$\displaystyle E_{y}\left[e^{c\hat{\Upsilon}^{A,r}_{j}(t)}\right]$
$\displaystyle=E_{y}\left[e^{c\mathcal{I}_{\left\\{\tau^{r,A}_{j}(t)=0\right\\}}\hat{\Upsilon}^{A}_{j}+c\mathcal{I}_{\left\\{\tau^{r,A}_{j}(t)>0\right\\}}\hat{\Upsilon}^{A,r}_{j}(t)}\right]$
(40) $\displaystyle\leq
e^{c\mathcal{I}_{\left\\{\tau^{r,A}_{j}(t)=0\right\\}}\hat{\Upsilon}^{A}_{j}}E_{y}\left[e^{\frac{c}{r}\mathcal{I}_{\left\\{\tau^{r,A}_{j}(t)>0\right\\}}u^{r}_{j}(\tau^{r,A}_{j}(t))}\right]$
$\displaystyle\leq
e^{c\hat{\Upsilon}^{A}_{j}}E_{y}\left[e^{c\mathcal{I}_{\left\\{\tau^{r,A}_{j}(t)>0\right\\}}u^{r}_{j}(\tau^{r,A}_{j}(t))}\right]\leq
e^{c\hat{\Upsilon}^{A}_{j}}B_{7}.$

Since $\left\\{\hat{\Upsilon}^{A,r}_{j}(t)\right\\}_{j=1}^{J}$ are
independent, for all $c\in(0,\delta)$, $t\geq 0$, and $r\geq R_{4}$ we then
have

$\displaystyle
E_{y}\left[e^{c\max_{j}\\{\hat{\Upsilon}^{A,r}_{j}(t)\\}}\right]\leq\sum_{j=1}^{J}E_{y}\left[e^{c\hat{\Upsilon}^{A,r}_{j}(t)}\right]\leq
e^{c\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}}JB_{7}.$ (41)

Note that if $t>\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}$ and $r\geq R_{4}\geq 1$
then $\tau^{r,A}_{j}(t)>0$ for all $j\in\mathbb{A}_{J}$. Consequently, a
similar estimate as for (41) shows that, for all
$t>\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}$, $c\in(0,\delta)$, and $r\geq R_{4}$
we have

$\displaystyle
E_{y}\left[e^{c\max_{j}\\{\hat{\Upsilon}^{A,r}_{j}(t)\\}}\right]\leq\sum_{j=1}^{J}E_{y}\left[e^{c\hat{\Upsilon}^{A,r}_{j}(t)}\right]\leq
JB_{7}.$ (42)

Let $\delta_{1}\doteq\min\left\\{B_{2}/2,\delta/(2C_{i})\right\\}$ and note
that for all $c\in(0,\delta_{1})$, $t\geq 0$, $r\geq R_{4}$, and
$i\in\mathbb{A}_{I}$ we have

$\displaystyle E_{y}\Big{[}e^{c\hat{W}^{r}_{i}(t)}\Big{]}$ $\displaystyle\leq
E_{y}\Big{[}\mathcal{I}_{\\{\hat{W}^{r}_{i}(t)>2C_{i}\max_{j}\\{\hat{\Upsilon}^{A,r}_{j}(t)\\}\\}}e^{2c\left(\hat{W}^{r}_{i}(t)-C_{i}\max_{j}\\{\hat{\Upsilon}^{A,r}_{j}(t)\\}\right)}$
$\displaystyle\quad+\mathcal{I}_{\\{\hat{W}^{r}_{i}(t)\leq
2C_{i}\max_{j}\\{\hat{\Upsilon}^{A,r}_{j}(t)\\}\\}}e^{2cC_{i}\max_{j}\\{\hat{\Upsilon}^{A,r}_{j}(t)\\}}\Big{]}$
$\displaystyle\leq
E_{y}\left[e^{2c\left(\hat{W}^{r}_{i}(t)-C_{i}\max_{j}\\{\hat{\Upsilon}^{A,r}_{j}(t)\\}\right)^{+}}\right]+E_{y}\left[e^{2cC_{i}\max_{j}\\{\hat{\Upsilon}^{A,r}_{j}(t)\\}}\right]$
$\displaystyle\leq
E_{y}\left[e^{B_{2}\left(\hat{W}^{r}_{i}(t)-C_{i}\max_{j}\\{\hat{\Upsilon}^{A,r}_{j}(t)\\}\right)^{+}}\right]+E_{y}\left[e^{2cC_{i}\max_{j}\\{\hat{\Upsilon}^{A,r}_{j}(t)\\}}\right]$
$\displaystyle\leq\frac{B_{4}}{B_{1}}e^{-\tilde{\delta}t+B_{5}(|\hat{q}|+|\hat{\Upsilon}|)}+\frac{B_{6}+B_{3}}{B_{1}}+E_{y}\left[e^{2cC_{i}\max_{j}\\{\hat{\Upsilon}^{A,r}_{j}(t)\\}}\right]$

where the last line used equation (39). Combining this with equations (41) and
(42) implies that for any $c\in(0,\delta_{1})$, $r\geq R_{4}$, and
$i\in\mathbb{A}_{I}$ if $t\in[0,\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}]$ we have

$\displaystyle
E_{y}\Big{[}e^{c\hat{W}^{r}_{i}(t)}\Big{]}\leq\frac{B_{4}}{B_{1}}e^{-\tilde{\delta}t+B_{5}(|\hat{q}|+|\hat{\Upsilon}|)}+\frac{B_{6}+B_{3}}{B_{1}}+e^{\delta_{1}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}}JB_{7}$

and if $t>\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}$ we have

$\displaystyle
E_{y}\Big{[}e^{c\hat{W}^{r}_{i}(t)}\Big{]}\leq\frac{B_{4}}{B_{1}}e^{-\tilde{\delta}t+B_{5}(|\hat{q}|+|\hat{\Upsilon}|)}+\frac{B_{6}+B_{3}}{B_{1}}+JB_{7}.$

Note that since $|\hat{\Upsilon}|\geq\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}$
this implies that for any $c\in(0,\delta_{1})$, $r\geq R_{4}$,
$i\in\mathbb{A}_{I}$, and $t\geq 0$ we have

$\displaystyle
E_{y}\Big{[}e^{c\hat{W}^{r}_{i}(t)}\Big{]}\leq\left(\frac{B_{4}}{B_{1}}+JB_{7}\right)e^{-\tilde{\delta}t+(B_{5}+\delta_{1}+\tilde{\delta})(|\hat{q}|+|\hat{\Upsilon}|)}+\frac{B_{6}+B_{3}}{B_{1}}+JB_{7}.$

Define $\delta_{2}\doteq I^{-\frac{1}{2}}\delta_{1}$ and note that for all
$c\in(0,\delta_{2})$, $t\geq 0$, and $r\geq R_{4}$ we have

$\displaystyle E_{y}\left[e^{c|\hat{W}^{r}(t)|_{2}}\right]$ $\displaystyle\leq
E_{y}\left[e^{cI^{\frac{1}{2}}\max_{i}\\{\hat{W}^{r}_{i}(t)\\}}\right]\leq\sum_{i=1}^{I}E_{y}\left[e^{cI^{\frac{1}{2}}\hat{W}^{r}_{i}(t)}\right]$
$\displaystyle\leq\sum_{i=1}^{I}\left(\left(\frac{B_{4}}{B_{1}}+JB_{7}\right)e^{-\tilde{\delta}t+(B_{5}+\delta_{1}+\tilde{\delta})(|\hat{q}|+|\hat{\Upsilon}|)}+\frac{B_{6}+B_{3}}{B_{1}}+JB_{7}\right)$
$\displaystyle\leq
I\left(\frac{B_{4}}{B_{1}}+JB_{7}\right)e^{-\tilde{\delta}t+(B_{5}+\delta_{1}+\tilde{\delta})(|\hat{q}|+|\hat{\Upsilon}|)}+I\left(\frac{B_{6}+B_{3}}{B_{1}}+JB_{7}\right).$

Since $y\in\mathcal{Y}^{r}$ was arbitrary this completes the proof. ∎

###  7.2 Proof of Proposition 3.5

We begin with some auxiliary results.

######  Lemma 7.4.

For all $s\geq 0$ if $\mathcal{Z}^{r}(s)\not\in\mathcal{M}$ then
$\tilde{d}\left(\left(\hat{Q}^{r}(s)-\tilde{c}_{2}r^{\kappa-1}\right)\vee
0\right)=0$.

###### Proof.

Let $s\geq 0$ be arbitrary and assume $\mathcal{Z}^{r}(s)\not\in\mathcal{M}$.
Note that for all $j\in\mathbb{A}_{J}$ we have
$\left(\hat{Q}_{j}^{r}(s)-\tilde{c}_{2}r^{\kappa-1}\right)^{+}=0$ if
$\mathcal{Z}^{r}_{j}(s)=1$. Consequently the result follows on applying
Proposition2.2 with
$q=\left(\hat{Q}^{r}(s)-\tilde{c}_{2}r^{\kappa-1}\right)^{+}$ and noting that
$z^{q}=\mathcal{Z}^{r}(s)$. ∎

The next result, which is analogous to Proposition 6.1, is proved in Section
9.1. Recall that, by convention, we take $v_{j}^{r}(0)=0$ for $r>0$ and
$j\in\mathbb{A}_{J}$.

######  Proposition 7.5 (Proof in Section 9.1).

Let $\delta$ be as in Condition 1. There exists $R<\infty$ and for any
$c<\delta$ a corresponding $K(c)<\infty$ such that for any
$j\in\mathbb{A}_{J}$, $y^{r}\in\mathcal{Y}^{r}$ and $t\geq 0$, we have

$\sup_{r\geq R}E_{y^{r}}\left[e^{cv^{r}_{j}(\tau^{r,S}_{j}(t))}\right]<K(c).$

The next two results are proved in Section 10.3. Recall the constant $\lambda$
from Section 2.4.

######  Proposition 7.6 (Proof in Section 10.3).

There exists a constant $B_{\hat{h}}\in(0,\infty)$ and $R\in(0,\infty)$ such
that for all $q^{1},q^{2}\in\mathbb{R}^{J}_{+}$ and $r\geq R$, we have

$|\lambda|(\tilde{d}(q^{2})-\tilde{d}(q^{1}))\leq
h\cdot\left(q^{2}-q^{1}\right)+B_{\hat{h}}|KM^{r}q^{2}-KM^{r}q^{1}|_{2}.$

######  Proposition 7.7 (Proof in Section 10.3).

There exists a constant $B_{\tilde{d}}<\infty$ such that for all
$q^{1},q^{2}\in\mathbb{R}^{J}_{+}$ we have

$|\tilde{d}(q^{1})-\tilde{d}(q^{2})|\leq B_{\tilde{d}}|q^{1}-q^{2}|_{2}$

We now proceed to the proof of Proposition 3.5.

Proof of Proposition 3.5. By adding and subtracting $\hat{h}(KM\hat{Q}^{r})$,
and using Proposition 2.1, Proposition 3.4, and the Lipschitz property of the
map $\hat{h}$, it is sufficient to show that for any $\epsilon_{0}\in(0,1)$
and $M<\infty$ there exists $T^{*},R\in(0,\infty)$ such that for all $r\geq
R$, $T\geq T^{*}$, $y^{r}\in\mathcal{Y}^{r}$ satisfying $\hat{q}^{r}\leq M$
and $\hat{\Upsilon}^{r}\leq r^{-1}M$, and $t\geq 0$ we have

$E_{y^{r}}\left[\frac{1}{T}\int_{0}^{T}\tilde{d}\left(\hat{Q}^{r}(t+s)\right)ds\right]\leq\epsilon_{0},\;\;E_{y^{r}}\left[\int_{0}^{\infty}e^{-\varsigma
s}\tilde{d}\left(\hat{Q}^{r}(t+s)\right)ds\right]\leq\epsilon_{0}.$

We now fix $\epsilon_{0}\in(0,1)$ and let $\epsilon>0$ be arbitrary which will
be chosen suitably later depending on $\epsilon_{0}$.

The main idea behind the proof is that, on average, the vectors
$v^{c}(\mathcal{Z}^{r}(t))$ reduce $\tilde{d}(\hat{Q}^{r}(t))$ faster than
other factors like randomness and the vectors $v^{b}(\mathcal{Z}^{r}(t))$
(which are used to keep the queues nonempty) increase
$\tilde{d}(\hat{Q}^{r}(t))$. This is used to show that for a given
$\epsilon>0$ there exists $k<\infty$ such that for all sufficiently large $r$
and all $t\geq 0$ we have

$E_{y^{r}}\left[\int_{t}^{t+kr^{-1}}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds\right]\leq\epsilon
3kr^{-1},$ (43)

and from this the proposition follows readily. Fix $\epsilon\in(0,1)$ and
$M<\infty$. Proposition 3.4 implies that there exist constants $B_{1}<\infty$
and $c>0$ such that for sufficiently large $r$, all $t\geq 0$, and any
$y^{r}\in\mathcal{Y}^{r}$ satisfying $\hat{q}^{r}\leq M$ and
$\hat{\Upsilon}^{r}\leq r^{-1}M$ we have
$E_{y^{r}}\left[e^{c|\hat{W}^{r}(t)|}\right]\leq B_{1}$. From the definition
of $\tilde{d}$ there exists a constant $B_{2}>0$ such that for all
$q\in\mathbb{R}^{J}_{+}$ and $r$ we have $\tilde{d}(q)\leq
B_{2}e^{\frac{c}{2}|KM^{r}q|}$ and $\tilde{d}(q)\leq B_{2}|KM^{r}q|$. For
notational convenience let
$\xi=\frac{\epsilon}{\left(1+4|\lambda||\tilde{\lambda}|^{-1}\right)B_{1}B_{2}}\wedge
1$, where $\tilde{\lambda}$ was defined in (5), and define for
$n\in\mathbb{N}$,

$\displaystyle\mathcal{A}^{r,n,t,j}$ $\displaystyle=$
$\displaystyle\left\\{\sup_{0\leq s\leq
rn}|A_{j}^{r,t}(s)-s\alpha^{r}_{j}|\geq\frac{1}{5}nr^{\frac{1}{2}}\right\\}\cup\left\\{\sup_{0\leq
s\leq
r\max_{i}\\{C_{i}\\}n}|S_{j}^{r,t}(s)-s\beta^{r}_{j}|\geq\frac{1}{5}nr^{\frac{1}{2}}\right\\}$
$\displaystyle\cup\left\\{r\bar{\Upsilon}^{A,r}_{j}(t)\geq\xi
n\right\\}\cup\left\\{\beta^{r}_{j}r^{\frac{3}{2}}\bar{\Upsilon}^{S,r}_{j}(t)\geq\frac{1}{5}\xi
n\right\\},$ $\mathcal{A}^{r,n,t}=\cup_{j=1}^{J}\mathcal{A}^{r,n,t,j},$

and

$\mathcal{B}^{r,n,t}\doteq\mathcal{A}^{r,n,t}\cup\left\\{\tilde{d}\left(\hat{Q}^{r}(t+\xi
nr^{-1})\right)>\xi
n\right\\}\cup\Big{\\{}\int_{r^{2}t}^{r^{2}t+rn}\sum_{j=1}^{J}\mathcal{I}_{\\{\mathcal{E}_{j}^{r}(s)=1\\}}ds\geq\xi
nr^{\frac{7}{4}\kappa}\Big{\\}}.$

Main proof idea. We will choose $k$ large enough that $P(\mathcal{B}^{r,k,t})$
is small and we will show that on $(\mathcal{B}^{r,k,t})^{c}$, because
$\hat{\Upsilon}^{r}(t)$ has a small impact after $t+\xi kr^{-1}$,
$\hat{Q}^{r}(s)$ has favorable behavior for $s\in[t+\xi kr^{-1},t+kr^{-1}]$.
When proving the key estimate (43) we break up the interval $[t,t+kr^{-1}]$
into the subintervals $[t,t+\xi k(1+4|\lambda||\tilde{\lambda}|^{-1})]$ and
$[t+\xi k(1+4|\lambda||\tilde{\lambda}|^{-1}),t+kr^{-1}]$. We can think of
$\xi$ as determining the size of the first subinterval over which we rely on a
crude bound on $E[\tilde{d}(\hat{Q}^{r}(s))]$ based on Proposition 3.4, but
because $\xi$ is small (recall its definition in terms of $\epsilon$) we can
use the length of the subinterval to show that

$E_{y^{r}}\left[\int_{t}^{t+\xi
k(1+4|\lambda||\tilde{\lambda}|^{-1})}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds\right]\leq\epsilon
kr^{-1}.$

For the second subinterval we demonstrate that on the set
$(\mathcal{B}^{r,k,t})^{c}$ $\tilde{d}\left(\hat{Q}^{r}(s)\right)$ is small
for all $s\in[t+\xi k(1+4|\lambda||\tilde{\lambda}|^{-1}),t+kr^{-1}]$, which
combined with the fact that $(\mathcal{B}^{r,k,t})^{c}$ occurs with high
probability, allows us to show

$E_{y^{r}}\left[\int_{t+\xi
k(1+4|\lambda||\tilde{\lambda}|^{-1})}^{t+kr^{-1}}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds\right]\leq
2\epsilon kr^{-1}$

and complete the proof of (43) from which the Proposition follows easily by
appealing to Proposition 2.1.

Note that for any $j\in\mathbb{A}_{J}$ we have

$\displaystyle
r\bar{\Upsilon}^{A,r}_{j}(t)\leq\mathcal{I}_{\\{\tau^{r,A}_{j}(t)=0\\}}\hat{\Upsilon}^{A,r}_{j}+r^{-1}\mathcal{I}_{\\{\tau^{r,A}_{j}(t)>0\\}}u^{r}_{j}(\tau^{r,A}_{j}(t))$
$\displaystyle\leq\mathcal{I}_{\\{\tau^{r,A}_{j}(t)=0\\}}r^{-1}M+r^{-1}\mathcal{I}_{\\{\tau^{r,A}_{j}(t)>0\\}}u^{r}_{j}(\tau^{r,A}_{j}(t))$

and, similarly

$\beta^{r}_{j}r^{\frac{3}{2}}\bar{\Upsilon}^{S,r}_{j}(t)\leq\mathcal{I}_{\\{\tau^{r,S}_{j}(t)=0\\}}\beta^{r}_{j}r^{-\frac{1}{2}}M+\beta^{r}_{j}r^{-\frac{1}{2}}\mathcal{I}_{\\{\tau^{r,S}_{j}(t)>0\\}}v^{r}_{j}(\tau^{r,S}_{j}(t)).$

This (recall that $\sup_{j}\beta^{r}_{j}<\infty$), combined with Propositions
6.2 ((14), (15), with $c_{1}=1/2$, $c_{2}=0$), 7.5, and 6.1 implies that there
exist constants $\tilde{B}_{1},\tilde{B}_{2}<\infty$ such that for all $t\geq
0$ and all sufficiently large $r$ we have
$P_{y^{r}}\left(\mathcal{A}^{r,n,t}\right)\leq\tilde{B}_{1}e^{-n\tilde{B}_{2}}$.
In addition, for $r\geq\frac{1}{\xi}$ we have

$\displaystyle
P_{y^{r}}\left(\int_{r^{2}t}^{r^{2}t+rn}\sum_{j=1}^{J}\mathcal{I}_{\\{\mathcal{E}_{j}^{r}(s)=1\\}}ds\geq\xi
nr^{\frac{7}{4}\kappa}\right)\leq P_{y^{r}}\left(\int_{r^{2}t}^{r^{2}(t+\xi
n)}\sum_{j=1}^{J}\mathcal{I}_{\\{\mathcal{E}_{j}^{r}(s)=1\\}}ds\geq\xi
nr^{\frac{7}{4}\kappa}\right)$
$\displaystyle\leq\sum_{j=1}^{J}P_{y^{r}}\left(\int_{r^{2}t}^{r^{2}(t+\xi
n)}\mathcal{I}_{\\{\mathcal{E}_{j}^{r}(s)=1\\}}ds\geq
r\hat{\Upsilon}^{A,r}_{j}(t)+\frac{\xi
n}{2J}r^{\frac{7}{4}\kappa}\right)+\sum_{j=1}^{J}P_{y^{r}}\left(r\hat{\Upsilon}^{A,r}_{j}(t)\geq\frac{n\xi}{2J}r^{\frac{7}{4}\kappa}\right).$

This, combined with Proposition 3.3, Proposition 6.1, and the fact that for
any $j\in\mathbb{A}_{J}$

$\displaystyle r\hat{\Upsilon}^{A,r}_{j}(t)$ $\displaystyle\leq$
$\displaystyle\mathcal{I}_{\\{\tau^{r,A}_{j}(t)=0\\}}r\hat{\Upsilon}^{A,r}_{j}+\mathcal{I}_{\\{\tau^{r,A}_{j}(t)>0\\}}u^{r}_{j}(\tau^{r,A}_{j}(t))$
$\displaystyle\leq$
$\displaystyle\mathcal{I}_{\\{\tau^{r,A}_{j}(t)=0\\}}M+\mathcal{I}_{\\{\tau^{r,A}_{j}(t)>0\\}}u^{r}_{j}(\tau^{r,A}_{j}(t))$

implies that there exist constant $\tilde{B}_{3},\tilde{B}_{4}<\infty$ such
that for sufficiently large $r$ and for all $t\geq 0$ and $n\geq 0$ we have

$P_{y^{r}}\left(\int_{r^{2}t}^{r^{2}t+rn}\sum_{j=1}^{J}\mathcal{I}_{\\{\mathcal{E}_{j}^{r}(s)=1\\}}ds\geq\xi
nr^{\frac{7}{4}\kappa}\right)\leq\tilde{B}_{3}e^{-n\tilde{B}_{4}}.$

Furthermore, since $\tilde{d}(\hat{Q}^{r}(t+\xi nr^{-1}))\leq
B_{2}|\hat{W}^{r}(t+\xi nr^{-1})|$ and $E_{y^{r}}\left[e^{c|\hat{W}^{r}(t+\xi
nr^{-1})|}\right]\leq B_{1}$ for sufficiently large $r$ we have

$P_{y^{r}}\left(\tilde{d}\left(\hat{Q}^{r}(t+\xi nr^{-1})\right)>\xi
n\right)\leq P_{y^{r}}\left(|\hat{W}^{r}(t+\xi nr^{-1})|>\frac{\xi
n}{B_{2}}\right)\leq B_{1}e^{-\frac{c\xi}{B_{2}}n}.$

Consequently we know there exist constants $B_{3},B_{4}<\infty$ such that for
sufficiently large $r$ we have for all $t\geq 0$ and $n\geq 0$
$P_{y^{r}}\left(\mathcal{B}^{r,n,t}\right)\leq B_{3}e^{-nB_{4}}$.

Our goal is to show that for any $\epsilon>0$ we can choose corresponding
$k,R<\infty$ sufficiently large such that for all $r\geq R$ (43) holds. Our
choices of $k,R<\infty$ will have to satisfy a large number of inequalities
used throughout the remainder of the proof, so for the ease of the reader we
list them all in one place below and reference them later when needed.

######  Summary 1.

Recall the definitions of $\mathcal{B}^{r,n,t}$, $B_{1}$, $B_{2}$, $B_{3}$,
$B_{4}$, $\xi$, and $c$. We choose the constants $k\geq\frac{1}{\xi}$ and
$R<\infty$ such that,
$B^{\frac{1}{2}}_{3}e^{-k\frac{B_{4}}{2}}B_{2}B^{\frac{1}{2}}_{1}\leq\epsilon$
and, for all $r\geq R$, the following hold:

  1. (a)

For all $t\geq 0$ we have $E_{y^{r}}\left[e^{c|\hat{W}^{r}(t)|}\right]\leq
B_{1}$

  2. (b)

For all $q\in\mathbb{R}^{J}_{+}$, we have $\tilde{d}(q)\leq
B_{2}e^{\frac{c}{2}|KM^{r}q|}$ and $\tilde{d}(q)\leq B_{2}|KM^{r}q|$

  3. (c)

For all $t\geq 0$ we have $P_{y^{r}}\left(\mathcal{B}^{r,k,t}\right)\leq
B_{3}e^{-kB_{4}}$.

  4. (d)

$\epsilon r^{\frac{1}{8}}\geq k$, $r\geq\frac{1}{\xi}$, $r\geq 25$, and
$\epsilon r^{\frac{1}{16}}\geq\xi k$.

