t}-\mu^{\star}(S_{t};t,v)\bigg{)}^{2}dt+\frac{1}{2\sigma^{2}}\int_{0}^{T}\Big{(}\mu^{\star}(S_{t};t,v)\Big{)}^{2}dt+\frac{1}{\sigma^{2}}\int_{0}^{T}{\frac{\partial\mu_{\theta}(t,v)}{\partial
t}\sigma^{\star}(S_{t};t,v)}dW_{t}.$

Given sufficient metaorder data, the sum of the log-likelihood can be
maximized to obtain an optimal value $\hat{\theta}_{\text{MLE}}$ (if exists),
which can be seen as an estimator for

$\displaystyle\theta^{\star}=$
$\displaystyle\mathop{\arg\max}_{\theta\in\Theta}\mathbb{E}\Big{[}l(\\{S_{t}\\}_{0\leq
t\leq T}|\theta)\Big{]}$ $\displaystyle=$
$\displaystyle\mathop{\arg\min}_{\theta\in\Theta}\int_{0}^{T}\mathbb{E}\bigg{[}\bigg{(}\frac{\partial\mu_{\theta}(t,v)}{\partial
t}-\mu^{\star}(S_{t};t,v)\bigg{)}^{2}\bigg{]}dt.$

Thus, in this example, under model misspecification, the $\theta^{\star}$
corresponds to a model whose drift term (i.e.,
$\frac{\partial\mu_{\theta}(t,v)}{\partial t}$, or the gradient of impact
function $\mu_{\theta}(t,v)$ w.r.t $t$,), on average, best matches true drift
term $\mu^{\star}(S_{t};t,v)$ of $S_{t}$ in mean squared error over $[0,T]$.
This example suggests the MLE from (1) seeks to recover an impact function
that best describes the shape of the impact function by matching gradients (or
how the gradient diminishes, in presence of concavity of impact function
[Zarinelli et al., 2015]), rather than the direct value the impact function.

More importantly, how does model misspecification affect our discussion,
largely based on the Fisher information, regarding the (asymptotic)
“efficiency” of MLE? When the model is correctly specified, as mentioned in
(16), the asymptotic covariance matrix of $\hat{\theta}_{\text{MLE}}$ scaling
with $\frac{1}{\sqrt{n}}$ (i.e., the number of metaorders) is the Fisher
information matrix evaluated at the true model parameter $\theta^{\star}$,

$\mathcal{I}(\theta^{\star})=\mathcal{A}(\theta^{\star})\triangleq\mathbb{E}\bigg{[}\Big{(}\frac{\partial
l(\theta|\bm{S})}{\partial\theta}\Big{)}\Big{(}\frac{\partial
l(\theta|\bm{S})}{\partial\theta}\Big{)}^{T}\bigg{]}$

which can be equivalently defined in Hessian form, known as the information
matrix equivalence theorem [White, 1982], as

$\mathcal{A}(\theta^{\star})=\mathcal{B}(\theta^{\star}),$ (54)

where
$[\mathcal{B}]_{ij}(\theta^{\star})\triangleq-\mathbb{E}\Big{[}\frac{\partial^{2}l(\theta|\bm{S})}{\partial\theta_{i}\partial\theta_{j}}\Big{]}$.
Note both $\mathbb{E}$ above refer to expectation taken w.r.t. the true
probability measure ($\mathbb{E}_{\theta^{\star}}$ when the model is correctly
specified and is generally unknown under model misspecification). However, as
shown in [White, 1982], under model misspecification, (54) no longer holds and
the (scaled) asymptotic variance as $\hat{\theta}_{\text{MLE}}$ approaches
$\theta^{\star}_{\text{KL}}$ becomes

$\mathcal{C}(\theta^{\star}_{\text{KL}})=\mathcal{B}^{-1}(\theta^{\star}_{\text{KL}})\mathcal{A}(\theta^{\star}_{\text{KL}})\mathcal{B}^{-1}(\theta^{\star}_{\text{KL}}).$

Moreover, [White, 1982] shows the discrepancies between empirical estimates of
$\mathcal{A}(\theta^{\star}_{\text{KL}})$ and
$\mathcal{B}(\theta^{\star}_{\text{KL}})$ can be used to test model
misspecification. However, this implies the conclusion drawn from our previous
theorems no longer holds, as they rely on the comparison of the Fisher
information of correctly specified models. In this case, it becomes less clear
how to choose the appropriate models or consider estimation efficiency of MLE
based on experiment design, as theoretical properties of
$\mathcal{A}(\theta^{\star}_{\text{KL}})$ or
$\mathcal{B}(\theta^{\star}_{\text{KL}})$ depends on the unknown, true
distribution of the price trajectory.

Thus, the presence of model misspecification will pose a limitation to the
results of previous theorems. To better discuss the extent of such limitations
and possible remedies, we revisit relevant concepts on model selection from
statistical learning/machine learning.

###  5.2 Simulation

In this section, we perform a simulation study for the model misspecification
case. The estimator model follows the power-law kernel $G(s)=s^{-\gamma}$ and
power-law impact $f(v)=v^{\delta}$ (so
$\mu_{\theta}(t,v)=\frac{v^{\delta}t^{1-\gamma}}{1-\gamma}$), where $\delta$
is the only parameter that needs to be estimated. However, the sample-
generating model does not have the same parametric format. The generating
model has $G(s)=s^{-\gamma}$ and $f(v)=1.5\ln(1+0.5v^{2}+0.7v)$ (so
$\mu^{\star}(t,v)=1.5\ln(1+0.5v^{2}+0.7v)\frac{t^{1-\gamma}}{1-\gamma}$). We
set $\sigma=0.2$, $\gamma=0.4$. Each estimation has 300 samples, and each
sample trajectory is assigned a random price trajectory length
$T\sim\mathrm{Uniform}(0.1,0.15)$ and a random
$v\sim\mathrm{Uniform}(0.3,0.4)$. The simulation is repeated 40,000 times for
each scenario. The diffusion process is discretized into $1024$ time bins. The
first model is based on the average cost of VWAP as in (39). Other models rely
on the price trajectory as discussed in section 3.2 with a subset or all
points at $t_{1}=\frac{T}{8},t_{2}=\frac{T}{2},t_{3}=\frac{5T}{8}$ and
$t_{4}=T$ relative to the corresponding path length $T$. The error of
estimation is the mean squared error of the cost across all sample paths.
Results are shown in Table 7. The conclusion is similar to the well-specified
model in section 4.3 and the simulation study therein that the cost-based
method is worse than the price trajectory-based methods if earlier price
trajectory points are selected (for example $t_{1}$ or $t_{2}$). In addition,
if more trajectory points are chosen or the earlier the points are selected,
the performance will be better. The performance is evaluated using the squared
error between the true cost and the estimated cost, and the parenthesis shows
the standard error of the mean error. $X=J$ is the model based on the average
cost of VWAP, others are based on the points along the price trajectory.
$t_{1}=\frac{T}{8},t_{2}=\frac{T}{2},t_{3}=\frac{5T}{8}$ and $t_{4}=T$. Bold
numbers indicate the method has a statistically significant better performance
than using $J$ based on two sample t-tests.

Estimator via $X$ | mean error $\times 10^{-7}$ (SD $\times 10^{-9}$)
---|---
$X=J$ |  $1.079$ ($4.831$)
$X=S_{t_{1}},S_{T}$ |  1.049 ($4.612$)
$X=S_{t_{2}},S_{T}$ |  1.077 ($4.822$)
$X=S_{t_{3}},S_{T}$ |  $1.144$ ($5.294$)
$X=S_{t_{1}},S_{t_{2}},S_{T}$ |  1.038 ($4.538$)
$X=S_{t_{1}},S_{t_{3}},S_{T}$ |  1.040 ($4.546$)
$X=S_{t_{2}},S_{t_{3}},S_{T}$ |  1.073 ($4.788$)
$X=S_{t_{1}},S_{t_{2}},S_{t_{3}},S_{T}$ |  1.034 ($4.509$)

Table 7: Numerical comparison of models with misspecification.

