how to compute this variance in Appendix C-B by applying (167)-(168). Fig. 4
shows the GMIs as the curve labeled “rGMI”, where we used the same power
levels as for the $I(A;Y)$ curve. The two curves are indistinguishable for
small $P$ but the “rGMI” rates are poor at large $P$. This example shows that
the forward model GMI with optimized powers can be substantially better than
the reverse model GMI with a reasonable, but suboptimal, power policy. Hence
it is useful to gain understanding via the forward models.

The forward model GMI (196) is simpler:

$\displaystyle
I_{1}(A;Y|H)=\frac{1}{2}\log\left(1+\text{SNR}\left(\sqrt{2}\right)\right)$
(258)

where $\text{SNR}\left(\sqrt{2}\right)$ is given by (197) with

$\displaystyle\tilde{P}_{T}\left(\sqrt{2}\right)=\left(\epsilon\sqrt{P(0)}+\bar{\epsilon}\sqrt{P(2)}\right)^{2}$
$\displaystyle{\rm
Var}\left[{\left.\sqrt{P(S_{T})}\right|H=h}\right]=1+2\,\epsilon\,\bar{\epsilon}\left(\sqrt{P(2)}-\sqrt{P(0)}\right)^{2}.$

Applying Remark 57, the optimal power control policy is

$\displaystyle\sqrt{P(s_{T})}$
$\displaystyle=\frac{p_{H|S_{T}}\left(\sqrt{2}\,\big{|}s_{T}\right)}{\gamma+\beta\,p_{H|S_{T}}\left(\sqrt{2}\,\big{|}s_{T}\right)}$
$\displaystyle=\left\\{\begin{array}[]{ll}\dfrac{\epsilon}{\gamma+\beta\,\epsilon},&s_{T}=0\vspace{0.2cm}\\\
\dfrac{\bar{\epsilon}}{\gamma+\beta\,\bar{\epsilon}},&s_{T}=2\end{array}\right.$
(261)

where

$\displaystyle\beta$
$\displaystyle=\frac{2\sqrt{\tilde{P}_{T}\left(\sqrt{2}\,\right)}}{{\rm
E}\left[{|Y|^{2}|H=\sqrt{2}\,}\right]}$ (262)

and $\gamma\geq 0$ is chosen so that $P(0)+P(2)=2P$. Fig. 4 shows the
resulting GMI as the curve labeled “GMI, K=1”. At low SNR, we achieve the rate
$\tilde{P}_{T}\left(\sqrt{2}\,\right)$ and the optimal power control has
$\beta\rightarrow 0$ so that

$\displaystyle
P(0)=\frac{2P\epsilon^{2}}{\epsilon^{2}+\bar{\epsilon}^{2}},\quad
P(2)=\frac{2P\bar{\epsilon}^{2}}{\epsilon^{2}+\bar{\epsilon}^{2}}$ (263)

and therefore

$\displaystyle\tilde{P}_{T}(\sqrt{2})=2\left(\epsilon^{2}+\bar{\epsilon}^{2}\right)P.$
(264)

We thus have $\dot{C}(0)=2\left(\epsilon^{2}+\bar{\epsilon}^{2}\right)$ and
lose a fraction of $(\epsilon^{2}+\bar{\epsilon}^{2})$ of the power as
compared to having full CSIT ($\epsilon=0$). For example, if $\epsilon=0.1$
then the minimal $E_{b}/N_{0}$ is approximately $-3.74$ dB.

We remark that both the $I(A;Y)$ and reverse model GMI curves lie above the
forward model curve if we choose the same power policy as for the forward
channel.

###  VII-C Partial CSIR, Full CSIT

This section studies $S_{T}=H$. The capacity with partial CSIR is given by
(150) for which we need to compute $p(y|a,s_{R})$ and $p(y|s_{R})$. We
consider two cases.

####  VII-C1 $S_{R}=1(G\geq t)$

Here we recover the case with full CSIR by choosing $t$ to satisfy $0<t\leq
2$.

####  VII-C2 $S_{R}=0$

The best power policy clearly has $P(0)=0$ and $P(\sqrt{2}\,)=2P$. The mutual
information is thus $I(A;Y)=I\big{(}X\big{(}\sqrt{2}\big{)};Y\big{)}$ and the
channel densities are (cf. (77) and (78))

$\displaystyle
p(y|a)=\frac{e^{-|y|^{2}}}{2\pi}+\frac{e^{-\left|y-2\sqrt{P}u\left(\sqrt{2}\,\right)\right|^{2}}}{2\pi}$
$\displaystyle
p(y)=\frac{e^{-|y|^{2}}}{2\pi}+\frac{e^{-|y|^{2}/(1+4P)}}{2\pi(1+4P)}.$

The information rates are shown in Fig. 5. Observe that at low SNR the rates
are larger than without fading; this is a consequence of the slightly bursty
nature of transmission.

Figure 5: Rates for on-off fading with $S_{T}=H$ and $S_{R}=0$. The GMI for
the $K=2$ partition uses the threshold $t_{R}=\sqrt{P}+3$.

The reverse model GMI (151) requires ${\rm Var}\left[{U|Y}\right]$. We compute
this variance in Appendix C-C by using the general expression (167) with (202)
and $\phi(s_{T})=0$. Fig. 5 shows the GMIs as the curve labeled “rGMI”.

Next, the TCP, TMF, TCI, and TMMSE power control policies are all the same for
$0<t\leq 2$, since they all sensibly use $P(0)=0$ and
$P\left(\sqrt{2}\,\right)=2P$. The resulting rate is given by (203)-(205) with
$\tilde{P}(0)=0$, $\tilde{P}(1)=P$, and ${\rm
Var}\left[{\left.\sqrt{GP(S_{T})}\right|S_{R}=1}\right]=P$ and

$\displaystyle I_{1}(A;Y)=\log\left(1+\frac{P}{1+P}\right).$ (265)

The rates are plotted in Fig. 5 as the curve labeled “GMI, K=1”. This example
again shows that choosing $K=1$ is a poor choice at high SNR.

To improve the auxiliary model at high SNR, consider the GMI (127) with $K=2$
and the subsets (67). We further choose the parameters $h_{1}=0$,
$\sigma_{1}^{2}=0$, $h_{2}=2$, $\sigma_{2}^{2}=1$, and adaptive coding with
$X(0)=0$, $X\left(\sqrt{2}\,\right)=\sqrt{2P}\,U$, $\bar{X}=\sqrt{P}\,U$,
where $U\sim\mathcal{CN}(0,1)$. The GMI (127) is

$\displaystyle I_{1}(A;Y)={\rm Pr}\left[{\mathcal{E}_{2}}\right]$
$\displaystyle\left[\log(1+4P)+\frac{{\rm
E}\left[{|Y|^{2}|\mathcal{E}_{2}}\right]}{1+4P}\right.$
$\displaystyle\quad\left.-{\rm
E}\left[{\left.\left|Y-\sqrt{4P}\,U\right|^{2}\right|\mathcal{E}_{2}}\right]\right].$
(266)

In Appendix B-B we show that choosing $t_{R}=P^{\lambda_{R}}+b$ where
$0<\lambda_{R}<1$ and $b$ is a real constant makes all terms behave as desired
as $P$ increases:

$\displaystyle\begin{array}[]{l}{\rm
Pr}\left[{\mathcal{E}_{2}}\right]\rightarrow 1/2\vspace{0.1cm}\\\ {\rm
E}\left[{|Y|^{2}|\mathcal{E}_{2}}\right]/(1+4P)\rightarrow 1\vspace{0.1cm}\\\
{\rm E}\left[{|Y-\sqrt{4P}U|^{2}|\mathcal{E}_{2}}\right]\rightarrow
1.\end{array}$ (270)

We thus have

$\displaystyle\lim_{P\rightarrow\infty}\;\frac{I_{1}(A;Y)}{\frac{1}{2}\log(1+4P)}=1.$
(271)

Fig. 5 shows the behavior of $I_{1}(X;Y)$ for $\lambda_{R}=1/2$ and $b=3$ as
the curve labeled “GMI, K=2”. As for the case without CSIT, at large SNR the
receiver can estimate $H$ accurately and one approaches the capacity with full
CSIR.

Finally, the rates of the large-$K$ forward model are computed using (72) but
where $\bar{X}$ replaces $X$. One may again use the results of Appendix C-C
and the simple relations

$\displaystyle{\rm E}\left[{\left.\bar{X}\right|Y=y}\right]$
$\displaystyle=\sqrt{P}\,{\rm E}\left[{U|Y=y}\right]$ $\displaystyle{\rm
E}\left[{\left.|\bar{X}|^{2}\right|Y=y}\right]$ $\displaystyle=P\,{\rm
E}\left[{|U|^{2}\big{|}Y=y}\right]$ $\displaystyle{\rm
Var}\left[{\bar{X}\big{|}Y=y}\right]$ $\displaystyle=P\,{\rm
Var}\left[{U|Y=y}\right].$

The rates are shown as the curve labeled “GMI, K=$\infty$” in Fig. 5. The
large-$K$ forward model is again good at high SNR but performs worse than the
best $K=1$ model at low SNR.

###  VII-D Partial CSIR, CSIT@R

Consider partial CSIR with $S_{T}=S_{R}$ and

$\displaystyle{\rm Pr}\left[{S_{R}=H}\right]=\bar{\epsilon},\quad{\rm
Pr}\left[{S_{R}\neq H}\right]=\epsilon$ (272)

where $0\leq\epsilon\leq\frac{1}{2}$. We thus have both CSIT@R and CSIR@T. To
compute $I(X;Y|S_{R})$ in (235), we write (236)-(237) as

$\displaystyle p_{Y|S_{R},X}(y|0,x)$
$\displaystyle=\bar{\epsilon}\,\frac{e^{-|y|^{2}}}{\pi}+\epsilon\,\frac{e^{-\left|y-\sqrt{2}\,x(0)\right|^{2}}}{\pi}$
$\displaystyle p_{Y|S_{R},X}(y|\sqrt{2},x)$
$\displaystyle=\bar{\epsilon}\,\frac{e^{-\left|y-\sqrt{2}\,x\left(\sqrt{2}\,\right)\right|^{2}}}{\pi}+\epsilon\,\frac{e^{-|y|^{2}}}{\pi}$
$\displaystyle p_{Y|S_{R}}(y|0)$
$\displaystyle=\bar{\epsilon}\,\frac{e^{-|y|^{2}}}{\pi}+\epsilon\,\frac{e^{-|y|^{2}/[1+2P(0)]}}{\pi[1+2P(0)]}$
$\displaystyle p_{Y|S_{R}}(y|\sqrt{2})$
$\displaystyle=\bar{\epsilon}\,\frac{e^{-|y|^{2}/\left[1+2P\left(\sqrt{2}\,\right)\right]}}{\pi\left[1+2P\left(\sqrt{2}\,\right)\right]}+\epsilon\,\frac{e^{-|y|^{2}}}{\pi}$

where $X(s_{T})$ is CSCG. We choose the transmit powers $P(0)$ and
$P\left(\sqrt{2}\,\right)$ as in (255) to compare with the best CSIR. Fig. 6
shows the resulting rates for $\epsilon=0.1$ as the curve labeled “Partial
CSIR, $I(X;Y|S_{R})$”. Observe that at high SNR the curve seems to approach
the best CSIR curve from Fig. 4 with $S_{R}=H\sqrt{P(S_{T})}$. We show that
this is indeed the case by studying a forward model GMI with $K=2$.

Figure 6: Rates for on-off fading with partial CSIR and CSIT@R. The curve
“Best CSIR” shows the capacity with $S_{R}=H\sqrt{P(S_{T})}$. The mutual
information $I(X;Y|S_{R})$ and the GMI are for ${\rm Pr}\left[{S_{R}\neq
H}\right]=0.1$ and with CSCG inputs $X(s_{T})$. The GMI for the $K=2$
partition uses $t_{R}=P^{0.4}$. The curve labeled ‘c-waterfill’ shows the
conventional waterfilling rates.

The reverse model GMI requires ${\rm Var}\left[{U|Y,S_{R}}\right]$ which can
be computed by simulation, see Appendix C-D. However, optimizing the powers
seems difficult. We instead focus on the forward model GMI of Theorem 3 for
which we compute

$\displaystyle\tilde{g}(0)=2\,\epsilon^{2},\quad\tilde{g}\left(\sqrt{2}\right)=2\,\bar{\epsilon}^{2}$
(273)
$\displaystyle\tilde{\sigma}^{2}(0)=\tilde{\sigma}^{2}\left(\sqrt{2}\right)=2\,\epsilon\,\bar{\epsilon}$
(274)

and therefore (242) is

$\displaystyle I_{1}(X;Y|S_{R})$
$\displaystyle=\frac{1}{2}\log\left(1+\frac{2\,\epsilon^{2}P(0)}{1+2\,\epsilon\,\bar{\epsilon}\,P(0)}\right)$
$\displaystyle\quad+\frac{1}{2}\log\left(1+\frac{2\,\bar{\epsilon}^{2}P\left(\sqrt{2}\,\right)}{1+2\,\epsilon\,\bar{\epsilon}\,P\left(\sqrt{2}\,\right)}\right).$
(275)

For CSIR@T the optimal power control policy is given by the quadratic
waterfilling specified by (244) or (250):

$\displaystyle P(0)$
$\displaystyle=\frac{1+\bar{\epsilon}}{4\,\epsilon\,\bar{\epsilon}}\left[\sqrt{1+8\,\epsilon\,\bar{\epsilon}\left(\frac{1}{\lambda}-\frac{1}{2\,\epsilon^{2}}\right)^{+}\frac{\epsilon}{(1+\bar{\epsilon})^{2}}}-1\right]$
$\displaystyle P\left(\sqrt{2}\right)$
$\displaystyle=\frac{1+\epsilon}{4\,\epsilon\,\bar{\epsilon}}\left[\sqrt{1+8\,\epsilon\,\bar{\epsilon}\left(\frac{1}{\lambda}-\frac{1}{2\,\bar{\epsilon}^{2}}\right)^{+}\frac{\bar{\epsilon}}{(1+\epsilon)^{2}}}-1\right].$

The rates are shown in Fig. 6 as the curve labeled “Partial CSIR, GMI, K=1”.
Observe that at high SNR the GMI (275) saturates at

$\displaystyle
I_{1}(X;Y|S_{R})\rightarrow\frac{1}{2}\log\left(1+\frac{\epsilon}{\bar{\epsilon}}\right)+\frac{1}{2}\log\left(1+\frac{\bar{\epsilon}}{\epsilon}\right).$
(276)

For example, for $\epsilon=0.1$ we approach $1.74$ bits at high SNR. On the
other hand, at low SNR the rate is maximized with $P(0)=0$ and
$P\left(\sqrt{2}\right)=2P$ so that $I_{1}(X;Y|S_{R})\approx
2\,\bar{\epsilon}^{2}P$. We thus achieve a fraction of $\bar{\epsilon}^{2}$ of
the power as compared to having full CSIT. For example, if $\epsilon=0.1$ then
the minimal $E_{b}/N_{0}$ is approximately $-3.69$ dB.

Fig. 6 also shows the conventional waterfilling rates as the curve labeled
“Partial CSIR, GMI, c-waterfill”. These rates are almost the same as the
quadratic waterfilling rates except for the range of $E_{b}/N_{0}$ between 9
to 13 dB shown in the inset.

To improve the auxiliary model at high SNR, we now use a $K=2$ GMI but with
(see Remark 66)

$\displaystyle h_{1}(s_{R})=0,\quad
h_{2}(s_{R})=\sqrt{2},\quad\sigma_{1}^{2}(s_{R})=\sigma_{2}^{2}(s_{R})=1$

for $s_{R}=0,\sqrt{2}$. The receiver chooses
$\bar{X}(s_{R})=\sqrt{P(s_{R})}\,U$ (see Remark 40) and we have (see Remark
41)

$\displaystyle I_{1}(X;Y|S_{R})=\frac{1}{2}{\rm
Pr}\left[{\mathcal{E}_{2}|S_{R}=0}\right]$
$\displaystyle\quad\left\\{\log\left(1+2P(0)\right)+\frac{{\rm
E}\left[{\left.|Y|^{2}\right|\mathcal{E}_{2},S_{R}=0}\right]}{1+2P(0)}\right.$
$\displaystyle\left.\qquad-{\rm
E}\left[{\left.\big{|}Y-\sqrt{2}\,X(0)\big{|}^{2}\right|\mathcal{E}_{2},S_{R}=0}\right]\right\\}$
$\displaystyle\quad+\frac{1}{2}{\rm
Pr}\left[{\mathcal{E}_{2}|S_{R}=\sqrt{2}\,}\right]$
$\displaystyle\quad\left\\{\log\left(1+2P\big{(}\sqrt{2}\,\big{)}\right)+\frac{{\rm
E}\left[{\left.|Y|^{2}\right|\mathcal{E}_{2},S_{R}=\sqrt{2}\,}\right]}{1+2P\left(\sqrt{2}\,\right)}\right.$
$\displaystyle\left.\qquad-{\rm
E}\left[{\left.|Y-\sqrt{2}\,X\big{(}\sqrt{2}\,\big{)}|^{2}\right|\mathcal{E}_{2},S_{R}=\sqrt{2}}\right]\right\\}$
(277)

where the $X(s_{T})$, $s_{T}\in\mathcal{S}_{T}$, are given by (134). We
consider $P(0)$ and $P\big{(}\sqrt{2}\,\big{)}$ that both scale in proportion
to $P$. In this case, Appendix B-C shows that choosing $t_{R}=P^{\lambda_{R}}$
where $0<\lambda_{R}<1$ gives the maximal GMI value for large $P$, which is
the rate specified in (254) :

$\displaystyle\lim_{P\rightarrow\infty}\frac{I_{1}(X;Y|S_{R})}{\frac{\epsilon}{2}\log\left(1+2P(0)\right)+\frac{\bar{\epsilon}}{2}\log\left(1+2P\left(\sqrt{2}\,\right)\right)}=1.$
(278)

In other words, by optimizing $P(0)$ and $P\left(\sqrt{2}\,\right)$, at high
SNR the $K=2$ GMI can approach the capacity of Proposition 2 in (254). This is
expected, since at high SNR the receiver can estimate $H\sqrt{P(S_{T})}$
reliably.