  5. (e)

$2|\theta|+|\frac{1}{\beta^{r}}|_{1}\leq r^{\frac{1}{8}}$

  6. (f)

For all $i\in\mathbb{A}_{I}$ we have
$r\left(\sum_{j=1}^{J}K_{i,j}\rho^{r}_{j}-C_{i}\right)\geq 2\theta_{i}$

  7. (g)

$J(c_{2}+1)r^{\kappa-1}+C_{i}\epsilon
r^{\frac{7}{4}\kappa-\frac{15}{16}}+|\frac{1}{\beta^{r}}|_{1}\epsilon
r^{-\frac{3}{8}}\leq\epsilon r^{-\frac{1}{4}}$

  8. (h)

For all $i\in\mathbb{A}_{I}$ we have
$\left(\sum_{j=1}^{J}K_{i,j}\rho^{r}_{j}-C_{i}\right)<0$

  9. (i)

$|h||\beta^{r}||\rho-\rho^{r}|+|h||\beta^{r}-\beta|\left(\max_{z\in\mathcal{M}}|v^{c}(z)|+\max_{z\in\left\\{0,1\right\\}^{J}}|v^{b}(z)|\right)\leq\frac{|\tilde{\lambda}|}{4}$

  10. (j)

$|\alpha^{r}|\leq 2|\alpha|$

  11. (k)

$\left(\frac{|\tilde{\lambda}|}{2}+2|h||\alpha|\right)\epsilon
r^{\frac{7}{4}\kappa-\frac{15}{16}}+|h|\epsilon r^{-\frac{3}{8}}\leq\epsilon
r^{-\frac{1}{4}}$

  12. (l)

$\frac{\epsilon}{|\lambda|}r^{-\frac{1}{4}}+\frac{B_{\hat{h}}}{|\lambda|}I^{\frac{1}{2}}\epsilon
r^{-\frac{1}{4}}\leq\xi k$

  13. (m)

$B_{\tilde{d}}J^{\frac{1}{2}}\tilde{c}_{2}r^{\kappa-1}+J^{\frac{1}{2}}B_{\tilde{d}}r^{-1}+\frac{\epsilon}{|\lambda|}r^{-\frac{1}{4}}+2B_{\hat{h}}I^{\frac{1}{2}}\frac{\epsilon}{|\lambda|}r^{-\frac{1}{4}}\leq\epsilon
r^{-\frac{1}{8}}$.

We now bound $\tilde{d}\left(\hat{Q}^{r}(\cdot)\right)$ from above on the set
$\left(\mathcal{B}^{r,k,t}\right)^{c}$. Using (9), on the set
$\left(\mathcal{B}^{r,k,t}\right)^{c}$ for all $r\geq R$, any $t+\xi
kr^{-1}\leq s_{1}<s_{2}\leq t+kr^{-1}$, and any $j\in\mathbb{A}_{J}$, we have

$\displaystyle\hat{Q}^{r}_{j}(s_{2})-\hat{Q}^{r}_{j}(s_{1})$ $\displaystyle=$
$\displaystyle\frac{1}{r}\mathcal{I}_{\\{s_{2}-t\geq\bar{\Upsilon}^{A,r}_{j}(t)>s_{1}-t\\}}-\frac{1}{r}\mathcal{I}_{\\{\bar{B}^{r}_{j}(s_{2})-\bar{B}^{r}_{j}(t)\geq\bar{\Upsilon}^{S,r}_{j}(t)>\bar{B}^{r}_{j}(s_{1})-\bar{B}^{r}_{j}(t)\\}}$
$\displaystyle+\frac{1}{r}A_{j}^{r,t}\left(r^{2}\left(s_{2}-t-\bar{\Upsilon}^{A,r}_{j}(t)\right)^{+}\right)-\frac{1}{r}A_{j}^{r,t}\left(r^{2}\left(s_{1}-t-\bar{\Upsilon}^{A,r}_{j}(t)\right)^{+}\right)$
$\displaystyle-\frac{1}{r}S_{j}^{r,t}\left(r^{2}\left(\bar{B}^{r}_{j}(s_{2})-\bar{B}^{r}_{j}(t)-\bar{\Upsilon}^{S,r}_{j}(t)\right)^{+}\right)+\frac{1}{r}S_{j}^{r,t}\left(r^{2}\left(\bar{B}^{r}_{j}(s_{1})-\bar{B}^{r}_{j}(t)-\bar{\Upsilon}^{S,r}_{j}(t)\right)^{+}\right)$
$\displaystyle\leq$
$\displaystyle\frac{1}{r}A_{j}^{r,t}\left(r^{2}\left(s_{2}-t-\bar{\Upsilon}^{A,r}_{j}(t)\right)\right)-\frac{1}{r}A_{j}^{r,t}\left(r^{2}\left(s_{1}-t-\bar{\Upsilon}^{A,r}_{j}(t)\right)\right)$
$\displaystyle-\frac{1}{r}S_{j}^{r,t}\left(r^{2}\left(\bar{B}^{r}_{j}(s_{2})-\bar{B}^{r}_{j}(t)-\bar{\Upsilon}^{S,r}_{j}(t)\right)^{+}\right)+\frac{1}{r}S_{j}^{r,t}\left(r^{2}\left(\bar{B}^{r}_{j}(s_{1})-\bar{B}^{r}_{j}(t)-\bar{\Upsilon}^{S,r}_{j}(t)\right)^{+}\right),$

since on the set $\left(\mathcal{B}^{r,k,t}\right)^{c}$ we have
$\bar{\Upsilon}^{A,r}_{j}(t)\leq\xi kr^{-1}$ meaning
$\mathcal{I}_{\\{s_{2}-t\geq\bar{\Upsilon}^{A,r}_{j}(t)>s_{1}-t\\}}=0$, and so

$\displaystyle\hat{Q}^{r}_{j}(s_{2})-\hat{Q}^{r}_{j}(s_{1})$
$\displaystyle\leq$
$\displaystyle\alpha^{r}_{j}r\left(s_{2}-s_{1}\right)+\frac{4}{5}kr^{-\frac{1}{2}}-\beta^{r}_{j}r\left(\left(\bar{B}^{r}_{j}(s_{2})-\bar{B}^{r}_{j}(t)-\bar{\Upsilon}^{S,r}_{j}(t)\right)^{+}\right.$
$\displaystyle\left.-\left(\bar{B}^{r}_{j}(s_{1})-\bar{B}^{r}_{j}(t)-\bar{\Upsilon}^{S,r}_{j}(t)\right)^{+}\right)$
$\displaystyle\leq$
$\displaystyle\alpha^{r}_{j}r\left(s_{2}-s_{1}\right)-\beta^{r}_{j}r\left(\bar{B}^{r}_{j}(s_{2})-\bar{B}^{r}_{j}(s_{1})\right)+\beta^{r}_{j}r\bar{\Upsilon}^{S,r}_{j}(t)+\frac{4}{5}kr^{-\frac{1}{2}}$
$\displaystyle\leq$
$\displaystyle\alpha^{r}_{j}r\left(s_{2}-s_{1}\right)-\beta^{r}_{j}r\left(\bar{B}^{r}_{j}(s_{2})-\bar{B}^{r}_{j}(s_{1})\right)+kr^{-\frac{1}{2}}$
$\displaystyle\leq$
$\displaystyle\alpha^{r}_{j}r\left(s_{2}-s_{1}\right)-\beta^{r}_{j}r\left(\bar{B}^{r}_{j}(s_{2})-\bar{B}^{r}_{j}(s_{1})\right)+\epsilon
r^{-\frac{3}{8}}$

due to Summary 1 part (d) and the fact that on the set
$\left(\mathcal{B}^{r,k,t}\right)^{c}$ we have
$\beta^{r}_{j}r^{\frac{3}{2}}\bar{\Upsilon}^{S,r}_{j}(t)\leq\frac{1}{5}\xi
k\leq\frac{1}{5}k$.

A similar calculation, using the inequalities,

$\left(\bar{B}^{r}_{j}(s_{2})-\bar{B}^{r}_{j}(t)-\bar{\Upsilon}^{S,r}_{j}(t)\right)^{+}-\left(\bar{B}^{r}_{j}(s_{1})-\bar{B}^{r}_{j}(t)-\bar{\Upsilon}^{S,r}_{j}(t)\right)^{+}\leq\bar{B}^{r}_{j}(s_{2})-\bar{B}^{r}_{j}(s_{1}),$

$-\frac{1}{r}\mathcal{I}_{\\{\bar{B}^{r}_{j}(s_{2})-\bar{B}^{r}_{j}(t)\geq\bar{\Upsilon}^{S,r}_{j}(t)>\bar{B}^{r}_{j}(s_{1})-\bar{B}^{r}_{j}(t)\\}}\geq-\frac{1}{r}$,
properties in Summary 1 part (d), and the fact that on the set
$\left(\mathcal{B}^{r,k,t}\right)^{c}$ we have
$\bar{\Upsilon}^{A,r}_{j}(t)\leq\xi kr^{-1}$ shows that on the set
$\left(\mathcal{B}^{r,k,t}\right)^{c}$ for any $t+\xi kr^{-1}\leq
s_{1}<s_{2}\leq t+kr^{-1}$, $r\geq R$, and $j\in\mathbb{A}_{J}$ we have

$\displaystyle\hat{Q}^{r}_{j}(s_{2})-\hat{Q}^{r}_{j}(s_{1})$
$\displaystyle\geq$
$\displaystyle\alpha^{r}_{j}r\left(s_{2}-s_{1}\right)-\beta^{r}_{j}r\left(\bar{B}^{r}_{j}(s_{2})-\bar{B}^{r}_{j}(s_{1})\right)-\epsilon
r^{-\frac{3}{8}}.$

Consequently on the set $\left(\mathcal{B}^{r,k,t}\right)^{c}$ for any $t+\xi
kr^{-1}\leq s_{1}<s_{2}\leq t+kr^{-1}$, $r\geq R$, and $j\in\mathbb{A}_{J}$ we
have

$|\hat{Q}^{r}_{j}(s_{2})-\hat{Q}^{r}_{j}(s_{1})-r\left(\alpha^{r}_{j}(s_{2}-s_{1})-\beta^{r}_{j}\left(\bar{B}^{r}_{j}(s_{2})-\bar{B}^{r}_{j}(s_{1})\right)\right)|\leq\epsilon
r^{-\frac{3}{8}}.$ (44)

We will next use Proposition 7.6 to bound $\tilde{d}(\hat{Q}^{r}(\cdot))$ from
above on the interval $[t+\xi kr^{-1},t+kr^{-1}]$. For this we will need to
bound $\sup_{s\in[t+\xi
kr^{-1},t+kr^{-1}]}|\hat{W}^{r}_{i}(s)-\hat{W}^{r}_{i}(t+\xi kr^{-1})|$ for
all $i\in\mathbb{A}_{I}$. Let $i\in\mathbb{A}_{I}$ be arbitrary. We will first
bound
$\inf_{s\in[t+\xi
kr^{-1},t+kr^{-1}]}\left(\hat{W}^{r}_{i}(s)-\hat{W}^{r}_{i}(t+\xi
kr^{-1})\right)$ from below. On the set $\left(\mathcal{B}^{r,k,t}\right)^{c}$
for all $r\geq R$ and $s\in[t+\xi kr^{-1},t+kr^{-1}]$ we have from (44)

$\displaystyle\hat{W}^{r}_{i}(s)$ $\displaystyle\geq$
$\displaystyle\hat{W}^{r}_{i}(t+\xi kr^{-1})-|\frac{1}{\beta^{r}}|_{1}\epsilon
r^{-\frac{3}{8}}$
$\displaystyle+r\sum_{j=1}^{J}K_{i,j}\frac{1}{\beta^{r}_{j}}\left(\alpha^{r}_{j}(s-t-\xi
kr^{-1})-\beta^{r}_{j}\left(\bar{B}^{r}_{j}(s)-\bar{B}^{r}_{j}(t+\xi
kr^{-1})\right)\right)$ $\displaystyle\geq$
$\displaystyle\hat{W}^{r}_{i}(t+\xi kr^{-1})+r(s-t-\xi
kr^{-1})\left(\sum_{j=1}^{J}K_{i,j}\rho^{r}_{j}-C_{i}\right)-|\frac{1}{\beta^{r}}|_{1}\epsilon
r^{-\frac{3}{8}}$ $\displaystyle\geq$ $\displaystyle\hat{W}^{r}_{i}(t+\xi
kr^{-1})+2\theta_{i}(s-t-\xi kr^{-1})-|\frac{1}{\beta^{r}}|_{1}\epsilon
r^{-\frac{3}{8}}$ $\displaystyle\geq$ $\displaystyle\hat{W}^{r}_{i}(t+\xi
kr^{-1})+2\theta_{i}kr^{-1}-|\frac{1}{\beta^{r}}|_{1}\epsilon
r^{-\frac{3}{8}}$ $\displaystyle\geq$ $\displaystyle\hat{W}^{r}_{i}(t+\xi
kr^{-1})+2\theta_{i}\epsilon
r^{-\frac{7}{8}}-|\frac{1}{\beta^{r}}|_{1}\epsilon r^{-\frac{3}{8}}$
$\displaystyle\geq$ $\displaystyle\hat{W}^{r}_{i}(t+\xi
kr^{-1})-\left(2|\theta_{i}|+|\frac{1}{\beta^{r}}|_{1}\right)\epsilon
r^{-\frac{3}{8}}\geq\hat{W}^{r}_{i}(t+\xi kr^{-1})-\epsilon r^{-\frac{1}{4}}$

due to equation (44) and Summary 1 parts (d), (e), and (f). Therefore on the
set $\left(\mathcal{B}^{r,k,t}\right)^{c}$ for all $r\geq R$ we have

$\inf_{s\in[t+\xi
kr^{-1},t+kr^{-1}]}\left(\hat{W}^{r}_{i}(s)-\hat{W}^{r}_{i}(t+\xi
kr^{-1})\right)\geq-\epsilon r^{-\frac{1}{4}}.$

Now we bound $\sup_{s\in[t+\xi
kr^{-1},t+kr^{-1}]}\left(\hat{W}^{r}_{i}(s)-\hat{W}^{r}_{i}(t+\xi
kr^{-1})\right)$ from above. For all $i\in\mathbb{A}_{I}$ define

$\gamma^{r,i}_{0}\doteq\inf\left\\{s\geq t+\xi kr^{-1}:\hat{W}^{r}_{i}(s)\geq
Jc_{2}r^{\kappa-1}\right\\}$

and for $l\geq 0$ define

$\gamma^{r,i}_{2l+1}\doteq\inf\left\\{s\geq\gamma^{r,i}_{2l}:\hat{W}^{r}_{i}(s)<Jc_{2}r^{\kappa-1}\right\\},\;\;\gamma^{r,i}_{2l+2}\doteq\inf\left\\{s\geq\gamma^{r,i}_{2l+1}:\hat{W}^{r}_{i}(s)\geq
Jc_{2}r^{\kappa-1}\right\\}.$

Using Proposition 2.6 part (d) and (44), for $r\geq R$, on the set
$\left(\mathcal{B}^{r,k,t}\right)^{c}$, for any $l\geq 0$ and all
$s\in[\gamma^{r,i}_{2l},\gamma^{r,i}_{2l+1})\cap[\gamma^{r,i}_{2l},t+kr^{-1}]$,

$\displaystyle\hat{W}^{r}_{i}(s)$
$\displaystyle\leq\hat{W}^{r}_{i}(\gamma^{r,i}_{2l})+r\sum_{j=1}^{J}K_{i,j}\frac{1}{\beta^{r}_{j}}\left(\alpha^{r}_{j}(s-\gamma^{r,i}_{2l})-\beta^{r}_{j}\left(\bar{B}^{r}(s)-\bar{B}^{r}(\gamma^{r,i}_{2l})\right)\right)+|\frac{1}{\beta^{r}}|_{1}\epsilon
r^{-\frac{3}{8}}$ $\displaystyle\leq\hat{W}^{r}_{i}(t+\xi
kr^{-1})+J(c_{2}+1)r^{\kappa-1}+r(s-\gamma^{r,i}_{2l})\left(\sum_{j=1}^{J}K_{i,j}\rho^{r}_{j}-C_{i}\right)$
$\displaystyle\quad+r^{-1}C_{i}\int_{r^{2}t}^{r^{2}t+rk}\sum_{j=1}^{J}\mathcal{I}_{\\{\mathcal{E}_{j}^{r}(u)=1\\}}du+|\frac{1}{\beta^{r}}|_{1}\epsilon
r^{-\frac{3}{8}}$ $\displaystyle\leq\hat{W}^{r}_{i}(t+\xi
kr^{-1})+J(c_{2}+1)r^{\kappa-1}+r(s-\gamma^{r,i}_{2l})\left(\sum_{j=1}^{J}K_{i,j}\rho^{r}_{j}-C_{i}\right)$
$\displaystyle\quad+C_{i}\xi
kr^{\frac{7}{4}\kappa-1}+|\frac{1}{\beta^{r}}|_{1}\epsilon r^{-\frac{3}{8}}$
$\displaystyle\leq\hat{W}^{r}_{i}(t+\xi
kr^{-1})+J(c_{2}+1)r^{\kappa-1}+r(s-\gamma^{r,i}_{2l})\left(\sum_{j=1}^{J}K_{i,j}\rho^{r}_{j}-C_{i}\right)$
$\displaystyle\quad+C_{i}\epsilon
r^{\frac{7}{4}\kappa-\frac{15}{16}}+|\frac{1}{\beta^{r}}|_{1}\epsilon
r^{-\frac{3}{8}}$ $\displaystyle\leq\hat{W}^{r}_{i}(t+\xi
kr^{-1})+r(s-\gamma^{r,i}_{2l})\left(\sum_{j=1}^{J}K_{i,j}\rho^{r}_{j}-C_{i}\right)+\epsilon
r^{-\frac{1}{4}}\leq\hat{W}^{r}_{i}(t+\xi kr^{-1})+\epsilon r^{-\frac{1}{4}}$

due to Summary 1 parts (d), (g) and (h), and the fact that on the set
$\left(\mathcal{B}^{r,k,t}\right)^{c}$ we have
$r^{-1}\int_{r^{2}t}^{r^{2}t+rk}\sum_{j=1}^{J}\mathcal{I}_{\\{\mathcal{E}_{j}^{r}(s)=1\\}}ds\leq\xi
kr^{\frac{7}{4}\kappa-1}$. In addition, for $r\geq R$ on the set
$\left(\mathcal{B}^{r,k,t}\right)^{c}$, for all $s\in[t+\xi
kr^{-1},\gamma^{r,i}_{0})\cap[t+\xi kr^{-1},t+kr^{-1}]$ and
$s\in[\gamma^{r,i}_{2l+1},\gamma^{r,i}_{2l+2})\cap[\gamma^{r,i}_{2l+1},t+kr^{-1}]$
for any $l\geq 0$ we have

$\displaystyle\hat{W}^{r}_{i}(s)$ $\displaystyle\leq$ $\displaystyle
J(c_{2}+1)r^{\kappa-1}\leq J(c_{2}+1)r^{\kappa-1}+\hat{W}^{r}_{i}(t+\xi
kr^{-1})\leq\hat{W}^{r}_{i}(t+\xi kr^{-1})+\epsilon r^{-\frac{1}{4}}$

due to Summary 1 part (g). Since $i\in\mathbb{A}_{I}$ was arbitrary we have
shown that on the set $\left(\mathcal{B}^{r,k,t}\right)^{c}$ for all $r\geq R$
and $i\in\mathbb{A}_{I}$ we have

$\sup_{s\in[t+\xi kr^{-1},t+kr^{-1}]}|\hat{W}^{r}_{i}(s)-\hat{W}^{r}_{i}(t+\xi
kr^{-1})|\leq\epsilon r^{-\frac{1}{4}}.$ (45)

Now we will apply Proposition 7.6 to bound $\tilde{d}(\hat{Q}^{r}(\cdot))$
from above on the interval $[t+\xi kr^{-1},t+kr^{-1}]$. From Proposition 7.7 ,
for any $q\in\mathbb{R}^{J}_{+}$,

$|\tilde{d}(q)-\tilde{d}(\left(q-\tilde{c}_{2}r^{\kappa}\right)\vee 0)|\leq
B_{\tilde{d}}|q-\left(q-\tilde{c}_{2}r^{\kappa}\right)\vee 0|_{2}\leq
B_{\tilde{d}}J^{\frac{1}{2}}\tilde{c}_{2}r^{\kappa}.$ (46)

Define

$\zeta^{r}_{0}\doteq\inf\left\\{s\geq t+\xi
kr^{-1}:\tilde{d}\left(Q^{r}(r^{2}s)\right)\leq
B_{\tilde{d}}J^{\frac{1}{2}}\tilde{c}_{2}r^{\kappa}\right\\}$

and for $l\geq 0$ define

$\displaystyle\zeta^{r}_{2l+1}$
$\displaystyle\doteq\inf\left\\{s\geq\zeta^{r}_{2l}:\tilde{d}\left(Q^{r}(r^{2}s)\right)>B_{\tilde{d}}J^{\frac{1}{2}}\tilde{c}_{2}r^{\kappa}\right\\},$
$\displaystyle\zeta^{r}_{2l+2}$
$\displaystyle\doteq\inf\left\\{s\geq\zeta^{r}_{2l+1}:\tilde{d}\left(Q^{r}(r^{2}s)\right)\leq
B_{\tilde{d}}J^{\frac{1}{2}}\tilde{c}_{2}r^{\kappa}\right\\}.$

Note that if $\mathcal{Z}^{r}(s)\not\in\mathcal{M}$ then due to Lemma 7.4 we
have $\tilde{d}\left(\left(Q^{r}(r^{2}s)-\tilde{c}_{2}r^{\kappa}\right)\vee
0\right)=0$ so, from (46),

$\displaystyle\tilde{d}\left(Q^{r}(r^{2}s)\right)$ $\displaystyle\leq$
$\displaystyle\tilde{d}\left(\left(Q^{r}(r^{2}s)-\tilde{c}_{2}r^{\kappa}\right)\vee
0\right)+|\tilde{d}(Q^{r}(r^{2}s))-\tilde{d}(\left(Q^{r}(r^{2}s)-\tilde{c}_{2}r^{\kappa}\right)\vee
0)|$ $\displaystyle\leq$ $\displaystyle
B_{\tilde{d}}J^{\frac{1}{2}}\tilde{c}_{2}r^{\kappa}.$

Consequently if
$\tilde{d}\left(\hat{Q}^{r}(s)\right)>B_{\tilde{d}}J^{\frac{1}{2}}\tilde{c}_{2}r^{\kappa-1}$
then $\mathcal{Z}^{r}(s)\in\mathcal{M}$ (here we have used the fact that for
any $r>0$ and $q\in\mathbb{R}^{J}_{+}$ we have
$\tilde{d}(\frac{1}{r}q)=\frac{1}{r}\tilde{d}(q)$). In particular,
$\mathcal{Z}^{r}(s)\in\mathcal{M}$ for all $s\in[t+\xi
kr^{-1},\zeta^{r}_{0})\cap[t+\xi kr^{-1},t+kr^{-1}]$ and all
$s\in[\zeta^{r}_{2l+1},\zeta^{r}_{2l+2})\cap[\zeta^{r}_{2l+1},t+kr^{-1}]$ for
any $l\geq 0$.