###  5.3 Model Selection

As in (52), given $(T,v)\sim G_{\text{order}}$, we use $F$ to denote the true
distribution of price trajectory $\bm{S}$ and $F(\theta)$ the one
characterized under $S_{t}=S_{0}+\mu_{\theta}(t,v)+\sigma W_{t}$ for
$S_{t}\in\bm{S}$ and $(T,v)\sim G_{\text{order}}$. Then, as discussed in the
previous section, if we measure the risk as

$R(\theta)=D_{\text{KL}}(F\lVert F(\theta)),$

which quantifies the discrepancy between distributions fitted by the model and
truth. Then, the model selection problem decomposes the risk as

$\displaystyle R(\theta)=$
$\displaystyle\underbrace{R(\theta)-\inf_{\theta\in\Theta}R(\theta)}_{\text{estimation
error}}+\underbrace{\inf_{\theta\in\Theta}R(\theta)}_{\text{approximation
error}}$ $\displaystyle=$
$\displaystyle\underbrace{R(\theta)-R(\theta^{\star}_{\text{KL}})}_{\text{estimation
error}}+\underbrace{R(\theta^{\star}_{\text{KL}})}_{\text{approximation
error}}$ (55)

In a correctly specified model, $\theta^{\star}_{\text{KL}}=\theta^{\star}$
and the estimation error is 0. The approximation error is directly related to
the efficiency for estimating $\theta^{\star}$ and the results of our previous
theorem, based on the Fisher information matrix, can be applied. When a model
misspecification occurs, different factors/concerns can come into play and the
landscape is less clear. For example,

  * •

If one is concerned with an accurate estimation of the price trajectory for
various values of $(T,v)$, then the estimation using the price trajectory
would increase the sample size, compared to using summary statistics, e.g.,
VWAP, which could lead to a positive effect on reducing estimation error. On
the other hand, depending on how the distribution is misspecified, e.g., heavy
tail versus light tail, the method using summary statistics could turn out to
be more robust than using the entire trajectory. In this case, the previous
comparison between using different summary statistics (VWAP or sampled price
points) may or may not hold.

  * •

If one is only concerned with an accurate estimation of summary statistics,
e.g., VWAP. Then, this could lead to a reduced approximation error because KL
divergence monotonically decreases under the non-invertible transformation (it
is well known that KL divergence is invariant under invertible
transformations, i.e., given random variable $X,Y$ and an invertible
transformation $g(\cdot)$, we have
$D_{\text{KL}}(X|Y)=D_{\text{KL}}(g(X)|g(Y))$. However, it can be shown that
$D_{\text{KL}}(X|Y)\geq D_{\text{KL}}(g(X)|g(Y))$ when $g$ is non-invertible
(see [Mena et al., 2018]). The transformation from the price trajectory
$\bm{S}$ to summary statistics are typically non-invertible.). However, it
could lead to an increase in approximation error due to reduced sample size or
inefficiencies of estimation.

During practical implementation, we should take these issues into
consideration. Typical methods include cross-validation upon different methods
(or some hybrid of them) under the specified loss function or Bayesian
inference based on carefully calibrated prior from previous data (or even
online data, where one can update the posterior as the order is executed and
adjust the execution). This is left for discussion and future research.

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##  Appendix A Appendix

###  A.1 Asymptotic Inference

####  A.1.1 Proof of Basic Results and Review

######  Definition 4 (Differentiability in Quadratic Mean).

A family of model $\\{\mathbb{P}_{\theta}\\}_{\Theta}$ with dominating measure
$m$ is differentiable in quadratic mean (q.m.d.) if there exists a vector of
measurable function
$\dot{l}_{\theta}=(\dot{l}_{\theta,1},...,\dot{l}_{\theta,K})$ such that

$\int\bigg{[}\sqrt{p_{\theta+h}}-\sqrt{p_{\theta}}-\frac{1}{2}h^{T}\dot{l}_{\theta}\sqrt{p_{\theta}}\bigg{]}dm=o(\|h^{2}\|)\text{
as }h\rightarrow 0.$

######  Definition 5 (Regular Estimator).

For a parametric family of probability measure
$\\{\mathbb{P}_{\theta}\\}_{\Theta}$, a sequence of estimators
$\\{T_{n}\\}_{n\geq 0}$ is called regular at $\theta$ for estimating
$\phi(\theta)$ if, for every $h$,

$\sqrt{n}\Big{(}T_{n}-\phi(\theta+\frac{h}{\sqrt{n}})\Big{)}\xrightarrow[\mathbb{P}_{\theta+\frac{h}{\sqrt{n}}}]{d}L_{\theta},$

i.e., the left hand side converges in distribution under
$\mathbb{P}_{\theta+\frac{h}{\sqrt{n}}}$ to some random variable $L_{\theta}$
which do not depend on $h$.

###### Proof of Proposition 2.

The proof mainly follows from basic results in [Van der Vaart, 2000] and
[Bickel et al., 1993]. Given assumption 3.3, we derive that
$\theta\rightarrow\sqrt{p_{\theta}(\bm{x})}$ is continuously for all $\bm{x}$.
Assumption 4 ensures that $I_{\bm{X}}(\theta)$ is well-defined and continuous
in $\theta$. Assumption 1 gives openness of $\Theta$. As as result, according
to Lemma 7.6 in [Van der Vaart, 2000] or Proposition 1 in [Bickel et al.,
1993], the model $\\{\mathbb{P}_{\theta}\\}_{\Theta}$ is q.m.d. with
$\dot{l}_{\theta}$ given by $\frac{\dot{p}_{\theta}}{p_{\theta}}$.

Then, we can use Assumptions 2 and 3.4 to derive

$|\log p_{\theta_{1}}(\bm{x})-\log p_{\theta_{2}}(\bm{x})|\leq
L\|\theta_{1}-\theta_{2}\|,$

for some $\theta_{1},\theta_{2}$ in the neighborhood of $\theta^{\star}$.
Assumption 3.1 gives consistency of $\hat{\theta}_{n}$ by Theorem 5.14 of [Van
der Vaart, 2000]. These two results, in addition to assumption 3.5 and q.m.d.
give (16), according to Theorem 7.12 in [Van der Vaart, 2000].

Finally, (17) follows by Theorem 8.11 in [Van der Vaart, 2000], using q.m.d.
and assumption 4. ∎

The locally asymptotic minimax theorem is only one of many ways to argue the
asymptotic optimality of MLE. For example, the Hájek-LeCam convolution theorem
and its variant show that for any regular estimator, the MLE attains the
“best” limiting distribution asymptotically and achieves the lowest possible
variance for any asymptotically regular sequence of the estimator. Moreover,
any improvement over this limit distribution can only be made on a Lebesgue
null set of parameters. For a comprehensive review, see section 8 of [Van der
Vaart, 2000].

####  A.1.2 Technical Lemma

The appendix mainly consists of technical proofs and additional lemmas. The
proof of the following support lemma can be found in [Zamir, 1998, Zegers,
2015].

######  Lemma 6 (Chain Rule for Fisher Information).

Let $X$ and $Y$ be two statistics of a given experiments
$(\bm{S},\mathcal{F},\\{\mathbb{P}_{\theta}\\}_{\theta\in\Theta})$, then

$\displaystyle\mathcal{I}_{X,Y}(\theta)=\mathcal{I}_{X}(\theta)+\mathcal{I}_{Y|X}(\theta),$

where $\mathcal{I}_{Y|X}(\theta)=\mathbb{E}_{X}[\mathcal{I}_{Y|X=x}(\theta)]$
is simply defined as

$\mathcal{I}_{Y|X=x}(\theta)=\mathbb{E}_{\theta}\bigg{[}\Big{(}\frac{\partial
l(Y|X=x,\theta)}{\partial\theta}\Big{)}\Big{(}\frac{\partial
l(Y|X=x,\theta)}{\partial\theta}\Big{)}^{T}\bigg{]}.$

More generally, if we have $n$ statistics $X_{1},X_{2},...,X_{n}$, then

$\mathcal{I}_{X_{1},X_{2},...,X_{n}}(\theta)=\mathcal{I}_{X_{1}}(\theta)+\sum_{i=1}^{n-1}\mathcal{I}_{X_{i+1}|X_{1},...,X_{i}}(\theta)$

where
$\mathcal{I}_{X_{i+1}|X_{1},...,X_{i}}(\theta)=\mathbb{E}_{X_{1},...,X_{i}}[\mathcal{I}_{X_{i+1}|X_{1},...,X_{i}=x_{1},...,x_{i}}(\theta)]$.