Fig. 6 shows the behavior of this GMI and $t_{R}=P^{0.4}$, and where we have
chosen $P(0)$ and $P\big{(}\sqrt{2}\,\big{)}$ according to (255). The
discontinuity at approximately 2.5 dB is because $P(0)$ becomes positive
beyond this $E_{b}/N_{0}$. Keeping $P(0)=0$ for $E_{b}/N_{0}$ up to about 12
dB gives better rates but for high SNR one should choose the powers according
to (255).

##  VIII Rayleigh Fading

Rayleigh fading has $H\sim\mathcal{CN}(0,1)$. The random variable $G=|H|^{2}$
thus has the density $p(g)=e^{-g}\cdot 1(g\geq 0)$. Sec. VIII-B reviews known
results.

###  VIII-A No CSIR, No CSIT

Suppose $S_{R}=S_{T}=0$ and $X\sim\mathcal{CN}(0,P)$. The densities to compute
$I(X;Y)$ for CSCG $X$ are

$\displaystyle p(y|x)$
$\displaystyle=\frac{e^{-|y|^{2}/(|x|^{2}+1)}}{\pi(|x|^{2}+1)}$ (279)
$\displaystyle p(y)$
$\displaystyle=\int_{0}^{\infty}\frac{e^{-g/P}}{P}\,\frac{e^{-|y|^{2}/(g+1)}}{\pi(g+1)}\,dg.$
(280)

One finds that the minimal $E_{b}/N_{0}$ is approximately 9.2 dB and the
forward model GMI (175) is zero. In fact, capacity is achieved by discrete and
finite $X$ [81] and at large SNR the capacity behaves as $\log\log P$ [82].
Further results can be found in [83, 12, 84, 85, 86].

###  VIII-B Full CSIR, CSIT@R

The capacity (176) for $B=0$ (no CSIT) is

$\displaystyle C(P)$
$\displaystyle=\int_{0}^{\infty}e^{-g}\log\left(1+g\,P\right)dg$
$\displaystyle=e^{1/P}E_{1}\left(1/P\right)\log(e)$ (281)

where the exponential integral $E_{1}(.)$ is given by (392) below. The
wideband values are given by (178):

$\displaystyle\left.\frac{E_{b}}{N_{0}}\right|_{\text{min}}=\log 2,\quad S=1.$

The minimal $E_{b}/N_{0}$ is $-1.59$ dB but the fading reduces the capacity
slope. At high SNR, we have

$\displaystyle C(P)\approx\log(P)-\gamma$

where $\gamma\approx 0.57721$ is Euler’s constant. The capacity thus behaves
as for the case without fading but with an SNR loss of approximately 2.5 dB.

The capacity (183) with $B=\infty$ (or $S_{T}=G$) is (see [80, Eq. (7)])

$\displaystyle C(P)$
$\displaystyle=\int_{\lambda}^{\infty}e^{-g}\log\left(g/\lambda\right)dg=E_{1}(\lambda).$
(282)

where $P(g)$ is given by (182) and $\lambda$ is chosen so that

$\displaystyle P$
$\displaystyle=\int_{\lambda}^{\infty}e^{-g}P(g)\,dg=\frac{e^{-\lambda}}{\lambda}-E_{1}(\lambda).$

At low SNR we have large $\lambda$ and using the approximation (395) below we
compute

$\displaystyle C(P)\approx e^{-\lambda}/\lambda\text{ and }P\approx
e^{-\lambda}/\lambda^{2}.$ (283)

We thus have $E_{b}/N_{0}\approx\log(2)/\lambda$ and the minimal $E_{b}/N_{0}$
is $-\infty$.

Consider now $B=1$ for which $P_{S_{T}}(3\Delta/2)=e^{-\Delta}$ and

$\displaystyle{\rm E}\left[{G|G\geq\Delta}\right]=1+\Delta$ (284)
$\displaystyle{\rm E}\left[{G^{2}|G\geq\Delta}\right]=2+2\Delta+\Delta^{2}.$
(285)

We thus have the wideband quantities in (187)-(188):

$\displaystyle\left.\frac{E_{b}}{N_{0}}\right|_{\text{min}}=\frac{\log
2}{1+\Delta}$ (286) $\displaystyle
S=\frac{2e^{-\Delta}(1+\Delta)^{2}}{2+2\Delta+\Delta^{2}}.$ (287)

Fig. 7 shows the capacities for $B=1$ and $\Delta=1,2,1/2$. The minimum
$E_{b}/N_{0}$ value is

$\displaystyle-1.59\text{ dB}-10\log_{10}\left(1+\Delta\right)$ (288)

and for $\Delta=1,2,1/2$ we gain 3 dB, 4.8 dB, 1.8 dB over no CSIT at low
power. Note that one bit of feedback allows one to closely approach the full
CSIT rates.

Figure 7: Capacities for Rayleigh fading with full CSIR, a one-bit quantizer
with threshold $\Delta$, and CSIT@R.

######  Remark 67.

For the scalar channel (161), knowing $H$ at both the transmitter and receiver
provides large gains at low SNR [68] but small gains at high SNR [80, Fig. 4]
as compared to knowing $H$ at the receiver only. Furthermore, the reliability
can be improved substantially [72, Fig. 5-7]. Large gains are also possible
for MIMO channels where beamforming and multiplexing play a role.

######  Remark 68.

An alternative way to derive (284)-(287) is as follows. Define
$\hat{P}=Pe^{\Delta}$ so for small $P$ the capacity is

$\displaystyle C(P)$
$\displaystyle=\int_{\Delta}^{\infty}e^{-g}\log\left(1+g\,\hat{P}\right)dg$
$\displaystyle=e^{1/\hat{P}}E_{1}\left(\frac{1}{\hat{P}}+\Delta\right)+e^{-\Delta}\log(1+\hat{P}\Delta)$
$\displaystyle\approx
P\,(1+\Delta)-\frac{1}{2}P^{2}e^{\Delta}\left(2+2\Delta+\Delta^{2}\right).$

###  VIII-C Full CSIR, Partial CSIT

Consider the noisy CSIT with

$\displaystyle{\rm Pr}\left[{S_{T}=1(G\geq\Delta)}\right]$
$\displaystyle=\bar{\epsilon}$ $\displaystyle{\rm Pr}\left[{S_{T}\neq
1(G\geq\Delta)}\right]$ $\displaystyle=\epsilon.$

We begin with the most informative CSIR.

####  VIII-C1 $S_{R}=\sqrt{P(S_{T})}H$

Proposition 2 gives the capacity

$\displaystyle
C(P)=\int_{0}^{\infty}e^{-g}\sum_{s_{T}}P(s_{T}|g)\log\left(1+g\,P(s_{T})\right)\,dg$
$\displaystyle=\int_{0}^{\Delta}e^{-g}\left[\bar{\epsilon}\,\log\left(1+g\,P(0)\right)+\epsilon\,\log\left(1+g\,P(1)\right)\right]\,dg$
$\displaystyle\quad+\int_{\Delta}^{\infty}e^{-g}\left[\bar{\epsilon}\,\log\left(1+g\,P(1)\right)+\epsilon\,\log\left(1+g\,P(0)\right)\right]\,dg.$
(289)

It remains to optimize $P(0)$, $P(1)$ and $\Delta$. There are two equations
for the Lagrange multiplier $\lambda$, namely

$\displaystyle\lambda\cdot P_{S_{T}}(0)$
$\displaystyle=\int_{0}^{\Delta}e^{-g}\cdot\dfrac{\bar{\epsilon}\,g}{1+gP(0)}dg$
$\displaystyle\qquad+\int_{\Delta}^{\infty}e^{-g}\cdot\dfrac{\epsilon\,g}{1+gP(0)}dg$
(290) $\displaystyle\lambda\cdot P_{S_{T}}(1)$
$\displaystyle=\int_{0}^{\Delta}e^{-g}\cdot\dfrac{\epsilon\,g}{1+gP(1)}dg$
$\displaystyle\qquad+\int_{\Delta}^{\infty}e^{-g}\cdot\dfrac{\bar{\epsilon}\,g}{1+gP(1)}dg$
(291)

where $P_{S_{T}}(0)=\bar{\epsilon}-(\bar{\epsilon}-\epsilon)e^{-\Delta}$ and
$P_{S_{T}}(1)=\epsilon+(\bar{\epsilon}-\epsilon)e^{-\Delta}$. The rates are
shown in Fig. 8

Figure 8: Capacities for Rayleigh fading, $S_{R}=\sqrt{P(S_{T})}H$, and a one-
bit quantizer with threshold $\Delta=1$, and various CSIT error probabilities
$\epsilon$.

For fixed $\Delta$ and large $P$, we have $1/\lambda\approx P(0)\approx
P(1)\approx P$ and approach the capacity (281) without CSIT. In contrast, for
small $P$ we may use similar steps as for (184)-(185). Observe the following
for (290) and (291):

  * •

both $P(0)$ and $P(1)$ decrease as $\lambda$ increases;

  * •

the maximal $\lambda$ in (290) is obtained with $P(0)=0$; this value is

$\displaystyle{\rm
E}\left[{G|S_{T}=0}\right]=\frac{\bar{\epsilon}-(\bar{\epsilon}-\epsilon)(1+\Delta)\,e^{-\Delta}}{P_{S_{T}}(0)}.$
(292)

  * •

the maximal $\lambda$ in (291) is obtained with $P(1)=0$; this value is

$\displaystyle{\rm
E}\left[{G|S_{T}=1}\right]=\frac{\epsilon+(\bar{\epsilon}-\epsilon)(1+\Delta)\,e^{-\Delta}}{P_{S_{T}}(1)}.$
(293)

Thus, if ${\rm E}\left[{G|S_{T}=0}\right]<{\rm E}\left[{G|S_{T}=1}\right]$ and
$0\leq\epsilon<1/2$ then for $P$ below some threshold we have $P(0)=0$,
$P(1)=P/P_{S_{T}}(1)$ and the capacity is

$\displaystyle C(P)$
$\displaystyle=\int_{0}^{\Delta}e^{-g}\,\epsilon\,\log\left(1+\frac{g\,P}{P_{S_{T}}(1)}\right)\,dg$
$\displaystyle\quad+\int_{\Delta}^{\infty}e^{-g}\,\bar{\epsilon}\,\log\left(1+\frac{g\,P}{P_{S_{T}}(1)}\right)\,dg.$
(294)

We compute $\dot{C}(0)={\rm E}\left[{G|S_{T}=1}\right]$ which is given by
(293) so that $1\leq\dot{C}(0)\leq 1+\Delta$, as expected from (286). For
example, for $\epsilon=0.1$ and $\Delta=1$ we have $\dot{C}(0)\approx 1.75$
and therefore the minimal $E_{b}/N_{0}$ is approximately $-4.01$ dB.

The best $\Delta$ is the unique solution $\hat{\Delta}$ of the equation

$\displaystyle e^{-\Delta}=\frac{\epsilon}{\bar{\epsilon}-\epsilon}(\Delta-1)$
(295)

and the result is $\dot{C}(0)=\hat{\Delta}\geq 1$. We have the simple bounds

$\displaystyle
1+\frac{1}{2}\log\left(\frac{1}{\epsilon}-2\right)\leq\dot{C}(0)\leq
1+\frac{1}{e}\left(\frac{1}{\epsilon}-2\right)$ (296)

where the left inequality follows by taking logarithms and using
$\log(\Delta-1)\leq\Delta-2$, and the right inequality follows by using
$e^{-\Delta}\leq e^{-1}$ in (295). For example, for $\epsilon\rightarrow 0$ we
have $\dot{C}(0)\rightarrow\infty$ and for $\epsilon\rightarrow 1/2$ we have
$\dot{C}(0)\rightarrow 1$.

####  VIII-C2 $S_{R}=H$

For the less informative CSIR, one may use (192) and (194) to compute
$I(A;Y|H)$. The reverse model GMI requires ${\rm Var}\left[{U|Y,S_{R}}\right]$
which can be computed by simulation, see Appendix C-B. Again, however,
optimizing the powers seems difficult. We instead focus on the forward model
GMI of Corollary 1, which is

$\displaystyle
I_{1}(A;Y|H)=\int_{0}^{\infty}e^{-g}\log\left(1+\text{SNR}(g)\right)\,dg$
(297)

where

$\displaystyle\text{SNR}(g)=\frac{g\tilde{P}_{T}(g)}{1+g\,\epsilon\,\bar{\epsilon}\left(\sqrt{P(0)}-\sqrt{P(1)}\right)^{2}}$
(298)

and

$\displaystyle\tilde{P}_{T}(g)$
$\displaystyle=\left\\{\begin{array}[]{ll}\left(\bar{\epsilon}\sqrt{P(0)}+\epsilon\sqrt{P(1)}\right)^{2},&g<\Delta\\\
\left(\epsilon\sqrt{P(0)}+\bar{\epsilon}\sqrt{P(1)}\right)^{2},&g\geq\Delta.\end{array}\right.$
(301)

It remains to optimize $P(0)$, $P(1)$ and $\Delta$. Computing the derivatives
seems complicated so we resort to numerical optimization for fixed $\Delta=1$
as in Fig. 8. The results are shown in Fig. 9. For fixed $\Delta$ and large
$P$, it is best to choose $P(0)\approx P(1)$ so that $SNR(g)\approx gP$ and we
approach the rate of no CSIT. For small $P$, however, the best $P(0)$ is no
longer zero and $\dot{C}(0)$ is smaller than (293).

Figure 9: Rates for Rayleigh fading, $S_{R}=H$ and $S_{R}=H\sqrt{P(S_{T})}$, a
one-bit quantizer with threshold $\Delta=1$, and various $\epsilon$. The
curves labeled “best CSIR” show the capacities with $S_{R}=H\sqrt{P(S_{T})}$.
The curves labeled “GMI” show the rates (297) for the optimal powers $P(0)$
and $P(1)$.

###  VIII-D Partial CSIR, Full CSIT

Consider $S_{T}=H$ and suppose we choose the $X(h)$ to be jointly CSCG with
variances ${\rm E}\left[{|X(h)|^{2}}\right]=P(h)$ and correlation coefficients

$\displaystyle\rho(h,h^{\prime})=\frac{{\rm
E}\left[{X(h)X(h^{\prime})^{\text{*}}}\right]}{\sqrt{P(h)P(h^{\prime})}}$

and where ${\rm E}\left[{P(H)}\right]\leq P$. We then have

$\displaystyle
p(y|s_{R})=\int_{\mathbb{C}}p(h|s_{R})\frac{e^{-|y|^{2}/(|h|^{2}P(h)+1)}}{\pi(|h|^{2}P(h)+1)}\,dh.$

As in (100), $p(y|s_{R})$ and therefore $h(Y|S_{R})$ depend only on the
marginals of $A$ and not on the $\rho(h,h^{\prime})$. We thus have the problem
of finding the $\rho(h,h^{\prime})$ that minimize

$\displaystyle h(Y|A,S_{R})=\int_{\mathcal{A}}p(a)\,h(Y|S_{R},A=a)\,da.$

We will use fully-correlated $X(h)$ as discussed in Sec. VII-C. We again
consider $S_{R}=0$ and $S_{R}=1(G\geq t)$.

####  VIII-D1 $S_{R}=0$

For the heuristic policies, the power (209) is

$\displaystyle\hat{P}=\frac{P}{\Gamma\left({1+a,t}\right)}$ (302)

and the rate (222) is

$\displaystyle I_{1}(A;Y)=\log\Bigg{(}1+$
$\displaystyle\left.\frac{P\,\Gamma\left({\frac{3+a}{2},t}\right)^{2}}{\Gamma\left({1+a,t}\right)+P\left[\Gamma\left({2+a,t}\right)-\Gamma\left({\frac{3+a}{2},t}\right)^{2}\right]}\right)$
(303)

where $\Gamma(s,x)$ is the upper incomplete gamma function, see Appendix A-C.
Moreover, the expression (223) is

$\displaystyle\left.\frac{E_{b}}{N_{0}}\right|_{\text{min}}$
$\displaystyle=\frac{\Gamma\left({1+a,t}\right)}{\Gamma\left({\frac{3+a}{2},t}\right)^{2}}\cdot\log
2.$ (304)

We remark that $\Gamma(s,0)=\Gamma(s)$ where $\Gamma(x)$ is the gamma
function. We further have

$\displaystyle\begin{array}[]{ll}\Gamma(0,t)=E_{1}(t),&\Gamma(1,t)=e^{-t}\\\
\Gamma(2,t)=e^{-t}(t+1),&\Gamma(3,t)=e^{-t}(t^{2}+2t+2).\end{array}$

For example, the TCP policy ($a=0$) has $\hat{P}=P\,e^{t}$. At low SNR, it
turns out that the best choice is $t=0.283$ for which we have
$\Gamma(1,t)/\Gamma(3/2,t)^{2}\approx 1.174$. The minimum $E_{b}/N_{0}$ in
(225) is thus $-0.90$ dB. At high SNR, the best choice is $t=0$ so that (303)
with $\Gamma(3/2,0)=\Gamma(3/2)=\sqrt{\pi}/2$ gives

$\displaystyle I_{1}(A;Y)=\log\left(1+\frac{P\,\pi/4}{1+P(1-\pi/4)}\right).$
(305)

The TCP rate thus saturates at $2.22$ bits per channel use, see the curve
labeled “TCP, GMI, K=1” in Fig. 10.