Next note that from Definition 2.3 for all $r\geq R$ if
$\mathcal{Z}^{r}(s)\in\mathcal{M}$ and
$\sum_{j=1}^{J}\mathcal{E}_{j}^{r}(s)=0$ we have
$\mathfrak{b}^{r}(s)=\rho-v^{c}(\mathcal{Z}^{r}(s))-v^{b}(\mathcal{Z}^{r}(s))$
and so

$\displaystyle
h\cdot\left(\alpha^{r}-\beta^{r}\mathfrak{b}^{r}(s)\right)=h\cdot\left(\alpha^{r}-\beta^{r}\left(\rho-v^{c}(\mathcal{Z}^{r}(s))-v^{b}(\mathcal{Z}^{r}(s))\right)\right)$
$\displaystyle=h\cdot\left(-\beta^{r}\left(\rho-\rho^{r}\right)+\beta
v^{c}(\mathcal{Z}^{r}(s))+\beta v^{b}(\mathcal{Z}^{r}(s))\right.$
$\displaystyle\quad\left.+(\beta^{r}-\beta)v^{c}(\mathcal{Z}^{r}(s))+(\beta^{r}-\beta)v^{b}(\mathcal{Z}^{r}(s))\right)$
$\displaystyle\leq|h||\beta^{r}||\rho-\rho^{r}|+|h||\beta^{r}-\beta|\left(\max_{z\in\mathcal{M}}|v^{c}(z)|+\max_{z\in\left\\{0,1\right\\}^{J}}|v^{b}(z)|\right)$
$\displaystyle\quad+h\beta\cdot v^{c}(\mathcal{Z}^{r}(s))+h\beta\cdot
v^{b}(\mathcal{Z}^{r}(s))$
$\displaystyle\leq\frac{|\tilde{\lambda}|}{4}-|\tilde{\lambda}|+|h||\beta|\frac{|\tilde{\lambda}|}{4|h|\beta|}\leq-\frac{|\tilde{\lambda}|}{2}$
(47)

where we have used Summary 1 part (i) and the fact that due to Definition 2.3
we have $\max_{z\in\mathcal{Z}}\left\\{h\beta\cdot
v^{c}(z)\right\\}\leq-|\tilde{\lambda}|$ and
$\max_{z\in\left\\{0,1\right\\}}|v^{b}(z)|\leq\frac{|\tilde{\lambda}|}{4|\beta||h|}$.
Next, on the set $\left(\mathcal{B}^{r,k,t}\right)^{c}$ for all $r\geq R$ and
$s\in[t+\xi kr^{-1},\zeta^{r}_{0})\cap[t+\xi kr^{-1},t+kr^{-1}]$ we have from
(44)

$\displaystyle h\cdot\left(\hat{Q}^{r}(s)-\hat{Q}^{r}(t+\xi
kr^{-1})\right)-|h|\epsilon r^{-\frac{3}{8}}$ $\displaystyle\leq
rh\cdot\left(\alpha^{r}\left(s-t-\xi
kr^{-1}\right)-\beta^{r}\left(\bar{B}^{r}(s)-\bar{B}^{r}(t+\xi
kr^{-1})\right)\right)$ $\displaystyle\leq\int_{t+\xi
kr^{-1}}^{s}\mathcal{I}_{\\{\sum_{j=1}^{J}\mathcal{E}_{j}^{r}(r^{2}z)=0\\}}rh\cdot\left(\alpha^{r}-\beta^{r}\mathfrak{b}^{r}(r^{2}z)\right)dz$
$\displaystyle\quad+\int_{t+\xi
kr^{-1}}^{s}\mathcal{I}_{\\{\sum_{j=1}^{J}\mathcal{E}_{j}^{r}(r^{2}z)>0\\}}rh\cdot\left(\alpha^{r}-\beta^{r}\mathfrak{b}^{r}(r^{2}z)\right)dz$
$\displaystyle\leq-r\left(s-t-\xi kr^{-1}-\int_{t+\xi
kr^{-1}}^{s}\mathcal{I}_{\\{\sum_{j=1}^{J}\mathcal{E}_{j}^{r}(r^{2}z)>0\\}}dz\right)\frac{|\tilde{\lambda}|}{2}$
$\displaystyle\quad+r(h\cdot\alpha^{r})\int_{t+\xi
kr^{-1}}^{s}\mathcal{I}_{\\{\sum_{j=1}^{J}\mathcal{E}_{j}^{r}(r^{2}z)>0\\}}dz$

where the last line is from (47). Thus

$\displaystyle h\cdot\left(\hat{Q}^{r}(s)-\hat{Q}^{r}(t+\xi kr^{-1})\right)$
$\displaystyle\leq-r\left(s-t-\xi
kr^{-1}\right)\frac{|\tilde{\lambda}|}{2}+\left(\frac{|\tilde{\lambda}|}{2}+2|h||\alpha|\right)\xi
kr^{\frac{7}{4}\kappa-1}+|h|\epsilon r^{-\frac{3}{8}}$
$\displaystyle\leq-r\left(s-t-\xi
kr^{-1}\right)\frac{|\tilde{\lambda}|}{2}+\left(\frac{|\tilde{\lambda}|}{2}+2|h||\alpha|\right)\epsilon
r^{\frac{7}{4}\kappa-\frac{15}{16}}+|h|\epsilon r^{-\frac{3}{8}}$
$\displaystyle\leq-r\left(s-t-\xi
kr^{-1}\right)\frac{|\tilde{\lambda}|}{2}+\epsilon r^{-\frac{1}{4}}$ (48)

due to Summary 1 parts (d), (j), and (k), and the fact that on the set
$\left(\mathcal{B}^{r,k,t}\right)^{c}$ we have
$\int_{t}^{t+kr^{-1}}\mathcal{I}_{\\{\sum_{j=1}^{J}\mathcal{E}_{j}^{r}(r^{2}z)>0\\}}dz\leq\xi
kr^{\frac{7}{4}\kappa-2}.$ An almost identical argument shows that on the set
$\left(\mathcal{B}^{r,k,t}\right)^{c}$ for all $r\geq R$, any $l\geq 0$, and
all $s\in[\zeta^{r}_{2l+1},\zeta^{r}_{2l+2})\cap[\zeta^{r}_{2l+1},t+kr^{-1}]$
we have

$h\cdot\left(\hat{Q}^{r}(s)-\hat{Q}^{r}(\zeta^{r}_{2l+1})\right)\leq-r\left(s-\zeta^{r}_{2l+1}\right)\frac{|\tilde{\lambda}|}{2}+\epsilon
r^{-\frac{1}{4}}.$ (49)

From Proposition 7.6, for $r\geq R$, on the set
$\left(\mathcal{B}^{r,k,t}\right)^{c}$ for all $s\in[t+\xi
kr^{-1},\zeta^{r}_{0})\cap[t+\xi kr^{-1},t+kr^{-1}]$ we have

$\displaystyle\tilde{d}\left(\hat{Q}^{r}(s)\right)$ $\displaystyle\leq$
$\displaystyle\tilde{d}\left(\hat{Q}^{r}(t+\xi
kr^{-1})\right)+\frac{1}{|\lambda|}h\cdot\left(\hat{Q}^{r}(s)-\hat{Q}^{r}(t+\xi
kr^{-1})\right)$
$\displaystyle+\frac{B_{\hat{h}}}{|\lambda|}|\hat{W}^{r}(s)-\hat{W}^{r}(t+\xi
kr^{-1})|_{2}$ $\displaystyle\leq$ $\displaystyle\xi k-r\left(s-t-\xi
kr^{-1}\right)\frac{|\tilde{\lambda}|}{2|\lambda|}+\frac{\epsilon}{|\lambda|}r^{-\frac{1}{4}}+\frac{B_{\hat{h}}}{|\lambda|}I^{\frac{1}{2}}\epsilon
r^{-\frac{1}{4}}$ $\displaystyle\leq$ $\displaystyle 2\xi k-r\left(s-t-\xi
kr^{-1}\right)\frac{|\tilde{\lambda}|}{2|\lambda|}$

where we have used (48), (45), Summary 1 part (l), and the fact that on the
set $\left(\mathcal{B}^{r,k,t}\right)^{c}$ we have
$\tilde{d}\left(\hat{Q}^{r}(t+\xi kr^{-1})\right)\leq\xi k$. This implies that
on the set $\left(\mathcal{B}^{r,k,t}\right)^{c}$ for $r\geq R$ if
$\zeta^{r}_{0}-t-\xi kr^{-1}>\frac{4\xi k|\lambda|}{|\tilde{\lambda}|}r^{-1}$
we have

$\tilde{d}\left(\hat{Q}^{r}\left(t+\xi kr^{-1}+\frac{4\xi
k|\lambda|}{|\tilde{\lambda}|}r^{-1}\right)\right)\leq 2\xi
k-r\left(\frac{4\xi
k|\lambda|}{|\tilde{\lambda}|}r^{-1}\right)\frac{|\tilde{\lambda}|}{2|\lambda|}\leq
0$

which contradicts the definition of $\zeta^{r}_{0}$. Consequently, on the set
$\left(\mathcal{B}^{r,k,t}\right)^{c}$ for $r\geq R$ we have

$\zeta^{r}_{0}\leq t+\xi kr^{-1}+\frac{4\xi
k|\lambda|}{|\tilde{\lambda}|}r^{-1}.$ (50)

Similarly, for $r\geq R$ on the set $\left(\mathcal{B}^{r,k,t}\right)^{c}$ for
all $s\in[\zeta^{r}_{2l+1},\zeta^{r}_{2l+2})\cap[\zeta^{r}_{2l+1},t+kr^{-1}]$
for any $l\geq 0$ we have

$\displaystyle\tilde{d}\left(\hat{Q}^{r}(s)\right)$ $\displaystyle\leq$
$\displaystyle\tilde{d}\left(\hat{Q}^{r}(\zeta^{r}_{2l+1})\right)+\frac{1}{|\lambda|}h\cdot\left(\hat{Q}^{r}(s)-\hat{Q}^{r}(\zeta^{r}_{2l+1})\right)+\frac{B_{\hat{h}}}{|\lambda|}|\hat{W}^{r}(s)-\hat{W}^{r}(\zeta^{r}_{2l+1})|_{2}$
$\displaystyle\leq$
$\displaystyle\tilde{d}\left(\hat{Q}^{r}(\zeta^{r}_{2l+1})\right)+\frac{1}{|\lambda|}h\cdot\left(\hat{Q}^{r}(s)-\hat{Q}^{r}(\zeta^{r}_{2l+1})\right)$
$\displaystyle+\frac{B_{\hat{h}}}{|\lambda|}|\hat{W}^{r}(s)-\hat{W}^{r}(t+\xi
kr^{-1})|_{2}+\frac{B_{\hat{h}}}{|\lambda|}|\hat{W}^{r}(\zeta^{r}_{2l+1})-\hat{W}^{r}(t+\xi
kr^{-1})|_{2}$ $\displaystyle\leq$ $\displaystyle
B_{\tilde{d}}J^{\frac{1}{2}}\tilde{c}_{2}r^{\kappa-1}+J^{\frac{1}{2}}B_{\tilde{d}}r^{-1}-r\left(s-\zeta^{r}_{2l+1}\right)\frac{|\tilde{\lambda}|}{2|\lambda|}+\frac{\epsilon}{|\lambda|}r^{-\frac{1}{4}}+2B_{\hat{h}}I^{\frac{1}{2}}\frac{\epsilon}{|\lambda|}r^{-\frac{1}{4}}$
$\displaystyle\leq$ $\displaystyle\epsilon
r^{-\frac{1}{8}}-r\left(s-\zeta^{r,\epsilon}_{2l+1}\right)\frac{|\tilde{\lambda}|}{2|\lambda|}$

where we have used (45), (49), Summary 1 part (m), and the fact that by the
definition of $\zeta^{r}_{2l+1}$ and Proposition 7.7 we have
$\tilde{d}\left(\hat{Q}^{r}(\zeta^{r}_{2l+1})\right)\leq
B_{\tilde{d}}J^{\frac{1}{2}}\tilde{c}_{2}r^{\kappa-1}+J^{\frac{1}{2}}B_{\tilde{d}}r^{-1}$.
This, combined with the fact that for $r\geq R$, any $l\geq 0$, and all
$s\in[\zeta^{r}_{2l},\zeta^{r}_{2l+1})$ we have
$\tilde{d}\left(\hat{Q}^{r}(s)\right)\leq
B_{\tilde{d}}J^{\frac{1}{2}}\tilde{c}_{2}r^{\kappa-1}\leq\epsilon
r^{-\frac{1}{8}}$ due to the definition of $\zeta^{r}_{2l+1}$ and Summary 1
part (m), implies that on the set $\left(\mathcal{B}^{r,k,t}\right)^{c}$ for
$r\geq R$ and all $s\in[\zeta^{r}_{0},t+kr^{-1}]$ we have
$\tilde{d}\left(\hat{Q}^{r}(s)\right)\leq\epsilon r^{-\frac{1}{8}}$.
Therefore, from (50), on the set $\left(\mathcal{B}^{r,k,t}\right)^{c}$ for
$r\geq R$ we have

$\tilde{d}\left(\hat{Q}^{r}(s)\right)\leq\epsilon r^{-\frac{1}{8}}\mbox{ for
all }s\in[t+(\xi k+4\xi k|\lambda||\tilde{\lambda}|^{-1})r^{-1},t+kr^{-1}].$
(51)

Note that for $r\geq R$ we have

$\displaystyle
E_{y^{r}}\int_{t}^{t+kr^{-1}}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds$
$\displaystyle\leq E_{y^{r}}\int_{t}^{t+\left(\xi k+\frac{4\xi
k|\lambda|}{|\tilde{\lambda}|}\right)r^{-1}}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds$
$\displaystyle+E_{y^{r}}\int_{t+\left(\xi k+\frac{4\xi
k|\lambda|}{|\tilde{\lambda}|}\right)r^{-1}}^{t+kr^{-1}}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds.$
(52)

and

$\displaystyle E_{y^{r}}\int_{t}^{t+\left(\xi k+\frac{4\xi
k|\lambda|}{|\tilde{\lambda}|}\right)r^{-1}}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds$
$\displaystyle\leq\int_{t}^{t+\left(\xi k+\frac{4\xi
k|\lambda|}{|\tilde{\lambda}|}\right)r^{-1}}B_{2}E_{y^{r}}\left[e^{c|\hat{W}^{r}(s)|}\right]ds$
(53) $\displaystyle\leq r^{-1}\left(\xi k+\frac{4\xi
k|\lambda|}{|\tilde{\lambda}|}\right)B_{2}B_{1}\leq r^{-1}k\epsilon$ (54)

due to our choice of $\xi$ and Summary 1 parts (a) and (b).

In addition, due to Summary 1 parts (a), (b), (c), and our choice of $k$, for
all $r\geq R$, $t\geq 0$, and $s\geq 0$ we have

$\displaystyle
E_{y^{r}}\left[\mathcal{I}_{\mathcal{B}^{r,k,t}}\tilde{d}\left(\hat{Q}^{r}(s)\right)\right]$
$\displaystyle\leq$ $\displaystyle
E_{y^{r}}\left[\mathcal{I}_{\mathcal{B}^{r,k,t}}B_{2}e^{\frac{c}{2}|\hat{W}^{r}(s)|}\right]$
$\displaystyle\leq$ $\displaystyle
P_{y^{r}}(\mathcal{B}^{r,k,t})^{\frac{1}{2}}B_{2}E_{y^{r}}\left[e^{c|\hat{W}^{r}(s)|}\right]^{\frac{1}{2}}\leq
B^{\frac{1}{2}}_{3}e^{-k\frac{B_{4}}{2}}B_{2}B^{\frac{1}{2}}_{1}\leq\epsilon.$

So for $r\geq R$ we have

$\displaystyle E_{y^{r}}\left[\int_{t+\left(\xi k+\frac{4\xi
k|\lambda|}{|\tilde{\lambda}|}\right)r^{-1}}^{t+kr^{-1}}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds\right]$
$\displaystyle\leq
E_{y^{r}}\left[\mathcal{I}_{\left(\mathcal{B}^{r,k,t}\right)^{c}}\int_{t+\left(\xi
k+\frac{4\xi
k|\lambda|}{|\tilde{\lambda}|}\right)r^{-1}}^{t+kr^{-1}}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds\right]$
$\displaystyle\quad+E_{y^{r}}\left[\mathcal{I}_{\mathcal{B}^{r,k,t}}\int_{t+\left(\xi
k+\frac{4\xi
k|\lambda|}{|\tilde{\lambda}|}\right)r^{-1}}^{t+kr^{-1}}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds\right]$
$\displaystyle\leq r^{-1}\left((1-\xi)k-\frac{4\xi
k|\lambda|}{|\tilde{\lambda}|}\right)\epsilon
r^{-\frac{1}{8}}+r^{-1}\left((1-\xi)k-\frac{4\xi
k|\lambda|}{|\tilde{\lambda}|}\right)\epsilon\leq r^{-1}k2\epsilon.$ (55)

Combining (52), (54), and (55), we have that for $r\geq R$ and any $t\geq 0$

$E_{y^{r}}\left[\int_{t}^{t+kr^{-1}}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds\right]\leq
r^{-1}3k\epsilon.$

We can now complete the proof. Consider first the discounted case. For $r\geq
R$ we have

$E_{y^{r}}\left[\int_{0}^{\infty}e^{-s\varsigma}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds\right]\leq\sum_{l=0}^{\infty}e^{-l\varsigma\left(kr^{-1}\right)}E_{y^{r}}\left[\int_{lkr^{-1}}^{(l+1)kr^{-1}}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds\right]\\\
\leq\sum_{l=0}^{\infty}e^{-l\varsigma\left(kr^{-1}\right)}r^{-1}3k\epsilon\leq\frac{1}{1-e^{-\varsigma\left(kr^{-1}\right)}}r^{-1}3k\epsilon\\\
\leq\frac{1}{e^{-\varsigma\left(kr^{-1}\right)}\varsigma\left(kr^{-1}\right)}3r^{-1}k\epsilon\leq\frac{3\epsilon
e^{\varsigma\left(kr^{-1}\right)}}{\varsigma}\leq\frac{3\epsilon
e^{\varsigma\epsilon r^{-\frac{7}{8}}}}{\varsigma}\leq\frac{3\epsilon
e^{\varsigma\epsilon}}{\varsigma}$

due to Summary 1 part (d). Taking $\epsilon=\frac{\varsigma
e^{-\varsigma}}{3}\epsilon_{0}\wedge 1$ proves the first statement in the
proposition.

Consider now the ergodic cost. Let $T^{*}=kR^{-1}$. For any $t\geq 0$, $T\geq
T^{*}$, and $r\geq R$ we have

$E_{y^{r}}\left[\frac{1}{T}\int_{t}^{t+T}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds\right]\leq\frac{1}{T}\sum_{l=1}^{\lceil\frac{T}{kr^{-1}}\rceil}E_{y^{r}}\left[\int_{t+(l-1)kr^{-1}}^{t+lkr^{-1}}\tilde{d}\left(\hat{Q}^{r}(s)\right)ds\right]\\\
\leq\frac{1}{T}\sum_{l=1}^{\lceil\frac{T}{kr^{-1}}\rceil}r^{-1}3k\epsilon\leq\frac{1}{T}\left(\frac{T}{kr^{-1}}+1\right)r^{-1}3k\epsilon\leq
3\epsilon+\frac{3kR^{-1}\epsilon}{T^{*}}\leq 6\epsilon.$

Taking $\epsilon=\epsilon_{0}/6$ completes the proof of the second statement
in the proposition. ∎

##  8 Proofs of Some Stability Results

In this section we prove the key stability results, namely Proposition 7.1,
Lemma 7.2, and Proposition 7.3, that were used in the proof of Proposition 3.4
in Section 7.1.

###  8.1 Proof of Proposition 7.1

We begin with the following auxiliary result which is proved in Section 9.3.

######  Proposition 8.1.

For any $c>0$ and $\epsilon>0$ there exist constants $B,R\in(0,\infty)$ such
that for all $T\geq 1$, $j\in\mathbb{A}_{J}$, and $r\geq R$ we have

$P\left(\sum_{l=1}^{\lceil r^{2}T\rceil}\mathcal{I}_{\\{v^{r}_{j}(l)\geq
rc\\}}v^{r}_{j}(l)\geq\epsilon T\right)\leq e^{-BT}$ (56)

and

$P\left(\sum_{l=1}^{\lceil r^{2}T\rceil}\mathcal{I}_{\\{u^{r}_{j}(l)\geq
rc\\}}u^{r}_{j}(l)\geq\epsilon T\right)\leq e^{-BT}.$ (57)

We now complete the proof of Proposition 7.1.

Proof of Proposition 7.1. Recall $\rho^{*}$ introduced below Proposition 2.3
and $\theta$ from Condition 2. Fix $i\in\mathbb{A}_{I}$ and
$y^{r}=(\hat{q}^{r},\hat{\Upsilon}^{r},\tilde{\mathcal{E}}^{r})\in\mathcal{Y}^{r}$.
Let $\tilde{\upsilon}>0$, $\xi\geq 0$ be given. Let
$\epsilon_{1}=\frac{\frac{|\theta_{i}|}{12}}{JC_{i}\left(1+\frac{4}{\rho^{*}}\right)}$,
$\epsilon_{2}=\frac{|\theta_{i}|}{24JC_{i}}$, and
$\epsilon_{3}=\frac{|\theta_{i}|\min_{j}\\{\beta_{j}\\}}{96J}$. Define the
sets

$\displaystyle\mathcal{A}^{r,T}_{j}$ $\displaystyle=$
$\displaystyle\left\\{\sup_{0\leq t\leq
r^{2}T}|A_{j}^{r}(t)-t\alpha^{r}_{j}|\leq\epsilon_{3}rT\right\\}\cap\left\\{\sup_{0\leq
t\leq
r^{2}\max_{i}\\{C_{i}\\}T}|S_{j}^{r}(t)-t\beta^{r}_{j}|\leq\epsilon_{3}rT\right\\}$
$\displaystyle\cap\left\\{\int_{0}^{r^{2}T}\mathcal{I}_{\\{\mathcal{E}_{j}^{r}(s)=1\\}}ds\leq
r\hat{\Upsilon}^{A,r}_{j}+\epsilon_{2}Tr\right\\}$
$\displaystyle\cap\left\\{\sum_{l=1}^{\lceil
2r^{2}\max_{i}\\{C_{i}\\}\beta^{r}_{j}T\rceil}\mathcal{I}_{\\{v^{r}_{j}(l)>r\frac{\tilde{\upsilon}}{2J+1}\\}}v^{r}_{j}(l)\leq
r\epsilon_{1}T\right\\}$ $\displaystyle\cap\left\\{\sum_{l=1}^{\lceil
2r^{2}\alpha^{r}_{j}T\rceil}\mathcal{I}_{\\{u^{r}_{j}(l)>r\frac{\tilde{\upsilon}}{2J+1}\\}}u^{r}_{j}(l)\leq
r\epsilon_{1}T\right\\}$

and $\mathcal{A}^{r,T}=\cap_{j=1}^{J}\mathcal{A}^{r,T}_{j}$. Due to
Propositions 6.2 (equations (14) and (15) with $c_{1}=1$ and $c_{2}=0$), 8.1 ,
and 3.3 (and since $\kappa<1/4$) we know there exist constants
$B_{1},B_{2},R\in(0,\infty)$ such that for all $r\geq R$,
$y^{r}\in\mathcal{Y}^{r}$, and $T\geq 1$ we have

$P_{y^{r}}\left(\left(\mathcal{A}^{r,T}\right)^{c}\right)\leq
B_{1}e^{-B_{2}T},\;\;\epsilon_{3}T\leq
r\max_{i}\\{C_{i}\\}\beta^{r}_{j}T,\;\;\mbox{ and }\epsilon_{3}T\leq
r\alpha^{r}_{j}T.$ (58)

In addition, we assume $R$ is sufficiently large that for all $r\geq R$ we
have $\frac{3\theta}{4}\geq r(K\rho^{r}-C)$, $\tilde{\upsilon}>Jr^{-1}$,
$\frac{\tilde{\upsilon}}{2J+1}\geq Jr^{\kappa-1}c_{2}$,
$2\min_{j}\\{\beta^{r}_{j}\\}\geq\min_{j}\\{\beta_{j}\\}$,and for all
$j\in\mathbb{A}_{J}$ and $t\geq 0$ we have
$x^{r}_{j}(t)\geq\frac{\rho^{*}}{4}$ (see Proposition 2.6 part (c)). For the
rest of the proof we will restrict ourselves to values of $r$ satisfying
$r\geq R$. Note that if
$\hat{W}^{r}_{i}(s)\geq\frac{\tilde{\upsilon}}{2J+1}\geq Jc_{2}r^{\kappa-1}$
and $\sum_{j=1}^{J}\mathcal{I}_{\\{\tilde{\mathcal{E}}^{r}_{j}(s)=1\\}}=0$
then due to Proposition 2.6 part (d) we have
$(K\mathfrak{b}^{r})_{i}(r^{2}s)=C_{i}$. This implies that if
$\sum_{j=1}^{J}\mathcal{I}_{\\{\tilde{\mathcal{E}}^{r}_{j}(s)=1\\}}=0$,