###  A.2 Proofs

####  A.2.1 The Proof of Lemma 3

###### Proof of Lemma 3.

The rigorous proof can be found in Theorem 7.2 of [Ibragimov and Has’ Minskii,
2013]. It can be checked that the assumptions in Theorem 7.2 are all
satisfied. For a more concise proof, similar to the proof in [Zamir, 1998], we
can use Lemma 6 to obtain:

$\mathcal{I}_{\bm{X}}(\theta)+\mathbb{E}_{\bm{X}}[\mathcal{I}_{\phi(\bm{X})|\bm{X}=\bm{x}}(\theta)]=\mathcal{I}_{\bm{X},\phi(\bm{X})}(\theta)=\mathcal{I}_{\phi(\bm{X})}(\theta)+\mathbb{E}_{\phi(\bm{X})}[\mathcal{I}_{\bm{X}|\phi(\bm{X})=\phi(\bm{x})}(\theta)].$
(56)

Since $\phi(\bm{X})$ is a statistic of $\bm{X}$, it is clear that the
conditional distribution of $\phi(\bm{X})$ given $\bm{X}$ is independent of
$\theta$, implying that $\mathcal{I}_{\phi(\bm{X})|\bm{X}}(\theta)=0$.
Consequently, (56) now becomes

$\mathcal{I}_{\bm{X}}(\theta)=\mathcal{I}_{\phi(\bm{X})}(\theta)+\mathbb{E}_{\phi(\bm{X})}[\mathcal{I}_{\bm{X}|\phi(\bm{X})=\phi(\bm{x})}(\theta)].$

Since $\mathcal{I}_{\bm{X}|\phi(\bm{X})}(\theta)\succcurlyeq{0}$ (any Fisher
information is positive semi-definite as it is the covariance matrix of the
score function), we have shown that

$\mathcal{I}_{\bm{X}}(\theta)\succcurlyeq\mathcal{I}_{\phi(\bm{X})}(\theta).$

Now, if $\phi(\bm{X})$ is a sufficient statistic, then the conditional
distribution of $\bm{X}$ given $\phi(\bm{X})$ is independent of $\theta$ which
leads to $\mathcal{I}_{\bm{X}|\phi(\bm{X})}(\theta)=0$ and the equality of
$\mathcal{I}_{\bm{X}}(\theta)=\mathcal{I}_{\phi(\bm{X})}(\theta)$. On the
other hand, if
$\mathcal{I}_{\bm{X}}(\theta)=\mathcal{I}_{\phi(\bm{X})}(\theta)$, we must
have
$E_{\phi(\bm{X})}[\mathcal{I}_{\bm{X}|\phi(\bm{X})=\phi(\bm{s})}(\theta)]=0$,
but this does not necessarily imply $\phi(\bm{X})$ is a sufficient statistic.
An interesting counter-example where an insufficient statistic preserves
Fisher information can be found in [Kagan and Shepp, 2005]. To complete the
proof, an additional condition on the continuous differentiability of density
is enough to establish the sufficiency of $\phi(\bm{X})$, the detailed proof
again is shown in [Kagan and Shepp, 2005]. ∎

####  A.2.2 The Proof of Lemma 4

###### Proof of Lemma 4.

The first claim follows directly from Lemma 6 since
$\mathcal{I}_{\bm{Y}|\bm{X}=\bm{x}}(\theta)\succcurlyeq{0}$ and
$\mathbb{E}_{\bm{X}}[\mathcal{I}_{\bm{Y}|\bm{X}=\bm{x}}(\theta)]\succcurlyeq{0}$.
For the second claim, notice by Lemma 6,

$\mathcal{I}_{\bm{Y}}(\theta)+\mathbb{E}_{\bm{Y}}[\mathcal{I}_{\bm{X}|\bm{Y}=\bm{y}}(\theta)]=\mathcal{I}_{\bm{X},\bm{Y}}(\theta)=\mathcal{I}_{\bm{X}}(\theta)+\mathbb{E}_{\bm{X}}[\mathcal{I}_{\bm{Y}|\bm{X}=\bm{x}}(\theta)].$

Since $\bm{Y}=\phi(\bm{X})$ and $\phi(\cdot)$ is a bijective mapping, it is
easy to check that
$\mathbb{E}_{\bm{X}}[\mathcal{I}_{\bm{Y}|\bm{X}=\bm{x}}(\theta)]=E_{\bm{Y}}[\mathcal{I}_{\bm{X}|\bm{Y}=\bm{y}}(\theta)]=0$
which implies $\mathcal{I}_{\bm{X}}(\theta)=\mathcal{I}_{\bm{Y}}(\theta)$. ∎

###  A.3 The Joint Distribution of $(I,J)$ in Almgren-Chriss Model

In this derivation, we fix $(v,T)$ as constant and derive the distribution of
$(I,J)$ conditional on $(v,T)$. Notice that, after the affine transformation
$P_{t}=\frac{S_{t}-S_{0}}{S_{0}}$, we have

$\displaystyle P_{t}=$ $\displaystyle g(v)t+h(v)+\sigma W_{t},\text{ when
$t\leq T$}$ $\displaystyle P_{t}=$ $\displaystyle g(v)T+\sigma W_{t},\text{
when $t>T$. }$

Following the notations in [Almgren et al., 2005], we let
$\bar{S}=\frac{\int_{0}^{t}S_{t}\dot{x}_{t}dt}{X}=\frac{\int_{0}^{T}S_{t}dt}{T}$.
First, following (3), we have

$\displaystyle I=P_{T_{\text{post}}}=Tg(v)+\sigma
W_{T_{\text{post}}},\implies$ $\displaystyle
I\sim\mathcal{N}(Tg(v),\sigma^{2}T_{\text{post}})$ $\displaystyle
J=\frac{\bar{S}-S_{0}}{S_{0}}=\frac{1}{T}\int_{0}^{T}P_{t}dt=\frac{T}{2}g(v)+h(v)+\sigma\frac{1}{T}\int_{0}^{T}W_{t}dt,\implies$
$\displaystyle
J\sim\mathcal{N}\big{(}\frac{T}{2}g(v)+h(v),\sigma^{2}\frac{T}{3}\big{)},$

where we know
$\int_{0}^{T}W_{t}dt=\int_{0}^{T}(T-t)dW_{t}\sim\mathcal{N}(0,\int_{0}^{T}(T-t)^{2}dt)=\mathcal{N}(0,\frac{T^{3}}{3})$
by Itô’s Lemma. Since $I$ and $J$ are clearly jointly normal and we have known
their mean and variance, we only need to calculate their correlation to
determine the joint distribution. First, we can calculate
$\text{Cov}(\int_{0}^{T}W_{t}dt,W_{T_{\text{post}}})$:

$\displaystyle\text{Cov}(\int_{0}^{T}W_{t}dt,W_{T_{\text{post}}})=$
$\displaystyle\text{Cov}(\int_{0}^{T}W_{t}dt,W_{T}+W_{T_{\text{post}}}-W_{T})$
$\displaystyle=$ $\displaystyle\text{Cov}(\int_{0}^{T}W_{t}dt,W_{T})$
$\displaystyle=$
$\displaystyle\mathbb{E}W_{T}\int_{0}^{T}W_{t}dt=\int_{0}^{T}\mathbb{E}[W_{T}W_{t}]dt=\int_{0}^{T}tdt=\frac{T^{2}}{2}.$

On the other hand, again using Itô’s Lemma, we have
$\text{Var}(\int_{0}^{T}W_{t}dt)=\frac{T^{3}}{3}$ and
$\text{Var}(W_{T_{\text{post}}})=T_{\text{post}}$. Thus, it can be checked
that
$\text{Corr}(\int_{0}^{T}W_{t}dt,W_{T_{\text{post}}})=\sqrt{\frac{3T}{4T_{\text{post}}}}$.
This allows characterizing the joint distribution of $I$ and $J$ completely,
equivalent as in equation (2) of [Almgren et al., 2005], by noting that there
exist two independent standard normal random variables $\xi_{1},\xi_{2}$ where

$\displaystyle W_{T_{\text{post}}}=$
$\displaystyle\sqrt{T_{\text{post}}}\xi_{1}$
$\displaystyle\int_{0}^{T}W_{t}dt=$
$\displaystyle\frac{T^{2}}{2\sqrt{T_{\text{post}}}}\xi_{1}+T\sqrt{\frac{T}{12}(4-3\frac{T}{T_{\text{post}}})}\cdot\xi_{2}$
(57)

which can then be shown to be equivalent as (3).