Figure 10: Rates for Rayleigh fading with $S_{T}=H$ and $S_{R}=0$. The
threshold $t$ was optimized for the $K=1$ curves, while $t=P^{-0.4}$ for the
$I(A;Y)$ and $K=2$ curves. The $K=2$ GMI uses $t_{R}=P^{0.4}$.

The TMF policy ($a=1$) has $\hat{P}=P\,e^{t}/(t+1)$. It turns out that the
best choice is $t=0$ for which we have $\Gamma(2)=1$ and $\Gamma(3)=2$ and
therefore (303) becomes

$\displaystyle I_{1}(A;Y)=\log\left(1+\frac{P}{1+P}\right).$ (306)

The minimum $E_{b}/N_{0}$ in (225) is thus $-1.59$ dB and at high SNR the TMF
rate saturates at $1$ bit per channel use. The rates are shown as the curve
labeled “TMF, GMI, K=1” in Fig. 10.

The TCI policy ($a=-1$) has $\hat{P}=P/E_{1}(t)$ and using
$\Gamma(0,t)=E_{1}(t)$ and $\Gamma(1,t)=e^{-t}$ gives

$\displaystyle
I_{1}(A;Y)=\log\left(1+\frac{P}{e^{2t}\,E_{1}(t)+P\left(e^{t}-1\right)}\right).$
(307)

The minimum $E_{b}/N_{0}$ in (304) is

$\displaystyle\left.\frac{E_{b}}{N_{0}}\right|_{\text{min}}=E_{1}(t)\,e^{2t}\cdot\log
2.$ (308)

Optimizing over $t$ by taking derivatives (see (393) below), the best $t$
satisfies the equation $2te^{t}E_{1}(t)=1$ which gives $t\approx 0.61$ and the
minimal $E_{b}/N_{0}$ is approximately 0.194 dB. On the other hand, for large
SNR we may choose $t=1/P$ and using $E_{1}(t)\approx\log(1/t)$ for small $t$
gives

$\displaystyle I_{1}(A;Y)\approx\log\left(1+\frac{P}{1+\log P}\right).$

Since the pre-log is at most 1, the capacity grows with pre-log 1 for large
$P$. We see that TMF is best at small $P$ while TCI is best at large $P$. The
rates are shown as the curve labeled “TCI, GMI, K=1” in Fig. 10.

The simple channel output of TCI permits further analysis. Using Remark 61 we
compute the mutual information $I(A;Y)$ by numerical integration, see the
curve labeled “TCI, $I(A;Y)$” in Fig. 10. We see that at high SNR the TCI
mutual information is larger than the GMI for TCP, TMF and (of course) TCI.
Moreover, the TCI mutual information can be made to work well at low SNR, as
we proceed to show.

Motivated by Sec. VII-C and Fig. 5, we again use the GMI (127) with $K=2$ and
(67). We further choose $h_{1}=0$, $\sigma_{1}^{2}=\sigma_{2}^{2}=1$, and

$\displaystyle\bar{X}=\frac{\sqrt{\hat{P}}}{h_{2}}\,U,\quad
U\sim\mathcal{CN}(0,1).$

The expression (127) simplifies to

$\displaystyle I_{1}(A;Y)={\rm Pr}\left[{\mathcal{E}_{2}}\right]$
$\displaystyle\left[\log\left(1+\hat{P}\right)+\frac{{\rm
E}\left[{|Y|^{2}|\mathcal{E}_{2}}\right]}{1+\hat{P}}\right.$
$\displaystyle\quad\left.-\;{\rm
E}\left[{\left.\left|Y-\sqrt{\hat{P}}\,U\right|^{2}\right|\mathcal{E}_{2}}\right]\right].$
(309)

The GMI (309) exhibits interesting scaling for both high and low SNR by
choosing the following thresholds $t,t_{R}$.

  * •

For high SNR, we choose

$\displaystyle t=P^{-\lambda}\text{ and }t_{R}=\hat{P}^{\lambda_{R}}$ (310)

where $0<\lambda<1$ and $0<\lambda_{R}<1$. As $P$ increases, $t$ decreases and
Appendix B-D shows that

$\displaystyle\begin{array}[]{l}{\rm
Pr}\left[{\mathcal{E}_{2}}\right]\rightarrow 1\vspace{0.1cm}\\\ {\rm
E}\left[{|Y|^{2}|\mathcal{E}_{2}}\right]\Big{/}\left(1+\hat{P}\right)\rightarrow
1\vspace{0.1cm}\\\ {\rm
E}\left[{\left|Y-\sqrt{\hat{P}}\,U\right|^{2}|\mathcal{E}_{2}}\right]\rightarrow
1.\end{array}$ (314)

Using (394), we have $\hat{P}=P/E_{1}(t)\approx P/(\lambda\log P)$ and

$\displaystyle\lim_{P\rightarrow\infty}\frac{I_{1}(A;Y)}{\log P}=1.$ (315)

The high SNR rates are shown as the curve labeled “TCI, GMI, K=2” in Fig. 10
for $\lambda=\lambda_{R}=0.4$.

  * •

For low SNR, we choose

$\displaystyle t=-\log(P/c)\text{ and }t_{R}=\hat{P}$ (316)

for a constant $c>0$. As $P$ decreases, both $t$ and $\hat{P}=P/E_{1}(t)$
increase and Appendix B-D shows that

$\displaystyle\begin{array}[]{l}{\rm Pr}\left[{\mathcal{E}_{2}}\right]\approx
e^{-t-1}=P/(c\,e)\vspace{0.1cm}\\\ {\rm
E}\left[{|Y|^{2}|\mathcal{E}_{2}}\right]\Big{/}\left(1+2\hat{P}\right)\rightarrow
1\vspace{0.1cm}\\\ {\rm
E}\left[{\left|Y-\sqrt{\hat{P}}\,U\right|^{2}|\mathcal{E}_{2}}\right]\rightarrow
1.\end{array}$ (320)

Using (395), we have $I_{1}(A;Y)\approx e^{-t-1}\log t$ which vanishes as $t$
grows. But we also have

$\displaystyle\frac{E_{b}}{N_{0}}=\frac{P}{R}\log 2$
$\displaystyle\approx\frac{c\,e^{-t}\log 2}{e^{-t-1}\log
t}\approx\frac{c\,e\log 2}{\log(-\log P)}$ (321)

which decreases (very slowly) as $P$ decreases. The minimal $E_{b}/N_{0}$ is
therefore $-\infty$. The low SNR rates are shown as the curve labeled “TCI,
GMI, K=2” in Fig. 11 for $c=1.4$.

Figure 11: Low-SNR rates for Rayleigh fading with $S_{T}=H$ and $S_{R}=0$. The
threshold $t$ was optimized for the $K=1$ curves, while $t=-\log(P/1.4)$ for
the $I(A;Y)$, rGMI, and $K=2$ curves. The $K=2$ GMI uses $t_{R}=\hat{P}$. The
TMF and TMMSE GMIs are indistinguishable for this range of rates.

Fig. 11 shows that the TCI mutual information achieves a minimal $E_{b}/N_{0}$
below $-1.59$ dB. The $K=2$ GMI decreases too quickly for this to be visible,
but at $E_{b}/N_{0}=-2$ dB we computed $I_{1}(A;Y)\approx 6\times 10^{-7}$ and
$I(A;Y)\approx 3\times 10^{-4}$. The $K=2$ partition is useful mainly to prove
that TCI can achieve $E_{b}/N_{0}$ arbitrarily close to zero. Fig. 11 also
shows the reverse model GMI as the curve labeled “TCI, rGMI” which has the
rate $I_{1}(A;Y)\approx 8\times 10^{-6}$ at $E_{b}/N_{0}=-2$ dB.

We compare with the full CSIR and full CSIT rates. At high SNR, the GMI for
$S_{R}=0$ achieves the same capacity pre-log as $S_{R}=H$. At low SNR, recall
from (283) that with full CSIR/CSIT we have
$E_{b}/N_{0}\approx\log(2)/\lambda$. To compare the rates for similar
$E_{b}/N_{0}$, we set $\lambda=\log t$, where $t$ is as in (316) and $c\approx
1$. The TCI $K=2$ GMI without CSIR is approximately $e^{-t}\log t$ while the
full CSIR rate (283) is approximately $e^{-\lambda}/\lambda\approx
1/(t\log(t))$. Thus, the $K=2$ GMI with no CSIR is a fraction
$te^{-t}\log(t)^{2}$ of the full CSIR capacity.

####  VIII-D2 $S_{R}=1(G\geq t)$

The power in (209) is again (302) and the rate (227) is

$\displaystyle I_{1}(A;Y|S_{R})=e^{-t}\cdot\log\Bigg{(}1+$
$\displaystyle\;\left.\frac{P\,e^{2t}\,\Gamma\left({\frac{3+a}{2},t}\right)^{2}}{\Gamma\left({1+a,t}\right)+P\left[e^{t}\,\Gamma\left({2+a,t}\right)-e^{2t}\,\Gamma\left({\frac{3+a}{2},t}\right)^{2}\right]}\right).$
(322)

Moreover, the expression (228) is

$\displaystyle\left.\frac{E_{b}}{N_{0}}\right|_{\text{min}}=\frac{\Gamma\left({1+a,t}\right)}{e^{t}\cdot\Gamma\left({\frac{3+a}{2},t}\right)^{2}}\cdot\log
2$ (323)

which is the same as (304) except for the factor $e^{t}$ in the denominator.
This implies that the minimal $E_{b}/N_{0}$ can be improved for $t>0$.

The TCP, TMF, and TCI rates (322) are the respective

$\displaystyle I_{1}(A;Y|S_{R})$
$\displaystyle\;=e^{-t}\log\left(1+\frac{P\,e^{2t}\,\Gamma\left({\frac{3}{2},t}\right)^{2}}{e^{-t}+P\left[t+1-e^{2t}\,\Gamma\left({\frac{3}{2},t}\right)^{2}\right]}\right)$
(324) $\displaystyle
I_{1}(A;Y|S_{R})=e^{-t}\log\left(1+\frac{P\,(t+1)^{2}}{e^{-t}(t+1)+P}\right)$
(325) $\displaystyle
I_{1}(A;Y|S_{R})=e^{-t}\log\left(1+\frac{P}{E_{1}(t)}\right).$ (326)

######  Remark 69.

As pointed out in Remark 64, the TCI GMI (326) is $I(A;Y|S_{R})$. One can also
understand this by observing that the receiver knows $\sqrt{GP(G)}$ for all
$G$. The mutual information is thus related to the rate (190) of Proposition
2.

Figure 12: Rates for Rayleigh fading with full CSIT and $S_{R}=1(G\geq t)$.

Next, the minimal $E_{b}/N_{0}$ in (323) are the respective

$\displaystyle\left.\frac{E_{b}}{N_{0}}\right|_{\text{min}}=\frac{1}{e^{2t}\cdot\Gamma\left({\frac{3}{2},t}\right)^{2}}\cdot\log
2$ (327)
$\displaystyle\left.\frac{E_{b}}{N_{0}}\right|_{\text{min}}=\frac{1}{t+1}\cdot\log
2$ (328)
$\displaystyle\left.\frac{E_{b}}{N_{0}}\right|_{\text{min}}=e^{t}\,E_{1}(t)\cdot\log
2.$ (329)

The above expressions mean that, for all three policies, we can make the
minimal $E_{b}/N_{0}$ as small as desired by increasing $t$. For example, for
TCI we can bound (see (397) below)

$\displaystyle\frac{1}{t+1}<e^{t}E_{1}(t)<\frac{1}{t}.$ (330)

TCI thus has a slightly larger (slightly worse) minimal $E_{b}/N_{0}$ than TMF
for the same $t$, as discussed above after (215).

For large $P$, the TCP rate (324) is optimized by $t\approx 0.163$ and the
rate saturates at $\approx 2.35$ bits per channel use. The TMF rate (325) is
optimized with $t=0$ and the rate saturates at 1 bit per channel use. For the
TCI rate (326), we again choose $t=1/P$ and use $E_{1}(t)\approx\log(1/t)$ for
small $t$ to show that the capacity grows with pre-log 1:

$\displaystyle I_{1}(A;Y|S_{R})\approx\log\left(1+\frac{P}{\log P}\right).$

Again, TMF is best at small $P$ while TCI is best at large $P$.

####  VIII-D3 Optimal Policy

Consider now the optimal power control policy. Suppose first that $S_{R}=0$
for which Theorem 2 gives the TMMSE policy with $t=0$:

$\displaystyle\sqrt{P(h)}$ $\displaystyle=\frac{\alpha|h|}{\beta+|h|^{2}}.$
(331)

For Rayleigh fading, we thus have (see (401) below)

$\displaystyle
P=\int_{0}^{\infty}e^{-g}\frac{\alpha^{2}g}{(\beta+g)^{2}}\,dg=\alpha^{2}\left[(\beta+1)e^{\beta}E_{1}(\beta)-1\right]$
(332)

with the two expressions (see (400) and (402) below)

$\displaystyle\tilde{P}=\int_{0}^{\infty}e^{-g}\frac{\alpha^{2}g}{\beta+g}\,dg=\alpha^{2}\left[1-\beta
e^{\beta}E_{1}(\beta)\right]^{2}$ (333) $\displaystyle{\rm
E}\left[{GP(H)}\right]$
$\displaystyle=\int_{0}^{\infty}e^{-g}\frac{\alpha^{2}g^{2}}{(\beta+g)^{2}}\,dg$
$\displaystyle=\alpha^{2}\left[1+\beta-\beta(\beta+2)e^{\beta}E_{1}(\beta)\right].$
(334)

Given $P$ and $\beta$, we may compute $\alpha^{2}$ from (332). We then search
for the optimal $\beta$ for fixed $P$. The rates are shown as the curve
labeled “TMMSE, GMI, K=1” in Figs. 10-11 and we see that the TMMSE strategy
has the best $K=1$ rates.

Consider next $S_{R}=1(G\geq t)$ and the TMMSE policy. We compute (see (401)
below)

$\displaystyle P$
$\displaystyle=\int_{t}^{\infty}e^{-g}\frac{\alpha^{2}g}{(\beta+g)^{2}}\,dg$
$\displaystyle=\alpha^{2}\left[(\beta+1)e^{\beta}E_{1}(t+\beta)-e^{-t}\frac{\beta}{t+\beta}\right]$
(335)

and (see (400) and (402) below)

$\displaystyle\sqrt{\tilde{P}(1)}=\int_{t}^{\infty}\frac{e^{-g}}{e^{-t}}\frac{\alpha
g}{\beta+g}\,dg$ $\displaystyle\quad=\alpha\left[1-\beta
e^{t+\beta}E_{1}(t+\beta)\right]$ (336) $\displaystyle{\rm
E}\left[{|Y|^{2}|S_{R}=1}\right]=\int_{t}^{\infty}\frac{e^{-g}}{e^{-t}}\left(1+\frac{\alpha^{2}g^{2}}{(\beta+g)^{2}}\right)\,dg$
$\displaystyle\quad=1+\alpha^{2}\left[1+\frac{\beta^{2}}{t+\beta}-\beta(\beta+2)e^{t+\beta}E_{1}(t+\beta)\right].$
(337)

We optimize as for the $S_{R}=0$ case: given $P$, $\beta$, $t$, we compute
$\alpha^{2}$ from (335). We then search for the optimal $\beta$ for fixed $P$
and $t$. The optimal $t$ turns out to be approximately a factor of 1.1 smaller
than for the TCI policy. The rates are shown in Fig. 12 as the curve labeled
“TMMSE, GMI”.

###  VIII-E Partial CSIR, CSIT@R

Suppose $S_{R}$ is defined by (see (173))

$\displaystyle H=\sqrt{\bar{\epsilon}}\,S_{R}+\sqrt{\epsilon}\,Z_{R}$

where $0\leq\epsilon\leq 1$ and $S_{R},Z_{R}$ are independent with
distribution $\mathcal{CN}(0,1)$. We further consider the CSIT
$S_{T}=|S_{R}|^{2}$.

The reverse model GMI again requires ${\rm Var}\left[{U|Y,S_{R}}\right]$ which
can be computed by simulation, see Appendix C-D. However, as in Sec. VII-D and
Sec. VIII-C, optimizing the powers seems difficult and we instead focus on the
forward model GMI. We compute the quantities in (240)-(241) in Sec. VI-F to be

$\displaystyle\tilde{g}(s_{R})=\bar{\epsilon}\,s_{T},\quad\tilde{\sigma}^{2}(s_{R})=\epsilon.$
(338)

The GMI (242) of Theorem 3 is

$\displaystyle I_{1}(X;Y|S_{R})$
$\displaystyle=\int_{\lambda/\bar{\epsilon}}^{\infty}e^{-s_{T}}\log\left(1+\frac{\bar{\epsilon}\,s_{T}P(s_{T})}{1+\epsilon\,P(s_{T})}\right)\,ds_{T}$
(339)

where the power control policy $P(s_{T})$ is given by (250). The parameter
$\lambda$ is chosen so that ${\rm E}\left[{P(S_{T})}\right]=P$. For example,
for $\epsilon\rightarrow 0$ we recover the waterfilling solution (182). Fig.
13 shows the quadratic and conventional waterfilling rates, which lie almost
on top of each other. For example, the inset shows the rates for
$\epsilon=0.2$ and a small range of $E_{b}/N_{0}$.

Figure 13: Rates for Rayleigh fading with partial CSIR and CSIT@R. The curves
labeled ‘q-waterfill’ and ‘c-waterfill’ are the quadratic and conventional
waterfilling rates, respectively.

##  IX Channels with In-Block Feedback

This section generalizes Shannon’s model of Sec. IV by including block fading
with in-block feedback. The model lets one include delay in the CSIT, for
example, and it permits many other generalizations for network models [22].