$\sum_{j=1}^{J}\mathcal{I}_{\\{\hat{\Upsilon}^{S,r}_{j}(s)>\frac{\tilde{\upsilon}}{2J+1}\\}}+\sum_{j=1}^{J}\mathcal{I}_{\\{\hat{\Upsilon}^{A,r}_{j}(s)>\frac{\tilde{\upsilon}}{2J+1}\\}}=0,\mbox{
and
}\hat{W}^{r}_{i}(s)+\sum_{j=1}^{J}\hat{\Upsilon}^{S,r}_{j}(s)+\sum_{j=1}^{J}\hat{\Upsilon}^{A,r}_{j}(s)\geq\tilde{\upsilon}$

then $(K\mathfrak{b}^{r})_{i}(r^{2}s)=C_{i}$. Define

$\hat{t}\doteq\sup\\{t\in[0,\xi]:\hat{W}^{r}_{i}(t)+\sum_{j=1}^{J}\hat{\Upsilon}^{S,r}_{j}(t)+\sum_{j=1}^{J}\hat{\Upsilon}^{A,r}_{j}(t)<\tilde{\upsilon}\\}$

with the convention that if

$\hat{W}^{r}_{i}(t)+\sum_{j=1}^{J}\hat{\Upsilon}^{S,r}_{j}(t)+\sum_{j=1}^{J}\hat{\Upsilon}^{A,r}_{j}(s)\geq\tilde{\upsilon}\mbox{
for all }t\in[0,\xi]\mbox{ then }\hat{t}\doteq 0.$

Define

$T^{*}\doteq
1+2\xi+\frac{4}{|\theta_{i}|}\left(\hat{w}^{r}_{i}+3\tilde{\upsilon}+\left(C_{i}+1\right)\sum_{j=1}^{J}\hat{\Upsilon}^{S,r}_{j}+3C_{i}\sum_{j=1}^{J}\hat{\Upsilon}^{A,r}_{j}\right)$

and let $T>T^{*}$. Then on the set
$\left\\{\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,\xi}>T\right\\}\cap\mathcal{A}^{r,T}$,
for all $s\in[\hat{t},T]$, we have

$\hat{W}^{r}_{i}(s)+\sum_{j=1}^{J}\hat{\Upsilon}^{S,r}_{j}(s)+\sum_{j=1}^{J}\hat{\Upsilon}^{A,r}_{j}(s)\geq\tilde{\upsilon}$

so for any $s\in[\hat{t},T]$ satisfying
$\sum_{j=1}^{J}\mathcal{I}_{\\{\mathcal{E}^{r}_{j}(s)=1\\}}=0,$ and

$\sum_{j=1}^{J}\mathcal{I}_{\\{\hat{\Upsilon}^{S,r}_{j}(s)>\frac{\tilde{\upsilon}}{2J+1}\\}}+\sum_{j=1}^{J}\mathcal{I}_{\\{\hat{\Upsilon}^{A,r}_{j}(s)>\frac{\tilde{\upsilon}}{2J+1}\\}}=0,$

we have $(K\mathfrak{b}^{r})_{i}(s)=C_{i}$. Consequently, on the set
$\left\\{\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,\xi}>T\right\\}\cap\mathcal{A}^{r,T}$,

$\displaystyle\sum_{j=1}^{J}K_{i,j}\left(\bar{B}^{r}_{j}(T)-\bar{B}^{r}_{j}(\hat{t})\right)=\int_{\hat{t}}^{T}\sum_{j=1}^{J}K_{i,j}\mathfrak{b}^{r}_{j}(r^{2}s)ds$
$\displaystyle\geq
C_{i}(T-\hat{t})-C_{i}\sum_{j=1}^{J}\int_{\hat{t}}^{T}\mathcal{I}_{\\{\tilde{\mathcal{E}}^{r}_{j}(s)=1\\}}ds$
$\displaystyle\quad-
C_{i}\sum_{j=1}^{J}\int_{\hat{t}}^{T}\mathcal{I}_{\\{\hat{\Upsilon}^{S,r}_{j}(s)>\frac{\tilde{\upsilon}}{2J+1}\\}}ds-
C_{i}\sum_{j=1}^{J}\int_{\hat{t}}^{T}\mathcal{I}_{\\{\hat{\Upsilon}^{A,r}_{j}(s)>\frac{\tilde{\upsilon}}{2J+1}\\}}ds$
$\displaystyle\geq
C_{i}(T-\hat{t})-C_{i}\sum_{j=1}^{J}\int_{0}^{T}\mathcal{I}_{\\{\tilde{\mathcal{E}}^{r}_{j}(s)=1\\}}ds$
$\displaystyle\quad-
C_{i}\sum_{j=1}^{J}\int_{0}^{T}\mathcal{I}_{\\{\hat{\Upsilon}^{S,r}_{j}(s)>\frac{\tilde{\upsilon}}{2J+1}\\}}ds-
C_{i}\sum_{j=1}^{J}\int_{0}^{T}\mathcal{I}_{\\{\hat{\Upsilon}^{A,r}_{j}(s)>\frac{\tilde{\upsilon}}{2J+1}\\}}ds.$
(59)

For all $j\in\mathbb{A}_{J}$ we have

$\displaystyle\int_{0}^{T}\mathcal{I}_{\left\\{\hat{\Upsilon}^{S,r}_{j}(s)>\frac{\tilde{\upsilon}}{2J+1}r\right\\}}ds$
$\displaystyle\leq
r^{-1}\hat{\Upsilon}^{S,r}_{j}+\sum_{l=1}^{\tau^{r,S}_{j}(T)}\mathcal{I}_{\\{v^{r}_{j}(l)>\frac{\tilde{\upsilon}}{2J+1}r\\}}\frac{1}{r^{2}}\int_{0}^{r^{2}T}\mathcal{I}_{\left\\{\sum_{k=1}^{l-1}v^{r}_{j}(k)\leq
B^{r}_{j}(s)\leq\sum_{k=1}^{l}v^{r}_{j}(k)\right\\}}ds.$

Recall that due to Proposition 2.6 part (c) and our assumption on the size of
$r$ we have $\mathfrak{b}^{r}_{j}(s)\geq\frac{\rho^{*}}{4}$ unless
$\mathcal{E}^{r}_{j}(s)=1$ so for all $l\in\\{1,...,\tau^{r,S}_{j}(T)\\}$ we
have

$\displaystyle\int_{0}^{r^{2}T}\mathcal{I}_{\left\\{\sum_{k=1}^{l-1}v^{r}_{j}(k)\leq
B^{r}_{j}(s)\leq\sum_{k=1}^{l}v^{r}_{j}(k)\right\\}}ds$ $\displaystyle\leq$
$\displaystyle\int_{0}^{r^{2}T}\mathcal{I}_{\\{\mathcal{E}^{r}_{j}(s)=0\\}}\mathcal{I}_{\left\\{\sum_{k=1}^{l-1}v^{r}_{j}(k)\leq
B^{r}_{j}(s)\leq\sum_{k=1}^{l}v^{r}_{j}(k)\right\\}}ds$
$\displaystyle+\int_{0}^{r^{2}T}\mathcal{I}_{\\{\mathcal{E}^{r}_{j}(s)=1\\}}\mathcal{I}_{\left\\{\sum_{k=1}^{l-1}v^{r}_{j}(k)\leq
B^{r}_{j}(s)\leq\sum_{k=1}^{l}v^{r}_{j}(k)\right\\}}ds$ $\displaystyle\leq$
$\displaystyle\frac{4}{\rho^{*}}v^{r}_{j}(l)+\int_{0}^{r^{2}T}\mathcal{I}_{\\{\mathcal{E}^{r}_{j}(s)=1\\}}\mathcal{I}_{\left\\{\sum_{k=1}^{l-1}v^{r}_{j}(k)\leq
B^{r}_{j}(s)\leq\sum_{k=1}^{l}v^{r}_{j}(k)\right\\}}ds.$

Combining these last two inequalities gives

$\displaystyle\int_{0}^{T}\mathcal{I}_{\left\\{\hat{\Upsilon}^{S,r}_{j}(s)>\frac{\tilde{\upsilon}}{2J+1}r\right\\}}ds\leq
r^{-1}\hat{\Upsilon}^{S,r}_{j}+\sum_{l=1}^{\tau^{r,S}_{j}(T)}\mathcal{I}_{\\{v^{r}_{j}(l)>\frac{\tilde{\upsilon}}{2J+1}r\\}}\frac{1}{r^{2}}\frac{4}{\rho^{*}}v^{r}_{j}(l)$
$\displaystyle\quad+\sum_{l=1}^{\tau^{r,S}_{j}(T)}\mathcal{I}_{\\{v^{r}_{j}(l)>\frac{\tilde{\upsilon}}{2J+1}r\\}}\frac{1}{r^{2}}\int_{0}^{r^{2}T}\mathcal{I}_{\\{\mathcal{E}^{r}_{j}(s)=1\\}}\mathcal{I}_{\left\\{\sum_{k=1}^{l-1}v^{r}_{j}(k)\leq
B^{r}_{j}(s)\leq\sum_{k=1}^{l}v^{r}_{j}(k)\right\\}}ds$ $\displaystyle\leq
r^{-1}\hat{\Upsilon}^{S,r}_{j}+\frac{4}{\rho^{*}}\frac{1}{r^{2}}\sum_{l=1}^{\tau^{r,S}_{j}(T)}\mathcal{I}_{\\{v^{r}_{j}(l)>\frac{\tilde{\upsilon}}{2J+1}r\\}}v^{r}_{j}(l)+\int_{0}^{T}\mathcal{I}_{\\{\tilde{\mathcal{E}}^{r}_{j}(s)=1\\}}ds.$

Note that on the set $\mathcal{A}^{r,T}$ we have $\tau^{r,S}_{j}(T)\leq\lceil
2r^{2}\max_{i}\\{C_{i}\\}\beta^{r}_{j}T\rceil$ and

$\frac{1}{r^{2}}\sum_{l=1}^{\lceil
2r^{2}\max_{i}\\{C_{i}\\}\beta^{r}_{j}T\rceil}\mathcal{I}_{\\{v^{r}_{j}(l)>\frac{\tilde{\upsilon}}{2J+1}r\\}}v^{r}_{j}(l)\leq
r^{-1}\epsilon_{1}T$

which gives

$\displaystyle\int_{0}^{T}\mathcal{I}_{\left\\{\hat{\Upsilon}^{S,r}_{j}(s)\geq\frac{\tilde{\upsilon}}{2J+1}r\right\\}}ds=r^{-1}\hat{\Upsilon}^{S,r}_{j}+r^{-1}\frac{4\epsilon_{1}}{\rho^{*}}T+\int_{0}^{T}\mathcal{I}_{\\{\tilde{\mathcal{E}}^{r}_{j}(s)=1\\}}ds.$

Also, on the set $\mathcal{A}^{r,T}$, we have $\tau^{r,A}_{j}(T)\leq\lceil
2r^{2}\alpha^{r}_{j}T\rceil$ and

$\frac{1}{r^{2}}\sum_{l=1}^{\lceil
2r^{2}\alpha^{r}_{j}T\rceil}\mathcal{I}_{\\{u^{r}_{j}(l)>\frac{\tilde{\upsilon}}{2J+1}r\\}}u^{r}_{j}(l)\leq
r^{-1}\epsilon_{1}T$

so

$\displaystyle\int_{0}^{T}\mathcal{I}_{\left\\{\hat{\Upsilon}^{A,r}_{j}(s)>\frac{\tilde{\upsilon}}{2J+1}r\right\\}}ds$
$\displaystyle\leq
r^{-1}\hat{\Upsilon}^{A,r}_{j}+\sum_{l=1}^{\tau^{r,A}_{j}(T)}\mathcal{I}_{\\{u^{r}_{j}(l)>\frac{\tilde{\upsilon}}{2J+1}r\\}}\frac{1}{r^{2}}\int_{0}^{r^{2}T}\mathcal{I}_{\left\\{\sum_{k=1}^{l-1}u^{r}_{j}(k)\leq
s\leq\sum_{k=1}^{l}u^{r}_{j}(k)\right\\}}ds$ $\displaystyle\leq
r^{-1}\hat{\Upsilon}^{A,r}_{j}+\frac{1}{r^{2}}\sum_{l=1}^{\tau^{r,A}_{j}(T)}\mathcal{I}_{\\{u^{r}_{j}(l)>\frac{\tilde{\upsilon}}{2J+1}r\\}}u^{r}_{j}(l)\leq
r^{-1}\hat{\Upsilon}^{A,r}_{j}+r^{-1}\epsilon_{1}T.$

Thus from (59), on
$\left\\{\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,\xi}>T\right\\}\cap\mathcal{A}^{r,T}$,

$\displaystyle\sum_{j=1}^{J}K_{i,j}\left(\bar{B}^{r}_{j}(T)-\bar{B}^{r}_{j}(\hat{t})\right)$
$\displaystyle\geq
C_{i}(T-\hat{t})-2C_{i}\sum_{j=1}^{J}\int_{0}^{T}\mathcal{I}_{\\{\mathcal{E}^{r}_{j}(r^{2}s)=1\\}}ds-r^{-1}JC_{i}\epsilon_{1}T\left(1+\frac{4}{\rho^{*}}\right)$
$\displaystyle\quad-r^{-1}C_{i}\sum_{j=1}^{J}\hat{\Upsilon}^{S,r}_{j}-r^{-1}C_{i}\sum_{j=1}^{J}\hat{\Upsilon}^{A,r}_{j}$
$\displaystyle\geq
C_{i}(T-\hat{t})-r^{-1}C_{i}J2\epsilon_{2}T-r^{-1}JC_{i}\epsilon_{1}T\left(1+\frac{4}{\rho^{*}}\right)$
$\displaystyle\quad-r^{-1}C_{i}\sum_{j=1}^{J}\hat{\Upsilon}^{S,r}_{j}-3r^{-1}C_{i}\sum_{j=1}^{J}\hat{\Upsilon}^{A,r}_{j}.$
(60)

Consequently
on$\left\\{\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,\xi}>T\right\\}\cap\mathcal{A}^{r,T}$
we have

$\displaystyle\hat{W}^{r}_{i}(T)$
$\displaystyle\leq\hat{W}^{r}_{i}(\hat{t})+Jr^{-1}+\sum_{j=1}^{J}K_{i,j}\frac{1}{\beta^{r}_{j}}\left(\hat{A}^{r}_{j}\left(\left(T-\bar{\Upsilon}^{A,r}_{j}\right)^{+}\right)-\hat{A}^{r}_{j}\left(\left(\hat{t}-\bar{\Upsilon}^{A,r}_{j}\right)^{+}\right)\right)$
$\displaystyle\quad-\sum_{j=1}^{J}K_{i,j}\frac{1}{\beta^{r}_{j}}\left(\hat{S}^{r}_{j}\left(\left(\bar{B}^{r}_{j}(T)-\bar{\Upsilon}^{S,r}_{j}\right)^{+}\right)-\hat{S}^{r}_{j}\left(\left(\bar{B}^{r}_{j}(\hat{t})-\bar{\Upsilon}^{S,r}_{j}\right)^{+}\right)\right)+\sum_{j=1}^{J}K_{i,j}\hat{\Upsilon}^{S,r}$
$\displaystyle\quad+(T-\hat{t})r((K\rho^{r})_{i}-C_{i})+r\left(C_{i}(T-\hat{t})-\sum_{j=1}^{J}K_{i,j}(\bar{B}^{r}(T)-\bar{B}^{r}(\hat{t}))\right)$
$\displaystyle\leq\hat{W}^{r}_{i}(\hat{t})+\tilde{\upsilon}+\frac{4J}{\min_{j}\\{\beta^{r}_{j}\\}}\epsilon_{3}T+\frac{3\theta_{i}}{4}(T-\hat{t})+C_{i}J2\epsilon_{2}T$
$\displaystyle\quad+JC_{i}\epsilon_{1}T\left(1+\frac{4}{\rho^{*}}\right)+(C_{i}+1)\sum_{j=1}^{J}\hat{\Upsilon}^{S,r}_{j}+3C_{i}\sum_{j=1}^{J}\hat{\Upsilon}^{A,r}_{j}.$

Note that
$\hat{W}^{r}_{i}(\hat{t})\leq\max\\{\hat{w}^{r}_{i},\tilde{\upsilon}+r^{-1}J\\}\leq\hat{w}^{r}_{i}+2\tilde{\upsilon}$
which, combined with our choices of $\epsilon_{i}$, gives

$\displaystyle\hat{W}^{r}_{i}(T)$ $\displaystyle\leq$
$\displaystyle\hat{w}^{r}_{i}+3\tilde{\upsilon}-\frac{\theta_{i}}{12}T+\frac{3\theta_{i}}{4}(T-\hat{t})-\frac{\theta_{i}}{12}T-\frac{\theta_{i}}{12}T+(C_{i}+1)\sum_{j=1}^{J}\hat{\Upsilon}^{S,r}_{j}+3C_{i}\sum_{j=1}^{J}\hat{\Upsilon}^{A,r}_{j}$
$\displaystyle\leq$
$\displaystyle\hat{w}^{r}_{i}+3\tilde{\upsilon}+\frac{3\theta_{i}}{4}(T-\xi)-\frac{\theta_{i}}{4}T+(C_{i}+1)\sum_{j=1}^{J}\hat{\Upsilon}^{S,r}_{j}+3C_{i}\sum_{j=1}^{J}\hat{\Upsilon}^{A,r}_{j}$
$\displaystyle\leq$
$\displaystyle\hat{w}^{r}_{i}+3\tilde{\upsilon}+(C_{i}+1)\sum_{j=1}^{J}\hat{\Upsilon}^{S,r}_{j}+3C_{i}\sum_{j=1}^{J}\hat{\Upsilon}^{A,r}_{j}+\frac{\theta_{i}}{4}(T-\xi),$

where in the last inequality we have used $T<2(T-\xi)$ which follows from our
choice of $T$. Since
$T>\xi+\frac{4}{|\theta_{i}|}\left(\hat{w}^{r}_{i}+3\tilde{\upsilon}+\left(C_{i}+1\right)\sum_{j=1}^{J}\hat{\Upsilon}^{S}_{j}+3C_{i}\sum_{j=1}^{J}\hat{\Upsilon}^{A}_{j}\right)$
and $\theta_{i}<0$ this implies

$\displaystyle\hat{W}^{r}_{i}(T)$ $\displaystyle<$
$\displaystyle\hat{w}^{r}_{i}+3\tilde{\upsilon}+\left(C_{i}+1\right)\sum_{j=1}^{J}\hat{\Upsilon}^{S}_{j}+3C_{i}\sum_{j=1}^{J}\hat{\Upsilon}^{A}_{j}$
$\displaystyle+\frac{1}{4}\theta_{i}\left(-\frac{4}{\theta_{i}}\left(w^{r}_{i}+3\tilde{\upsilon}+\left(C_{i}+1\right)\sum_{j=1}^{J}\hat{\Upsilon}^{S}_{j}+3C_{i}\sum_{j=1}^{J}\hat{\Upsilon}^{A}_{j}\right)\right)=0$

which contradicts the fact that $\hat{W}^{r}_{i}(t)\geq 0$ for all $t\geq 0$.
Therefore for all $T>T^{*}$ we have
$\left\\{\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,\xi}>T\right\\}\cap\mathcal{A}^{r,T}=\emptyset$
which says that
$\left\\{\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,\xi}>T\right\\}\subset\left(\mathcal{A}^{r,T}\right)^{c}$
and consequently from (58)
$P_{y^{r}}\left(\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,\xi}>T\right)\leq
P_{y^{r}}\left(\left(\mathcal{A}^{r,T}\right)^{c}\right)\leq
B_{1}e^{-B_{2}T}$. Finally, for any $c\in(0,\frac{1}{2}B_{2}]$ and $r\geq R$
we have

$\displaystyle E_{y^{r}}[e^{c\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,\xi}}]$
$\displaystyle\leq$ $\displaystyle
e^{cT^{*}}+\int_{\left(e^{cT^{*}},\infty\right)}P_{y^{r}}\left(\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,\xi}>\frac{\ln(x)}{c}\right)dx$
$\displaystyle\leq$ $\displaystyle
e^{cT^{*}}+\int_{\left(e^{cT^{*}},\infty\right)}B_{1}e^{-\frac{B_{2}\ln(x)}{c}}dx$
$\displaystyle\leq$ $\displaystyle e^{\frac{1}{2}B_{2}T^{*}}+B_{1}.$

The result follows on recalling the definition of $T^{*}$. ∎

###  8.2 Proof of Lemma 7.2

Since $\hat{Y}^{r}(\cdot)$ is a $\mathcal{G}^{r}(t)$ Markov process, we have,
for $0\leq s<t$,

$\displaystyle
E_{y}\left[V^{r,\tilde{\upsilon}}_{i}\left(\hat{Y}^{r}(t)\right)\right]$
$\displaystyle=$ $\displaystyle
E_{y}\left[e^{\tilde{\delta}(\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,t}-t)}\right]=E_{y}\left[E_{y}\left[e^{\tilde{\delta}(\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,t}-t)}|\mathcal{G}^{r}(s)\right]\right]$
$\displaystyle=$ $\displaystyle
E_{y}\left[E_{\hat{Y}^{r}(s)}\left[e^{\tilde{\delta}(\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,t-s}-(t-s))}\right]\right]=E_{y}\left[E_{\hat{Y}^{r}(s)}\left[V^{r,\tilde{\upsilon}}_{i}\left(\hat{Y}^{r}(t-s)\right)\right]\right].$

######  Proposition 8.2.

For $\tilde{\upsilon}>0$ let $\tilde{\delta}$ be as in Proposition 7.1. There
exist constants $\tilde{R},\tilde{B}\in(0,\infty)$ such that for any
$y=(\hat{q},\hat{\Upsilon},\tilde{\mathcal{E}})\in\mathcal{Y}^{r}$,
$i\in\mathbb{A}_{I}$, $s\in[0,1]$, and $r\geq\tilde{R}$ we have

$E_{y}\left[V^{r,\tilde{\upsilon}}_{i}\left(\hat{Y}^{r}(s)\right)\right]\leq
e^{-\tilde{\delta}s}V^{r,\tilde{\upsilon}}_{i}\left(y\right)+\tilde{B}.$

###### Proof.

Let $\tilde{\upsilon}>0$ be fixed, let $B_{1},B_{2},B_{3},R$ be as in Theorem
7.1, and let $y\in\mathcal{Y}^{r}$, $i\in\mathbb{A}_{I}$, and $s\in[0,1]$ be
arbitrary. Due to the fact that $\hat{Y}^{r}(t)$ is a $\mathcal{G}^{r}(t)$
strong Markov process and $\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,0}\wedge s$
is a bounded $\mathcal{G}^{r}(t)$ stopping time, for all $r\geq R$ we have,

$\displaystyle
E_{y}\left[V^{r,\tilde{\upsilon}}_{i}\left(\hat{Y}^{r}(s)\right)\right]$
$\displaystyle\leq$ $\displaystyle
E_{y}\left[e^{\tilde{\delta}(\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,s}-s)}\right]$
$\displaystyle=$ $\displaystyle
E_{y}\left[\mathcal{I}_{\\{\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,0}\geq
s\\}}e^{\tilde{\delta}(\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,s}-s)}\right]+E_{y}\left[\mathcal{I}_{\\{\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,0}<s\\}}e^{\tilde{\delta}(\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,s}-s)}\right]$
$\displaystyle=$ $\displaystyle
E_{y}\left[e^{\tilde{\delta}(\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,0}-s)}\right]+E_{y}\left[E_{y}\left[\mathcal{I}_{\\{\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,0}<s\\}}e^{\tilde{\delta}(\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,s}-s)}|\mathcal{G}^{r}(\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,0}\wedge
s)\right]\right]$ $\displaystyle=$ $\displaystyle
e^{-\tilde{\delta}s}E_{y}\left[e^{\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,0}}\right]+\sup_{(y,z):w_{i}+|\hat{\Upsilon}|\leq\tilde{\upsilon},z\in[0,s]}\left\\{E_{y}\left[e^{\tilde{\delta}\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,z}}\right]\right\\}$
$\displaystyle\leq$ $\displaystyle
e^{-\tilde{\delta}s}V^{r,\tilde{\upsilon}}_{i}(y)+B_{1}e^{B_{2}\left(1+2\tilde{\upsilon}\right)}+B_{3}$

where the last line uses Proposition 7.1. This completes the proof. ∎

We now complete the proof of Lemma 7.2.