####  A.3.1 Proof of Lemma 5

Based on assumption 1, 2 on the lower and upper boundedness of $v$ and $T$,
and the specific form of (3), one can routinely check assumption 3.1,3.3,3.4
are satisfied. For assumption 3.4, we note that, by (3) the $\Sigma_{I,J}(T)$
in Gaussian variable (5) is given by

$\Sigma_{I,J}(T)\triangleq\sigma^{2}\begin{bmatrix}T_{\text{post}}&-\frac{T_{\text{post}}-T}{2}\\\
-\frac{T_{\text{post}}-T}{2}&\frac{T_{\text{post}}}{4}-\frac{T}{6}\end{bmatrix}.$

Define the ratio $\tau=\frac{T_{\text{post}}}{T}-1$, we note $\tau$ is random
but $\tau>0$. Then, one can write

$\Sigma_{I,J}(T)\triangleq\sigma^{2}T\begin{bmatrix}1+\tau&-\frac{\tau}{2}\\\
-\frac{\tau}{2}&\frac{1+\tau}{4}-\frac{1}{6}\end{bmatrix}\triangleq\sigma^{2}T\Sigma_{\tau}.$

The smallest eigenvalue $\lambda_{1}(\tau)$ of $\Sigma_{\tau}$ can be
explicitly computed by $a-\sqrt{a^{2}-b}$ where
$a=\frac{5(1+\tau)}{8}-\frac{1}{12}$ is the mean of diagonal elements and
$b=\frac{(1+\tau)^{2}}{4}-\frac{(1+\tau)}{6}-\frac{\tau^{2}}{4}$ is the
determinant. For $\tau>0$, $\lambda_{1}(\tau)$ is lower bounded by
$\lambda_{1}(0)$ which is a positive constant. Thus, by assumption 2, we have

$\Sigma_{I,J}(T)=\sigma^{2}T\Sigma_{\tau}\succcurlyeq\sigma^{2}T_{L}\lambda_{1}(0)\textbf{I}.$

For assumption 3.5, the uniqueness of $\theta^{\star}$ can be violated only if
there exists
$\theta^{\prime}=(\gamma^{\prime},\eta^{\prime},\alpha^{\prime},\beta^{\prime})$
such that

$\displaystyle\gamma^{\prime}v^{\alpha^{\prime}}T=$
$\displaystyle\gamma^{\star}v^{\alpha^{\star}}T$
$\displaystyle\eta^{\prime}v^{\beta^{\prime}}=$
$\displaystyle\eta^{\star}v^{\beta^{\star}},$ (58)

for all $(v,T)$, $G_{\text{order}}$ almost surely. This is impossible when the
support of the marginal distribution of $v$ from $G_{\text{order}}$ has
infinite cardinality (or finite and sufficiently large). For assumption,
recall from (3) that,

$\displaystyle\mathcal{I}_{I,J}(\theta)=$ $\displaystyle\mathbb{E}_{(v,T)\sim
G_{\text{order}}}\bigg{[}\mathcal{J}^{T}_{I,J}(\theta,T,v)\cdot\sigma^{-2}\begin{bmatrix}\frac{1}{T}\frac{T_{\text{post}}/4-T/6}{T_{\text{post}}/3-T/4}&\frac{1}{T}\frac{T_{\text{post}}/2-T/2}{T_{\text{post}}/3-T/4}\\\
\frac{1}{T}\frac{T_{\text{post}}/2-T/2}{T_{\text{post}}/3-T/4}&\frac{1}{T}\frac{T_{\text{post}}}{T_{\text{post}}/3-T/4}\end{bmatrix}\cdot\mathcal{J}_{I,J}(\theta,T,v)\bigg{]},$
$\displaystyle\mathcal{J}_{I,J}(\theta,T,v)=$
$\displaystyle\begin{bmatrix}Tv^{\alpha}&0&T\gamma v^{\alpha}\ln(v)&0\\\
0&v^{\beta}&0&\eta v^{\beta}\ln(v)\end{bmatrix}.$

In order to show $\mathcal{I}_{I,J}(\theta)\succ 0$, it suffices to show that,
for any $\bm{x}\in\mathbb{R}^{K}$ with $\|\bm{x}\|=1$, we have

$\bm{x}^{T}\mathcal{I}_{I,J}(\theta)\bm{x}>0.$

The non-singularity of $\Sigma_{I,J}$ and form of $\mathcal{I}_{I,J}(\theta)$
ensures $\bm{x}^{T}\mathcal{I}_{I,J}(\theta)\bm{x}\geq 0$ for all such
$\bm{x}$. If there exists one such $\bm{x}$ that
$\bm{x}^{T}\mathcal{I}_{I,J}(\theta)\bm{x}=0$, we must have

$\bm{x}^{T}\mathcal{J}_{I,J}(\theta,T,v)\Sigma_{I,J}^{-1}(T)\mathcal{J}_{I,J}(\theta,T,v)\bm{x}=0$

for all $(v,T)\sim G_{\text{order}}$ almost surely. To see why, simply note
that, for a random variable $X\geq 0$, $\mathbb{E}[X]=0$ if and only if
$\mathbb{P}(X=0)=1$. Due to the non-singularity of $\Sigma_{I,J}(T)$, this
requires $\mathcal{J}_{I,J}(\theta,T,v)\bm{x}=0$ for all $(v,T)\sim
G_{\text{order}}$ almost surely:

$\displaystyle x_{1}Tv^{\alpha}+x_{3}T\gamma v^{\alpha}\ln(v)=$ $\displaystyle
0$ $\displaystyle x_{2}v^{\beta}+x_{4}\eta v^{\beta}\ln(v)=0$

for all $(v,T)\sim G_{\text{order}}$ almost surely. This is not possible for
any $\bm{x}=1$, when the support of marginal distribution of $v$ from
$G_{\text{order}}$ has infinite cardinality (e.g., $x_{1}+x_{3}\ln(v)=0$ for
all $v$ iff $x_{1}=x_{3}=0$).

####  A.3.2 Derivation of $\mathcal{I}_{P_{T},P_{T_{\text{post}}}}(\theta)$ in
Almgren-Chriss Model

In this derivation, we fix $(v,T)$ as constant. By Lemma 4, we have
$\mathcal{I}_{P_{T},P_{T_{\text{post}}}}(\theta)=\mathcal{I}_{P_{T_{\text{post}}},P_{T_{\text{post}}}-P_{T}}(\theta)$.
To calculate
$\mathcal{I}_{P_{T_{\text{post}}},P_{T_{\text{post}}}-P_{T}}(\theta)$, note
that:

$\displaystyle\begin{pmatrix}P_{T_{\text{post}}}\\\
P_{T}-P_{T_{\text{post}}}\end{pmatrix}\sim\mathcal{N}\begin{pmatrix}\begin{pmatrix}Tg(v;\theta)\\\
h(v;\theta)\end{pmatrix},\sigma^{2}\begin{bmatrix}T_{\text{post}}&-(T_{\text{post}}-T)\\\
-(T_{\text{post}}-T)&T_{\text{post}}-T\end{bmatrix}\end{pmatrix}.$

The rest follows from standard calculation based on Lemma 1 and let $(v,T)\sim
G_{\text{order}}$.

####  A.3.3 Proof of Theorem 1

###### Proof of Theorem 1.

Throughout the proof, we assume $(v,T)$ is fixed, i.e., $G_{\text{order}}$ is
a point mass. Results for generic $G_{\text{order}}$ follow trivially since,
for any random variable, $X\geq Y$ implies $\mathbb{E}[X]\geq\mathbb{E}[Y]$.

Following (28), we can see that,
$(P_{T},\frac{P_{t^{\prime}_{1}}}{t^{\prime}_{1}}-\frac{P_{T_{\text{post}}}-P_{T}}{T_{\text{post}-T}})$
is sufficient for experiment based on $\bm{S}$, thus we also have

$\mathcal{I}_{\bm{S}}(\theta)=\mathcal{I}_{P_{T},\phi_{\tau_{1}}(P_{t_{1}},P_{T},P_{T_{\text{post}}})}(\theta).$
(59)

where
$\phi_{\tau_{1}}(P_{t_{1}},P_{T},P_{T_{\text{post}}})=\frac{P_{t_{1}}}{t_{1}}-\frac{P_{T_{\text{post}}}-P_{T}}{T_{\text{post}-T}}$
and $\tau_{1}$ is fixed for the experiment. Together with (29), we can see
that, for $(P_{t},P_{T},P_{T_{\text{post}}})$,

$\mathcal{I}_{P_{t},P_{T},P_{T_{\text{post}}}}(\theta)=\mathcal{I}_{P_{T},\phi_{\tau}(P_{t},P_{T},P_{T_{\text{post}}})}(\theta).$
(60)