The problem is specified by the FDG shown in Fig. 14. The model has a message
$M$ and the channel input and output block strings

$\displaystyle X_{i}^{L}$ $\displaystyle=(X_{i1},\ldots,X_{iL})$
$\displaystyle Y_{i}^{L}$ $\displaystyle=(Y_{i1},\ldots,Y_{iL})$

with $L$ symbols each for block $i=1,\ldots,n$. The channel is specified by a
string $S_{H}^{n}=(S_{H1},\ldots,S_{Hn})$ of i.i.d. hidden channel states. The
CSIR $S_{Ri\ell}$ is a (possibly noisy) function of $S_{Hi}$ for all $i$ and
$\ell$. The receiver sees the channel outputs (see (161))

$\displaystyle(Y_{i\ell},S_{Ri\ell})=\left(f_{\ell}\left(X_{i}^{\ell},S_{Hi},Z_{i}^{L}\right),S_{Ri\ell}\right)$
(340)

for some functions $f_{\ell}(\cdot)$, $\ell=1,\dots,L$. Observe that the
$X_{i}^{\ell}$ influence the $Y_{i\ell}$ in a causal fashion. The random
symbols $M,S_{H1},\ldots,S_{Hn},Z_{1}^{L},\ldots,Z_{n}^{L}$ are mutually
independent.

Figure 14: FDG for a block fading model with $n=2$ blocks of length $L=2$ and
in-block feedback. Across-block dependence via past $S_{Ti\ell}$ is not shown.

We now permit past channel symbols to influence the CSIT, see Sec. I-B.
Suppose the CSIT has the form

$\displaystyle
S_{Ti\ell}=f_{T\ell}\left(S_{Hi},X_{i}^{\ell-1},Y_{i}^{\ell-1}\right)$ (341)

for some function $f_{T\ell}(.)$ and for all $i$ and $\ell$. The motivation
for (341) is that useful CSIR may not be available until the end of a block,
or even much later. But in the mean time, the receiver can, e.g., quantize the
$Y_{i}^{\ell-1}$ and transmit the quantization bits via feedback. This lets
one study fast power control and/or beamforming without precise knowledge of
the channel coefficients.

Define the string of past and current states as

$\displaystyle
s_{T}^{i\ell}=\left(s_{T1}^{L},\dots,s_{T(i-1)}^{L},s_{Ti}^{\ell}\right).$
(342)

The channel input at time $i\ell$ is $X(s_{T}^{i\ell})$ and the adaptive
codeword $A^{nL}$ is defined by the ordered lists

$\displaystyle A_{i\ell}$
$\displaystyle=\left[X(s_{T}^{i\ell}),\;\forall\;s_{T}^{i\ell}\right]$ (343)

for $1\leq i\leq n$ and $1\leq\ell\leq L$. The adaptive codeword $A^{nL}$ is a
function of $M$ and is thus independent of $S_{H}^{n}$ and $S_{T}^{nL}$.

######  Remark 70.

The model under consideration is a special case of the channels introduced in
[22, Sec. V]. However, the model in [22] has transmission and reception begin
at time $\ell=2$ rather than $\ell=1$. To compare the theory, one must thus
shift the time indexes by 1 unit and increase $L$ to $L+1$, see Remark 71
below. The capacity for our model is given by [22, Thm. 2] which we write as

$\displaystyle C$
$\displaystyle\overset{(a)}{=}\max_{A^{L}}\frac{1}{L}I(A^{L};Y^{L},S_{R}^{L})$
$\displaystyle\overset{(b)}{=}\max_{A^{L}}\frac{1}{L}I(A^{L};Y^{L}\big{|}S_{R}^{L}).$
(344)

where $(a)$ follows by normalizing by $L$ rather than $L+1$, and step $(b)$
follows by the independence of $A^{L}$ and $S_{R}^{L}$.

###  IX-A GMI for Scalar Channels

We will study scalar block fading channels; extensions to vector channels
follow as described in Sec. IV-D. Let $\underline{Y}=[Y_{1},\dots,Y_{L}]^{T}$
be the vector form of $Y^{L}$ and similarly for other strings with $L$
symbols. The GMI with parameter $s$ is

$\displaystyle I_{s}(A^{L};Y^{L}\big{|}S_{R}^{L})={\rm
E}\left[{\log\frac{q(\underline{Y}\big{|}\underline{A},\underline{S}_{R})^{s}}{q(\underline{Y}\big{|}\underline{S}_{R})}}\right]$
(345)

####  IX-A1 Reverse Model

For the reverse model, let $\underline{A}$ be a column vector that stacks the
$X_{\ell}(s_{T}^{\ell})$ for all $s_{T}^{\ell}$ and $\ell$. Consider a reverse
density as in (107):

$\displaystyle
q\big{(}a^{L}|y^{L}\big{)}=\frac{\exp\left(-\underline{z}(\underline{y},\underline{s}_{R})^{\dagger}\,Q_{\underline{A}|\underline{Y}=\underline{y},\underline{S}_{R}=\underline{s}_{R}}^{-1}\,\underline{z}(\underline{y},\underline{s}_{R})\right)}{\pi^{N}\det
Q_{\underline{A}|\underline{Y}=\underline{y},\underline{S}_{R}=\underline{s}_{R}}}$

where

$\displaystyle\underline{z}(\underline{y},\underline{s}_{R})=\underline{a}-{\rm
E}\left[{\underline{A}|\underline{Y}=\underline{y},\underline{S}_{R}=\underline{s}_{R}}\right].$

Using the forward model $q(y^{L}|a^{L})=q(a^{L}|y^{L})/p(a^{L})$, the GMI with
$s=1$ becomes

$\displaystyle I_{1}(A^{L};Y^{L},S_{R}^{L})={\rm E}\left[{\log\frac{\det
Q_{\underline{A}}}{\det
Q_{\underline{A}|\underline{Y},\underline{S}_{R}}}}\right].$ (346)

To simplify, consider adaptive symbols as in (93) (cf. (109)):

$\displaystyle
X_{\ell}(S_{T}^{\ell})=\sqrt{P_{\ell}(S_{T}^{\ell})}\,e^{j\phi_{\ell}(S_{T}^{\ell})}\,U_{\ell}$
(347)

where $\underline{U}\sim\mathcal{CN}(\underline{0},{\bf I})$. In other words,
consider a conventional codebook represented by the $U_{\ell}$ and adapt the
power and phase based on past CSIT. The mutual information becomes
$I(A^{L};Y^{L},S_{R}^{L})=I(U^{L};Y^{L},S_{R}^{L})$ (cf. (98)) and the GMI
with $s=1$ is (cf. (110))

$\displaystyle I_{1}(A^{L};Y^{L}\big{|}S_{R}^{L})={\rm E}\left[{-\log\det{\bf
Q}_{\underline{U}\big{|}\underline{Y},\underline{S}_{R}}}\right].$ (348)

In fact, one may also consider choosing $U_{\ell}=U$ for all $\ell$ in which
case we compute (cf. (151))

$\displaystyle I_{1}(A^{L};Y^{L}\big{|}S_{R}^{L})={\rm E}\left[{-\log{\rm
Var}\left[{U\big{|}\underline{Y},\underline{S}_{R}}\right]}\right].$ (349)

####  IX-A2 Forward Model

Consider the following forward model (cf. (113) and (153)):

$\displaystyle
q(\underline{y}\big{|}\underline{a},\underline{s}_{R})=\frac{\exp\left(-\underline{z}(\underline{s}_{R})^{\dagger}{\bf
Q}_{\underline{Z}}(\underline{s}_{R})^{-1}\underline{z}(\underline{s}_{R})\right)}{\pi^{L}\det{\bf
Q}_{\underline{Z}}(\underline{s}_{R})}.$ (350)

with

$\displaystyle\underline{z}(\underline{s}_{R})=\underline{y}-{\bf
H}(\underline{s}_{R})\,\underline{\bar{x}}(\underline{s}_{R})$

and where similar to (154) we define

$\displaystyle\underline{\bar{X}}(\underline{s}_{R})=\sum_{\underline{s}_{T}}{\bf
W}(\underline{s}_{T},\underline{s}_{R})\,\underline{X}(\underline{s}_{T})$
(351)

where the ${\bf W}(\underline{s}_{T},\underline{s}_{R})$ are $L\times L$
complex matrices. Note that

$\displaystyle\underline{X}(\underline{s}_{T})=[X_{1}(s_{T1}),X_{2}(s_{T}^{2}),\dots,X_{2}(s_{T}^{L})]^{T}$
(352)

so $X_{\ell}$ is a function of $A^{L}$ and $S_{T}^{\ell}$, $\ell=1,\dots,L$.

We have the following generalization of Lemma 5 (see also Theorem 1) where the
novelty is that $S_{T}$ is replaced with $\underline{S}_{T}$. Define
$\underline{U}(\underline{s}_{T})\sim\mathcal{CN}(\underline{0},{\bf I})$ and
$\underline{X}(\underline{s}_{T})={\bf
Q}_{\underline{X}(\underline{s}_{T})}^{1/2}\,\underline{U}(\underline{s}_{T})$
for all $\underline{s}_{T}$.

######  Theorem 4.

A GMI (345) for the scalar block fading channel $p(y^{L}|a^{L},s_{R}^{L})$, an
adaptive codeword $A^{L}$ with CSCG entries, the auxiliary model (350), and
with fixed ${\bf Q}_{X(\underline{s}_{T})}$ is

$\displaystyle I_{1}(A^{L};Y^{L}|S_{R}^{L})$ $\displaystyle={\rm
E}\left[{\log\left(\frac{\det{\bf
Q}_{\underline{Y}}(\underline{S}_{R})}{\det\left({\bf
Q}_{\underline{Y}}(\underline{S}_{R})-\tilde{\bf
D}(\underline{S}_{R})\,\tilde{\bf
D}(\underline{S}_{R})^{\dagger}\right)}\right)}\right].$ (353)

where

$\displaystyle{\bf Q}_{\underline{Y}}(\underline{s}_{R})$ $\displaystyle={\rm
E}\left[{\left.\underline{Y}\,\underline{Y}^{\dagger}\right|\underline{S}_{R}=\underline{s}_{R}}\right]$
(354)

and for $M\times M$ unitary ${\bf V}_{R}(\underline{s}_{T},\underline{s}_{R})$
the matrix $\tilde{\bf D}(\underline{s}_{R})$ is

$\displaystyle{\rm E}\left[{\left.{\bf
U}_{T}(\underline{S}_{T},\underline{s}_{R})\,{\bf\Sigma}(\underline{S}_{T},\underline{s}_{R})\,{\bf
V}_{R}(\underline{S}_{T},\underline{s}_{R})^{\dagger}\right|\underline{S}_{R}=\underline{s}_{R}}\right]$
(355)

and ${\bf U}_{T}(\underline{s}_{T},\underline{s}_{R})$ and
${\bf\Sigma}(\underline{s}_{T},\underline{s}_{R})$ are $N\times N$ unitary and
$N\times M$ rectangular diagonal matrices, respectively, of the SVD

$\displaystyle{\rm
E}\left[{\left.\underline{Y}\,{\underline{U}(\underline{s}_{T})}^{\dagger}\right|\underline{S}_{T}=\underline{s}_{T},\underline{S}_{R}=\underline{s}_{R}}\right]$
$\displaystyle\quad={\bf
U}_{T}(\underline{s}_{T},\underline{s}_{R})\,{\bf\Sigma}(\underline{s}_{T},\underline{s}_{R})\,{\bf
V}_{T}(\underline{s}_{T},\underline{s}_{R})^{\dagger}$ (356)

for all $\underline{s}_{T}$, $\underline{s}_{R}$ and the ${\bf
V}_{T}(\underline{s}_{T},\underline{s}_{R})$ are $M\times M$ unitary matrices.
One may maximize (353) over the unitary ${\bf
V}_{R}(\underline{s}_{T},\underline{s}_{R})$.

Suppose next that the true channel is
$\underline{Y}=H\underline{X}+\underline{Z}$ where
$\underline{Z}\sim\mathcal{CN}(\underline{0},{\bf I})$. The extension of (148)
to block fading channels with CSIR is (see (169))

$\displaystyle I_{1}(A^{L};Y^{L}|S_{R}^{L})$
$\displaystyle=\sum_{\ell=1}^{L}{\rm
E}\left[{\log\left(1+\frac{\tilde{P}_{\ell}(\underline{S}_{R})}{1+{\rm
E}\left[{GP_{\ell}(S_{T}^{\ell})\big{|}\underline{S}_{R}}\right]-\tilde{P}_{\ell}(\underline{S}_{R})}\right)}\right]$
(357)

where

$\displaystyle\tilde{P}_{\ell}(\underline{s}_{R})={\rm E}\left[{\,\left|{\rm
E}\left[{\left.H\sqrt{P_{\ell}(S_{T}^{\ell})}\,\right|S_{T}^{\ell},\underline{S}_{R}=\underline{s}_{R}}\right]\right|\,}\right]^{2}$
$\displaystyle{\rm
E}\left[{|Y_{\ell}|^{2}\big{|}\underline{S}_{R}=\underline{s}_{R}}\right]=1+{\rm
E}\left[{GP_{\ell}(S_{T}^{\ell})\big{|}\underline{S}_{R}=\underline{s}_{R}}\right].$

###  IX-B CSIT@R

Continuing as in Sec. V-B, suppose the CSIT in (341) can be written by
replacing $S_{Hi}$ with $S_{Ri}^{\ell}$ for all $i$ and $\ell$:

$\displaystyle
S_{Ti\ell}=f_{T\ell}\left(S_{Ri}^{\ell},X_{i}^{\ell-1},Y_{i}^{\ell-1}\right).$
(358)

The capacity (344) then simplifies to a directed information. To see this,
expand the mutual information in (344) as

$\displaystyle I(A^{L};Y^{L}\big{|}S_{R}^{L})$
$\displaystyle\overset{(a)}{=}\sum_{\ell=1}^{L}I\left(A^{L},X^{\ell};Y_{\ell}\big{|}S_{R}^{L},Y^{\ell-1}\right)$
$\displaystyle\overset{(b)}{=}\sum_{\ell=1}^{L}I(X^{\ell};Y_{\ell}\big{|}S_{R}^{L},Y^{\ell-1})$
(359)

where step $(a)$ follows because $X^{\ell}$ is a function of $A^{L}$ and
$S_{T}^{\ell}$ in (358), and step $(b)$ follows by the Markov chains

$\displaystyle A^{L}-[S_{R}^{L},X^{\ell},Y^{\ell-1}]-Y_{\ell}.$ (360)

The capacity is therefore (see the definition (26))

$\displaystyle
C=\max_{X_{\ell}(S_{T}^{\ell}),\;\ell=1,\ldots,L}\frac{1}{L}I(X^{L}\rightarrow
Y^{L}\big{|}S_{R}^{L}).$ (361)

The maximization in (361) under a cost constraint becomes a constrained
maximization so that ${\rm E}\left[{c(X^{L},Y^{L})}\right]\leq LP$ for some
cost function $c(\cdot)$.

######  Remark 71.

As outlined in Remark 70, the capacity (361) is a special case of the theory
in [22, eq. (48)]. To see this, define the extended and time-shifted strings

$\displaystyle\hat{A}^{L+1}=(0,A^{L}),\;\;\hat{X}^{L+1}=(0,X^{L}),\;\;\hat{Y}^{L+1}=(0,Y^{L}).$

Now expand (359) as

$\displaystyle
I(A^{L};Y^{L}\big{|}S_{R}^{L})=I(A^{L}\,;\,(S_{R2},\dots,S_{RL},0),Y^{L}\big{|}S_{R1})$
$\displaystyle\quad\overset{(a)}{=}\sum_{\ell=1}^{L}I(A^{L},X^{\ell}\,;\,S_{R(\ell+1)},Y_{\ell}\big{|}S_{R}^{\ell},Y^{\ell-1})$
$\displaystyle\quad\overset{(b)}{=}\sum_{\ell=1}^{L}I(X^{\ell}\,;\,S_{R(\ell+1)},Y_{\ell}\big{|}S_{R}^{\ell},Y^{\ell-1})$
$\displaystyle\quad=\sum_{\ell=2}^{L+1}I(\hat{X}^{\ell}\,;\,S_{R\ell},\hat{Y}_{\ell}\big{|}S_{R}^{\ell-1},\hat{Y}^{\ell-1})$
(362)

where $S_{R(L+1)}=0$, step $(a)$ follows because $X^{\ell}$ is a function of
$A^{L}$ and $S_{T}^{\ell}$ in (358), and step $(b)$ follows by the Markov
chains

$\displaystyle
A^{L}-[X^{\ell},Y^{\ell-1},S_{R}^{\ell}]-[Y_{\ell},S_{R(\ell+1)}].$ (363)

The expression (362) is the desired directed information

$\displaystyle
I(A^{L};Y^{L},S_{R}^{L})=I(\hat{X}^{L+1}\rightarrow\hat{Y}^{L+1},S_{R}^{L+1}).$
(364)

######  Remark 72.

Consider the basic CSIT model

$\displaystyle S_{Ti\ell}=f_{T}(S_{Ri\ell})$ (365)

for some function $f_{T}(\cdot)$ and for $\ell=1,\ldots,L$ and $i=1,\ldots,n$.
This model was studied in [87, Sec. III.C] and its capacity is derived as (see
[87, eq. (35) with eq. (13)])

$\displaystyle C$
$\displaystyle=\max_{X_{\ell}(S_{T}^{\ell}),\;\ell=1,\ldots,L}\frac{1}{L}I(X^{L};Y^{L}\big{|}S_{R}^{L},S_{T}^{L}).$
(366)

To see that (366) is a special case of (361), observe that

$\displaystyle I(X^{L}\rightarrow Y^{L}|S_{R}^{L})$
$\displaystyle\overset{(a)}{=}\sum_{\ell=1}^{L}I(X^{\ell};Y_{\ell}\big{|}S_{R}^{L},S_{T}^{L},Y^{\ell-1})$
$\displaystyle\overset{(b)}{=}\sum_{\ell=1}^{L}I(X^{L};Y_{\ell}\big{|}S_{R}^{L},S_{T}^{L},Y^{\ell-1})$
(367)

where step $(a)$ follows by (359), and step $(b)$ follows by the Markov chain

$\displaystyle[X_{\ell+1},\ldots,X_{L}]-[S_{R}^{L},S_{T}^{L},Y^{\ell-1},X^{\ell}]-Y_{\ell}.$
(368)

The expression (367) gives (366). Related results are available in [10, Sec.
III] and [88, 89].