Proof of Lemma 7.2. Let $\tilde{\upsilon}>0$ be fixed and note that, with
$\tilde{\delta}$ as in Proposition 7.1, due to Proposition 8.2, there exist
constants $B_{1},R\in(0,\infty)$ such that for all $y\in\mathcal{Y}^{r}$,
$r\geq R$, $i\in\mathbb{A}_{I}$, and $t\in[0,1]$ we have

$E_{y}\left[V^{r,\tilde{\upsilon}}_{i}\left(\hat{Y}^{r}(t)\right)\right]\leq
e^{-\tilde{\delta}t}V^{r,\tilde{\upsilon}}_{i}\left(y\right)+B_{1}.$

Consequently for $T\in[0,1]$ we already have the result (with $B=B_{1}$). Let
$T>1$ be arbitrary and note that there exists $t\in\left[\frac{1}{2},1\right]$
and $n\in\mathbb{N}$ such that $T=nt$. Using the Markov property we have

$\displaystyle
E_{y}\left[V^{r,\tilde{\upsilon}}_{i}\left(\hat{Y}^{r}(T)\right)\right]$
$\displaystyle=E_{y}\left[V^{r,\tilde{\upsilon}}_{i}\left(\hat{Y}^{r}(nt)\right)\right]=E_{y}\left[E\left[V^{r,\tilde{\upsilon}}_{i}\left(\hat{Y}^{r}(nt)\right)|\mathcal{G}^{r}\left((n-1)t\right)\right]\right]$
$\displaystyle=E_{y}\left[E_{\hat{Y}^{r}((n-1)t)}\left[V^{r,\tilde{\upsilon}}_{i}\left(\hat{Y}^{r}(t)\right)\right]\right]\leq
e^{-\tilde{\delta}t}E_{y}\left[V^{r,\tilde{\upsilon}}_{i}\left(\hat{Y}^{r}((n-1)t)\right)\right]+B_{1}.$

Now a standard recursive argument shows that

$\displaystyle
E_{y}\left[V^{r,\tilde{\upsilon}}_{i}\left(\hat{Y}^{r}(T)\right)\right]$
$\displaystyle\leq$ $\displaystyle
e^{-\tilde{\delta}nt}E_{y}\left[V^{r,\tilde{\upsilon}}_{i}\left(\hat{Y}^{r}(0)\right)\right]+B_{1}\sum_{l=0}^{n-1}e^{-\tilde{\delta}lt}$
$\displaystyle\leq$ $\displaystyle
e^{-\tilde{\delta}T}V^{r,\tilde{\upsilon}}_{i}\left(y\right)+B_{1}\sum_{l=0}^{\infty}e^{-\tilde{\delta}l\frac{1}{2}}\leq
e^{-\tilde{\delta}T}V^{r,\tilde{\upsilon}}_{i}\left(y\right)+\frac{B_{1}}{1-e^{-\frac{1}{2}\tilde{\delta}}}.$

This completes the proof on taking
$B=B_{1}/(1-e^{-\frac{1}{2}\tilde{\delta}})$. ∎

###  8.3 Proof of Proposition 7.3

Let $\tilde{\upsilon}>0$ be fixed and let
$y=(\hat{q},\hat{\Upsilon},\tilde{\mathcal{E}})\in\mathcal{Y}^{r}$ and
$i\in\mathbb{A}_{I}$ be arbitrary. For $T\geq 1$ define

$\displaystyle\mathcal{A}^{r,T}_{j}$ $\displaystyle=$
$\displaystyle\left\\{\sup_{0\leq t\leq
r^{2}T}|A_{j}^{r}(t)-t\alpha^{r}_{j}|\leq\frac{|\theta_{i}|}{4J}rT\right\\}\cap\left\\{\sup_{0\leq
t\leq
r^{2}\max_{i}\\{C_{i}\\}T}|S_{j}^{r}(t)-t\beta^{r}_{j}|\leq\frac{|\theta_{i}|}{4J}rT\right\\}$

and $\mathcal{A}^{r,T}=\cap_{j=1}^{J}\mathcal{A}^{r,T}_{j}$. From Proposition
6.2 (equations (14) and (15) with $c_{1}=1$, $c_{2}=0$) we know there exist
constants $B_{1},B_{2},R\in(0,\infty)$ such that for all $r\geq R$ and $T\geq
1$ we have

$P\left(\mathcal{A}^{r,T}\right)\geq
1-B_{1}e^{-B_{2}T},\;\;r^{-1}\sum_{j=1}^{J}\frac{1}{\beta^{r}_{j}}\leq\tilde{\upsilon},\;\;\mbox{
and,
}r\left(\sum_{j=1}^{J}K_{i,j}\rho^{r}_{j}-C_{i}\right)\geq\frac{3\theta_{i}}{2}.$

Choose $L\in[1,\infty)$ such that $1-B_{1}e^{-B_{2}L}\geq\frac{1}{2}$. If

$\hat{w}_{i}-C_{i}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}\geq
2|\theta_{i}|L+2\tilde{\upsilon},\mbox{ let
}\tilde{T}\doteq\frac{1}{2|\theta_{i}|}\left(\hat{w}_{i}-C_{i}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}-2\tilde{\upsilon}\right)$

and note that $\tilde{T}\geq L$. For all $r\geq R$ and $s\in[0,\tilde{T})$ on
the set $\mathcal{A}^{r,\tilde{T}}$ we have

$\displaystyle\hat{W}^{r}_{i}(s)$ $\displaystyle=$
$\displaystyle\hat{w}_{i}+\sum_{j=1}^{J}K_{i,j}\frac{1}{r\beta^{r}_{j}}A^{r}_{j}\left(r^{2}\left(s-\bar{\Upsilon}_{j}^{A}\right)^{+}\right)+\sum_{j=1}^{J}K_{i,j}\frac{1}{r\beta^{r}_{j}}\mathcal{I}_{\\{s\geq\bar{\Upsilon}_{j}^{A}>0\\}}$
$\displaystyle-\sum_{j=1}^{J}K_{i,j}\frac{1}{r\beta^{r}_{j}}S^{r}_{j}\left(r^{2}\left(\bar{B}^{r}(s)-\bar{\Upsilon}_{j}^{S}\right)^{+}\right)-\sum_{j=1}^{J}K_{i,j}\frac{1}{r\beta^{r}_{j}}\mathcal{I}_{\\{\bar{B}^{r}(s)\geq\bar{\Upsilon}_{j}^{S}>0\\}}$
$\displaystyle\geq$
$\displaystyle\hat{w}_{i}+\sum_{j=1}^{J}K_{i,j}\frac{1}{r\beta^{r}_{j}}A^{r}_{j}\left(r^{2}\left(s-\bar{\Upsilon}^{A}_{j}\right)^{+}\right)-\sum_{j=1}^{J}K_{i,j}\frac{1}{r\beta^{r}_{j}}S^{r}_{j}\left(r^{2}\bar{B}^{r}_{j}(s)\right)-r^{-1}\sum_{j=1}^{J}\frac{1}{\beta^{r}_{j}}$
$\displaystyle\geq$
$\displaystyle\hat{w}_{i}+r\sum_{j=1}^{J}K_{i,j}\rho^{r}_{j}\left(s-\bar{\Upsilon}^{A}_{j}\right)^{+}-r\sum_{j=1}^{J}K_{i,j}\bar{B}^{r}_{j}(s)+\frac{\theta_{i}}{2}\tilde{T}-\tilde{\upsilon}$
$\displaystyle\geq$
$\displaystyle\hat{w}_{i}+r\sum_{j=1}^{J}K_{i,j}\rho^{r}_{j}\left(s-\max_{j}\\{\bar{\Upsilon}^{A}_{j}\\}\right)^{+}-rsC_{i}+\frac{\theta_{i}}{2}\tilde{T}-\tilde{\upsilon}$
$\displaystyle\geq$
$\displaystyle\hat{w}_{i}-C_{i}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}+sr\left(\sum_{j=1}^{J}K_{i,j}\rho^{r}_{j}-C_{i}\right)+\frac{\theta_{i}}{2}\tilde{T}-\tilde{\upsilon}$
$\displaystyle\geq$
$\displaystyle\hat{w}_{i}-C_{i}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}+\frac{3\theta_{i}}{2}s+\frac{\theta_{i}}{2}\tilde{T}-\tilde{\upsilon}>\hat{w}_{i}-C_{i}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}-\tilde{\upsilon}+2\theta_{i}\tilde{T}.$

Since
$\hat{w}_{i}-C_{i}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}-\tilde{\upsilon}+2\theta_{i}\tilde{T}=\tilde{\upsilon}$
this means that for $r\geq R$ and $s\in[0,\tilde{T})$ on the set
$\mathcal{A}^{r,\tilde{T}}$ we have $\hat{W}^{r}_{i}(s)>\tilde{\upsilon}$ and
consequently
$\\{\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,0}\geq\tilde{T}\\}\supset\mathcal{A}^{r,\tilde{T}}$.
Therefore for $r\geq R$ if
$\hat{w}_{i}-C_{i}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}\geq
2|\theta_{i}|L+2\tilde{\upsilon}$ we have

$\displaystyle V^{r,\tilde{\upsilon}}_{i}(y)$
$\displaystyle=E_{y}\left[e^{\tilde{\delta}\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,0}}\right]\geq
P\left(\mathcal{A}^{r,\tilde{T}}\right)e^{\tilde{\delta}\tilde{T}}$
$\displaystyle\geq\left(1-B_{1}e^{-B_{2}L}\right)e^{\tilde{\delta}\tilde{T}}\geq\frac{e^{-\frac{\tilde{\delta}\tilde{\upsilon}}{|\theta_{i}|}}}{2}e^{\frac{\tilde{\delta}}{2|\theta_{i}|}\left(\hat{w}_{i}-C_{i}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}\right)}.$
(61)

If
$\hat{w}_{i}-C_{i}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}<2|\theta_{i}|L+2\tilde{\upsilon}$
by definition we have
$V^{r,\tilde{\upsilon}}_{i}(y)=E_{y}\left[e^{\tilde{\delta}\tilde{\gamma}^{r,\tilde{\upsilon}}_{i,0}}\right]\geq
1$. Let

$\tilde{B}_{1}\doteq\min\left\\{e^{-\frac{\tilde{\delta}}{2|\theta_{i}|}\left(2|\theta_{i}|L+2\tilde{\upsilon}\right)},\frac{e^{-\frac{\tilde{\delta}\tilde{\upsilon}}{|\theta_{i}|}}}{2}\right\\},\mbox{
and, }\tilde{B}_{2}\doteq\frac{\tilde{\delta}}{2|\theta_{i}|}.$

Then if $\hat{w}_{i}-C_{i}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}\leq
2|\theta_{i}|L+2\tilde{\upsilon}$ we have
$\tilde{B}_{1}e^{\tilde{B}_{2}\left(\hat{w}_{i}-C_{i}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}\right)^{+}}\leq
1\leq V^{r,\tilde{\upsilon}}_{i}(y)$ and if
$\hat{w}_{i}-C_{i}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}\geq
2|\theta_{i}|L+2\tilde{\upsilon}$ we have from (61)

$\tilde{B}_{1}e^{\tilde{B}_{2}\left(\hat{w}_{i}-C_{i}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}\right)^{+}}\leq\frac{e^{-\frac{\tilde{\delta}\tilde{\upsilon}}{|\theta_{i}|}}}{2}e^{\frac{\tilde{\delta}}{2|\theta_{i}|}\left(\hat{w}_{i}-C_{i}\max_{j}\\{\hat{\Upsilon}^{A}_{j}\\}\right)}\leq
V^{r,\tilde{\upsilon}}_{i}(y).$

The result follows. ∎

##  9 Proofs of Some Exponential Estimates

The following lemma gives a key independence property.

######  Lemma 9.1.

Let $t\in\mathbb{R}_{+}$ and $r\in\mathbb{N}$. Then
$(\hat{A}^{r,t}(\cdot),\hat{S}^{r,t}(\cdot))$ is independent of
$\mathcal{G}^{r}(t)$ and $(\hat{A}^{r,t}(\cdot),\hat{S}^{r,t}(\cdot))$ has the
same distribution as $(\hat{A}^{r}(\cdot),\hat{S}^{r}(\cdot))$.

###### Proof.

For $a,b\in\mathbb{N}^{J}$ define

$\tilde{u}^{r}_{a}(b)=\left(u^{r}_{1}(a_{1}+1)...,u^{r}_{1}(a_{1}+b_{1}),u^{r}_{2}(a_{2}+1)...,u^{r}_{2}(a_{2}+b_{2})...,u^{r}_{J}(a_{J}+1)...,u^{r}_{J}(a_{J}+b_{J})\right)$

and

$\tilde{v}^{r}_{a}(b)=\left(v^{r}_{1}(a_{1}+1)...,v^{r}_{1}(a_{1}+b_{1}),v^{r}_{2}(a_{2}+1)...,v^{r}_{2}(a_{2}+b_{2}),...,v^{r}_{J}(a_{J}+1)...,v^{r}_{J}(a_{J}+b_{J})\right).$

It suffices to show that for all $(\tilde{n},\tilde{m})\geq(0,0)$
$(\tilde{u}^{r}_{\tau^{r,A}(t)}(\tilde{n}),\tilde{v}^{r}_{\tau^{r,S}(t)}(\tilde{m}))$
is independent of $\mathcal{G}^{r}(t)=\mathcal{F}^{r}(\tau^{r}(t))$ and
$(\tilde{u}^{r}_{\tau^{r,A}(t)}(\tilde{n}),\tilde{v}^{r}_{\tau^{r,S}(t)}(\tilde{m}))$
has the same distribution as
$(\tilde{u}^{r}_{0}(\tilde{n}),\tilde{v}^{r}_{0}(\tilde{m}))$. Let
$(n,m)\geq(0,0)$,
$f:\mathbb{R}_{+}^{|\tilde{n}|_{1}+|\tilde{m}|_{1}}\rightarrow\mathbb{R}$, and
$G\in\mathcal{F}^{r}(\tau^{r}(t))$ be arbitrary. Then

$\displaystyle
E\left[\mathcal{I}_{G}f\left(\tilde{u}^{r}_{\tau^{r,A}(t)}(\tilde{n}),\tilde{v}^{r}_{\tau^{r,S}(t)}(\tilde{m})\right)\right]$
$\displaystyle=\sum_{(n,m)\geq(0,0)}E\left[\mathcal{I}_{\\{\tau^{r}(t)=(n,m)\\}}\mathcal{I}_{G}f\left(\tilde{u}^{r}_{\tau^{r,A}(t)}(\tilde{n}),\tilde{v}^{r}_{\tau^{r,S}(t)}(\tilde{m})\right)\right]$
$\displaystyle=\sum_{(n,m)\geq(0,0)}E\left[\mathcal{I}_{G\cap\\{\tau^{r}(t)=(n,m)\\}}f\left(\tilde{u}^{r}_{n}(\tilde{n}),\tilde{v}^{r}_{m}(\tilde{m})\right)\right]$
$\displaystyle=\sum_{(n,m)\geq(0,0)}E\left[\mathcal{I}_{G\cap\\{\tau^{r}(t)=(n,m)\\}}\right]E\left[f\left(\tilde{u}^{r}_{n}(\tilde{n}),\tilde{v}^{r}_{m}(\tilde{m})\right)\right]$
$\displaystyle=E\left[f\left(\tilde{u}^{r}_{0}(\tilde{n}),\tilde{v}^{r}_{0}(\tilde{m})\right)\right]\sum_{(n,m)\geq(0,0)}E\left[\mathcal{I}_{G\cap\\{\tau^{r}(t)=(n,m)\\}}\right]=P\left(G\right)\left[f\left(\tilde{u}^{r}_{0}(\tilde{n}),\tilde{v}^{r}_{0}(\tilde{m})\right)\right]$

where the third equality is from the fact that $G\cap\\{\tau^{r}(t)=(n,m)\\}$
is $\mathcal{F}^{r}(n,m)$-measurable and
$(\tilde{u}^{r}_{n}(\tilde{n}),\tilde{v}^{r}_{m}(\tilde{m}))$ is independent
of $\mathcal{F}^{r}(n,m)$ and the fourth equality uses the fact that
$(\tilde{u}^{r}_{n}(\tilde{n}),\tilde{v}^{r}_{m}(\tilde{m}))$ has the same
distribution as $(\tilde{u}^{r}_{0}(\tilde{n}),\tilde{v}^{r}_{0}(\tilde{m}))$.
The result follows. ∎

###  9.1 Proof of Proposition 7.5

Fix $j\in\mathbb{A}_{J}$. Define

$\mathcal{F}^{r,S}_{j}(k)\doteq\sigma\left\\{u^{r}_{l}(m^{u}_{l}),v^{r}_{l^{\prime}}(m^{v}_{l^{\prime}}):m^{u}_{l}\in\mathbb{N},l\in\mathbb{A}_{J},m^{v}_{l^{\prime}}\in\mathbb{N},l^{\prime}\in\mathbb{A}_{J}\setminus\left\\{j\right\\},\text{
and }m^{v}_{j}\leq k\right\\}$

which is the filtration that contains the information from all inter-arrival
times, all service times from queues other than the $j$-th queue, and the
first $k$ service times from queue $j$. Note that $\tau^{r,S}_{j}(t)$ is a
$\mathcal{F}^{r,S}_{j}(k)$ stopping time. For all $n\geq 1$ define
$\tilde{L}^{r}_{j}(n)=\sup\left\\{s\geq
0:B^{r}_{j}(s)<\sum_{l=1}^{n}v^{r}_{j}(l)\right\\}$ and since
$B^{r}_{j}(\cdot)$ is continuous we have
$B^{r}_{j}\left(\tilde{L}^{r}_{j}(n)\right)=\sum_{l=1}^{n}v^{r}_{j}(l)$.

Define

$o^{r}_{j}(n)=\inf\left\\{s\geq\tilde{L}^{r}_{j}(n):\mathcal{E}^{r}_{j}(s)=0\right\\}$

and note that due to property (a) of Proposition 2.6 for all
$s\in[\tilde{L}^{r}_{j}(n),o^{r}_{j}(n)]$ we have
$B^{r}_{j}\left(s\right)=\sum_{l=1}^{n}v^{r}_{j}(l)$. Define
$\tilde{a}=\frac{\rho^{*}}{4}$ and note that due to properties (b) and (c) of
Proposition 2.6 for all $r$ sufficiently large we have
$\mathfrak{b}^{r}_{j}(s)\geq\tilde{a}>0$ for all
$s\in[o^{r}_{j}(n),\tilde{L}^{r}_{j}(n+1)]$.

Let $n\geq 2$ be arbitrary and assume that
$\tau^{r,S}_{j}\left(\left(t-\frac{v^{r}_{j}(n)}{\tilde{a}r^{2}}\right)^{+}\right)>n-1$
and $\tau^{r,S}_{j}(t)=n$. Note that in this case we must have
$t>\frac{v^{r}_{j}(n)}{\tilde{a}r^{2}}$. Since
$\tau^{r,S}_{j}\left(t-\frac{v^{r}_{j}(n)}{\tilde{a}r^{2}}\right)>n-1$ implies
$B^{r}_{j}\left(r^{2}\left(t-\frac{v^{r}_{j}(n)}{\tilde{a}r^{2}}\right)\right)>\sum_{l=1}^{n-1}v^{r}_{j}(l)$
we have
$o^{r}_{j}(n-1)<r^{2}\left(t-\frac{v^{r}_{j}(n)}{\tilde{a}r^{2}}\right)$
because $o^{r}_{j}(n-1)\geq
r^{2}\left(t-\frac{v^{r}_{j}(n)}{\tilde{a}r^{2}}\right)$ implies
$B^{r}_{j}\left(r^{2}\left(t-\frac{v^{r}_{j}(n)}{\tilde{a}r^{2}}\right)\right)=\sum_{l=1}^{n-1}v^{r}_{j}(l)$.
In addition, $\tau^{r,S}_{j}(t)=n$ implies $\tilde{L}^{r}_{j}(n)\geq tr^{2}$
so
$[r^{2}\left(t-\frac{v^{r}_{j}(n)}{r^{2}\tilde{a}}\right),r^{2}t]\subset[o^{r}_{j}(n-1),\tilde{L}^{r}_{j}(n)]$
and consequently $\mathfrak{b}^{r}_{j}(s)\geq\tilde{a}$ for all
$s\in[r^{2}\left(t-\frac{v^{r}_{j}(n)}{r^{2}\tilde{a}}\right),r^{2}t]$ and
thus in particular
$B^{r}_{j}(r^{2}t)-B^{r}_{j}\left(r^{2}\left(t-\frac{v^{r}_{j}(n)}{r^{2}\tilde{a}}\right)\right)\geq
v^{r}_{j}(n)$. Therefore

$\sum_{l=1}^{n}v^{r}_{j}(l)=\sum_{l=1}^{n-1}v^{r}_{j}(l)+v^{r}_{j}(n)<B^{r}_{j}\left(r^{2}\left(t-\frac{v^{r}_{j}(n)}{r^{2}\tilde{a}}\right)\right)+v^{r}_{j}(n)\leq
B^{r}_{j}\left(tr^{2}\right)$

which contradicts the assumption that $\tau^{r,S}_{j}(t)=n$. Consequently

$\left\\{\tau^{r,S}_{j}\left(\left(t-\frac{v^{r}_{j}(n)}{\tilde{a}r^{2}}\right)^{+}\right)>n-1\right\\}\cap\left\\{\tau^{r,S}_{j}(t)=n\right\\}=\emptyset$

and so

$\left\\{\tau^{r,S}_{j}(t)=n\right\\}=\left\\{\tau^{r,S}_{j}\left(\left(t-\frac{v^{r}_{j}(n)}{\tilde{a}r^{2}}\right)^{+}\right)\leq
n-1\right\\}\cap\left\\{\tau^{r,S}_{j}(t)=n\right\\}.$

Thus we have

$\displaystyle\left\\{\tau^{r,S}_{j}(t)=n\right\\}$ $\displaystyle=$
$\displaystyle\left\\{\tau^{r,S}_{j}(t)>n-1\right\\}\cap\left\\{\tau^{r,S}_{j}(t)=n\right\\}$
$\displaystyle=$
$\displaystyle\left\\{\tau^{r,S}_{j}(t)>n-1\right\\}\cap\left\\{\tau^{r,S}_{j}\left(\left(t-\frac{v^{r}_{j}(n)}{\tilde{a}r^{2}}\right)^{+}\right)\leq
n-1\right\\}\cap\left\\{\tau^{r,S}_{j}(t)=n\right\\}$ $\displaystyle\subset$
$\displaystyle\left\\{\tau^{r,S}_{j}(t)>n-1\right\\}\cap\left\\{\tau^{r,S}_{j}\left(\left(t-\frac{v^{r}_{j}(n)}{\tilde{a}r^{2}}\right)^{+}\right)\leq
n-1\right\\}.$

Note that for any $z\geq 0$, $t\geq 0$, and $n\geq 2$ both
$\left\\{\tau^{r,S}_{j}(t)>n-1\right\\}$ and
$\left\\{\tau^{r,S}_{j}\left(\left(t-\frac{z}{\tilde{a}r^{2}}\right)^{+}\right)\leq
n-1\right\\}$ are $\mathcal{F}^{r,S}_{j}(n-1)$-measurable and $v^{r}_{j}(n)$
is independent of $\mathcal{F}^{r,S}_{j}(n-1)$.