Now, we can check

$\displaystyle\begin{pmatrix}P_{T}\\\
t\cdot\phi_{\tau}(P_{t},P_{T},P_{T_{\text{post}}})\end{pmatrix}\sim\mathcal{N}\begin{pmatrix}\bar{\mu}(\theta),\bar{\Sigma},\end{pmatrix}$
(61)

where

$\displaystyle\bar{\mu}(\theta)\triangleq\begin{pmatrix}Tg(v;\theta)+h(v;\theta)\\\
tg(v;\theta)+(1+\frac{t}{T_{\text{post}}-T})h(v;\theta)\end{pmatrix},\bar{\Sigma}\triangleq\sigma^{2}\begin{bmatrix}T&t\\\
t&t+\frac{t^{2}}{T_{\text{post}}-T}\end{bmatrix}.$ (62)

Recall the definition in (3) where

$\displaystyle{\mu}(\theta)\triangleq\begin{pmatrix}Tg(v;\theta)\\\
h(v;\theta)\end{pmatrix}.$

Now define the matrix

$\displaystyle A\triangleq\begin{bmatrix}1&1\\\
\frac{t}{T}&1+\frac{t}{T_{\text{post}}-T}\end{bmatrix}.$

It is straightforward to check that for all $t\in(0,T)$ and
$T_{\text{post}}>T$, $A$ is non-singular and

$\bar{\mu}(\theta)=A\mu(\theta).$

Thus, if we define $U\in\mathbb{R}^{2}$ as

$U\triangleq A^{-1}\begin{bmatrix}P_{T}\\\
t\cdot\phi_{\tau}(P_{t},P_{T},P_{T_{\text{post}}})\end{bmatrix},$ (63)

we can check that

$U\sim\mathcal{N}\big{(}\mu(\theta),A^{-1}\bar{\Sigma}(A^{-1})^{T}\big{)}$
(64)

Since $A$ is non-singular, we know from Lemma 4 and (60) that

$\displaystyle\mathcal{I}_{U}(\theta)=\mathcal{I}_{P_{T},t\cdot\phi_{\tau}(P_{t},P_{T},P_{T_{\text{post}}})}(\theta)=\mathcal{I}_{P_{t},P_{T},P_{T_{\text{post}}}}(\theta).$
(65)

Thus, it suffices to calculate $\mathcal{I}_{U}(\theta)$, for which we use
Lemma 1 on (64) to obtain:

$\mathcal{I}_{P_{t},P_{T},P_{T_{\text{post}}}}(\theta)=\mathcal{I}_{U}(\theta)=\mathcal{J}^{T}_{I,J}(\theta)\big{(}A^{T}\bar{\Sigma}^{-1}A\big{)}\mathcal{J}_{I,J}(\theta),$

where we can carry out direct calculations to get

$A^{T}\bar{\Sigma}^{-1}A=\sigma^{-2}\begin{bmatrix}\frac{1}{T}&\frac{1}{T}\\\
\frac{1}{T}&\frac{1}{t}+\frac{1}{T_{\text{post}}-T}\end{bmatrix}.$

Recall from equation (3) that

$\mathcal{I}_{I,J}(\theta)=\mathcal{J}^{T}_{I,J}(\theta)\cdot\sigma^{-2}\begin{bmatrix}\frac{1}{T}\frac{T_{\text{post}}/4-T/6}{T_{\text{post}}/3-T/4}&\frac{1}{T}\frac{T_{\text{post}}/2-T/2}{T_{\text{post}}/3-T/4}\\\
\frac{1}{T}\frac{T_{\text{post}}/2-T/2}{T_{\text{post}}/3-T/4}&\frac{1}{T}\frac{T_{\text{post}}}{T_{\text{post}}/3-T/4}\end{bmatrix}\cdot\mathcal{J}_{I,J}(\theta).$

In order to show
$\mathcal{I}_{U}(\theta)\succcurlyeq\mathcal{I}_{I,J}(\theta)$ for generic
form of $h,g$, it suffices to show that

$\begin{bmatrix}\frac{1}{T}&\frac{1}{T}\\\
\frac{1}{T}&\frac{1}{t}+\frac{1}{T_{\text{post}}-T}\end{bmatrix}\succcurlyeq\begin{bmatrix}\frac{1}{T}\frac{T_{\text{post}}/4-T/6}{T_{\text{post}}/3-T/4}&\frac{1}{T}\frac{T_{\text{post}}/2-T/2}{T_{\text{post}}/3-T/4}\\\
\frac{1}{T}\frac{T_{\text{post}}/2-T/2}{T_{\text{post}}/3-T/4}&\frac{1}{T}\frac{T_{\text{post}}}{T_{\text{post}}/3-T/4}\end{bmatrix}.$
(66)

Define $\tau_{1}\triangleq\frac{T_{\text{post}}}{T}$ and
$\tau_{2}\triangleq\frac{T}{t}$ (notice $\tau_{1},\tau_{2}>1$), showing (66)
is equivalent as showing:

$B\triangleq\begin{bmatrix}1-\frac{\tau_{1}/4-1/6}{\tau_{1}/3-1/4}&1-\frac{\tau_{1}/2-1/2}{\tau_{1}/3-1/4}\\\
1-\frac{\tau_{1}/2-1/2}{\tau_{1}/3-1/4}&\tau_{2}+\frac{1}{\tau_{1}-1}-\frac{\tau_{1}}{\tau_{1}/3-1/4}\end{bmatrix}\succcurlyeq{0}.$

To check the positive semi-definiteness of the above matrix, we use
Sylvester’s criterion, which reduces to checking:

$\begin{cases}[B]_{11}=1-\frac{\tau_{1}/4-1/6}{\tau_{1}/3-1/4}\geq
0,&\text{}\\\
[B]_{22}=\tau_{2}+\frac{1}{\tau_{1}-1}-\frac{\tau_{1}}{\tau_{1}/3-1/4}\geq
0&\text{}\\\ \text{det}(B)\geq{0}&\text{}\end{cases}.$ (67)

For $[B]_{11}$, it is easy to check that $[B]_{11}>0$ whenever $\tau_{1}>1$,
which is satisfied by definition of
$\tau_{1}\triangleq\frac{T_{\text{post}}}{T}$. For $\text{det}(B)$, we wanted
to check for

$\text{det}(B)=(1-\frac{\tau_{1}/4-1/6}{\tau_{1}/3-1/4})(\tau_{2}+\frac{1}{\tau_{1}-1}-\frac{\tau_{1}}{\tau_{1}/3-1/4})-(1-\frac{\tau_{1}/2-1/2}{\tau_{1}/3-1/4})^{2}\geq
0.$ (68)

By multiplying a strictly positive term $12^{2}(\tau_{1}/3-1/4)^{2}$, one can
check (68) is greater than 0 if and only if

$(\tau_{2}(\tau_{1}-1)+1)(4\tau_{1}-3)-12\tau_{1}(\tau_{1}-1)-(2\tau_{1}-3)^{2}\geq
0$

which simplifies to

$(\tau_{2}-4)(4\tau_{1}^{2}-7\tau_{1}+3)\geq 0.$

One can check $4\tau_{1}^{2}-7\tau_{1}+3>0$ for all $\tau_{1}>1$. Thus,
$\text{det}(B)\geq 0$ if and only if $\tau_{2}\geq 4$. So far, the validity of
(67) requires $\tau_{1}>1$ and $\tau_{2}\geq 4$. Finally, for $[B]_{22}$, by
multiplying a strictly positive term $12(\tau_{1}-1)(\tau_{1}/3-1/4)$, it can
be checked that $[B]_{22}\geq 0$ if and only if

$4(\tau_{2}-3)\tau_{1}^{2}+(16-7\tau_{2})\tau_{1}+3(\tau_{2}-1)\geq 0.$ (69)

For $\tau_{2}\geq 4$, (69) is a quadratic function in $\tau_{1}$ with positive
quadratic coefficient. The discriminant of (69) is

$\Delta=(16-7\tau_{2})^{2}-48(\tau_{2}-3)(\tau_{2}-1),$ (70)

which satisfies $\Delta<0$ (no roots) if and only if $4<\tau_{2}<28$. Thus,
for $4<\tau_{2}<28$, $[B]_{22}>0$ for all $\tau_{1}>1$. When $\tau_{2}=4$, we
have $\Delta=0$ and $[B]_{22}\geq 0$ for all $\tau_{1}>1$. Thus, On the other
hand, when $\tau_{2}\geq 28$, (69) has two roots (possibly with multiplicity):

$\frac{7\tau_{2}-16\pm\sqrt{\Delta}}{8\tau_{2}-24}.$ (71)

However, it can be checked that, when $\tau_{2}\geq 28$,
$(\tau_{2}-8)^{2}-(\tau_{2}-4)(\tau_{2}-28)=16(\tau_{2}-3)>0$, thus we have