######  Remark 73.

The capacity (361) has only $S_{R}^{L}$ in the conditioning while (366) has
both $S_{R}^{L}$ and $S_{T}^{L}$ in the conditioning. This subtle difference
is due to permitting $X^{\ell-1}$ to influence the $S_{T\ell}$ in (358) and it
complicates the analysis. On the other hand, if we remove only $X^{\ell-1}$
from (358) then the receiver knows $S_{T\ell}$ at time $\ell$ and the capacity
(361) can be written as (see the definition (27))

$\displaystyle
C=\max_{X_{\ell}(S_{T}^{\ell}),\;\ell=1,\ldots,L}\frac{1}{L}I(X^{L}\rightarrow
Y^{L}\big{\|}S_{T}^{L}\big{|}S_{R}^{L}).$ (369)

We treat such a model with quantized channel output feedback in Sec. IX-F
below.

###  IX-C Fading Channels with AWGN

The expression (361) is valid for general statistics. We next specialize to
the block-fading AWGN model

$\displaystyle Y_{\ell}=HX_{\ell}+Z_{\ell}$ (370)

where $\ell=1,\dots,L$, $Z^{L}\sim\mathcal{CN}(\underline{0},{\bf I})$, and
$H,A^{L},Z^{L}$ are mutually independent. Consider the power constraint

$\displaystyle\sum_{\ell=1}^{L}{\rm
E}\left[{P_{\ell}\left(S_{T}^{\ell}\right)}\right]\leq LP$ (371)

where $P_{\ell}(s_{T}^{\ell})={\rm
E}\left[{|X_{\ell}(s_{T}^{\ell})|^{2}}\right]$. The optimization of (361)
under the constraint (371) is usually intractable and we again desire
expressions with $\log(1+\text{SNR})$ terms to obtain insight.

A capacity upper bound with such terms uses similar steps as in (166) in
Remark 45:

$\displaystyle
I(A^{L};Y^{L}|S_{R}^{L})=\sum_{\ell=1}^{L}I\left(A^{L};Y_{\ell}\,\big{|}\,S_{R}^{L},Y^{\ell-1}\right)$
$\displaystyle\leq\sum_{\ell=1}^{L}\left[h(Y_{\ell}|S_{R}^{L},Y^{\ell-1})-h(Z_{\ell})\right]$
$\displaystyle\overset{(a)}{\leq}\sum_{\ell=1}^{L}{\rm
E}\left[{\log\left(1+{\rm
E}\left[{GP_{\ell}(S_{T}^{\ell})\big{|}S_{R}^{L},Y^{\ell-1}}\right]\,\right)}\right]$
(372)

where $G=|H|^{2}$ and step $(a)$ follows by the entropy bound

$\displaystyle h(Y_{\ell}\big{|}S_{R}^{L}=s_{R}^{L},Y^{\ell-1}=y^{\ell-1})$
$\displaystyle\quad\leq\log\left(\pi e\,{\rm
Var}\left[{Y_{\ell}\big{|}S_{R}^{L}=s_{R}^{L},Y^{\ell-1}=y^{\ell-1}}\right]\right).$
(373)

It remains to maximize (372) over the power control policies satisfying the
power constraint (371).

######  Remark 74.

A bound of the type (163) of Lemma 6 seems more difficult to obtain unless we
simplify the CSIT, see Proposition 3 below.

Deriving lower bounds is now more subtle than in Sec. VI. Consider the CSIT
model (358) where for each block we have

$\displaystyle S_{T\ell}=f_{T\ell}(H,X^{\ell-1},Y^{\ell-1})$

for all $\ell$. The capacity (361) is

$\displaystyle C(P)$
$\displaystyle=\max_{X_{\ell}(S_{T}^{\ell}),\;\ell=1,\ldots,L}\frac{1}{L}I(X^{L}\rightarrow
Y^{L}\big{|}H)$ (374)
$\displaystyle=\max_{X_{\ell}(S_{T}^{\ell}),\;\ell=1,\ldots,L}\left[\frac{1}{L}h(Y^{L}\big{|}H)\right]-\log(\pi
e).$ (375)

However, CSCG inputs are not necessarily optimal since $X^{\ell-1}$ and
$Y^{\ell-1}$ affect the CSIT.

Instead of trying to optimize the input, consider $X_{\ell}$ that are CSCG. We
can now write

$\displaystyle\frac{1}{L}I(X^{L}\rightarrow
Y^{L}|H)=\frac{1}{L}\sum_{\ell=1}^{L}{\rm
E}\left[{\log\left(1+GP_{\ell}(S_{T}^{\ell})\right)}\right]$ (376)

and the Lagrangians to maximize (376) are

$\displaystyle\sum_{\ell=1}^{L}{\rm
E}\left[{\log\left(1+GP_{\ell}(S_{T}^{\ell})\right)}\right]+\lambda\left(LP-\sum_{\ell=1}^{L}{\rm
E}\left[{P_{\ell}(S_{T}^{\ell})}\right]\right).$ (377)

Suppose the $S_{T\ell}$ are discrete random variables. Taking the derivative
with respect to $P_{\ell}(s_{T}^{\ell})$, we obtain

$\displaystyle\lambda=\int_{0}^{\infty}p(g|s_{T}^{\ell})\frac{g}{1+gP_{\ell}(s_{T}^{\ell})}dg$
$\displaystyle\;+\sum_{k=\ell+1}^{L}\sum_{s_{T}^{k}}\int_{0}^{\infty}p(g)\frac{dP_{S_{T}^{k}|G}(s_{T}^{k}|g)}{dP_{\ell}(s_{T}^{\ell})}\frac{\log\left(1+gP_{k}(s_{T}^{k})\right)}{P_{S_{T}^{\ell}}(s_{T}^{\ell})}dg$
(378)

as long as $P_{\ell}(s_{T}^{\ell})>0$. This expression is complicated because
the choice of transmit powers $P_{\ell}(s_{T}^{\ell})$ influences the
statistics of the CSIT. As before, if (378) cannot be satisfied then choose
$P_{\ell}(s_{T}^{\ell})=0$. Finally, set $\lambda$ so that
$\sum_{\ell=1}^{L}{\rm E}\left[{P_{\ell}(S_{T}^{\ell})}\right]=LP$.

Instead of the above, consider the simpler CSIT model with
$S_{T\ell}=f_{T\ell}(H)$ for all $\ell$, cf. (365). The capacity (366) is now
given by (376) with CSCG inputs and (378) simplifies because the derivatives
with respect to $P_{\ell}(s_{T}^{\ell})$ are zero, i.e., for all $\ell$ and
$s_{T}^{\ell}$ we have

$\displaystyle\lambda=\int_{0}^{\infty}p(g|s_{T}^{\ell})\frac{g}{1+gP_{\ell}(s_{T}^{\ell})}dg.$
(379)

We use (379) for (383)-(385) in Sec. IX-F below.

###  IX-D Full CSIR, Partial CSIT

We next generalize Proposition 2 in Sec. VI-D to the block-fading AWGN model
(370) with the CSIR $S_{R\ell}=H\sqrt{P(S_{T}^{\ell})}$, $\ell=1,\dots,L$, and
where $S_{T\ell}=f_{T\ell}(S_{H})$, i.e., we have discarded $X_{i}^{\ell-1}$
and $Y_{i}^{\ell-1}$ in (341). We then have the following capacity result that
implies that this CSIR is the best possible since one achieves a capacity
upper bound similar to (163) in Lemma 6.

######  Proposition 3.

The capacity of the channel (370) with

$\displaystyle
S_{R}^{L}=H\left[\sqrt{P_{1}(S_{T1})},\sqrt{P_{2}(S_{T}^{2})},\dots,\sqrt{P_{L}(S_{T}^{L})}\right]$
(380)

and $S_{T\ell}=f_{T\ell}(S_{H})$ for $\ell=1,\dots,L$ is

$\displaystyle C(P)$ $\displaystyle=\max\frac{1}{L}\sum_{\ell=1}^{L}{\rm
E}\left[{\log\left(1+GP_{\ell}(S_{T}^{\ell})\right)}\right]$ (381)

where the maximization is over the power control policies
$P_{\ell}(S_{T}^{\ell})$ such that $\sum_{\ell=1}^{L}{\rm
E}\left[{P_{\ell}(S_{T}^{\ell})}\right]\leq LP$. One may use (379) to compute
the $P_{\ell}(S_{T}^{\ell})$.

###### Proof.

For achievability, apply (357) with

$\displaystyle\tilde{P}_{\ell}(\underline{S}_{R})=GP_{\ell}(S_{T}^{\ell})\text{
and }{\rm
E}\left[{|Y_{\ell}|^{2}|\underline{S}_{R}}\right]=1+\tilde{P}_{\ell}(\underline{S}_{R}).$

The converse follows by applying similar steps as in the proof of Lemma 6:

$\displaystyle I(A^{L};Y^{L}|S_{R}^{L})\leq
I(A^{L};Y^{L},S_{T}^{L},H|S_{R}^{L})$
$\displaystyle=\sum_{\ell=1}^{L}I\left(A^{L};Y_{\ell}\,\big{|}\,S_{R}^{L},S_{T}^{L},H,Y^{\ell-1}\right)$
$\displaystyle\leq\sum_{\ell=1}^{L}\left[h(Y_{\ell}|S_{R}^{L},S_{T}^{L},H,Y^{\ell-1})-h(Z_{\ell})\right]$
$\displaystyle\overset{(a)}{\leq}\sum_{\ell=1}^{L}{\rm E}\left[{\log{\rm
Var}\left[{Y_{\ell}|S_{R}^{L},S_{T}^{L},H,Y^{\ell-1}}\right]}\right].$ (382)

Finally, insert ${\rm
Var}\left[{Y_{\ell}|S_{R}^{L},S_{T}^{L},H,Y^{\ell-1}}\right]=1+GP_{\ell}(S_{T}^{\ell})$.
∎

As in Remark 45, the RHS of (382) is at most the RHS of (372) and hence (382)
gives a better bound. However, the bound (382) is valid only for the CSIT
$S_{T\ell}=f_{T\ell}(S_{H})$.

###  IX-E On-Off Fading with Delayed CSIT

Consider on-off fading where the CSIT is delayed by $D$ symbols, i.e., we have
$S_{T\ell}=0$ for $\ell=1,\dots,D$ and $S_{T(D+1)}=H$. Define the transmit
powers as $P_{\ell}(s_{T}^{\ell})={\rm E}\left[{|X(s_{T}^{\ell})|^{2}}\right]$
for $\ell=1,\dots,L$. The capacity is

$\displaystyle C(P)$
$\displaystyle=\frac{D}{2L}\log\left(1+2P_{1}\right)+\frac{L-D}{2L}\log\left(1+2P_{D+1}\right)$

where we write $P_{D+1}=P_{D+1}\big{(}s_{T}^{D+1}\big{)}$. Optimizing the
powers we obtain

$\displaystyle\left\\{\begin{array}[]{l}P_{1}=P-\frac{L-D}{4L}\\\
P_{D+1}=2P+\frac{D}{2L}\end{array}\right\\}\quad\text{if }P\geq\frac{L-D}{4L}$
$\displaystyle\left\\{\begin{array}[]{l}P_{1}=0\\\
P_{D+1}=\frac{2LP}{L-D}\end{array}\right\\}\quad\text{else.}$

For large $P$, we thus have $C(P)\approx\frac{1}{2}\log(P)$ for all $0\leq
D\leq L$. For small $P$, we have

$\displaystyle C(P)$
$\displaystyle=\left\\{\begin{array}[]{ll}\frac{L-D}{2L}\log\left(1+\frac{4LP}{L-D}\right),&\text{if
$0\leq D<L$}\\\ \log(1+2P)/2,&\text{if $D=L$}\end{array}\right.$
$\displaystyle\approx\left\\{\begin{array}[]{ll}\left(2P-\frac{4L}{L-D}P^{2}\right)\log(e),&\text{if
$0\leq D<L$}\\\ \left(P-P^{2}\right)\log(e),&\text{if
$D=L$}.\end{array}\right.$

Feedback thus gives a 3 dB power gain at low SNR since $C(P)\approx 2P\log(e)$
for $0\leq D<L$ and $C(P)\approx P\log(e)$ for $D=L$. Furthermore, using [68,
Thm. 9] the slope of the capacity versus $E_{b}/N_{0}$ in bits/s/Hz/(3 dB) is

$\displaystyle\begin{array}[]{ll}1-D/L&\text{if $0\leq D<L$}\\\ 1&\text{if
$D=L$.}\end{array}$

In other words, the delay reduces the low-SNR rate by a factor of $1-D/L$ for
$0\leq D<L$.

###  IX-F Rayleigh Fading and One-Bit Feedback

Let $q_{u}(.)$ be the one-bit quantizer in Sec. II-I. We study Rayleigh fading
for two scenarios.

  * •

For the CSIT (365), we study the rates of delayed feedback where $S_{T\ell}=0$
for $\ell=1,\dots,L-1$ and $S_{TL}=q_{u}(G)$. The delay is thus $D=L-1$ in the
sense of Sec. IX-E.

  * •

For the CSIT (358), we study the rates where $S_{T1}=0$,
$S_{T2}=q_{u}(|Y_{1}|)$, and $S_{T\ell}=0$ for $\ell=3,\dots,L$. The delay is
thus $D=1$ in the sense of Sec. IX-E.

For both cases, we consider $S_{R}^{L}=H$, i.e., the receiver knows $H$
perfectly after the $L$ transmissions of each block.

####  IX-F1 Delayed Quantized CSIR Feedback

Consider $S_{T\ell}=0$ for $\ell=1,\dots,L-1$ and $S_{TL}=q_{u}(G)$. CSCG
inputs are optimal and (367) has the same form as (381). The Lagrangians are
given by (377) and we again obtain (379). For the case at hand, we have $L+1$
equations for $\lambda$, namely

$\displaystyle\lambda$
$\displaystyle=\int_{0}^{\infty}e^{-g}\frac{g}{1+gP_{\ell}}\,dg,\quad\ell=1,\dots,L-1$
(383) $\displaystyle\lambda$
$\displaystyle=\int_{0}^{\Delta}\frac{e^{-g}}{1-e^{-\Delta}}\frac{g}{1+gP_{L}(\Delta/2)}\,dg$
(384) $\displaystyle\lambda$
$\displaystyle=\int_{\Delta}^{\infty}\frac{e^{-g}}{e^{-\Delta}}\dfrac{g}{1+gP_{L}(3\Delta/2)}\,dg$
(385)

where we used (40)-(43) and abused notation by writing $P_{L}(s_{TL})$ for
$P_{L}(s_{T}^{L})$. We thus have $P_{1}=\dots=P_{L-1}$ and obtain three
equations. We now search for $\lambda$ such that

$\displaystyle(L-1)P_{1}+\sum_{s}P_{S_{TL}}(s)P_{L}(s)=LP$

and the capacity (374) is

$\displaystyle C(P)$
$\displaystyle=\frac{L-1}{L}e^{1/P_{1}}E_{1}\left(1/P_{1}\right)$
$\displaystyle\quad+\frac{1}{L}\sum_{s}\int_{\mathcal{I}(s)}e^{-g}\log\left(1+gP_{L}(s)\right)dg$
(386)

where the sums are over $s=\Delta/2,3\Delta/2$ and

$\displaystyle\mathcal{I}(\Delta/2)=[0,\Delta),\quad\mathcal{I}(3\Delta/2)=[\Delta,\infty).$

We remark that if $P_{1}=0$ then we set
$e^{1/P_{1}}E_{1}\left(1/P_{1}\right)=0$ since
$\lim_{x\rightarrow\infty}e^{x}E_{1}(x)=0$.

Fig. 15 shows these capacities for $L=1,2,3$ and $\Delta=1$. At low SNR (e.g.
for $L=3$ below -2.97 dB) we have $P_{1}=0$ and $P_{L}(\Delta/2)=0$, i.e., the
transmitter is silent unless $S_{TL}=3\Delta/2$ and it uses power at time
$\ell=L$ only. Observe that as in Sec. IX-E a delay of $L$ steps reduces the
low-SNR slope, and therefore the low-SNR rates, by a factor of $L$. Delay can
thus be costly at low SNR.

Figure 15: Capacities for Rayleigh block fading with $L=1,2,3$ and a CSIT
delay of $D=L-1$. The CSIT at symbol $L$ is $S_{TL}=q_{u}(G)$.

####  IX-F2 Quantized Channel Output Feedback

Consider $S_{T1}=0$, $S_{T2}=q_{u}(|Y_{1}|)$, and $S_{T\ell}=0$ for
$\ell=3,\dots,L$. As discussed in Remark 73, the capacity is given by the
directed information expression (369). However, optimizing the input
statistics seems difficult, i.e., CSCG inputs are not necessarily optimal.
Instead, we compute achievable rates for a strategy where one symbol partially
acts as a pilot.