Let $\gamma^{r}_{j}(dz)$ denote the pdf of $v^{r}_{j}(1)$ (recall that the
$\\{v^{r}_{j}(l)\\}_{l=1}^{\infty}$ are i.i.d.)and let $c\in(0,\delta)$ be
arbitrary where $\delta$ is as in Condition 1. Since
$P(\tau^{r,S}_{j}(t)<\infty)=1$, we have, recalling the convention
$v_{j}^{r}(0)=0$,

$\displaystyle E\left[e^{cv^{r}_{j}(\tau^{r,S}_{j}(t))}\right]$
$\displaystyle\leq
1+E\left[e^{cv^{r}_{j}(1)}\right]+\sum_{n=2}^{\infty}E\left[e^{cv^{r}_{j}(n)}\mathbb{I}_{\\{\tau^{r,S}_{j}(t)=n\\}}\right]$
$\displaystyle\leq
1+E\left[e^{cv^{r}_{j}(1)}\right]+\sum_{n=2}^{\infty}E\left[e^{cv^{r}_{j}(n)}\mathbb{I}_{\left\\{\tau^{r,S}_{j}(t)>n-1\right\\}\cap\left\\{\tau^{r,S}_{j}\left(\left(t-\frac{v^{r}_{j}(n)}{r^{2}\tilde{a}}\right)^{+}\right)\leq
n-1\right\\}}\right]$
$\displaystyle=1+\int_{(0,\infty)}e^{cz}E\left[1+\sum_{n=1}^{\infty}\mathbb{I}_{\left\\{\tau^{r,S}_{j}\left(\left(t-\frac{z}{\tilde{a}r^{2}}\right)^{+}\right)\leq
n\right\\}\cap\left\\{\tau^{r,S}_{j}(t)>n\right\\}}\right]\gamma^{r}_{j}(dz)$
$\displaystyle=1+\int_{(0,\infty)}e^{cz}\left(1+E\left[\tau^{r,S}_{j}(t)-\tau^{r,S}_{j}\left(\left(t-\frac{z}{\tilde{a}r^{2}}\right)^{+}\right)\right]\right)\gamma^{r}_{j}(dz).$
(62)

Let

$\hat{\tau}^{r,S,t}_{j}(z)=\min\left\\{n\geq
1:\sum_{l=\tau^{r,S}_{j}\left(\left(t-\frac{z}{\tilde{a}r^{2}}\right)^{+}\right)+1}^{\tau^{r,S}_{j}\left(\left(t-\frac{z}{\tilde{a}r^{2}}\right)^{+}\right)+n}v^{r}_{j}(l)\geq\frac{\max_{i}\\{C_{i}\\}z}{\tilde{a}}\right\\}$

and note that since
$B^{r}_{j}(r^{2}t)-B^{r}_{j}\left(r^{2}\left(t-\frac{z}{\tilde{a}r^{2}}\right)^{+}\right)\leq\frac{\max_{i}\\{C_{i}\\}z}{\tilde{a}}$
and
$B^{r}_{j}\left(r^{2}\left(t-\frac{z}{\tilde{a}r^{2}}\right)^{+}\right)\leq\sum_{l=1}^{\tau^{r,S}_{j}\left(\left(t-\frac{z}{\tilde{a}r^{2}}\right)^{+}\right)}v^{r}_{j}(l)$
we have
$\hat{\tau}^{r,S,t}_{j}(z)\geq\tau^{r,S}_{j}(t)-\tau^{r,S}_{j}\left(\left(t-\frac{z}{\tilde{a}r^{2}}\right)^{+}\right).$
In addition, if we define $\hat{\tau}^{r,S}_{j}(z)=\min\left\\{n\geq
1:\sum_{l=1}^{n}v^{r}_{j}(l)\geq\frac{\max_{i}\\{C_{i}\\}z}{\tilde{a}}\right\\}$
then we have $\hat{\tau}^{r,S,t}_{j}(z)\overset{d}{=}\hat{\tau}^{r,S}_{j}(z)$
due to Lemma 9.1. Consequently, from (62),

$E\left[e^{cv^{r}_{j}(\tau^{r,S}_{j}(t))}\right]\leq
1+\int_{(0,\infty)}e^{cz}\left(1+E\left[\hat{\tau}^{r,S}_{j}(z)\right]\right)\gamma^{r}_{j}(dz).$
(63)

We will now bound $E\left[\hat{\tau}^{r,S}_{j}(z)\right]$ from above. For all
$r$ sufficiently large we have
$\frac{2}{\beta_{j}}\geq\frac{1}{\beta^{r}_{j}}\geq\frac{3}{4\beta_{j}}$ and
$2\sigma^{v}_{j}\geq\sigma^{v,r}_{j}\geq\frac{1}{2}\sigma^{v}_{j}$. Due to
Condition 1 there exists $1\leq K<\infty$ such that
$\sup_{r}\left\\{E\left[v^{r}_{j}(1)\mathbb{I}_{\\{v^{r}_{j}(1)>K\\}}\right]\right\\}\leq\frac{1}{4\beta_{j}}$.
Therefore for all $r$ sufficiently large we have

$\displaystyle\frac{3}{4\beta_{j}}\leq\frac{1}{\beta^{r}_{j}}\leq
E\left[v^{r}_{j}(1)\mathbb{I}_{\\{v^{r}_{j}(1)\leq
K\\}}\right]+E\left[v^{r}_{j}(1)\mathbb{I}_{\\{v^{r}_{j}(1)>K\\}}\right]$

and so

$\displaystyle E\left[v^{r}_{j}(1)\mathbb{I}_{\\{v^{r}_{j}(1)\leq
K\\}}\right]\geq\frac{3}{4\beta_{j}}-E\left[v^{r}_{j}(1)\mathbb{I}_{\\{v^{r}_{j}(1)>K\\}}\right]\geq\frac{1}{2\beta_{j}}.$

In addition,

$E\left[v^{r}_{j}(1)\mathbb{I}_{\\{v^{r}_{j}(1)\leq K\\}}\right]\leq
E\left[v^{r}_{j}(1)\mathbb{I}_{\\{v^{r}_{j}(1)<\frac{1}{4\beta_{j}}\\}}\right]+E\left[v^{r}_{j}(1)\mathbb{I}_{\\{\frac{1}{4\beta_{j}}\leq
v^{r}_{j}(1)\leq K\\}}\right]\\\
\leq\frac{1}{4\beta_{j}}P\left(v^{r}_{j}(1)<\frac{1}{4\beta_{j}}\right)+KP\left(\frac{1}{4\beta_{i}}\leq
v^{r}_{j}(1)\leq
K\right)\leq\frac{1}{4\beta_{j}}+KP\left(\frac{1}{4\beta_{j}}\leq
v^{r}_{j}(1)\right)$

so for all $r$ sufficiently large we have

$P\left(\frac{1}{4\beta_{j}}\leq
v^{r}_{j}(1)\right)\geq\frac{1}{K}E\left[v^{r}_{j}(1)\mathbb{I}_{\\{v^{r}_{j}(1)\leq
K\\}}\right]-\frac{1}{4K\beta_{j}}\geq\frac{1}{2K\beta_{j}}-\frac{1}{4K\beta_{j}}\geq\frac{1}{4K\beta_{j}}.$
(64)

Define
$C^{r}_{j}(n)=\sum_{l=1}^{n}\mathbb{I}_{\\{v^{r}_{j}(l)\geq\frac{1}{4K\beta_{j}}\\}}$
and $\zeta^{r}_{j}(z)=\min\\{n\geq 0:C^{r}_{j}(n)=\lceil
4\frac{z}{\tilde{a}}K\max_{i}\\{C_{i}\\}\beta_{j}\rceil\\}$ and note that
$E\left[\hat{\tau}^{r,S}_{j}(z)\right]\leq E\left[\zeta^{r}_{j}(z)\right].$
However, because the $\\{v^{r}_{j}(l)\\}_{l=1}^{\infty}$ are i.i.d. it follows
that $\zeta^{r}_{j}(z)$ is just the sum of $\lceil
4\frac{z}{\tilde{a}}K\max_{i}\\{C_{i}\\}\beta_{j}\rceil$ independent geometric
distributions with probability of success $p\geq\frac{1}{4K\beta_{j}}$ for all
$r$ sufficiently large which gives

$\displaystyle E\left[\hat{\tau}^{r,S}_{j}(z)\right]\leq
E\left[\zeta^{r}_{j}(z)\right]\leq(1+4\frac{z}{\tilde{a}}K\max_{i}\\{C_{i}\\}\beta_{j})4K\beta_{j}\leq
4K\beta_{j}+16\frac{z}{\tilde{a}}K^{2}\max_{i}\\{C_{i}\\}\beta^{2}_{j}.$

Thus, from (63) we have, for all $r$ sufficiently large,

$\displaystyle E\left[e^{cv^{r}_{j}(\tau^{r,S}_{j}(t))}\right]\leq
1+\int_{(0,\infty)}e^{cz}\left(1+E\left[\hat{\tau}^{r,S}_{j}(z)\right]\right)\gamma^{r}_{j}(dz)$
$\displaystyle\leq
1+\int_{(0,\infty)}e^{cz}(1+4K\beta_{j}+16\frac{z}{\tilde{a}}K^{2}\max_{i}\\{C_{i}\\}\beta^{2}_{j})\gamma^{r}_{j}(dz)$
$\displaystyle\leq
1+\left(1+4K\beta_{j}\right)\int_{(0,\infty)}e^{cz}\gamma^{r}_{j}(dz)+\frac{16K^{2}\beta^{2}_{j}}{\tilde{a}}\max_{i}\\{C_{i}\\}\int_{(0,\infty)}ze^{cz}\gamma^{r}_{j}(dz).$

The result follows on using Condition 1. ∎

###  9.2 Proof of Proposition 6.2.

In this section we prove the key large deviation estimates given in
Proposition 6.2 which have been used on several occasions.

Fix $j\in\mathbb{A}_{J}$. Since the proof is identical for $A^{r}_{j}$ and
$S^{r}_{j}$ we will only present it for $A^{r}_{j}$. Throughout in the proof
we suppress the subscript $j$. Define $C^{r}(n)\doteq\sum_{l=1}^{n}u^{r}(1)$
and $\Lambda^{r}(y)=\log(E[e^{y(u^{r}(1)-\frac{1}{\alpha^{r}})}]).$ Note that
$y\mapsto\Lambda^{r}(y)$ is infinitely differentiable for $|y|<\delta$ and due
to Jensen’s inequality $\Lambda^{r}(y)\geq 0$ for all $|y|<\delta$. Due to
Condition 1 there exists $K_{\Lambda}<\infty$ such that

$\displaystyle\sup_{r,|y|\leq\frac{\delta}{2}}\Big{|}\frac{d^{3}\Lambda^{r}}{dy^{3}}(y)\Big{|}\leq
K_{\Lambda}.$

Since, $\Lambda^{r}(0)=0$, $\frac{d\Lambda^{r}}{dy}(0)=0$, and
$\frac{d^{2}\Lambda^{r}}{dy^{2}}(0)=\left(\sigma^{u,r}\right)^{2}$ we have for
$|y|\leq\delta/2$,

$\left|\Lambda^{r}(y)-\frac{y^{2}}{2}\left(\sigma^{u,r}\right)^{2}\right|\leq\frac{|y|^{3}}{6}K_{\Lambda}.$
(65)

We will first prove that, for any $c\geq 0$ there exist
$B_{1},B_{2},R\in(0,\infty)$ such that for all $r\geq R$ and $T\geq 0$ we have

$P\left(\sup_{0\leq t\leq r^{2c}T}|A^{r}(t)-t\alpha^{r}|\geq\epsilon
r^{c}T\right)=P\left(\sup_{0\leq t\leq
T}|A^{r}(r^{2c}t)-r^{2c}t\alpha^{r}|\geq\epsilon r^{c}T\right)\leq
B_{1}e^{-TB_{2}}.$ (66)

Due to the fact that $A^{r}$ is integer-valued and the definition of $C^{r}$,
for all $t\in[0,T]$ we have

$\displaystyle\left\\{A^{r}(r^{2c}t)-\alpha^{r}r^{2c}t\geq
r^{c}T\epsilon\right\\}$ $\displaystyle=$
$\displaystyle\left\\{A^{r}(r^{2c}t)\geq\lceil\alpha^{r}r^{2c}t+r^{c}T\epsilon\rceil\right\\}$
$\displaystyle=$
$\displaystyle\left\\{C^{r}(\lceil\alpha^{r}r^{2c}t+r^{c}T\epsilon\rceil)\leq
r^{2c}t\right\\}$

and

$\displaystyle\left\\{A^{r}(r^{2c}t)-\alpha^{r}r^{2c}t\leq-r^{c}T\epsilon\right\\}$
$\displaystyle=$
$\displaystyle\left\\{A^{r}(r^{2c}t)\leq\lfloor\alpha^{r}r^{2c}t-r^{c}T\epsilon\rfloor\right\\}$
$\displaystyle=$
$\displaystyle\left\\{C^{r}(\lfloor\alpha^{r}r^{2c}t-r^{c}T\epsilon\rfloor+1)>r^{2c}t\right\\}.$

In addition, since for all $t\in[0,T]$ we have

$\frac{1}{\alpha^{r}}\lceil\alpha^{r}r^{2c}t+r^{c}T\epsilon\rceil-\frac{1}{\alpha^{r}}\left(r^{c}T\epsilon\right)\geq
r^{2c}t$

and

$\frac{1}{\alpha^{r}}\left(\lfloor\alpha^{r}r^{2c}t-r^{c}T\epsilon\rfloor+1\right)+\frac{1}{\alpha^{r}}\left(r^{c}T\epsilon-1\right)\leq
r^{2c}t$

it follows that for all $t\in[0,T]$ we have

$\displaystyle\left\\{C^{r}(\lceil\alpha^{r}r^{2c}t+r^{c}T\epsilon\rceil)\leq
r^{2c}t\right\\}$
$\displaystyle\subset\left\\{\frac{1}{\alpha^{r}}\lceil\alpha^{r}r^{2c}t+r^{c}T\epsilon\rceil-C^{r}(\lceil\alpha^{r}r^{2c}t+r^{c}T\epsilon\rceil)\geq\frac{1}{\alpha^{r}}\left(r^{c}T\epsilon\right)\right\\}$
$\displaystyle\subset\left\\{\frac{1}{\alpha^{r}}\lceil\alpha^{r}r^{2c}t+r^{c}T\epsilon\rceil-C^{r}(\lceil\alpha^{r}r^{2c}t+r^{c}T\epsilon\rceil)>\frac{1}{\alpha^{r}}\left(r^{c}T\epsilon-1\right)\right\\}$

and

$\displaystyle\left\\{C^{r}(\lfloor\alpha^{r}r^{2c}t-r^{c}T\epsilon\rfloor+1)>r^{2c}t\right\\}$
$\displaystyle\subset\left\\{\frac{1}{\alpha^{r}}\left(\lfloor\alpha^{r}r^{2c}t-r^{c}T\epsilon\rfloor+1\right)-C^{r}(\lfloor\alpha^{r}r^{2c}t-r^{c}T\epsilon\rfloor+1)<-\frac{1}{\alpha^{r}}\left(r^{c}T\epsilon-1\right)\right\\}$
$\displaystyle=\left\\{C^{r}(\lfloor\alpha^{r}r^{2c}t-r^{c}T\epsilon\rfloor+1)-\frac{1}{\alpha^{r}}\left(\lfloor\alpha^{r}r^{2c}t-r^{c}T\epsilon\rfloor+1\right)>\frac{1}{\alpha^{r}}\left(r^{c}T\epsilon-1\right)\right\\}.$

Using these observations, we have, with
$p_{c}^{r}\doteq\lceil\alpha^{r}r^{2c}T+r^{c}\epsilon T\rceil$

$\left\\{\sup_{0\leq t\leq T}\left(A^{r}(r^{2c}t)-\alpha^{r}r^{2c}t\right)\geq
r^{c}T\epsilon\right\\}\subset\left\\{\sup_{0\leq n\leq
p_{c}^{r}}\left(\frac{1}{\alpha^{r}}n-C^{r}(n)\right)>\frac{1}{\alpha^{r}}\left(r^{c}\epsilon
T-1\right)\right\\}$ (67)

and with $q_{c}^{r}=\lfloor\alpha^{r}r^{2c}T-r^{c}\epsilon T\rfloor$

$\left\\{\inf_{0\leq t\leq
T}\left(A^{r}(r^{2c}t)-\alpha^{r}r^{2c}t\right)\leq-r^{c}T\epsilon\right\\}\subset\left\\{\sup_{0\leq
n\leq
q_{c}^{r}+1}\left(C^{r}(n)-\frac{1}{\alpha^{r}}n\right)>\frac{1}{\alpha^{r}}\left(r^{c}\epsilon
T-1\right)\right\\}.$ (68)

For $0<y<\frac{\delta}{4}$

$P\Big{(}\sup_{0\leq n\leq
p_{c}^{r}}\Big{(}\frac{1}{\alpha^{r}}n-C^{r}(n)\Big{)}>\frac{1}{\alpha^{r}}\Big{(}r^{c}\epsilon
T-1\Big{)}\Big{)}e^{\frac{y}{\alpha^{r}}\Big{(}r^{c}\epsilon T-1\Big{)}}\leq
E\Big{[}\sup_{0\leq n\leq
p_{c}^{r}}e^{y\Big{(}\frac{1}{\alpha^{r}}n-C^{r}(n)\Big{)}}\Big{]}.$ (69)

Since $\Lambda^{r}(-\frac{y}{2})\geq 0$ we have

$\displaystyle E\left[\sup_{0\leq n\leq
p_{c}^{r}}\left(e^{y\left(\frac{1}{\alpha^{r}}n-C^{r}(n)\right)}\right)\right]$
$\displaystyle=E\left[\sup_{0\leq n\leq
p_{c}^{r}}\left(e^{y\left(\frac{1}{\alpha^{r}}n-C^{r}(n)\right)-n2\Lambda^{r}(-\frac{y}{2})}e^{n2\Lambda^{r}(-\frac{y}{2})}\right)\right]$
$\displaystyle\leq e^{p_{c}^{r}2\Lambda^{r}(-\frac{y}{2})}E\left[\sup_{0\leq
n\leq
p_{c}^{r}}\left(e^{y\left(\frac{1}{\alpha^{r}}n-C^{r}(n)\right)-n2\Lambda^{r}\left(-\frac{y}{2}\right)}\right)\right]$
(70)

Since
$e^{\frac{y}{2}\left(\frac{1}{\alpha^{r}}n-C^{r}(n)\right)-n\Lambda^{r}(-\frac{y}{2})}$
is a nonnegative martingale and
$Ee^{\frac{y}{2}\left(\frac{1}{\alpha^{r}}n-C^{r}(n)\right)-n\Lambda^{r}(-\frac{y}{2})}<\infty$
for all $n\in\mathbb{N}$ the Doob-Kolmogorov inequality gives

$\displaystyle E\left[\sup_{0\leq n\leq
p_{c}^{r}}\left(e^{y\left(\frac{1}{\alpha^{r}}n-C^{r}(n)\right)-2n\Lambda^{r}(-\frac{y}{2})}\right)\right]$
$\displaystyle\leq$ $\displaystyle
4E\left[e^{y\left(\frac{1}{\alpha^{r}}p_{c}^{r}-C^{r}(p_{c}^{r})\right)-2p_{c}^{r}\Lambda^{r}(-\frac{y}{2})}\right]$
$\displaystyle\leq$ $\displaystyle
4e^{-2p_{c}^{r}\Lambda^{r}(-\frac{y}{2})}e^{p_{c}^{r}\Lambda^{r}(-y)}.$

Consequently, from (67), (69) and (70),

$\displaystyle P\left(\sup_{0\leq t\leq
T}\left(A^{r}(r^{2c}t)-\alpha^{r}r^{2c}t\right)\geq r^{c}T\epsilon\right)$
$\displaystyle\leq 4e^{-\frac{y}{\alpha^{r}}\left(r^{c}\epsilon
T-1\right)+p_{c}^{r}\Lambda^{r}(-y)}\leq
4e^{-\frac{y}{\alpha^{r}}\left(r^{c}\epsilon
T-1\right)+p_{c}^{r}\left(\frac{y^{2}}{2}\left(\sigma^{u,r}\right)^{2}+\frac{|y|^{3}}{6}K_{\Lambda}\right)}$
$\displaystyle\leq 4e^{-\frac{y}{\alpha^{r}}\left(r^{c}\epsilon
T-1\right)+\left(\alpha^{r}r^{2c}T+r^{c}\epsilon
T+1\right)\left(\frac{y^{2}}{2}\left(\sigma^{u,r}\right)^{2}+\frac{|y|^{3}}{6}K_{\Lambda}\right)},$
(71)

where the second inequality is from (65).

Choose $R<\infty$ sufficiently large that for all $r\geq R$ we have
$\alpha/2\leq\alpha^{r}\leq 2\alpha$, $\left(\sigma^{u,r}\right)^{2}\leq
2\left(\sigma^{u}\right)^{2}$.

Consider first the case where $c>0$. Define
$k=\epsilon/(16\left(\alpha\sigma^{u}\right)^{2})$, and assume without loss
that $R$ is large enough that for all $r\geq R$

$r^{-c}\frac{1}{6}\alpha^{r}k^{3}K_{\Lambda}+r^{-c}\frac{1}{2}\epsilon
k^{2}\left(\sigma^{u,r}\right)^{2}+r^{-2c}\frac{1}{6}\epsilon
k^{3}K_{\Lambda}\leq\frac{k\epsilon}{8\alpha},$
$r^{-c}\frac{k}{\alpha^{r}}+r^{-2c}\frac{1}{2}k^{2}\left(\sigma^{u,r}\right)^{2}+r^{-3c}\frac{1}{6}k^{3}K_{\Lambda}\leq
1,\;\;\mbox{ and }kr^{-c}<\frac{\delta}{4}.$

Then for all $r\geq R$, the exponent in (71) with $y=kr^{-c}$ satisfies

$\displaystyle-\frac{kr^{-c}}{\alpha^{r}}\left(r^{c}\epsilon
T-1\right)+\left(\alpha^{r}r^{2c}T+r^{c}\epsilon
T\right)\left(\frac{\left(kr^{-c}\right)^{2}}{2}\left(\sigma^{u,r}\right)^{2}+\frac{\left(kr^{-c}\right)^{3}}{6}K_{\Lambda}\right)$
$\displaystyle+\left(\frac{\left(kr^{-c}\right)^{2}}{2}\left(\sigma^{u,r}\right)^{2}+\frac{\left(kr^{-c}\right)^{3}}{6}K_{\Lambda}\right)$
$\displaystyle=-\frac{k\epsilon
T}{\alpha^{r}}+\frac{k}{r^{c}\alpha^{r}}+\frac{1}{2}\alpha^{r}Tk^{2}\left(\sigma^{u,r}\right)^{2}+\frac{1}{6r^{c}}\alpha^{r}Tk^{3}K_{\Lambda}+\frac{1}{2r^{c}}\epsilon
Tk^{2}\left(\sigma^{u,r}\right)^{2}+\frac{1}{6r^{2c}}\epsilon
Tk^{3}K_{\Lambda}$
$\displaystyle\quad+\frac{1}{2r^{2c}}k^{2}\left(\sigma^{u,r}\right)^{2}+\frac{1}{6r^{3c}}k^{3}K_{\Lambda}$
$\displaystyle=T\left(\frac{1}{2}\alpha^{r}k^{2}\left(\sigma^{u,r}\right)^{2}-\frac{k\epsilon}{\alpha^{r}}\right)+T\left(r^{-c}\frac{1}{6}\alpha^{r}k^{3}K_{\Lambda}+r^{-c}\frac{1}{2}\epsilon
k^{2}\left(\sigma^{u,r}\right)^{2}+r^{-2c}\frac{1}{6}\epsilon
k^{3}K_{\Lambda}\right)$
$\displaystyle\quad+r^{-c}\frac{k}{\alpha^{r}}+r^{-2c}\frac{1}{2}k^{2}\left(\sigma^{u,r}\right)^{2}+r^{-3c}\frac{1}{6}k^{3}K_{\Lambda}$
$\displaystyle\leq T\left(2\alpha
k^{2}\left(\sigma^{u}\right)^{2}-\frac{k\epsilon}{2\alpha}\right)+T\left(\frac{k\epsilon}{8\alpha}\right)+1=T\left(\frac{k\epsilon}{8\alpha}-\frac{k\epsilon}{2\alpha}\right)+T\left(\frac{k\epsilon}{8\alpha}\right)+1=-T\frac{k\epsilon}{4\alpha}+1.$

Consequently for $r\geq R$, substituting $y=kr^{-c}$ in (71) gives

$P\left(\sup_{0\leq t\leq T}\left(A^{r}(r^{2c}t)-\alpha^{r}r^{2c}t\right)\geq
r^{c}T\epsilon\right)\leq 4e^{-T\frac{k\epsilon}{4\alpha}+1}\leq
4ee^{-T\frac{k\epsilon}{4\alpha}}.$

Similarly, using the inequality (68), an almost identical argument (which we
omit for the sake of brevity) provides a similar bound on $P\left(\inf_{0\leq
t\leq
T}\left(A^{r}(r^{2c}t)-\alpha^{r}r^{2c}t\right)\leq-r^{c}T\epsilon\right)$ and
combining the two bounds we have the estimate in (66) for $c>0$.