$\sqrt{\Delta}=\sqrt{(\tau_{2}-4)(\tau_{2}-28)}\leq\sqrt{(\tau_{2}-8)^{2}}=\tau_{2}-8.$

Consequently, for $\tau_{2}\geq 28$, we have for (71) that

$\frac{7\tau_{2}-16+\sqrt{\Delta}}{8\tau_{2}-24}\leq 1,$

which implies the roots of (69) is no greater than 1 and $[B]_{22}>0$ for
$\tau_{1}>1$. Thus, in summary, (67) is satisfied as long as $\tau_{1}>1$ and
$\tau_{2}\geq 4$. The rest follows by letting $(v,T)\sim G_{\text{order}}$. ∎

####  A.3.4 Technical Proof in Theorem 2

Fix $(v,T)$. We prove the case for the kernel $G(s)\propto s^{-\gamma}$ and
the case for $G(s)\propto l_{0}(l_{0}+s)^{-\gamma}$ or $G(s)\propto e^{-\rho
s}$ is similar. For simplicity, we assume $G(s)=s^{-\gamma}$ so that
$\eta_{i}(\gamma)=\int_{t_{i-1}}^{t_{i}}G(s;\gamma)ds=\frac{1}{1-\gamma}(t_{i}^{1-\gamma}-t_{i-1}^{1-\gamma})=\frac{(\Delta
t)^{1-\gamma}}{1-\gamma}(i^{1-\gamma}-(i-1)^{1-\gamma})$. We want to show that
the convex hull of $\\{\eta(\gamma)\\}_{\gamma\in\Gamma}$ has dimension $N$,
or equivalently: there exists
$\\{\gamma_{0},\gamma_{1},...,\gamma_{N}\\}\subseteq\Theta$ such that
$\\{\eta(\gamma_{i})-\eta(\gamma_{0})\\}_{1\leq i\leq N}$ is linearly
independent. Define $a_{i}(\gamma)\triangleq\frac{(\Delta
t)^{1-\gamma}}{1-\gamma}(i^{1-\gamma}-(i-1)^{1-\gamma})$, we then want to show
there exists $(\gamma_{0},\gamma_{1},...,\gamma_{N})\in\Gamma^{N+1}$, such
that the matrix

$A(\gamma_{0},\gamma_{1},...,\gamma_{N})\triangleq\begin{bmatrix}a_{1}(\gamma_{1})-a_{1}(\gamma_{0})&\dots&a_{1}(\gamma_{N})-a_{N}(\gamma_{0})\\\
\vdots&\ddots&\vdots\\\
a_{N}(\gamma_{1})-a_{N}(\gamma_{0})&\dots&a_{N}(\gamma_{N})-a_{N}(\gamma_{0})\end{bmatrix}$
(72)

has full rank $N$. To show this, it suffices to show there exists
$(\gamma_{0},\gamma_{1},...,\gamma_{N})\in\Gamma^{N+1}$ such that

$\textup{det}(A(\gamma_{0},\gamma_{1},...,\gamma_{N}))\neq 0.$ (73)

To show this, note that the dimension of the convex hull of
$\\{\eta(\gamma)\\}_{\gamma\in\Gamma}$ is also the dimension of the smallest
affine set (affine hull) that contains $\\{\eta(\gamma)\\}_{\gamma\in\Gamma}$.
There are two scenarios:

Scenario 1 If $\\{\eta(\gamma)\\}_{\gamma\in\Gamma}$ cannot be contained in a
linear hyper-plane of dimension $N-1$, then the convex hull of
$\\{\eta(\gamma)\\}_{\gamma\in\Gamma}$ has dimension $N$, therefore there must
exist $(\gamma_{0},\gamma_{1},...,\gamma_{N})\in\Gamma^{N+1}$ that satisfies
(73).

Scenario 2 If $\eta(\gamma)$ can be contained in a linear hyper-plane of
dimension $N-1$, then there exists some
$\bm{w}\in\mathbb{R}^{N+1}=(w_{0},w_{1},...,w_{N})$ such that
$\bm{w}\neq\bm{0}$ and

$b(\gamma)\triangleq\sum_{i=1}^{N}w_{i}a_{i}(\gamma)-w_{0}=0$ (74)

for all $\gamma\in\Gamma$. First, we note that $a_{i}(\gamma)$ is an analytic
function on not just $\Gamma$, but also $(-\infty,1)$. It follows from
[Mityagin, 2020] that the zero set of $b(\gamma)$:

$B=\\{\gamma\in(-\infty,1):b(\gamma)=0\\}$

either has Lebesgue measure 0 or is the entire $(-\infty,1)$. First, we
assume, for the sake of contradiction, that $B=(-\infty,1)$, then

$b(\gamma)=0\Leftrightarrow\sum_{i=1}^{N}w_{i}(t_{i}^{1-\gamma}-t_{i-1}^{1-\gamma})=w_{0}(1-\gamma),$
(75)

for all $\gamma\in(-\infty,1)$. Then we can define $i^{\star}=\max\\{i:0\leq
i\leq N,w_{i}\neq 0\\}$. Note that $i^{\star}$ is well-defined since
$\bm{w}\neq\bm{0}$. Moreover, $i^{\star}\geq 1$ since otherwise we would would
have

$0=\sum_{i=1}^{N}w_{i}(t_{i}^{1-\gamma}-t_{i-1}^{1-\gamma})\neq
w_{0}(1-\gamma),$

which contradicts (75). Thus, we can rewrite (75) as

$\sum_{i=1}^{i^{\star}}w_{i}(t_{i}^{1-\gamma}-t_{i-1}^{1-\gamma})=w_{0}(1-\gamma),$
(76)

for all $\gamma\in(-\infty,1)$.Divide both side by $t_{i^{\star}}^{1-\gamma}$
and let $\gamma\downarrow-\infty$, the left side of (76) goes to
$w_{i^{\star}}$ while the right side of (76) becomes

$\lim_{\gamma\downarrow-\infty}\frac{w_{0}(1-\gamma)}{t_{i^{\star}}^{1-\gamma}},$

which cannot converge to a constant other than 0 (i.e., not $w_{i^{\star}}$),
regardless of the value of $t_{i^{\star}}$. Thus, we arrive at a contradiction
and we conclude that $B$ cannot be $(-\infty,1)$. Thus, $B$ must have Lebesgue
measure 0. However, we assumed that $\Gamma$ contains an open set, which
implies the Lebesgue of $\Gamma$ on $\mathbb{R}$ is greater than 0. As a
result, there exists $\gamma\in\Gamma$ (in fact there exists a set of positive
measures with such points) such that $b(\gamma)\neq 0$, which contradicts
(74). Thus, Scenario 2 is not possible, we must only have Scenario 1, which
completes the proof.

####  A.3.5 Proof of Theorem 3

Fix $(v,T)$. Denote $\Delta S_{i}\triangleq S_{t_{i}}-S_{t_{i-1}}$ and $\Delta
G_{i}=\int_{t_{i-1}}^{t_{i}}G(s)ds$. It follows from (4.1) and Theorem 2 that

$\phi(\bm{S}_{\text{full}})\triangleq\sum_{i=1}^{N}(S_{t_{i}}-S_{t_{i-1}})\int_{t_{i-1}}^{t_{i}}G(s)ds=\sum_{i=1}^{N}\Delta
S_{i}\Delta G_{i},$

is a sufficient statistic. Since
$\phi(\bm{S}_{\text{full}})\sim\mathcal{N}(f(v;\theta)\sum_{i=1}^{N}(\Delta
G_{i})^{2},\sigma^{2}\Delta t\sum_{i=1}^{N}(\Delta G_{i})^{2})$, as $\Delta
t\rightarrow 0$, it follows from Lemma 1 that

$\mathcal{I}_{\bm{S}_{\text{full}}}(\theta)=\mathcal{I}_{\phi(\bm{S}_{\text{full}})}(\theta)=\frac{(\frac{\partial
f}{\partial\theta})^{2}}{\sigma^{2}}\sum_{i=1}^{N}\frac{\Delta
G^{2}_{i}}{\Delta t}\rightarrow\frac{(\frac{\partial
f}{\partial\theta})^{2}}{\sigma^{2}}\int_{0}^{T}G^{2}(t)dt,$ (77)

which proves (43). Now, based on definition 3, we have

$J=\sum_{i=1}^{N}S_{t_{i}}\Delta t=\sum_{i=1}^{N}\Delta S_{i}(T-t_{i}+\Delta
t).$ (78)