Suppose the transmitter sends $X_{1}=\sqrt{P_{1}}e^{j\Phi}$ as the first
symbol of each block, where $\Phi$ is uniformly distributed in $[0,2\pi)$. The
idea is that $|X_{1}|=\sqrt{P_{1}}$ is known at the receiver and thus $X_{1}$
acts as a pilot to test the channel amplitude. Next, we choose a variation of
flash signaling. Define the event
$\mathcal{E}=\\{|Y_{1}|\geq\Delta\\}=\\{S_{T2}=3\Delta/2\\}$. If this event
does not occur, then the transmitter sends $X_{\ell}=0$ for $\ell=2,\dots,L$.
Otherwise, the transmitter sends independent CSCG $X_{\ell}$ with variance
$P_{2}/{\rm Pr}\left[{\mathcal{E}}\right]$ for $\ell=2,\dots,L$. Define
$P_{\ell}(s_{T}^{\ell})={\rm E}\left[{|X(s_{T}^{\ell})|^{2}}\right]$. We have
$P_{\ell}=P_{2}$ for $\ell\geq 2$ and the power constraint is
$P_{1}+(L-1)P_{2}\leq LP$.

We use (367) to write

$\displaystyle
C(P)\geq\frac{1}{L}I(X_{1};Y_{1}|H)+\frac{L-1}{L}I(X^{2};Y_{2}|H,Y_{1}).$
(387)

The first mutual information in (387) is

$\displaystyle I(X_{1};Y_{1}|H)$ $\displaystyle=h(Y_{1}|H)-\log(\pi e)$

and we compute (see [50, App. A])

$\displaystyle
p(y_{1}|h)=\frac{1}{\pi}e^{-(|y_{1}|^{2}+P_{1}|h|^{2})}I_{0}\left(2\,|y_{1}|\,|h|\,\sqrt{P_{1}}\right)$

where $I_{0}(.)$ is the modified Bessel function of the first kind of order
zero. The Jacobian of the mapping from Cartesian coordinates
$[\Re(y_{1}),\Im(y_{1})]$ to polar coordinates $[|y_{1}|,\arg y_{1}]$ is
$|y_{1}|$, so we have

$\displaystyle
h(Y_{1}|H=h)=\int_{0}^{\infty}-\,p(y_{1}|h)\log(p(y_{1}|h))\,2\pi|y_{1}|\,d|y_{1}|.$

We further compute

$\displaystyle I\left(X^{2};Y_{2}|H,Y_{1}\right)$
$\displaystyle=\int_{0}^{\infty}e^{-g}{\rm
Pr}\left[{\mathcal{E}|G=g}\right]\log\left(1+\frac{gP_{2}}{{\rm
Pr}\left[{\mathcal{E}}\right]}\right)dg.$ (388)

The conditional probability of a high-energy $Y_{1}$ is

$\displaystyle{\rm
Pr}\left[{\mathcal{E}|G=g}\right]=Q_{1}\left(\sqrt{2gP_{1}},\sqrt{2}\Delta\right)$

where $Q_{1}(.)$ is the Marcum Q-function of order 1, see (391) in Appendix
A-A. For Rayleigh fading we compute

$\displaystyle{\rm Pr}\left[{\mathcal{E}}\right]={\rm
Pr}\left[{\left|H\sqrt{P_{1}}e^{j\Phi}+Z_{1}\right|^{2}\geq\Delta^{2}}\right]=e^{-\Delta^{2}/(P_{1}+1)}.$

The resulting rates are shown in Fig. 16 for the block lengths $L=10,20,100$.
Observe that each curve turns back on itself, which reflects the non-concavity
of $C(P)$ in $P$, see [90]. All rates below the curves are achievable by
“time-wasting”, i.e., by transmitting for some fraction of the time only. This
suggests that the usual flash signaling [68] will improve the rates further,
since one sends information in the choice of whether to transmit or not.

Figure 16: Rates for Rayleigh block fading with block lengths $L=10,20,100$.
The CSIT at symbol 2 is $S_{T2}=q_{u}(|Y_{1}|)$.

##  X Conclusions

This paper reviewed and developed information rates for channels with CSIR,
CSIT, and for channels with block fading and in-block feedback. Various types
of CSIT and CSIR were studied. GMI calculations were developed for reverse
channel models, as well as for adaptive codewords by using CSCG forward models
with inputs that are linear functions of the adaptive codewords’ symbols. For
scalar channels, the input distribution that maximizes the GMI was shown to
generate a conventional codebook, and have the codeword symbols multiplied by
a complex number that depends on the CSIT. The GMI was increased by
partitioning the channel output alphabet and by modifying the auxiliary model
parameters for each subset of the partition. The partitioning helps to
determine the capacity scaling at high SNR. A class of power control policies
was developed for full CSIT and partial CSIR, including a TMMSE policy for
full CSIT. The theory was applied to channels with on-off fading and Rayleigh
fading, and to block fading channels with in-block feedback.

There are many possible extensions of this work. For example, a practical
problem is to develop better auxiliary models for adaptive codewords. It
should also be interesting to apply the GMI to multi-user channels such as
multi-access and broadcast channels, to channels with beamforming [91] or
intelligent reflecting surfaces, and more generally to information networks
with in-block memory and in-block feedback, see [22].

## Acknowledgements

The author wishes to thank W. Zhang for sending his recent paper [48].

Appendix A

Special Functions

This appendix reviews three classes of functions that we use to analyze
information rates for fading channels: the non-central chi-squared
distribution, the exponential integral, and gamma functions.

###  A-A Non-Central Chi-Squared Distribution

The non-central chi-squared distribution with two degrees of freedom is the
probability distribution of $Y=|x+Z|^{2}$ where $x\in\mathbb{C}$ and
$Z\sim\mathcal{CN}(0,2)$. The density is

$\displaystyle p(y)=\frac{1}{2}e^{-(y+|x|^{2})/2}I_{0}(|x|\sqrt{y})\cdot
1(y\geq 0)$ (389)

where $I_{0}(.)$ is the modified Bessel function of the first kind of order
zero. The cumulative distribution function is

$\displaystyle{\rm Pr}\left[{Y\leq t}\right]=1-Q_{1}\left(|x|,\sqrt{t}\right)$
(390)

where $Q_{1}(.)$ is the Marcum Q-function of order 1. Observe that if we
change $Z$ to $Z\sim\mathcal{CN}(0,\sigma^{2})$ then for $Y=|x+Z|^{2}$ we
instead have

$\displaystyle{\rm Pr}\left[{Y\leq
t}\right]=1-Q_{1}\left(\sqrt{2|x|^{2}/\sigma^{2}},\sqrt{2t/\sigma^{2}}\right).$
(391)

###  A-B Exponential Integral

The exponential integral is defined for $x>0$ as

$\displaystyle E_{1}(x)=\int_{x}^{\infty}\frac{e^{-t}}{t}dt.$ (392)

The derivative of $E_{1}(x)$ is

$\displaystyle\frac{dE_{1}(x)}{dx}=\frac{-e^{-x}}{x}.$ (393)

For small $x$ one may apply [92, Eq. (3)]

$\displaystyle E_{1}(x)\approx-\gamma-\ln(x)+x$ (394)

where $\gamma\approx 0.57721$ is Euler’s constant. For large $x$ we have

$\displaystyle
E_{1}(x)\approx\frac{e^{-x}}{x}\left(1-\frac{1}{x}+\frac{2}{x^{2}}-\frac{6}{x^{3}}\right).$
(395)

We have the bounds [93]

$\displaystyle\frac{1}{2}\log\left(1+\frac{2}{x}\right)$
$\displaystyle<e^{x}E_{1}(x)<\log\left(1+\frac{1}{x}\right)$ (396)
$\displaystyle\frac{1}{x+1}$ $\displaystyle<e^{x}E_{1}(x)<\frac{x+1}{x(x+2)}.$
(397)

Using integration by parts, for $x>0$ we have

$\displaystyle\int_{x}^{\infty}e^{-t}\log t\;dt=E_{1}(x)+e^{-x}\log(x)$ (398)
$\displaystyle\int_{x}^{\infty}e^{-t}\frac{1}{t^{2}}\;dt=\frac{e^{-x}}{x}-E_{1}(x).$
(399)

Using the translation $\tilde{t}=t+y$ we also have

$\displaystyle\int_{x}^{\infty}e^{-t}\frac{t}{t+y}dt=e^{-x}-y\,e^{y}\,E_{1}(x+y)$
(400)
$\displaystyle\int_{x}^{\infty}e^{-t}\frac{t}{(t+y)^{2}}dt=-e^{-x}\frac{y}{x+y}+(y+1)\,e^{y}\,E_{1}(x+y)$
(401)
$\displaystyle\int_{x}^{\infty}e^{-t}\frac{t^{2}}{(t+y)^{2}}dt=e^{-x}\left(1+\frac{y^{2}}{x+y}\right)$
$\displaystyle\qquad\qquad\qquad\qquad\qquad-y(y+2)\,e^{y}\,E_{1}(x+y).$ (402)

###  A-C Gamma Functions

The upper and lower incomplete gamma functions are the respective

$\displaystyle\Gamma(s,t)=\int_{t}^{\infty}e^{-g}\,g^{s-1}\,dg$ (403)
$\displaystyle\gamma(s,t)=\int_{0}^{t}e^{-g}\,g^{s-1}\,dg.$ (404)

For instance, we have $\Gamma(1,t)=e^{-t}$ and $\gamma(1,t)=1-e^{-t}$. We
further have $\Gamma(0,t)=E_{1}(t)$ where $E_{1}(x)$ is the exponential
integral defined in Appendix A-B.

The Gamma function is $\Gamma(s)=\Gamma(s,0)=\gamma(s,\infty)$ and for
positive integers $n$ we have

$\displaystyle\Gamma(n)$ $\displaystyle=(n-1)!$
$\displaystyle\Gamma\left(n-\frac{1}{2}\right)$
$\displaystyle=\frac{(2n-2)!}{4^{n-1}\,(n-1)!}\sqrt{\pi}.$

For example, the following cases are used in Sec. VIII-D:

$\displaystyle\Gamma(1)=\Gamma(2)=1,$
$\displaystyle\Gamma\left(\frac{1}{2}\right)=\sqrt{\pi},\quad\Gamma\left(\frac{3}{2}\right)=\frac{\sqrt{\pi}}{2},\quad\Gamma\left(\frac{5}{2}\right)=\frac{3}{4}\sqrt{\pi}.$

The value $\Gamma(0)$ is undefined but we have $\lim_{x\rightarrow
0^{+}}\Gamma(x)=\infty$.

Appendix B

Forward Model GMIs with $K=2$

This appendix studies $K=2$ GMIs to develop capacity scaling results at high
and low SNR.

###  B-A On-Off Fading

Consider on-off fading as in Sec. III-C and the $K=2$ partition in Remark 11
with $h_{2}=\sqrt{2}$. We compute

$\displaystyle{\rm Pr}\left[{\mathcal{E}_{2}}\right]$
$\displaystyle=\sum_{h=0,\sqrt{2}}{\rm Pr}\left[{H=h}\right]{\rm
Pr}\left[{\mathcal{E}_{2}|H=h}\right]$
$\displaystyle=\frac{1}{2}e^{-t_{R}}+\frac{1}{2}e^{-t_{R}/(2P+1)}.$ (405)

For example, if we choose $t_{R}=P^{\lambda_{R}}+b$ where $0<\lambda_{R}<1$
and $b$ is a real constant then ${\rm
Pr}\left[{\mathcal{E}_{2}}\right]\rightarrow 1/2$ as $P\rightarrow\infty$, as
desired. We further have

$\displaystyle{\rm Pr}\left[{\left.H=0\,\right|\mathcal{E}_{2}}\right]$
$\displaystyle=\frac{e^{-t_{R}}}{2{\rm Pr}\left[{\mathcal{E}_{2}}\right]}$
(406) $\displaystyle{\rm
Pr}\left[{\left.H=\sqrt{2}\,\right|\mathcal{E}_{2}}\right]$
$\displaystyle=\frac{e^{-t_{R}/(2P+1)}}{2{\rm
Pr}\left[{\mathcal{E}_{2}}\right]}.$ (407)

The choice $t_{R}=P^{\lambda_{R}}+b$ gives ${\rm
Pr}\left[{\left.H=\sqrt{2}\,\right|\mathcal{E}_{2}}\right]\rightarrow 1$ as
$P\rightarrow\infty$. In other words, the receiver can reliably determine $H$
by choosing $t_{R}$ to grow with $P$, but not too fast.

We next compute

$\displaystyle{\rm
E}\left[{|Y|^{2}|\mathcal{E}_{2}}\right]=\sum_{h=0,\sqrt{2}}{\rm
Pr}\left[{H=h|\mathcal{E}_{2}}\right]{\rm
E}\left[{|Y|^{2}|\mathcal{E}_{2},H=h}\right]$
$\displaystyle\quad=\frac{e^{-t_{R}}(t_{R}+1)+e^{-t_{R}/(2P+1)}(t_{R}+1+2P)}{2{\rm
Pr}\left[{\mathcal{E}_{2}}\right]}$
$\displaystyle\quad\geq\frac{e^{-t_{R}}(t_{R}+1)+(1+2P)-t_{R}^{2}/(1+2P)}{2{\rm
Pr}\left[{\mathcal{E}_{2}}\right]}$ (408)

where the last step uses $e^{-x}\geq 1-x$. The choice $t_{R}=P^{\lambda_{R}}$
makes ${\rm E}\left[{|Y|^{2}|\mathcal{E}_{2}}\right]/(1+2P)\rightarrow 1$ as
$P\rightarrow\infty$.

Finally, we compute

$\displaystyle{\rm E}\left[{|Y-\sqrt{2}X|^{2}|\mathcal{E}_{2}}\right]$
$\displaystyle=\sum_{h=0,\sqrt{2}}{\rm
Pr}\left[{H=h|\mathcal{E}_{2}}\right]{\rm
E}\left[{\left.\left|Y-\sqrt{2}X\right|^{2}\right|\mathcal{E}_{2},H=h}\right]$
$\displaystyle=\frac{1}{2{\rm
Pr}\left[{\mathcal{E}_{2}}\right]}\bigg{\\{}e^{-t_{R}}(t_{R}+1+2P)$
$\displaystyle\qquad\left.+\,e^{-t_{R}/(2P+1)}{\rm
E}\left[{|Z|^{2}\left|\left|\sqrt{2}X+Z\right|^{2}\geq
t_{R}\right.}\right]\right\\}$
$\displaystyle\leq\frac{e^{-t_{R}}(t_{R}+1+2P)+1}{2{\rm
Pr}\left[{\mathcal{E}_{2}}\right]}$ (409)

where we have used the law of total expectation to write ${\rm
Pr}\left[{\mathcal{E}}\right]{\rm E}\left[{|Z|^{2}|\mathcal{E}}\right]\leq{\rm
E}\left[{|Z|^{2}}\right]$ for any event $\mathcal{E}$. The choice
$t_{R}=P^{\lambda_{R}}$ makes ${\rm
E}\left[{|Y-\sqrt{2}X|^{2}|\mathcal{E}_{2}}\right]\rightarrow a$ where $a\leq
1$ as $P\rightarrow\infty$. Thus, all terms in (85) behave as desired.

To calculate $I_{1}(X;Y)$ for the plots, we use

$\displaystyle{\rm Pr}\left[{\left|\sqrt{2}X+Z\right|^{2}\geq
t_{R}}\right]{\rm E}\left[{|Z|^{2}\left|\left|\sqrt{2}X+Z\right|^{2}\geq
t_{R}\right.}\right]$
$\displaystyle\quad=\int_{\mathbb{C}}\frac{e^{-|z|^{2}}}{\pi}\,|z|^{2}\,{\rm
Pr}\left[{\left|\sqrt{2}X+z\right|^{2}\geq t_{R}}\right]\,dz$
$\displaystyle\quad=\int_{0}^{\infty}e^{-\tilde{g}}\,\tilde{g}\,Q_{1}\left(\sqrt{\frac{\tilde{g}}{P}},\sqrt{\frac{t_{R}}{P}}\right)\,d\tilde{g}$
(410)

where the last step follows by applying (391) and substituting
$\tilde{g}=|z|^{2}$.

###  B-B On-Off Fading, Partial CSIR, and Full CSIT

The analysis for Sec. VII-C is similar to that of Appendix B-A. Consider the
GMI (266) and observe that we can simply replace $2P$ with $4P$ in
(405)-(408). We also have

$\displaystyle{\rm
E}\left[{\left.|Y-\sqrt{4P}\,U|^{2}\right|\mathcal{E}_{2}}\right]$
$\displaystyle=\sum_{h=0,\sqrt{2}}{\rm
Pr}\left[{H=h|\mathcal{E}_{2}}\right]{\rm
E}\left[{\left.|Y-\sqrt{4P}\,U|^{2}\right.|\mathcal{E}_{2},H=h}\right]$
$\displaystyle=\frac{1}{2{\rm
Pr}\left[{\mathcal{E}_{2}}\right]}\bigg{\\{}e^{-t_{R}}(t_{R}+1+4P)$
$\displaystyle\qquad\left.+\,e^{-t_{R}/(4P+1)}{\rm
E}\left[{\left.|Z|^{2}\right|\left|\sqrt{2}X(\sqrt{2})+Z\right|^{2}\geq
t_{R}}\right]\right\\}$
$\displaystyle\leq\frac{e^{-t_{R}}(t_{R}+1+4P)+1}{2{\rm
Pr}\left[{\mathcal{E}_{2}}\right]}$ (411)

where we have again used ${\rm Pr}\left[{\mathcal{E}}\right]{\rm
E}\left[{|Z|^{2}|\mathcal{E}}\right]\leq{\rm E}\left[{|Z|^{2}}\right]$. The
choice $t_{R}=P^{\lambda_{R}}+b$ as as in Appendix B-A makes all terms in
(266) behave as desired.