Now consider the case $c=0$ and define

$q=\min\left\\{1,\frac{\delta}{4},\frac{\epsilon}{8\alpha\left(\alpha\left(\sigma^{u}\right)^{2}+\frac{\alpha}{6}K_{\Lambda}+\epsilon\left(\sigma^{u}\right)^{2}+\frac{\epsilon}{6}K_{\Lambda}\right)}\right\\}.$

Then for all $r\geq R$ we have, for the exponent in (71) with $y=q$,

$\displaystyle-\frac{q}{\alpha^{r}}\left(\epsilon
T-1\right)+\left(\alpha^{r}T+\epsilon
T\right)\left(\frac{q^{2}}{2}\left(\sigma^{u,r}\right)^{2}+\frac{q^{3}}{6}K_{\Lambda}\right)+\left(\frac{q^{2}}{2}\left(\sigma^{u,r}\right)^{2}+\frac{q^{3}}{6}K_{\Lambda}\right)$
$\displaystyle\leq-\frac{\epsilon
Tq}{2\alpha}+\frac{2q}{\alpha}+\alpha^{r}Tq^{2}\left(\sigma^{u}\right)^{2}+\alpha^{r}T\frac{q^{3}}{6}K_{\Lambda}+\epsilon
Tq^{2}\left(\sigma^{u}\right)^{2}+\epsilon
T\frac{q^{3}}{6}K_{\Lambda}+q^{2}\left(\sigma^{u}\right)^{2}+\frac{q^{3}}{6}K_{\Lambda}$
$\displaystyle\leq-\frac{\epsilon
Tq}{2\alpha}+2q^{2}T\left(\alpha\left(\sigma^{u}\right)^{2}+\frac{\alpha}{6}K_{\Lambda}+\epsilon\left(\sigma^{u}\right)^{2}+\frac{\epsilon}{6}K_{\Lambda}\right)+q^{2}\left(\sigma^{u}\right)^{2}+\frac{2q}{\alpha}+\frac{q^{3}}{6}K_{\Lambda}$
$\displaystyle\leq-\frac{\epsilon Tq}{2\alpha}+\frac{\epsilon
Tq}{4\alpha}+q^{2}\left(\sigma^{u}\right)^{2}+\frac{2q}{\alpha}+\frac{q^{3}}{6}K_{\Lambda}=-T\frac{\epsilon
q}{4\alpha}+q^{2}\left(\sigma^{u}\right)^{2}+\frac{2q}{\alpha}+\frac{q^{3}}{6}K_{\Lambda}.$

Consequently for $r\geq R$, substituting $y=q$ in (71) with $c=0$, we have

$\displaystyle P\left(\sup_{0\leq t\leq
T}\left(A^{r}(t)-\alpha^{r}t\right)\geq T\epsilon\right)$ $\displaystyle\leq$
$\displaystyle\left(4e^{q^{2}\left(\sigma^{u}\right)^{2}+\frac{2q}{\alpha}+\frac{q^{3}}{6}K_{\Lambda}}\right)e^{-T\frac{\epsilon
q}{4\alpha}}.$

Finally, using the inclusion in (68) with $c=0$ we have, by an almost
identical argument (which is omitted), a similar upper bound on
$P\left(\inf_{0\leq t\leq
T}\left(A^{r}(t)-\alpha^{r}t\right)\leq-T\epsilon\right)$. Combining the two
bounds we now have the estimate in (66) for $c=0$. We have thus proved (14)
with $c_{2}=0$. Now (14) for general $c_{1},c_{2}\geq 0$ (and $T$ replaced by
$S\geq 0$) follows on taking $c=c_{1}$ and $T=Sr^{c_{2}}$ in (66). Finally,
the inequality in (16) follows on taking $c_{1}=c_{2}=\frac{\kappa}{2}$ in
(14). ∎

###  9.3 Proof of Proposition 8.1.

The proof for (57) is the same as that for (56) and so we will only show the
latter. Let $c>0$, $\epsilon>0$, and $j\in\mathbb{A}_{J}$ be arbitrary. Once
again we suppress $j$ from the notation. Due to Condition 1 we have
$\sup_{r>0}\\{Ee^{\frac{3\delta}{4}v^{r}(1)}\\}=\tilde{K}<\infty$ where
$\delta$ is as in Condition 1. Choose $R<\infty$ such that for all $r\geq R$
we have
$\left(r^{2}+1\right)\tilde{K}e^{-r\frac{\delta}{4}c}\leq\frac{\epsilon\delta}{4}.$
Consequently

$\displaystyle E\left[e^{\frac{\delta}{2}\mathcal{I}_{\\{v^{r}(1)\geq
rc\\}}v^{r}(1)}\right]$ $\displaystyle=$ $\displaystyle
1+E\left[\mathcal{I}_{\\{v^{r}(1)\geq
rc\\}}e^{\frac{\delta}{2}v^{r}(1)}\right]\leq
1+E\left[\mathcal{I}_{\\{v^{r}(1)\geq
rc\\}}e^{-r\frac{\delta}{4}c}e^{\frac{3\delta}{4}v^{r}(1)}\right]$
$\displaystyle\leq$ $\displaystyle
1+e^{-r\frac{\delta}{4}c}E\left[e^{\frac{3\delta}{4}v^{r}(1)}\right]\leq
1+\tilde{K}e^{-r\frac{\delta}{4}c}\leq e^{\tilde{K}e^{-r\frac{\delta}{4}c}}.$

So for all $T\geq 1$ and $r\geq R$ we have

$E\left[e^{\frac{\delta}{2}\sum_{l=1}^{\lceil
r^{2}T\rceil}\mathcal{I}_{\\{v^{r}(l)\geq rc\\}}v^{r}_{l}}\right]\leq
e^{T\left(r^{2}+1\right)\tilde{K}e^{-r\frac{\delta}{4}c}}\leq
e^{T\frac{\epsilon\delta}{4}}$

which implies

$P\left(\sum_{l=1}^{\lceil r^{2}T\rceil}\mathcal{I}_{\\{v^{r}(l)\geq
rc\\}}v^{r}(l)\geq\epsilon T\right)\leq
e^{-\frac{\epsilon\delta}{2}T}e^{T\frac{\epsilon\delta}{4}}\leq
e^{-\frac{\epsilon\delta}{4}T}.$

∎

##  10 General Network and Cost Properties

###  10.1 Proof of Proposition 2.1.

From the discussion above Proposition 2.1, the result is clearly true when
$\mathcal{C}_{K}^{h}=\text{ker}(K)$. Consider now the complementary case,
namely, $\text{dim}(\mathcal{C}_{K}^{h})=\text{dim}(\text{ker}(K))-1=J-I-1$.
Let $q\in\mathbb{R}_{+}^{J}$ be arbitrary. For
$\tilde{q}\in\Lambda\left(KMq\right)$ define $\tilde{v}=(\tilde{q}-q)/\beta$.
Note that
$0\leq\tilde{q}_{j}=q_{j}+(\tilde{q}_{j}-q_{j})=q_{j}+\tilde{v}_{j}\beta_{j}$
for all $j\in\mathbb{A}_{J}$ and
$K\tilde{v}=KM(\tilde{q}-q)=KM\tilde{q}-KMq=0$ and so $\tilde{v}\in\Xi(q)$. In
addition, for any $\hat{v}\in\Xi(q)$ define $\hat{q}=q+\beta\hat{v}$ and note
that by the definition of $\Xi(q)$ we have $\hat{q}\in\mathbb{R}^{J}_{+}$ and
$KM\hat{q}=KM(q+\beta\hat{v})=KMq+K\hat{v}=KMq$ so
$\hat{q}\in\Lambda\left(KMq\right)$. Consequently
$\inf_{\tilde{q}\in\Lambda(KMq)}(h\cdot\tilde{q})=\inf_{\tilde{v}\in\Xi(q)}(h\cdot\left(q+\beta\tilde{v}\right))$.
Therefore

$\displaystyle h\cdot q-\hat{h}\left(KMq\right)=h\cdot
q-\inf_{\tilde{q}\in\Lambda(KMq)}(h\cdot\tilde{q})=h\cdot
q-\inf_{\tilde{v}\in\Xi(q)}(h\cdot\left(q+\beta\tilde{v}\right))$
$\displaystyle=\sup_{\tilde{v}\in\Xi(q)}(-h\cdot\left(\beta\tilde{v}\right))=\sup_{\tilde{v}\in\Xi(q)}\left(-\beta
h\cdot\sum_{j=1}^{J-I}u_{j}(u_{j}\cdot\tilde{v})\right)=\sup_{\tilde{v}\in\Xi(q)}\left(-\beta
h\cdot u_{J-I}(u_{J-I}\cdot\tilde{v})\right)$
$\displaystyle=|\lambda|\sup_{\tilde{v}\in\Xi(q)}\left(u_{J-I}\cdot\tilde{v}\right)=|\lambda|\tilde{d}(q),$

where the last equality on the second line uses $\beta h\cdot u_{j}=0$ for
$j=1,\ldots J-I-1$, and the last line follows on recalling the definition of
$\lambda$ and $\tilde{d}$. The result follows. ∎

###  10.2 Proof of Proposition 2.3

Fix $z\in\chi_{J}$. Due to the local traffic assumption (Condition 3) for each
$l\in\mathcal{A}_{z}$ we can choose $s_{l}\in\mathbb{A}_{J}$ such that
$K_{l,s_{l}}=1$ and $\sum_{i\in\mathbb{A}_{I}}K_{i,s_{l}}=1$. Define
$S=\\{s_{l}:l\in\mathcal{A}_{z}\\}$,
$P=\\{j\in\mathbb{A}_{J}:z_{j}=1\\}\setminus S$ and
$N=\\{j\in\mathbb{A}_{J}:z_{j}=0\\}\setminus S$. For $j\in N$ define
$v_{j}=-J$, for $j\in P$ define $v_{j}=1$, and for all $l\in\mathcal{A}_{z}$
define $v_{s_{l}}=-\sum_{j\neq s_{l}}K_{l,j}v_{j}$. For any
$l\in\mathcal{A}_{z}$ we have

$\sum_{j\in\mathbb{A}_{J}}K_{l,j}v_{j}=\sum_{j\neq
s_{l}}K_{l,j}v_{j}+v_{s_{l}}=\sum_{j\neq s_{l}}K_{l,j}v_{j}-\sum_{j\neq
s_{l}}K_{l,j}v_{j}=0.$

This verifies the first statement in the proposition with the above choice of
$v$. Next consider $j\in\mathbb{N}_{j}$ such that $z_{j}=1$. If $j\in P$, then
by definition $v_{j}>0$. To complete the proof consider now
$l\in\mathcal{A}_{z}$ with $z_{s_{l}}=1$. Due to the definition of
$\mathcal{A}_{z}$ we have $\sum_{j:z_{j}=0}K_{l,j}\geq 1$ and we also know
that $K_{l,s_{i}}=0$ for all $i\in\mathcal{A}_{z}\setminus\\{l\\}$. This
implies that there exists $j^{*}\in N$ such that $K_{l,j^{*}}=1$ and for all
$j\in\mathbb{N_{J}}\setminus\\{j^{*},s_{l}\\}$ such that $K_{l,j}=1$, either
$v_{j}=1$ or $v_{j}=-J$. Since $v_{j^{*}}=-J$ we have

$v_{s_{l}}=-\sum_{j\neq
s_{l}}K_{l,j}v_{j}=J-\sum_{j\in\mathbb{N}\setminus\\{j^{*},s_{l}\\}}K_{l,j}v_{j}\geq
J-(J-2)\geq 2.$

Thus we have shown that for all $j\in\mathbb{A}_{J}$ such that $z_{j}=1$ we
have $v_{j}>0$. This proves the second statement in the proposition. ∎

###  10.3 Proofs of Propositions 7.6 and 7.7.

We begin with two auxiliary results.

######  Lemma 10.1.

There exists a $\hat{B}_{\hat{h}}^{-}\in(0,\infty)$ such that for all
$w^{1},w^{2}\in\mathbb{N}^{I}_{+}$ satisfying $w^{1}\geq w^{2}$, we have

$\hat{h}(w^{2})\leq\hat{h}(w^{1})+\hat{B}_{\hat{h}}^{-}|w^{1}-w^{2}|_{2}.$

###### Proof.

For $i\in\mathbb{A}_{I}$ let
$\mathcal{C}_{i}=\left\\{j\in\mathbb{A}_{J}:K_{i,j}=1\right\\}$. For
$j\in\mathcal{C}_{i}$ define
$\mathcal{O}^{j}_{i}=\left\\{l\in\mathbb{A}_{I}:K_{l,j}=1\text{ and }l\neq
i\right\\}$. Thus the set $\mathcal{O}^{j}_{i}$ consists of all the resources
that job type $j$ impacts aside from resource $i$. For each
$i\in\mathbb{A}_{I}$ let $s_{i}\in\mathbb{A}_{J}$ be such that
$\sum_{l=1}^{I}K_{l,s_{i}}=K_{i,s_{i}}=1$. For any $j\in\mathbb{A}_{J}$ let
$e^{j}\in\mathbb{R}^{J}_{+}$ be the unit vector with one on the $j$th
coordinate so $e^{j}_{j}=1$ and $e^{j}_{l}=0$ for $l\neq j$. Note that for any
$c>0$ and $j\in\mathbb{A}_{J}$ if we define
$\tilde{w}^{1}=KMc\beta_{j}e^{j}=cK_{\cdot,j}$ (here $K_{\cdot,j}$ is the
$j$th column vector of the matrix $K$) and
$\tilde{w}^{2}=KM\left(\sum_{l\in\mathcal{O}^{j}_{i}}c\beta_{s_{l}}e^{s_{l}}\right)$
then $\tilde{w}^{2}_{l}=\tilde{w}^{1}_{l}$ for $l\neq i$ but
$\tilde{w}^{2}_{i}=0$ and $\tilde{w}^{1}_{i}=c$. In other words, by replacing
$c\beta_{j}e^{j}$ with $\sum_{l\in\mathcal{O}^{i}_{j}}c\beta_{s_{l}}e^{s_{l}}$
in the queue length vector we’ve reduced the workload for server $i$ by $c$
and we’ve changed the cost by
$c\left(\sum_{l\in\mathcal{O}^{j}_{i}}h_{s_{l}}\beta_{s_{l}}-h_{j}\beta_{j}\right)$.
This is the key idea in the proof.

Define

$\tilde{R_{i}}=\max_{j\in\mathcal{C}_{i}}\left(\sum_{l\in\mathcal{O}^{j}_{i}}h_{s_{l}}\beta_{s_{l}}-h_{j}\beta_{j}\right)^{+}\;\;\mbox{
and }\tilde{R}=\max_{i\in\mathbb{A}_{I}}\left\\{\tilde{R_{i}}\right\\}.$

Now let $w^{1},w^{2}\in\mathbb{N}^{I}_{+}$ be such that $w^{1}\geq w^{2}$ and
let $\tilde{q}^{0}\in\mathbb{R}^{J}_{+}$ satisfy $w^{1}=KM\tilde{q}^{0}$ and
$\hat{h}(w^{1})=h\cdot\tilde{q}^{0}$. Since
$\sum_{j\in\mathcal{C}_{1}}\frac{1}{\beta_{j}}\tilde{q}^{0}_{j}=\sum_{j\in\mathbb{A}_{J}}K_{1,j}\frac{1}{\beta_{j}}\tilde{q}^{0}_{j}=w^{1}_{1}\geq
w^{1}_{1}-w^{2}_{1}$ we can choose
$\tilde{c}^{1}_{j}\in[0,\tilde{q}^{0}_{j}/\beta_{j}]$ for all
$j\in\mathcal{C}_{1}$ such that
$\sum_{j\in\mathcal{C}_{1}}\tilde{c}^{1}_{j}=w^{1}_{1}-w^{2}_{1}$. Define

$\tilde{q}^{1}=\tilde{q}^{0}+\sum_{j\in\mathcal{C}_{1}}\left(\sum_{l\in\mathcal{O}^{j}_{1}}\tilde{c}^{1}_{j}\beta_{s_{l}}e^{s_{l}}-\tilde{c}^{1}_{j}\beta_{j}e^{j}\right)$

and note that $\tilde{q}^{1}\in\mathbb{R}^{J}_{+}$,

$KM\tilde{q}^{1}=\left(w^{1}_{1}-\sum_{j\in\mathcal{C}_{1}}\tilde{c}^{1}_{j},w^{1}_{2},w^{1}_{3},...,w^{1}_{I}\right)=\left(w^{2}_{1},w^{1}_{2},w^{1}_{3},...,w^{1}_{I}\right),$

and

$\displaystyle h\cdot\tilde{q}^{1}$ $\displaystyle=$ $\displaystyle
h\cdot\tilde{q}^{0}+\sum_{j\in\mathcal{C}_{1}}\left(\sum_{l\in\mathcal{O}^{j}_{1}}\tilde{c}^{1}_{j}h_{s_{l}}\beta_{s_{l}}-\tilde{c}^{1}_{j}h_{j}\beta_{j}\right)$
$\displaystyle\leq$
$\displaystyle\hat{h}(w^{1})+\sum_{j\in\mathcal{C}_{1}}\tilde{c}^{1}_{j}\left(\sum_{l\in\mathcal{O}^{j}_{1}}h_{s_{l}}\beta_{s_{l}}-h_{j}\beta_{j}\right)^{+}\leq\hat{h}(w^{1})+\tilde{R}|w^{1}_{1}-w^{2}_{1}|.$

Now assume that for $k\in\\{1,...,I-1\\}$ there exists
$\tilde{q}^{k}\in\mathbb{R}^{J}_{+}$ such that

$KM\tilde{q}^{k}=(w^{2}_{1},w^{2}_{2},...,w^{2}_{k},w^{1}_{k+1},w^{1}_{k+2},...,w^{1}_{I}),\;\;\mbox{
and
}h\cdot\tilde{q}^{k}\leq\hat{h}(w^{1})+\tilde{R}\sum_{i=1}^{k}|w^{1}_{i}-w^{2}_{i}|.$
(72)

Since
$\sum_{j\in\mathcal{C}_{k+1}}\tilde{q}^{k}_{j}/\beta_{j}=\sum_{j\in\mathbb{A}_{J}}K_{k+1,j}\tilde{q}^{k}_{j}/\beta_{j}=w^{1}_{k+1}\geq
w^{2}_{k+1}$ we can choose
$\tilde{c}^{k+1}_{j}\in[0,\tilde{q}^{k}_{j}/\beta_{j}]$ for all
$j\in\mathcal{C}_{k+1}$ such that
$\sum_{j\in\mathcal{C}_{k+1}}\tilde{c}^{k+1}_{j}=w^{1}_{k+1}-w^{2}_{k+1}$.
Define

$\tilde{q}^{k+1}=\tilde{q}^{k}+\sum_{j\in\mathcal{C}_{k+1}}\left(\sum_{l\in\mathcal{O}^{j}_{k+1}}\tilde{c}^{k+1}_{j}\beta_{s_{l}}e^{s_{l}}-\tilde{c}^{k+1}_{j}\beta_{j}e^{j}\right).$

Then, as before $\tilde{q}^{k+1}\in\mathbb{R}_{+}^{J}$, and from (72),

$\displaystyle KM\tilde{q}^{k+1}$ $\displaystyle=$
$\displaystyle\left(w^{2}_{1},w^{2}_{2},...,w^{2}_{k},w^{1}_{k+1}-\sum_{j\in\mathcal{C}_{k+1}}\tilde{c}^{k+1}_{j},w^{1}_{k+2},...,w^{1}_{I}\right)$
$\displaystyle=$
$\displaystyle\left(w^{2}_{1},w^{2}_{2},...,w^{2}_{k},w^{2}_{k+1},w^{1}_{k+2},...,w^{1}_{I}\right)$

and

$\displaystyle h\cdot\tilde{q}^{k+1}$ $\displaystyle=$ $\displaystyle
h\cdot\tilde{q}^{k}+\sum_{j\in\mathcal{C}_{k+1}}\left(\sum_{l\in\mathcal{O}^{j}_{k+1}}\tilde{c}^{k+1}_{j}h_{s_{l}}\beta_{s_{l}}-\tilde{c}^{k+1}_{j}h_{j}\beta_{j}\right)$
$\displaystyle\leq$
$\displaystyle\hat{h}(w^{1})+\tilde{R}\sum_{i=1}^{k}|w^{1}_{i}-w^{2}_{i}|+\sum_{j\in\mathcal{C}_{k+1}}\tilde{c}^{k+1}_{j}\left(\sum_{l\in\mathcal{O}^{j}_{k+1}}h_{s_{l}}\beta_{s_{l}}-h_{j}\beta_{j}\right)^{+}$
$\displaystyle\leq$
$\displaystyle\hat{h}(w^{1})+\tilde{R}\sum_{i=1}^{k}|w^{1}_{i}-w^{2}_{i}|+\tilde{R}|w^{1}_{k+1}-w^{2}_{k+1}|=\hat{h}(w^{1})+\tilde{R}\sum_{i=1}^{k+1}|w^{1}_{i}-w^{2}_{i}|.$

By induction this implies that there exists
$\tilde{q}^{I}\in\mathbb{R}^{J}_{+}$ such that

$KM\tilde{q}^{I}=w^{2},\mbox{ and,
}h\cdot\tilde{q}^{I}\leq\hat{h}(w^{1})+\tilde{R}\sum_{i=1}^{I}|w^{1}_{i}-w^{2}_{i}|.$

Since $\sum_{i=1}^{I}|w^{1}_{i}-w^{2}_{i}|\leq\sqrt{J}|w^{1}-w^{2}|_{2}$ and
$\hat{h}(w^{2})\leq h\cdot\tilde{q}^{I}$ due to the fact that
$KM\tilde{q}^{I}=w^{2}$, we have

$\hat{h}(w^{2})\leq\hat{h}(w^{1})+\tilde{R}\sqrt{J}|w^{1}-w^{2}|_{2}.$

This completes the proof. ∎

######  Lemma 10.2.

There exists $\hat{B}_{\hat{h}}\in(0,\infty)$ such that for all
$w^{1},w^{2}\in\mathbb{R}^{I}_{+}$ we have

$|\hat{h}(w^{1})-\hat{h}(w^{2})|\leq\hat{B}_{\hat{h}}|w^{1}-w^{2}|_{2}.$

###### Proof.

Let $w^{1},w^{2}\in\mathbb{R}^{I}_{+}$ be arbitrary. Let
$q^{*}\in\mathbb{R}_{+}^{J}$ satisfy $KMq^{*}=w^{1}\wedge w^{2}$ and $h\cdot
q^{*}=\hat{h}\left(w^{1}\wedge w^{2}\right)$. Note that for every
$i\in\mathbb{A}_{I}$ we have $|w^{1}_{i}-w^{2}_{i}|=|w^{1}_{i}-w^{1}_{i}\wedge
w^{2}_{i}|+|w^{2}_{i}-w^{1}_{i}\wedge w^{2}_{i}|$ so

$|w^{1}-w^{1}\wedge w^{2}|_{2}\leq|w^{1}-w^{2}|_{2},\mbox{ and,
}|w^{2}-w^{1}\wedge w^{2}|_{2}\leq|w^{1}-w^{2}|_{2},$

and

$|w^{1}-w^{2}|_{2}\leq|w^{1}-w^{1}\wedge w^{2}|_{2}+|w^{2}-w^{1}\wedge
w^{2}|_{2}.$

Due to Lemma 10.1 we have

$\hat{h}(w^{1}\wedge
w^{2})\leq\hat{h}(w^{1})+\hat{B}^{-}_{\hat{h}}|w^{1}-w^{1}\wedge
w^{2}|_{2}\leq\hat{h}(w^{1})+\hat{B}^{-}_{\hat{h}}|w^{1}-w^{2}|_{2}$

and

$\hat{h}(w^{1}\wedge
w^{2})\leq\hat{h}(w^{2})+\hat{B}^{-}_{\hat{h}}|w^{2}-w^{1}\wedge
w^{2}|_{2}\leq\hat{h}(w^{2})+\hat{B}^{-}_{\hat{h}}|w^{1}-w^{2}|_{2}.$

Let $\hat{B}_{\hat{h}}^{+}\in(0,\infty)$ be such that for all
$w\in\mathbb{R}^{I}_{+}$ we have $\hat{h}(w)\leq\hat{B}_{\hat{h}}^{+}|w|_{2}$.
Then, using the definition of $\hat{h}$,

$\displaystyle\hat{h}(w^{1})$ $\displaystyle\leq$
$\displaystyle\hat{h}(w^{1}\wedge w^{2})+\hat{h}(w^{1}-w^{1}\wedge
w^{2})\leq\hat{h}(w^{1}\wedge w^{2})+\hat{B}^{+}_{\hat{h}}|w^{1}-w^{1}\wedge
w^{2}|_{2}$ $\displaystyle\leq$ $\displaystyle\hat{h}(w^{1}\wedge
w^{2})+\hat{B}^{+}_{\hat{h}}|w^{1}-w^{2}|_{2}$

and similarly

$\hat{h}(w^{2})\leq\hat{h}(w^{1}\wedge
w^{2})+\hat{B}^{+}_{\hat{h}}|w^{1}-w^{2}|_{2}.$

Consequently

$|\hat{h}(w^{1})-\hat{h}(w^{1}\wedge
w^{2})|\leq\max\\{\hat{B}^{+}_{\hat{h}},\hat{B}^{-}_{\hat{h}}\\}|w^{1}-w^{2}|_{2}$

and

$|\hat{h}(w^{2})-\hat{h}(w^{1}\wedge
w^{2})|\leq\max\\{\hat{B}^{+}_{\hat{h}},\hat{B}^{-}_{\hat{h}}\\}|w^{1}-w^{2}|_{2}$

and so

$\displaystyle|\hat{h}(w^{1})-\hat{h}(w^{2})|$ $\displaystyle\leq$
$\displaystyle|\hat{h}(w^{1})-\hat{h}(w^{1}\wedge
w^{2})|+|\hat{h}(w^{2})-\hat{h}(w^{1}\wedge w^{2})|$ $\displaystyle\leq$
$\displaystyle
2\max\\{\hat{B}^{+}_{\hat{h}},\hat{B}^{-}_{\hat{h}}\\}|w^{1}-w^{2}|_{2}.$

This completes the proof. ∎

We can now complete the proof of Proposition 7.6.