Now, using Lemma 6, we have

$\mathcal{I}_{J}(\theta)=\mathcal{I}_{\phi(\bm{S}_{\text{full}})}(\theta)-\mathbb{E}_{x\sim
J}[\mathcal{I}_{\phi(\bm{S}_{\text{full}})|J=x}(\theta)].$ (79)

However, based on (78), the joint distribution of
$(\phi(\bm{S}_{\text{full}}),J)$ is

$\displaystyle\begin{pmatrix}\phi(\bm{S}_{\text{full}})\\\ J\end{pmatrix}=$
$\displaystyle\begin{pmatrix}\sum_{i=1}^{N}\Delta S_{i}\Delta G_{i}\\\
\sum_{i=1}^{N}\Delta S_{i}(T-t_{i}+\Delta t)\end{pmatrix}$ $\displaystyle\sim$
$\displaystyle\mathcal{N}\begin{pmatrix}f(v;\theta)\begin{pmatrix}\sum_{i=1}^{N}(\Delta
G_{i})^{2}\\\ \sum_{i=1}^{N}(T-t_{i}+\Delta t)\Delta
G_{i}\end{pmatrix},\sigma^{2}\Delta t\begin{bmatrix}\sum_{i=1}^{N}(\Delta
G_{i})^{2}&\sum_{i=1}^{N}(T-t_{i}+\Delta t)\Delta G_{i}\\\
\sum_{i=1}^{N}(T-t_{i}+\Delta t)\Delta G_{i}&\sum_{i=1}^{N}(T-t_{i}+\Delta
t)^{2}\end{bmatrix}\end{pmatrix}.$

Then, standard Gaussian calculation gives conditional distribution:

$\displaystyle\phi(\bm{S}_{\text{full}})|J=x\sim$
$\displaystyle\mathcal{N}\Bigg{(}f(v;\theta)\sum_{i=1}^{N}(\Delta
G_{i})^{2}+\frac{\sum_{i=1}^{N}(T-t_{i}+\Delta t)\Delta
G_{i}}{\sum_{i=1}^{N}(T-t_{i}+\Delta
t)^{2}}\Big{(}x-f(v;\theta)\cdot\sum_{i=1}^{N}(T-t_{i}+\Delta t)\Delta
G_{i}\Big{)},$ $\displaystyle\sum_{i=1}^{N}(\Delta
G_{i})^{2}-\frac{(\sum_{i=1}^{N}(T-t_{i}+\Delta t)\Delta
G_{i})^{2}}{\sum_{i=1}^{N}(T-t_{i}+\Delta t)^{2}}\Bigg{)}.$

Now, based on Lemma 1, we have, as $\Delta t\rightarrow 0$,

$\displaystyle\mathcal{I}_{\phi(\bm{S}_{\text{full}})|J}(\theta)=$
$\displaystyle\frac{(\frac{\partial
f}{\partial\theta})^{2}}{\sigma^{2}}\Big{(}\sum_{i=1}^{N}\frac{(\Delta
G_{i})^{2}}{\Delta t}-\frac{(\sum_{i=1}^{N}(T-t_{i}+\Delta t)\Delta
G_{i})^{2}}{\sum_{i=1}^{N}(T-t_{i}+\Delta t)^{2}\Delta t}\Big{)}$
$\displaystyle\rightarrow$ $\displaystyle\frac{(\frac{\partial
f}{\partial\theta})^{2}}{\sigma^{2}}\Big{(}\int_{0}^{T}G^{2}(t)dt-\frac{(\int_{0}^{T}(T-t)G(t)dt)^{2}}{\int_{0}^{t}(T-t)^{2}dt}\Big{)}$
$\displaystyle=$ $\displaystyle\frac{(\frac{\partial
f}{\partial\theta})^{2}}{\sigma^{2}}\Big{(}\int_{0}^{T}G^{2}(t)dt-\frac{3}{T^{3}}(\int_{0}^{T}(T-t)G(t)dt)^{2}\Big{)},$

which, combined with (79) and (43), gives (44):

$\mathcal{I}_{J}(\theta)\rightarrow\frac{(\frac{\partial
f}{\partial\theta})^{2}}{\sigma^{2}}\cdot\frac{3}{T^{3}}\Big{(}\int_{0}^{T}G(t)(T-t)dt\Big{)}^{2}.$

The fact that $\mathcal{I}_{J}(\theta)=\frac{(\frac{\partial
f}{\partial\theta})^{2}}{\sigma^{2}}\cdot\frac{3}{T^{3}}\Big{(}\int_{0}^{T}G(t)(T-t)dt\Big{)}^{2}$
can be directly calculated using Proposition 1, and note that by Fubini
theorem that $\int_{0}^{T}\int_{0}^{t}G(s)ds=\int_{0}^{T}G(t)(T-t)dt$.

Finally, to prove (45), we first write $\bm{S}_{\text{partial}}$ as $\bm{S}$
for ease of notation. Note that it follows from Lemma 4 that

$\mathcal{I}_{\bm{S}}(\theta)=\mathcal{I}_{S_{t_{1}^{\mathcal{K}}}-S_{t_{0}^{\mathcal{K}}},S_{t_{2}^{\mathcal{K}}}-S_{t_{1}^{\mathcal{K}}},...,S_{t_{N_{\mathcal{K}}}^{\mathcal{K}}}-S_{t_{{N_{\mathcal{K}}}-1}^{\mathcal{K}}}}(\theta).$
(80)

Now denote $\Delta
S_{i}^{\mathcal{K}}=S_{t_{i}^{\mathcal{K}}}-S_{t_{i-1}^{\mathcal{K}}}$ for
$1\leq i\leq{N_{\mathcal{K}}}$. Since $\Delta S_{i}^{\mathcal{K}}$ is
independent with $\Delta S_{j}^{\mathcal{K}}$ for $i\neq j$, it follows that
(80) becomes

$\mathcal{I}_{\bm{S}}(\theta)=\mathcal{I}_{\\{\Delta
S_{i}^{\mathcal{K}}\\}_{1\leq i\leq
n}}(\theta)=\sum_{i=1}^{N_{\mathcal{K}}}\mathcal{I}_{\Delta
S_{i}^{\mathcal{K}}}(\theta).$ (81)

On the other hand, fix $1\leq i\leq{N_{\mathcal{K}}}$, define
$T_{i}=t_{i}^{\mathcal{K}}-t_{i-1}^{\mathcal{K}}$,
$\Delta_{i}=\frac{T_{i}}{N}$ and $\Delta S_{ij}=j\Delta_{i}$ so that we have

$\Delta S_{i}^{\mathcal{K}}=\sum_{j=1}^{N}\Delta S_{ij}.$

We further define similarly $\Delta G_{ij}$ and

$\phi^{\mathcal{K}}_{i}(\bm{S})\triangleq\sum_{j=1}^{N}\Delta S_{j}\Delta
G_{j},$

which, similar as (77), can be shown to satisfy as $N\rightarrow\infty$,

$\mathcal{I}_{\phi^{\mathcal{K}}_{i}(\bm{S})}(\theta)=\frac{(\frac{\partial
f}{\partial\theta})^{2}}{\sigma^{2}}\int_{t_{i-1}^{\mathcal{K}}}^{t_{i}^{\mathcal{K}}}G^{2}(s)ds.$
(82)

Now, fix $1\leq i\leq n$, as in (79), base on Lemma 6, we have

$\mathcal{I}_{\Delta
S_{i}^{\mathcal{K}}}(\theta)=\mathcal{I}_{\phi^{\mathcal{K}}_{i}(\bm{S})}-\mathbb{E}_{x\sim\Delta
S_{i}^{\mathcal{K}}}[\mathcal{I}_{\phi^{\mathcal{K}}_{i}(\bm{S})|\Delta
S_{i}^{\mathcal{K}}=x}(\theta)].$ (83)