To calculate $I_{1}(A;Y)$ for the plots, we use (see (410))

$\displaystyle{\rm Pr}\left[{\left|\sqrt{2}X(\sqrt{2})+Z\right|^{2}\geq
t_{R}}\right]$ $\displaystyle\quad\cdot\,{\rm
E}\left[{\left.|Z|^{2}\right|\left|\sqrt{2}X(\sqrt{2})+Z\right|^{2}\geq
t_{R}}\right]$
$\displaystyle=\int_{0}^{\infty}e^{-\tilde{g}}\,\tilde{g}\,Q_{1}\left(\sqrt{\frac{\tilde{g}}{2P}},\sqrt{\frac{t_{R}}{2P}}\right)\,d\tilde{g}.$
(412)

###  B-C On-Off Fading, Partial CSIR, and CSIT@R

The analysis for Sec. VII-D is similar to that of Appendices B-A and B-B. We
compute

$\displaystyle{\rm Pr}\left[{\mathcal{E}_{2}|S_{R}=0}\right]$
$\displaystyle=\bar{\epsilon}\,e^{-t_{R}}+\epsilon\,e^{-t_{R}/[2P(0)+1]}$
(413) $\displaystyle{\rm Pr}\left[{\mathcal{E}_{2}|S_{R}=\sqrt{2}}\right]$
$\displaystyle=\epsilon\,e^{-t_{R}}+\bar{\epsilon}\,e^{-t_{R}/[2P(\sqrt{2})+1]}\;.$
(414)

Suppose $P(0)$ and $P(\sqrt{2})$ both scale in proportion to $P$. If we choose
$t_{R}=P^{\lambda_{R}}+b$ as in Appendix B-A then ${\rm
Pr}\left[{\mathcal{E}_{2}|S_{R}=0}\right]\rightarrow\epsilon$ and ${\rm
Pr}\left[{\mathcal{E}_{2}|S_{R}=\sqrt{2}}\right]\rightarrow\bar{\epsilon}$ as
$P\rightarrow\infty$. We also have

$\displaystyle{\rm
Pr}\left[{\left.H=0\,\right|\mathcal{E}_{2},S_{R}=0}\right]$
$\displaystyle=\frac{\bar{\epsilon}\,e^{-t_{R}}}{{\rm
Pr}\left[{\mathcal{E}_{2}|S_{R}=0}\right]}$ (415) $\displaystyle{\rm
Pr}\left[{\left.H=\sqrt{2}\,\right|\mathcal{E}_{2},S_{R}=0}\right]$
$\displaystyle=\frac{\epsilon\,e^{-t_{R}/[2P(0)+1]}}{{\rm
Pr}\left[{\mathcal{E}_{2}|S_{R}=0}\right]}$ (416)

and similarly for the probabilities ${\rm
Pr}\left[{H=0\,|\mathcal{E}_{2},S_{R}=\sqrt{2}}\right]$ and ${\rm
Pr}\left[{H=\sqrt{2}\,|\mathcal{E}_{2},S_{R}=\sqrt{2}}\right]$. Choosing
$t_{R}=P^{\lambda_{R}}+b$ gives the desired values ${\rm
Pr}\left[{H=\sqrt{2}\,|\mathcal{E}_{2},S_{R}=0}\right]\rightarrow 1$ and ${\rm
Pr}\left[{H=\sqrt{2}\,|\mathcal{E}_{2},S_{R}=\sqrt{2}}\right]\rightarrow 1$ as
$P\rightarrow\infty$. Again, the receiver can reliably determine $H$ by
choosing $t_{R}$ to grow with $P$, but not too fast.

We next have

$\displaystyle{\rm E}\left[{|Y|^{2}|\mathcal{E}_{2},S_{R}=0}\right]$
$\displaystyle=\frac{\bar{\epsilon}\,e^{-t_{R}}(t_{R}+1)+\epsilon\,e^{-t_{R}/[2P(0)+1]}(t_{R}+1+2P(0))}{{\rm
Pr}\left[{\mathcal{E}_{2}|S_{R}=0}\right]}.$ (417)

The expression for ${\rm
E}\left[{|Y|^{2}|\mathcal{E}_{2},S_{R}=\sqrt{2}}\right]$ is similar but
$\epsilon$ and $\bar{\epsilon}$ are swapped and $P(0)$ is replaced with
$P(\sqrt{2})$. We also have

$\displaystyle{\rm
E}\left[{\left.|Y-\sqrt{2}\,X(0)|^{2}\right|\mathcal{E}_{2},S_{R}=0}\right]$
$\displaystyle=\frac{1}{{\rm
Pr}\left[{\mathcal{E}_{2}|S_{R}=0}\right]}\bigg{\\{}\bar{\epsilon}\,e^{-t_{R}}(t_{R}+1+2P(0))+$
$\displaystyle\qquad\left.\epsilon\,e^{-t_{R}/[2P(0)+1]}{\rm
E}\left[{|Z|^{2}\left|\left|\sqrt{2}X(0)+Z\right|^{2}\geq
t_{R}\right.}\right]\right\\}$
$\displaystyle\leq\frac{\bar{\epsilon}\,e^{-t_{R}}(t_{R}+1+2P(0))+\epsilon}{{\rm
Pr}\left[{\mathcal{E}_{2}|S_{R}=0}\right]}$ (418)

where again ${\rm Pr}\left[{\mathcal{E}}\right]{\rm
E}\left[{|Z|^{2}|\mathcal{E}}\right]\leq{\rm E}\left[{|Z|^{2}}\right]$. The
expression for ${\rm
E}\left[{\left.|Y-\sqrt{2}\,X(0)|^{2}\right|\mathcal{E}_{2},S_{R}=\sqrt{2}}\right]$
is similar: swap $\epsilon$ and $\bar{\epsilon}$ and replace $P(0)$ with
$P(\sqrt{2})$. The choice $t=P^{\lambda_{R}}+b$ makes all terms in (277)
behave as desired and we obtain (278).

To calculate $I_{1}(X;Y|S_{R})$ for the plots, we use (see (410))

$\displaystyle{\rm Pr}\left[{\left|\sqrt{2}X(s_{T})+Z\right|^{2}\geq
t_{R}}\right]$ $\displaystyle\quad\cdot\,{\rm
E}\left[{|Z|^{2}\left|\left|\sqrt{2}X(s_{T})+Z\right|^{2}\geq
t_{R}\right.}\right]$
$\displaystyle=\int_{0}^{\infty}e^{-\tilde{g}}\,\tilde{g}\,Q_{1}\left(\sqrt{\frac{\tilde{g}}{P(s_{T})}},\sqrt{\frac{t_{R}}{P(s_{T})}}\right)\,d\tilde{g}$
(419)

for $s_{T}=0,\sqrt{2}$.

###  B-D Rayleigh Fading, No CSIR, full CSIT, and TCI

The analysis for Sec. VII-D is similar to that of Appendices B-A to B-C, but
we now have a continuous $H$. Recall that $\mathcal{E}_{2}=\\{|Y|^{2}\geq
t_{R}\\}$ and $Y=\sqrt{P(h)}\,U+Z$ where $P(h)=0$ for $g<t$ and $P(h)=\hat{P}$
otherwise.

We compute

$\displaystyle{\rm Pr}\left[{\mathcal{E}_{2}}\right]$ $\displaystyle={\rm
Pr}\left[{G<t}\right]{\rm Pr}\left[{\mathcal{E}_{2}|G<t}\right]$
$\displaystyle\quad+{\rm Pr}\left[{G\geq t}\right]{\rm
Pr}\left[{\mathcal{E}_{2}|G\geq t}\right]$
$\displaystyle=(1-e^{-t})e^{-t_{R}}+e^{-t}e^{-t_{R}/(1+\hat{P})}$ (420)

where we used ${\rm Pr}\left[{\mathcal{E}_{2}|G<t}\right]={\rm
Pr}\left[{|Z|^{2}\geq t_{R}}\right]$ and similarly for ${\rm
Pr}\left[{\mathcal{E}_{2}|G\geq t}\right]$. For example, for the $t$ and
$t_{R}$ in (310) we find that ${\rm
Pr}\left[{\mathcal{E}_{2}}\right]\rightarrow 1$ as $P$ grows. Similarly, for
the $t$ and $t_{R}$ in (316) we find that ${\rm
Pr}\left[{\mathcal{E}_{2}}\right]\approx e^{-t-1}$ as $P$ decreases.

Next, consider

$\displaystyle{\rm E}\left[{|Y|^{2}|\mathcal{E}_{2}}\right]$
$\displaystyle={\rm Pr}\left[{G<t|\mathcal{E}_{2}}\right]{\rm
E}\left[{\left.|Z|^{2}\right|\mathcal{E}_{2},G<t}\right]$
$\displaystyle\quad+{\rm Pr}\left[{G\geq t|\mathcal{E}_{2}}\right]{\rm
E}\left[{\left.\left|\sqrt{\hat{P}}\,U+Z\right|^{2}\right|\mathcal{E}_{2},G\geq
t}\right]$
$\displaystyle=\frac{(1-e^{-t})e^{-t_{R}}(t_{R}+1)+e^{-t}e^{-t_{R}/(1+\hat{P})}(t_{R}+1+\hat{P})}{{\rm
Pr}\left[{\mathcal{E}_{2}}\right]}.$ (421)

For the $t$ and $t_{R}$ in (310) we have ${\rm
E}\left[{|Y|^{2}|\mathcal{E}_{2}}\right]/(1+\hat{P})\rightarrow 1$ as $P$
grows. Similarly, for the $t$ and $t_{R}$ in (316) we find that ${\rm
E}\left[{|Y|^{2}|\mathcal{E}_{2}}\right]/(1+2\hat{P})\rightarrow 1$ as $P$
decreases.

We also have

$\displaystyle{\rm
E}\left[{\left.\left|Y-\sqrt{\hat{P}}\,U\right|^{2}\right|\mathcal{E}_{2}}\right]$
$\displaystyle={\rm Pr}\left[{G<t|\mathcal{E}_{2}}\right]{\rm
E}\left[{\left.\left|Z-\sqrt{\hat{P}}\,U\right|^{2}\right||Z|^{2}\geq
t_{R}}\right]$ $\displaystyle\quad+{\rm Pr}\left[{G\geq
t|\mathcal{E}_{2}}\right]{\rm
E}\left[{|Z|^{2}\left|\left|\sqrt{\hat{P}}\,U+Z\right|^{2}\geq
t_{R}\right.}\right]$ $\displaystyle=\frac{1}{{\rm
Pr}\left[{\mathcal{E}_{2}}\right]}\bigg{\\{}(1-e^{-t})e^{-t_{R}}(t_{R}+1+\hat{P})+$
$\displaystyle\qquad\left.e^{-t}e^{-t_{R}/(1+\hat{P})}{\rm
E}\left[{|Z|^{2}\left|\left|\sqrt{\hat{P}}\,U+Z\right|^{2}\geq
t_{R}\right.}\right]\right\\}$
$\displaystyle\leq\frac{(1-e^{-t})e^{-t_{R}}(t_{R}+1+\hat{P})+e^{-t}}{{\rm
Pr}\left[{\mathcal{E}_{2}}\right]}$ (422)

where we again used ${\rm Pr}\left[{\mathcal{E}}\right]{\rm
E}\left[{|Z|^{2}|\mathcal{E}}\right]\leq{\rm E}\left[{|Z|^{2}}\right]$. For
the $t$ and $t_{R}$ in (310) we find that (422) approaches (at most) 1 as $P$
grows. Similarly, for the $t$ and $t_{R}$ in (316) we find that (422)
approaches (at most) $1$ as $P$ decreases.

To calculate $I_{1}(A;Y)$ for the plots, we use (see (410))

$\displaystyle{\rm Pr}\left[{\left|\sqrt{\hat{P}}\,U+Z\right|^{2}\geq
t_{R}}\right]{\rm
E}\left[{|Z|^{2}\left|\left|\sqrt{\hat{P}}\,U+Z\right|^{2}\geq
t_{R}\right.}\right]$
$\displaystyle\quad=\int_{\mathbb{C}}\frac{e^{-|z|^{2}}}{\pi}\,|z|^{2}\,{\rm
Pr}\left[{\left|\sqrt{\hat{P}}\,U+z\right|^{2}\geq t_{R}}\right]\,dz$
$\displaystyle\quad=\int_{0}^{\infty}e^{-\tilde{g}}\,\tilde{g}\,Q_{1}\left(\sqrt{\frac{2\tilde{g}}{\hat{P}}},\sqrt{\frac{2t_{R}}{\hat{P}}}\right)\,d\tilde{g}.$
(423)

Appendix C

Conditional Second Order Statistics

This appendix shows how to compute conditional second orders statistics for
the reverse model GMIs and the forward model GMIs with $K=\infty$. To begin,
suppose that $U,Y$ are jointly CSCG given $H=h$. Using (24)-(25) we have

$\displaystyle{\rm E}\left[{U|Y=y,H=h}\right]=\frac{{\rm
E}\left[{UY^{\text{*}}\big{|}H=h}\right]}{{\rm
E}\left[{|Y|^{2}\big{|}H=h}\right]}\cdot y$ (424) $\displaystyle{\rm
Var}\left[{U|Y=y,H=h}\right]$ $\displaystyle\quad={\rm
E}\left[{|U|^{2}\big{|}H=h}\right]-\frac{\left|{\rm
E}\left[{UY^{\text{*}}\big{|}H=h}\right]\right|^{2}}{{\rm
E}\left[{|Y|^{2}\big{|}H=h}\right]}.$ (425)

Now consider the channel $Y=HX+Z$ where $H,Z$ and $X,Z$ are independent, and
where $X=\sqrt{P(S_{T})}e^{j\phi(S_{T})}U$ with $U\sim\mathcal{CN}(0,1)$. We
may write

$\displaystyle{\rm E}\left[{U\big{|}Y=y,S_{R}=s_{R}}\right]$
$\displaystyle=\int_{\mathbb{C}}p(h,s_{T}|y,s_{R})\,\frac{h^{\text{*}}\sqrt{P(s_{T})}e^{j\phi(s_{T})}y}{1+|h|^{2}P(s_{T})}\,ds_{T}\,dh$
(426)

and

$\displaystyle{\rm
E}\left[{|U|^{2}\big{|}Y=y,S_{R}=s_{R}}\right]=\int_{\mathbb{C}}p(h,s_{T}|y,s_{R})$
$\displaystyle\quad\left(\frac{1}{1+|h|^{2}P(s_{T})}+\frac{|h|^{2}P(s_{T})|y|^{2}}{\left(1+|h|^{2}P(s_{T})\right)^{2}}\right)\,ds_{T}\,dh.$
(427)

###  C-A No CSIR, No CSIT

Consider $S_{R}=S_{T}=0$. The expectations in (426)-(427) are computed via

$\displaystyle p(h|y)=\frac{p(h)\,p(y|h)}{p(y)}.$ (428)

The expression (426) with $\phi(0)=0$ gives

$\displaystyle{\rm
E}\left[{U|Y=y}\right]=\int_{\mathbb{C}}p(h|y)\,\frac{h^{\text{*}}\sqrt{P}\,y}{1+|h|^{2}P}\,dh.$
(429)

Similarly, the expression (427) gives

$\displaystyle{\rm E}\left[{|U|^{2}\big{|}Y=y}\right]$
$\displaystyle=\int_{\mathbb{C}}p(h|y)\,{\rm
E}\left[{\left.|X|^{2}\right|Y=y,H=h}\right]\,dh$
$\displaystyle=\int_{\mathbb{C}}p(h|y)\left(\frac{1}{1+|h|^{2}P}+\frac{|h|^{2}P|y|^{2}}{\left(1+|h|^{2}P\right)^{2}}\right)dh$
(430)

We may now compute ${\rm Var}\left[{U|Y=y}\right]$ using (429) and (430). For
the expressions (71) and (72), one may simply use

$\displaystyle{\rm E}\left[{X|Y=y}\right]$ $\displaystyle=\sqrt{P}\,{\rm
E}\left[{U|Y=y}\right]$ $\displaystyle{\rm E}\left[{|X|^{2}\big{|}Y=y}\right]$
$\displaystyle=P\,{\rm E}\left[{|U|^{2}\big{|}Y=y}\right].$

For example, for on-off fading as in Sec. III-C we compute

$\displaystyle{\rm
E}\left[{X|Y=y}\right]=P_{H|Y}\left(\left.\sqrt{2}\,\right|y\right)\frac{\sqrt{2}P}{1+2P}\cdot
y$ (431) $\displaystyle{\rm
E}\left[{\left.|X|^{2}\right|Y=y}\right]=P_{H|Y}(0|y)\,P$
$\displaystyle\qquad+P_{H|Y}\left(\left.\sqrt{2}\,\right|y\right)\left(\frac{P}{1+2P}+\frac{2P^{2}|y|^{2}}{(1+2P)^{2}}\right)$
(432)

and therefore

$\displaystyle{\rm Var}\left[{X|Y=y}\right]=P_{H|Y}(0|y)\,P$
$\displaystyle+P_{H|Y}\left(\left.\sqrt{2}\,\right|y\right)\left(\frac{P}{1+2P}+\frac{2P^{2}|y|^{2}}{(1+2P)^{2}}\,P_{H|Y}(0|y)\right)$
(433)

where $P_{H|Y}\left(\left.\sqrt{2}\,\right|y\right)=1-P_{H|Y}(0|y)$ and

$\displaystyle
P_{H|Y}(0|y)=\frac{e^{-|y|^{2}}}{e^{-|y|^{2}}+\frac{1}{1+2P}e^{-|y|^{2}/(1+2P)}}.$
(434)

For Rayleigh fading as in Sec. VIII-A the density (428) is

$\displaystyle
p(h|y)=\frac{e^{-g}\,e^{-|y|^{2}/(1+gP)}}{\pi^{2}(1+gP)}\cdot\frac{1}{p(y)}$

where $g=|h|^{2}$. Moreover, $p(y)$ in (280) depends on $g$ only. We thus have
${\rm E}\left[{U|Y=y}\right]=0$ and the integrand in (430) depends on $g$ and
$|y|^{2}$ only.