Proof of Proposition 7.6 For any $q^{1},q^{2}\in\mathbb{R}^{J}_{+}$, from
Proposition 2.1, we have

$\displaystyle|\lambda|(\tilde{d}(q^{2})-\tilde{d}(q^{1}))=h\cdot
q^{2}-\hat{h}(KMq^{2})-\left(h\cdot q^{1}-\hat{h}(KMq^{1})\right)$
$\displaystyle=h\cdot q^{2}-h\cdot
q^{1}-\left(\hat{h}(KMq^{2})-\hat{h}(KMq^{1})\right)\leq
h\cdot\left(q^{2}-q^{1}\right)+|\hat{h}(KMq^{2})-\hat{h}(KMq^{1})|.$

The result now follows from Lemma 10.2 on observing that we can find
$R\in(0,\infty)$ such that for all $x\in\mathbb{R}^{J}$ and $r\geq R$,
$|KMx|_{2}\leq 2|KM^{r}x|_{2}$. ∎

Finally we complete the proof of Proposition 7.7.

Proof of Proposition 7.7 Let $q^{1},q^{2}\in\mathbb{R}^{J}_{+}$ be arbitrary.
Due to Proposition 7.6 we have

$\displaystyle|\lambda||\tilde{d}(q^{2})-\tilde{d}(q^{1})|$
$\displaystyle\leq$
$\displaystyle|h\cdot\left(q^{2}-q^{1}\right)|+B_{\hat{h}}|KMq^{2}-KMq^{1}|_{2}$
$\displaystyle\leq$
$\displaystyle|h|_{2}|q^{2}-q^{1}|_{2}+B_{\hat{h}}|KM||q^{2}-q^{1}|_{2}\leq(|h|_{2}+B_{\hat{h}}|KM|)|q^{2}-q^{1}|_{2}$

which completes the proof. ∎

##  11 Proofs of Propositions 3.6 and 3.7

Results analogous to Propositions 3.6 and 3.7 for exponential primitives were
studied in [6] and therefore we only give proof sketches here.

###  11.1 Proof of Proposition 3.6.

The fact that $J^{r}_{E}(B^{r},y^{r})<\infty$ for $r$ sufficiently large
follows from Proposition 3.4 and the fact that there exists a constant
$B<\infty$ such that $h\cdot q\leq B|w|$ for all $q\in\Lambda(w)$ and
$w\in\mathbb{R}^{I}_{+}$. To prove tightness of $\\{\nu^{r}\\}$ it is
sufficient to show that the two marginals are tight. The tightness of
$\nu^{r}_{(1)}$ follows immediately from Proposition 3.4. To show the
tightness of the second marginal, $\nu^{r}_{(2)}(dx)$, since $x(0)=0$
$\nu_{(2)}^{r}$-a.s. it is sufficient to show that for any
$\epsilon_{1},\epsilon_{2}>0$ there exists $\delta>0$ and $R<\infty$ such that
for all $r\geq R$ we have

$E\left[\nu^{r}_{(2)}\left(\left\\{\sup_{s,t\in[0,1],|s-t|<\delta}\|x(s)-x(t)\|>\epsilon_{1}\right\\}\right)\right]<\epsilon_{2}.$
(73)

Note that the left side above equals

$\frac{1}{T_{r}}\int_{0}^{T_{r}}P\left(\sup_{s,t\in[u,u+1],|s-t|<\delta}\|\hat{X}^{r}(s)-\hat{X}^{r}(t)\|>\epsilon_{1}\right)du$

and for any $t,u\geq 0$ we have

$\displaystyle\hat{X}^{r}(u+t)=$
$\displaystyle\hat{X}^{r}(u)+KM^{r}\hat{A}^{r,u}\left(\left(t-\bar{\Upsilon}^{A,r}(u)\right)^{+}\right)+\frac{1}{r}KM^{r}\mathcal{I}_{\\{t\geq\bar{\Upsilon}^{A,r}(u)>0\\}}$
$\displaystyle-
KM^{r}\hat{S}^{r,u}\left(\left(\bar{B}^{r}(t+u)-\bar{B}^{r}(u)-\bar{\Upsilon}^{S,r}(u)\right)^{+}\right)$
$\displaystyle-\frac{1}{r}KM^{r}\mathcal{I}_{\\{\bar{B}^{r}(t+u)-\bar{B}^{r}(u)\geq\bar{\Upsilon}^{S,r}(u)>0\\}}+rtK(\rho^{r}-\rho)$
$\displaystyle-
rK\rho^{r}\left(t\wedge\bar{\Upsilon}^{A,r}(u)\right)+rK\left((\bar{B}^{r}(t+u)-\bar{B}^{r}(u))\wedge\bar{\Upsilon}^{S,r}(u)\right).$
(74)

From Proposition 6.2 and Lemma 9.1 it follows that for any
$\epsilon_{1},\epsilon_{2}>0$ there exists $\delta>0$ and $R<\infty$ such that
for all $r\geq R$ and $j\in\mathbb{A}_{J}$ we have

$\frac{1}{T_{r}}\int_{0}^{T_{r}}P\left(\sup_{s,t\in[u,u+1],|s-t|<\delta}\|\hat{A}_{j}^{r,u}(s)-\hat{A}_{j}^{r,u}(t)\|>\epsilon_{1}\right)du<\epsilon_{2}$

and

$\frac{1}{T_{r}}\int_{0}^{T_{r}}P\left(\sup_{s,t\in[u,u+\max_{i}\\{C_{i}\\}],|s-t|<\max_{i}\\{C_{i}\\}\delta}\|\hat{S}_{j}^{r,u}(s)-\hat{S}_{j}^{r,u}(t)\|>\epsilon_{1}\right)du<\epsilon_{2}.$

In addition, due to Proposition 6.1, Proposition 7.5, and the assumption that
$\sup_{r}r\hat{\Upsilon}^{r}<\infty$ it follows that for any
$\epsilon_{1},\epsilon_{2}>0$ there exists $\delta>0$ and $R<\infty$ such that
for all $r\geq R$ we have

$\frac{1}{T_{r}}\int_{0}^{T_{r}}P\left(\|rK\rho^{r}\bar{\Upsilon}^{A,r}(u)\|>\epsilon_{1}\right)du<\epsilon_{2}$

and

$\frac{1}{T_{r}}\int_{0}^{T_{r}}P\left(\|rK\bar{\Upsilon}^{S,r}(u)\|>\epsilon_{1}\right)du<\epsilon_{2}.$

Also note that $rK(\rho^{r}-\rho)\rightarrow\theta$ (due to Condition 2 and
the paragraph that follows) and for all $s,t\geq 0$ and $j\in\mathbb{A}_{J}$
we have $|\bar{B}_{j}^{r}(s)-\bar{B}_{j}^{r}(t)|\leq\max_{i}\\{C_{i}\\}(t-s)$.
These observations, together with the form of
$\|\hat{X}^{r}(s)-\hat{X}^{r}(t)\|$ for $s,t\geq u$ given by (74), give (73)
and complete the proof. ∎

###  11.2 Proof of Proposition 3.7.

Let $\nu^{*}$ and $(w,x)$ be as in the statement of the proposition. In what
follows we let $P_{\nu^{*}(\omega)}$ and $E_{\nu^{*}(\omega)}$ denote
probability and expectation under $\nu^{*}(\omega)$. The proof that for a.e.
$\omega$, $P_{\nu^{*}(\omega)}(x\in\mathcal{C}([0,1]:\mathbb{R}^{I})=1$ is the
same as the proof of [6, Theorem 15 part 1]. To complete the proof of part (a)
it suffices to show that for any $f\in\mathcal{C}^{2}_{c}(\mathbb{R}^{I})$
(which is the space of continuous, real-valued functions on $\mathbb{R}^{I}$
with compact support and continuous first and second derivatives)
$f(x(t))-\int_{0}^{t}\mathcal{L}f(x(s))ds$ is a $\sigma(w,x(s):s\leq
t)$-martingale under $\nu^{*}(\omega)$ for a.e $\omega$ where

$\mathcal{L}f(y)=\sum_{i=1}^{I}\theta_{i}\frac{\partial f}{\partial
y_{i}}(y)+\frac{1}{2}\sum_{i=1}^{I}\sum_{j=1}^{I}\Sigma_{i,j}\frac{\partial^{2}f}{\partial
y_{i}\partial y_{j}}(y),\;y\in\mathbb{R}^{I}.$

Let $0\leq s<t\leq 1$ be fixed and let
$g_{s}\in\mathcal{C}_{b}(\mathbb{R}_{+}^{I}\times\mathcal{D}([0,s]))$ and
$f\in\mathcal{C}^{2}_{c}(\mathbb{R}^{I})$ be arbitrary (here
$\mathcal{C}_{b}(\mathbb{R}_{+}^{I}\times\mathcal{D}([0,s])$ denotes the space
of continuous, bounded real-valued functions on
$\mathbb{R}_{+}^{I}\times\mathcal{D}([0,s])$). Denoting, for $s\in[0,1]$, by
$x_{s}$ the restriction of $x$ on $[0,s]$, we see that

$g_{s}(w,x_{s})\left(f(x(t))-f(x(s))-\int_{s}^{t}\mathcal{L}f(x(z))dz\right)$

is a bounded, continuous function on
$\mathbb{R}_{+}^{I}\times\mathcal{D}([0,1]:\mathbb{R}^{I})$ so

$\displaystyle
E\left[E_{\nu^{*}(\omega)}\left[g_{s}(w,x_{s})\left(f(x(t))-f(x(s))-\int_{s}^{t}\mathcal{L}f(x(u))du\right)\right]^{2}\right]$
$\displaystyle=\lim_{m\rightarrow\infty}E\left[E_{\nu^{r_{m}}(\omega)}\left[g_{s}(w,x_{s})\left(f(x(t))-f(x(s))-\int_{s}^{t}\mathcal{L}f(x(u))du\right)\right]^{2}\right].$
(75)

For $(w,y)\in\mathbb{R}^{I}_{+}\times\mathcal{D}^{I}$ and $0\leq s\leq 1$,
define

$G^{u}_{s}(w,y)=g_{s}(w,[y(u+\cdot)-y(u)]_{s}),$

where $[y(u+\cdot)-y(u)]_{s}$ is the restriction of $y(u+\cdot)-y(u)$ on
$[0,s]$ and for $0\leq s\leq t\leq 1$,

$F^{u}_{s,t}(y)=f(y(u+t)-y(u))-f(y(u+s)-y(u))-\int_{s}^{t}\mathcal{L}f(y(u+z)-y(u))dz.$

Then the expectation in (75) can be written as

$\displaystyle
E\left[\left(\frac{1}{T_{r_{m}}}\int_{0}^{T_{r_{m}}}G^{u}_{s}(\hat{W}^{r_{m}}(u),\hat{X}^{r_{m}})F^{u}_{s,t}(\hat{X}^{r_{m}})du\right)^{2}\right]$
$\displaystyle=E\left[\frac{2}{T^{2}_{r_{m}}}\int_{0}^{T_{r_{m}}}\int_{0}^{u}G^{u}_{s}(\hat{W}^{r_{m}}(u),\hat{X}^{r_{m}})F^{u}_{s,t}(\hat{X}^{r_{m}})G^{v}_{s}(\hat{W}^{r_{m}}(v),\hat{X}^{r_{m}})F^{v}_{s,t}(\hat{X}^{r_{m}})dvdu\right]$

It can be shown using Proposition 3.4, Proposition 6.2, Lemma 9.1,
Propositions 6.1 and 7.5, and the assumptions that
$\sup_{r}\hat{q}^{r}<\infty$ and $\sup_{r}r\hat{\Upsilon}^{r}<\infty$, that

$\lim_{m\rightarrow\infty}\sup_{u\geq
0}P\left(\sup_{t\in[0,1]}\|\bar{B}^{r}(t+u)-\bar{B}^{r}(u)-t\rho\|>\epsilon\right)=0.$
(76)

The proof of the above assertion is very similar to that of [6, Theorem 15
part 2 ] and is therefore omitted.

For any $u\geq 0$ define

$\tilde{X}^{r,u}(t)=KM^{r}\hat{A}^{r,u}\left(t\right)-KM^{r}\hat{S}^{r,u}\left(\rho
t\right)+rtK(\rho^{r}-\rho).$

and

$\breve{X}^{r_{m},u}(t)\doteq\begin{cases}\hat{X}^{r_{m}}(t)&\mbox{ if }u\geq
t\\\ \hat{X}^{r_{m}}(u)+\tilde{X}^{r,u}(t-u)&\mbox{ otherwise.}\end{cases}$

Note that due to Lemma 9.1 the distribution of $\tilde{X}^{r_{m},u}$ does not
depend on $u$, meaning $\tilde{X}^{r_{m},u}\overset{d}{=}\tilde{X}^{r_{m},0}$
for all $u\geq 0$, and from the central limit theorem for renewal processes
(see e.g. [4, Theorem 14.6]) $\tilde{X}^{r,0}\rightarrow\hat{X}$ in
distribution on $\mathcal{D}^{I}$ where $\hat{X}(\cdot)$ is as introduced
above (10). Since $\hat{X}\in\mathcal{C}^{I}$ a.s., it follows from (76),
Propositions 6.1 and 7.5, and the assumption that
$\sup_{r}\hat{\Upsilon}^{r}<\infty$ that, for any $\epsilon>0$,

$\lim_{m\rightarrow\infty}\sup_{u\geq
0}P\left(\sup_{z\in[0,1]}\|\hat{X}^{r_{m}}(u+z)-\breve{X}^{r_{m},u}(u+z)\|>\epsilon\right)=0.$

Since $f$ and it’s first and second derivatives are bounded and uniformly
continuous it follows that

$\lim_{m\rightarrow\infty}\sup_{u\geq
0}E\left[\|F^{u}_{s,t}(\hat{X}^{r_{m}})-F^{u}_{s,t}(\breve{X}^{r_{m},u+s})\|\right]=0.$
(77)

and using the fact that the distribution of $\tilde{X}^{r_{m},u}$ does not
depend on $u$, that $f\in\mathcal{C}^{2}_{c}(\mathbb{R}^{I})$ (in particular
it has compact support), and that

$E\left[f(\hat{X}(t-s)+x)-f(x)-\int_{s}^{t}\mathcal{L}f(\hat{X}(z-s)+x)dz\right]=0$

for all $x\in\mathbb{R}^{I}$ we have

$\lim_{m\rightarrow\infty}\sup_{x\in\mathbb{R}^{I},u\geq
0}\left\\{E\left[f(\tilde{X}^{r_{m},u+s}(t-s)+x)-f(x)-\int_{s}^{t}\mathcal{L}f(\tilde{X}^{r_{m},u+s}(z-s)+x)dz\right]\right\\}=0.$
(78)

Recall that

$\displaystyle F^{u}_{s,t}(\breve{X}^{r_{m},u+s})=$ $\displaystyle
f(\tilde{X}^{r_{m},u+s}(t-s)+\hat{X}^{r_{m}}(u+s)-\hat{X}^{r_{m}}(u))-f(\hat{X}^{r_{m}}(u+s)-\hat{X}^{r_{m}}(u))$
$\displaystyle-\int_{s}^{t}\mathcal{L}f(\tilde{X}^{r_{m},u+s}(z-s)+\hat{X}^{r_{m}}(u+s)-\hat{X}^{r_{m}}(u))dz$

and note that since $\tilde{X}^{r_{m},u+s}$ is independent of
$\mathcal{G}^{r_{m}}(u+s)$ and $\hat{X}^{r_{m}}(u+s)-\hat{X}^{r_{m}}(u)$ is
$\mathcal{G}^{r_{m}}(u+s)$-measurable we have

$\displaystyle\sup_{u\geq
0}E\left[F^{u}_{s,t}(\breve{X}^{r_{m},u+s})|\mathcal{G}^{r_{m}}(u)\right]$
$\displaystyle\leq\sup_{x\in\mathbb{R}^{I},u\geq
0}E\left[f(\tilde{X}^{r_{m},u+s}(t-s)+x)-f(x)-\int_{s}^{t}\mathcal{L}f(\tilde{X}^{r_{m},u+s}(z-s)+x)dz\right].$
(79)

The fact that $F^{u}_{s,t}$ and $G^{u}_{s}$ are uniformly bounded in $u$ gives

$\displaystyle\lim_{m\rightarrow\infty}E\frac{2}{T^{2}_{r_{m}}}\int_{0}^{T_{r_{m}}}\int_{0}^{u}G^{u}_{s}(\hat{W}^{r_{m}}(u),\hat{X}^{r_{m}})F^{u}_{s,t}(\hat{X}^{r_{m}})G^{v}_{s}(\hat{W}^{r_{m}}(v),\hat{X}^{r_{m}})F^{v}_{s,t}(\hat{X}^{r_{m}})dvdu$
$\displaystyle=\lim_{m\rightarrow\infty}E\frac{2}{T^{2}_{r_{m}}}\int_{0}^{T_{r_{m}}}\int_{0}^{u-1}G^{u}_{s}(\hat{W}^{r_{m}}(u),\hat{X}^{r_{m}})F^{u}_{s,t}(\breve{X}^{r_{m},u+s})$
$\displaystyle\hskip
126.47249ptG^{v}_{s}(\hat{W}^{r_{m}}(v),\hat{X}^{r_{m}})F^{v}_{s,t}(\hat{X}^{r_{m}})dvdu$
$\displaystyle=\lim_{m\rightarrow\infty}E\frac{2}{T^{2}_{r_{m}}}\int_{0}^{T_{r_{m}}}\\!\\!\\!\\!\int_{0}^{u-1}\\!\\!E\left[F^{u}_{s,t}(\breve{X}^{r_{m},u+s})|\mathcal{G}^{r_{m}}(u+s)\right]\\!G^{u}_{s}(\hat{W}^{r_{m}}(u),\hat{X}^{r_{m}})$
$\displaystyle\hskip
126.47249ptG^{v}_{s}(\hat{W}^{r_{m}}(v),\hat{X}^{r_{m}})F^{v}_{s,t}(\hat{X}^{r_{m}})dvdu=0$

where the first equality comes from (77), and the second comes from the fact
that for $u-1\geq v$
$G^{u}_{s}(\hat{W}^{r_{m}}(u),\hat{X}^{r_{m}})G^{v}_{s}(\hat{W}^{r_{m}}(v),\hat{X}^{r_{m}})F^{v}_{s,t}(\hat{X}^{r_{m}})$
and $\hat{X}^{r_{m}}(u+s)-\hat{X}^{r_{m}}(u)$ are
$\mathcal{G}^{r_{m}}(u+s)$-measurable, and the third comes from (79) and (78).
Putting this all together gives

$E\left[E_{\nu^{*}(\omega)}\left[g_{s}(w,x(\cdot))(f(x(t))-f(x(s))-\int_{s}^{t}\mathcal{L}f(x(u))du\right]^{2}\right]=0.$

Proof of (a) now follows by a standard separability argument.

We will now prove part (b). Let $f\in\mathcal{C}_{c}(\mathbb{R}^{I})$ (the
space of continuous functions on $\mathbb{R}^{I}$ with compact support)) and
$s\in[0,1]$ be arbitrary From part (a),
$P_{\nu^{*}(\omega)}(x\in\mathcal{C}([0,1]:\mathbb{R}^{I})=1$ for a.e.
$\omega$ which implies

$E\left|E_{\nu^{*}(\omega)}\left[f(w)-f(\Gamma(w+x(\cdot))(s))\right]\right|=\lim_{m\rightarrow\infty}E\left|E_{\nu^{r_{m}}(\omega)}\left[f(w)-f(\Gamma(w+x(\cdot))(s))\right]\right|.$

The expectation on the right side is bounded above by

$\displaystyle
E\left[\left|\frac{1}{T_{r_{m}}}\int_{0}^{T_{r_{m}}}f(\hat{W}^{r_{m}}(u+s))-f(\Gamma(\hat{W}^{r_{m}}(u)+\hat{X}^{r_{m}}(u+\cdot)-\hat{X}^{r_{m}}(u))(s))du\right|\right]$
$\displaystyle+E\left[\left|\frac{1}{T_{r_{m}}}\int_{0}^{T_{r_{m}}}f(\hat{W}^{r_{m}}(u))-f(\hat{W}^{r_{m}}(u+s))du\right|\right].$

Due to Proposition 3.2, Proposition 3.3, Propositions 6.1 and 7.5, the
assumption that $\sup_{r}\hat{\Upsilon}^{r}<\infty$, and the fact that $f$ is
uniformly continuous and bounded we have

$\lim_{m\rightarrow\infty}E\left|\frac{1}{T_{r_{m}}}\int_{0}^{T_{r_{m}}}f(\hat{W}^{r_{m}}(u+s))-f(\Gamma(\hat{W}^{r_{m}}(u)+\hat{X}^{r_{m}}(u+\cdot)-\hat{X}^{r_{m}}(u))(s))du\right|=0.$

In addition, since $f$ is bounded and $T_{r_{m}}\rightarrow\infty$ we have

$\lim_{m\rightarrow\infty}E\left[\left|\frac{1}{T_{r_{m}}}\int_{0}^{T_{r_{m}}}[f(\hat{W}^{r_{m}}(u))-f(\hat{W}^{r_{m}}(u+s))]du\right|\right]=0.$

Together these observations show
$E\left[\left|E_{\nu^{*}(\omega)}\left[f(w)-f(\Gamma(w+x(\cdot))(s))\right]\right|\right]=0$.
By standard separability arguments it now follows that
$w\overset{d}{=}\Gamma(w+x(\cdot))(s)$ for all $s\in[0,1]$ under
$\nu^{*}(\omega)$ for a.e. $\omega$ which proves part (b). ∎

[Acknowledgments] We would like to thank Prof. Mike Harrison whose many
questions and suggestions significantly improved the exposition of the work.
Research of the first author was supported in part by awards DMS-2134107,
DMS-2152577 from the NSF.

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