Again, we calculate the joint distribution

$\displaystyle\begin{pmatrix}\phi_{i}^{\mathcal{K}}(\bm{S})\\\ \Delta
S_{i}^{\mathcal{K}}\end{pmatrix}=$
$\displaystyle\begin{pmatrix}\sum_{j\in[N]}\Delta S_{ij}\Delta G_{ij}\\\
\sum_{j\in[N]}\Delta S_{ij}\end{pmatrix}$ $\displaystyle\sim$
$\displaystyle\mathcal{N}\begin{pmatrix}f(v;\theta)\begin{pmatrix}\sum_{j\in[N]}(\Delta
G_{ij})^{2}\\\ \sum_{j\in[N]}\Delta G_{ij}\end{pmatrix},\sigma^{2}\Delta
t\begin{bmatrix}\sum_{j\in[N]}(\Delta G_{ij})^{2}&\sum_{j\in[N]}\Delta
G_{ij}\\\ \sum_{j\in[N]}\Delta
G_{ij}&\sum_{j\in[N]}1\end{bmatrix}\end{pmatrix}.$

which further gives a conditional distribution

$\displaystyle\phi_{i}^{\mathcal{K}}(\bm{S})|\Delta S_{i}^{\mathcal{K}}=x\sim$
$\displaystyle\mathcal{N}\Bigg{(}f(v;\theta)\sum_{j\in[N]}(\Delta
G_{ij})^{2}+\frac{\sum_{j\in[N]}\Delta
G_{ij}}{\sum_{j\in[N]}1}\Big{(}x-f(v;\theta)\cdot\sum_{j\in[N]}\Delta
G_{ij}\Big{)},$ $\displaystyle\sum_{j\in[N]}(\Delta
G_{ij})^{2}-\frac{(\sum_{j\in[N]}\Delta
G_{ij})^{2}}{\sum_{j\in[N]}1}\Bigg{)}.$

Again, following Lemma 1, we have, as $N\rightarrow\infty$,

$\displaystyle\mathcal{I}_{\phi_{i}^{\mathcal{K}}(\bm{S})|\Delta
S_{i}^{\mathcal{K}}}(\theta)=$ $\displaystyle\frac{(\frac{\partial
f}{\partial\theta})^{2}}{\sigma^{2}}\Big{(}\sum_{j\in[N]}\frac{(\Delta
G_{ij})^{2}}{\Delta t}-\frac{(\sum_{j\in[N]}\Delta
G_{ij})^{2}}{\sum_{j\in[N]}\Delta t}\Big{)}$ $\displaystyle\rightarrow$
$\displaystyle\frac{(\frac{\partial
f}{\partial\theta})^{2}}{\sigma^{2}}\Big{(}\int_{t_{i-1}^{\mathcal{K}}}^{t_{i}^{\mathcal{K}}}G^{2}(t)dt-\frac{(\int_{t_{i-1}^{\mathcal{K}}}^{t_{i}^{\mathcal{K}}}G(t)dt)^{2}}{t_{i}^{\mathcal{K}}-t_{i-1}^{\mathcal{K}}}\Big{)}.$
(84)

Now, since $\mathcal{I}_{\bm{S}_{\text{partial}}}(\theta)$ does not depend on
$N$, we can combine (82), (83),(A.3.5) and (81) to show (45):

$\mathcal{I}_{\bm{S}_{\text{partial}}}(\theta)=\frac{(\frac{\partial
f}{\partial\theta})^{2}}{\sigma^{2}}\Big{(}\sum_{i=1}^{n}\frac{(\int_{t_{i-1}^{\mathcal{K}}}^{t_{i}^{\mathcal{K}}}G(t)dt)^{2}}{t_{i}^{\mathcal{K}}-t_{i-1}^{\mathcal{K}}}\Big{)}.$

####  A.3.6 Derivation of Example 9

To calculate (47), notice when $G(s)=e^{-\rho s}$, we have

$\frac{(\int_{0}^{t}G(t)dt)^{2}}{t}+\frac{(\int_{t}^{T}G(t)dt)^{2}}{T-t}=\frac{1}{t}\Big{(}\frac{1-e^{-\rho
t}}{\rho}\Big{)}^{2}+\frac{1}{T-t}\Big{(}\frac{e^{-\rho t}-e^{-\rho
T}}{\rho}\Big{)}^{2}$ (85)

and

$\Big{(}\int_{0}^{T}G(t)(T-t)dt\Big{)}^{2}=\frac{3}{T^{3}}\Big{(}\frac{\rho
T+e^{-\rho T}-1}{\rho^{2}}\Big{)}^{2}.$ (86)

To show (85) is greater than (86), it suffices to show

$\frac{(\frac{1-e^{-\rho t}}{\rho})^{2}}{t/T}+\frac{(\frac{e^{-\rho
t}-e^{-\rho T}}{\rho})^{2}}{1-t/T}\geq\frac{3}{T^{2}}\Big{(}\frac{\rho
T+e^{-\rho T}-1}{\rho^{2}}\Big{)}^{2}$ (87)

Now, let $\frac{t}{T}\rightarrow\tau$ while $t,T\rightarrow\infty$, (87)
becomes

$\frac{1}{\tau\rho^{2}}\geq\frac{3}{\rho^{2}}.$

or simply $\tau\leq\frac{1}{3}$.

####  A.3.7 Proof of Corollary 3

Under the setting of Corollary 3, we have

$\mu(T,v)=\frac{c}{\delta}(vT)^{\delta}$

and

$\frac{J}{X}\sim\mathcal{N}(\frac{c}{\delta(1+\delta)}(vT)^{\delta},\frac{\sigma^{2}T}{3})$

for some $c$. Under the setting of Corollary 3, for $S_{t},S_{T}$, we have

$\displaystyle\mu(\delta)=\left[\begin{array}[]{c}c\frac{1}{\delta}(vt)^{\delta}\\\
c\frac{1}{\delta}(vT)^{\delta}\end{array}\right]$ $\displaystyle
S_{t}\sim\mathcal{N}\left(c\frac{1}{\delta}(vt)^{\delta},\sigma^{2}t\right)$
$\displaystyle
S_{T}\sim\mathcal{N}\left(c\frac{1}{\delta}(vT)^{\delta},\sigma^{2}T\right)$
$\displaystyle\Sigma=\frac{1}{\sigma^{2}t(T-t)}\left[\begin{array}[]{cc}T&-t\\\
-t&t\end{array}\right]$

According to Lemma 1, the Fisher information with respect to $\delta$ is,

$\displaystyle\mathcal{I}_{S_{t},S_{T}}(\delta)=\mathcal{J}(\delta)^{T}\Sigma^{-1}\mathcal{J}(\delta)$
(88) $\displaystyle=$
$\displaystyle\frac{c^{2}}{\sigma^{2}t(T-t)}\Bigg{(}T\Big{(}-\frac{1}{\delta^{2}}(vt)^{\delta}+\frac{\ln(vt)}{\delta}(vt)^{\delta}\Big{)}^{2}+t\Big{(}-\frac{1}{\delta^{2}}(vT)^{\delta}+\frac{\ln(vT)}{\delta}(vT)^{\delta}\Big{)}^{2}$
(89)
$\displaystyle-2t\Big{(}-\frac{1}{\delta^{2}}(vt)^{\delta}+\frac{\ln(vt)}{\delta}(vt)^{\delta}\Big{)}\Big{(}-\frac{1}{\delta^{2}}(vT)^{\delta}+\frac{\ln(vT)}{\delta}(vT)^{\delta}\Big{)}\Bigg{)}$
(90)

where the Jacobian matrix is,

$\mathcal{J}(\delta)=\left[\begin{array}[]{c}\frac{\partial}{\partial\delta}c\frac{1}{\delta}(vt)^{\delta}\\\
\frac{\partial}{\partial\delta}c\frac{1}{\delta}(vT)^{\delta}\end{array}\right]=c\left[\begin{array}[]{c}-\frac{1}{\delta^{2}}(vt)^{\delta}+\frac{\ln(vt)}{\delta}(vt)^{\delta}\\\
-\frac{1}{\delta^{2}}(vT)^{\delta}+\frac{\ln(vT)}{\delta}(vT)^{\delta}\end{array}\right]$
(91)

For $J$, consider

$\displaystyle\frac{J}{X}\sim\mathcal{N}\left(c\frac{1}{\delta(1+\delta)}(vT)^{\delta},\frac{\sigma^{2}T}{3}\right)$

Applying Lemma 1, the Fisher information with respect to $\delta$ is,

$\mathcal{I}_{J}(\delta)=\frac{3c^{2}}{\sigma^{2}T}\left(-\frac{2\delta+1}{\delta^{2}(1+\delta)^{2}}(vT)^{\delta}+\frac{\ln(vT)}{\delta(1+\delta)}(vT)^{\delta}\right)^{2}.$

Finally, it follows from an ad-hoc numerical analysis, which does not depend
on $c$ (as long as it is not 0), that for $0.005\leq vT\leq 1$ and $\delta\leq
0.8$, we have

$\mathcal{I}_{J}(\delta)\leq\mathcal{I}_{S_{t},S_{T}}$

as long as $1.2\%\leq t/T\leq 22.2\%$.