###  C-B Full CSIR, Partial CSIT

Consider $S_{R}=H$ and partial $S_{T}$. The expectations in (426)-(427) are
computed via (195) that we repeat here:

$\displaystyle
p(h,s_{T}|y,s_{R})=\delta(h-s_{R})\,\frac{p(s_{T}|h)\,p(y|h,s_{T})}{p(y|h)}.$

For on-off fading as in Sec. VII-B, the expression (426) with $\phi(0)=0$
gives ${\rm E}\left[{U|Y=y,H=0}\right]=0$ and

$\displaystyle{\rm E}\left[{U\big{|}Y=y,H=\sqrt{2}}\right]$
$\displaystyle=\sum_{s_{T}=0,2}P_{S_{T}|Y,H}(s_{T}|y,\sqrt{2}\,)\,\frac{\sqrt{2P(s_{T})}\,y}{1+2P(s_{T})}$

and, similarly, (427) gives ${\rm E}\left[{|U|^{2}|Y=y,H=0}\right]=1$ and

$\displaystyle{\rm
E}\left[{|U|^{2}\big{|}Y=y,H=\sqrt{2}}\right]=\sum_{s_{T}=0,2}$ $\displaystyle
P_{S_{T}|Y,H}(s_{T}|y,\sqrt{2}\,)\,\left(\frac{1}{1+2P(s_{T})}+\frac{2P(s_{T})|y|^{2}}{\left(1+2P(s_{T})\right)^{2}}\right)$

where $P_{S_{T}|Y,H}(2|y,\sqrt{2}\,)=1-P_{S_{T}|Y,H}(0|y,\sqrt{2}\,)$ and

$\displaystyle P_{S_{T}|Y,H}(0|y,\sqrt{2}\,)$
$\displaystyle=\frac{\frac{\epsilon}{1+2P(0)}\,e^{-|y|^{2}/(1+2P(0))}}{\frac{\epsilon}{1+2P(0)}\,e^{-|y|^{2}/(1+2P(0))}+\frac{\bar{\epsilon}}{1+2P(2)}\,e^{-|y|^{2}/(1+2P(2))}}.$

For Rayleigh fading as in Sec. VIII-C the sums over $s_{T}=0,2$ become sums
over $s_{T}=0,1$ and the probabilities $P(s_{T}|y,h)$ take on similar forms as
above.

###  C-C Partial CSIR, Full CSIT

Consider $S_{T}=H$ and partial $S_{R}$. The expectation in (426)-(427) are
computed via (202) that we repeat here:

$\displaystyle
p(h,s_{T}|y,s_{R})=\delta(s_{T}-h)\,\frac{p(h|s_{R})\,p(y|h,s_{R})}{p(y|s_{R})}.$

For on-off fading with $S_{R}=0$ as in Sec. VII-C, the expression (426) with
$\phi(0)=0$ gives

$\displaystyle{\rm
E}\left[{U\big{|}Y=y}\right]=P_{H|Y}\left(\left.\sqrt{2}\,\right|y\right)\,\frac{\sqrt{4P}\,y}{1+4P}$

and (427) gives

$\displaystyle{\rm E}\left[{|U|^{2}\big{|}Y=y}\right]=P_{H|Y}(0|y)$
$\displaystyle\qquad\left.+\>P_{H|Y}\left(\left.\sqrt{2}\,\right|y\right)\,\left(\frac{1}{1+4P}+\frac{4P|y|^{2}}{(1+4P)^{2}}\right)\right]$

where $P_{H|Y}\left(\left.\sqrt{2}\,\right|y\right)=1-P_{H|Y}(0|y)$ and

$\displaystyle
P_{H|Y}(0|y)=\frac{e^{-|y|^{2}}}{e^{-|y|^{2}}+\frac{1}{1+4P}e^{-|y|^{2}/(1+4P)}}.$

For Rayleigh fading with $S_{R}=0$ and TCI as in Sec. VIII-D, the expressions
(426)-(427) give (cf. (431)-(432))

$\displaystyle{\rm E}\left[{U\big{|}Y=y}\right]={\rm Pr}\left[{G\geq
t|Y=y}\right]\frac{\sqrt{\hat{P}}\,y}{1+\hat{P}}$ $\displaystyle{\rm
E}\left[{|U|^{2}\big{|}Y=y}\right]={\rm Pr}\left[{G<t|Y=y}\right]$
$\displaystyle\qquad\left.+\>{\rm Pr}\left[{G\geq
t|Y=y}\right]\left(\frac{1}{1+\hat{P}}+\frac{\hat{P}\,|y|^{2}}{(1+\hat{P})^{2}}\right)\right]$

and therefore (cf. (433))

$\displaystyle{\rm Var}\left[{U|Y=y}\right]$ $\displaystyle={\rm
Pr}\left[{G<t|Y=y}\right]+{\rm Pr}\left[{G\geq t|Y=y}\right]$
$\displaystyle\quad\cdot\left(\frac{1}{1+\hat{P}}+\frac{\hat{P}\,|y|^{2}}{(1+\hat{P})^{2}}\,{\rm
Pr}\left[{G<t|Y=y}\right]\right)$

where (cf. (434))

$\displaystyle{\rm
Pr}\left[{G<t|Y=y}\right]=\frac{\big{(}1-e^{-t}\big{)}\,e^{-|y|^{2}}}{\big{(}1-e^{-t}\big{)}e^{-|y|^{2}}+e^{-t}\frac{1}{1+\hat{P}}e^{-|y|^{2}/(1+\hat{P})}}.$

###  C-D Partial CSIR, CSIT@R

Consider $S_{T}=S_{R}$ and partial $S_{R}$. The expectation in (426)-(427) are
computed via (239) that we repeat here:

$\displaystyle
p(h,s_{T}|y,s_{R})=\delta\big{(}s_{T}-f(s_{R})\big{)}\,\frac{p(h|s_{R})\,p(y|h,s_{R})}{p(y|s_{R})}.$

For on-off fading as in Sec. VII-C, the expression (426) with $\phi(0)=0$
gives

$\displaystyle{\rm E}\left[{U\big{|}Y=y,S_{R}=0}\right]$
$\displaystyle=P_{H|Y,S_{R}}(1|y,0)\,\frac{\sqrt{2P(0)}\,y}{1+2P(0)}$
$\displaystyle{\rm E}\left[{U\big{|}Y=y,S_{R}=\sqrt{2}}\right]$
$\displaystyle=P_{H|Y,S_{R}}(1|y,\sqrt{2}\,)\,\frac{\sqrt{2P(\sqrt{2}\,)}\,y}{1+2P(\sqrt{2}\,)}$

and (427) gives

$\displaystyle{\rm
E}\left[{|U|^{2}\big{|}Y=y,S_{R}=0}\right]=P_{H|Y,S_{R}}(0|y,0)$
$\displaystyle\qquad+P_{H|Y,S_{R}}(1|y,0)\left(\frac{1}{1+2P(0))}+\frac{2P(0)|y|^{2}}{\left(1+2P(0)\right)^{2}}\right)$
$\displaystyle{\rm
E}\left[{|U|^{2}\big{|}Y=y,S_{R}=\sqrt{2}}\right]=P_{H|Y,S_{R}}(0|y,\sqrt{2}\,)$
$\displaystyle\qquad+P_{H|Y,S_{R}}(1|y,\sqrt{2}\,)\left(\frac{1}{1+2P(\sqrt{2}\,)}+\frac{2P(\sqrt{2}\,)|y|^{2}}{\left(1+2P(\sqrt{2}\,)\right)^{2}}\right)$

where $P_{H|Y,S_{R}}(\sqrt{2}\,|y,s_{R})=1-P_{H|Y,S_{R}}(0|y,s_{R})$ and

$\displaystyle
P_{H|Y,S_{R}}(0|y,0)=\frac{\bar{\epsilon}\,e^{-|y|^{2}}}{\bar{\epsilon}\,e^{-|y|^{2}}+\frac{\epsilon}{1+2P(0)}\,e^{-|y|^{2}/(1+2P(0))}}$
$\displaystyle
P_{H|Y,S_{R}}(0|y,\sqrt{2}\,)=\frac{\epsilon\,e^{-|y|^{2}}}{\epsilon\,e^{-|y|^{2}}+\frac{\bar{\epsilon}}{1+2P(\sqrt{2}\,)}\,e^{-|y|^{2}/(1+2P(\sqrt{2}\,))}}.$

For Rayleigh fading as in Sec. VIII-E the probabilities $P(h|y,s_{R})$ take on
similar forms as above.

Appendix D

Proof of Lemma 2 and (121)

We prove Lemma 2 by using the same steps as in the proof of Proposition 1. The
GMI (104) with a vector $\underline{Y}$ is

$\displaystyle I_{s}(A;\underline{Y})=\log\det\left({\bf I}+\left({\bf
Q}_{\underline{Z}}/s\right)^{-1}{\bf H}{\bf Q}_{\underline{\bar{X}}}{\bf
H}^{\dagger}\right)$ $\displaystyle\quad+{\rm
E}\left[{{\underline{Y}}^{\dagger}\left({\bf Q}_{\underline{Z}}/s+{\bf H}{\bf
Q}_{\underline{\bar{X}}}{\bf H}^{\dagger}\right)^{-1}\underline{Y}}\right]$
$\displaystyle\quad-{\rm E}\left[{\left(\underline{Y}-{\bf
H}\,\underline{\bar{X}}\right)^{\dagger}\left({\bf
Q}_{\underline{Z}}/s\right)^{-1}\left(\underline{Y}-{\bf
H}\,\underline{\bar{X}}\right)}\right].$ (435)

One can again set $s=1$. Choosing ${\bf H}=\tilde{\bf H}$ and ${\bf
Q}_{\underline{Z}}=\tilde{\bf Q}_{\underline{\tilde{Z}}}$ then gives (114).

Next, consider the channel $\underline{Y}_{a}=\tilde{\bf
H}\underline{\bar{X}}+\underline{\tilde{Z}}$ where $\underline{\tilde{Z}}$ is
CSCG with covariance matrix ${\bf Q}_{\underline{\tilde{Z}}}$ and
$\underline{\tilde{Z}}$ is independent of $\underline{\bar{X}}$. Generalizing
(52)-(53), we compute ${\bf Q}_{\underline{Y}_{a}}={\bf Q}_{\underline{Y}}$
and

$\displaystyle{\rm E}\left[{\left(\underline{Y}_{a}-\tilde{\bf
H}\,\underline{\bar{X}}\right)\left(\underline{Y}_{a}-\tilde{\bf
H}\,\underline{\bar{X}}\right)^{\dagger}}\right]$ $\displaystyle\qquad={\rm
E}\left[{\left(\underline{Y}-{\bf
H}\,\underline{\bar{X}}\right)\left(\underline{Y}-{\bf
H}\,\underline{\bar{X}}\right)^{\dagger}}\right].$ (436)

In other words, the second-order statistics for the two channels with outputs
$\underline{Y}$ (the true channel output) and $\underline{Y}_{a}$ are the
same. Moreover, the GMI (114) is the mutual information
$I(A;\underline{Y}_{a})$. Using (106) and (435), for any $s$, $\bf H$ and
${\bf Q}_{\underline{Z}}$ we have

$\displaystyle I(A;\underline{Y}_{a})$ $\displaystyle=\log\det\left({\bf
I}+{\bf Q}_{\underline{\tilde{Z}}}^{-1}\tilde{\bf H}{\bf
Q}_{\underline{\bar{X}}}\tilde{\bf H}^{\dagger}\right)$ $\displaystyle\geq
I_{s}(A;\underline{Y}_{a})=I_{s}(A;\underline{Y})$ (437)

and equality holds if ${\bf H}=\tilde{\bf H}$ and ${\bf
Q}_{\underline{Z}}/s={\bf Q}_{\underline{\tilde{Z}}}$.

To prove (121), recall that ${\rm tr}\left({\bf AB}\right)={\rm tr}\left({\bf
BA}\right)$ for matrices $\bf A$ and $\bf B$ with appropriate dimensions.
Furthermore, for Hermitian matrices $\bf A,B,C$ with the same dimensions we
have

$\displaystyle{\rm tr}\left({\bf ABC}\right)={\rm tr}\left({({\bf
ABC})^{\dagger}}\right)={\rm tr}\left({\bf CBA}\right)={\rm tr}\left({\bf
ACB}\right).$ (438)

For notational convenience, consider the covariance matrix (119) with $s=1$
and use

$\displaystyle{\bf A}$ $\displaystyle={\bf Q}_{\underline{\bar{Z}}},\quad{\bf
B}=\left({\bf H}{\bf Q}_{\underline{\bar{X}}}{\bf
H}^{\dagger}\right)^{-1/2}\left({\bf Q}_{\underline{Y}}-{\bf
Q}_{\underline{\bar{Z}}}\right)^{1/2}$ $\displaystyle{\bf C}$
$\displaystyle={\bf Q}_{\underline{\bar{Z}}}^{-1}\left({\bf
Q}_{\underline{Y}}-{\bf Q}_{\underline{\bar{Z}}}\right)^{1/2}\left({\bf H}{\bf
Q}_{\underline{\bar{X}}}{\bf H}^{\dagger}\right)^{-1/2}$

to compute (cf. (435))

$\displaystyle{\rm E}\left[{\left(\underline{Y}-{\bf
H}\,\underline{\bar{X}}\right)^{\dagger}{\bf
Q}_{\underline{Z}}^{-1}\left(\underline{Y}-{\bf
H}\,\underline{\bar{X}}\right)}\right]={\rm tr}\left({{\bf
Q}_{\underline{\bar{Z}}}{\bf Q}_{\underline{Z}}^{-1}}\right)$
$\displaystyle\overset{(a)}{=}{\rm tr}\left({\left({\bf
Q}_{\underline{Y}}-{\bf Q}_{\underline{\bar{Z}}}\right)\left({\bf H}{\bf
Q}_{\underline{\bar{X}}}{\bf H}^{\dagger}\right)^{-1}}\right)$ (439)

where step $(a)$ follows by (438). Next, by using (119) we have

$\displaystyle\left({\bf Q}_{\underline{Z}}+{\bf H}{\bf
Q}_{\underline{\bar{X}}}{\bf H}^{\dagger}\right)^{-1}$
$\displaystyle\quad=\left({\bf Q}_{\underline{Y}}-{\bf
Q}_{\underline{\bar{Z}}}\right)^{1/2}\left({\bf H}{\bf
Q}_{\underline{\bar{X}}}{\bf H}^{\dagger}\right)^{-1/2}{\bf
Q}_{\underline{Y}}^{-1}$ $\displaystyle\qquad\cdot\left({\bf H}{\bf
Q}_{\underline{\bar{X}}}{\bf H}^{\dagger}\right)^{-1/2}\left({\bf
Q}_{\underline{Y}}-{\bf Q}_{\underline{\bar{Z}}}\right)^{1/2}$ (440)

and therefore (cf. (435))

$\displaystyle{\rm E}\left[{{\underline{Y}}^{\dagger}\left({\bf
Q}_{\underline{Z}}+{\bf H}{\bf Q}_{\underline{\bar{X}}}{\bf
H}^{\dagger}\right)^{-1}\underline{Y}}\right]$ $\displaystyle\quad={\rm
tr}\left({{\bf Q}_{\underline{Y}}\left({\bf Q}_{\underline{Z}}+{\bf H}{\bf
Q}_{\underline{\bar{X}}}{\bf H}^{\dagger}\right)^{-1}}\right)$
$\displaystyle\quad\overset{(a)}{=}{\rm tr}\left({\left({\bf
Q}_{\underline{Y}}-{\bf Q}_{\underline{\bar{Z}}}\right)\left({\bf H}{\bf
Q}_{\underline{\bar{X}}}{\bf H}^{\dagger}\right)^{-1}}\right)$ (441)

where step $(a)$ again follows by (438). We are thus left with the logarithm
term in (435). Finally, the determinant in (435) is

$\displaystyle\det\left({\bf I}+{\bf Q}_{\underline{Z}}^{-1}\,{\bf H}{\bf
Q}_{\underline{\bar{X}}}{\bf H}^{\dagger}\right)=\det\left({\bf
Q}_{\underline{\bar{Z}}}^{-1}{\bf Q}_{\underline{Y}}\right)$ (442)

where we applied (119) and Sylvester’s identity (32).

Appendix E

Large $K$ for Sec. IV-E

We complete Remark 30 by proceeding as in Appendix C-A. The main issue to deal
with to generalize (72) is that unit-rank matrices
$\underline{y}\,\underline{y}^{\dagger}$ do not have inverses. Consider first
finite $K$. Conditioned on the event $\mathcal{E}_{k}$, we may write

$\displaystyle\underline{Y}=\underline{y}_{k}+\epsilon^{1/2}\underline{\tilde{Z}}_{k}$
(443)

where $\underline{y}_{k}={\rm E}\left[{\underline{Y}|\mathcal{E}_{k}}\right]$
and ${\rm
E}\left[{\left.\underline{\tilde{Z}}_{k}\right|\mathcal{E}_{k}}\right]=\underline{0}$.
We abuse notation and write the conditional covariance matrix of
$\underline{\tilde{Z}}_{k}$ as ${\bf Q}_{\underline{\tilde{Z}}_{k}}$, and we
assume that ${\bf Q}_{\underline{\tilde{Z}}_{k}}$ is invertible. Define
$\underline{\tilde{y}}_{k}={\bf
Q}_{\underline{\tilde{Z}}_{k}}^{-1/2}\underline{y}_{k}$ and compute

$\displaystyle{\bf Q}_{\underline{Y}}^{(k)}$ $\displaystyle=\epsilon\,{\bf
Q}_{\underline{\tilde{Z}}_{k}}^{1/2}\left[{\bf
I}+\frac{1}{\epsilon}\,\underline{\tilde{y}}_{k}\underline{\tilde{y}}_{k}^{\dagger}\right]{\bf
Q}_{\underline{\tilde{Z}}_{k}}^{1/2}$ (444) $\displaystyle\left({\bf
Q}_{\underline{Y}}^{(k)}\right)^{-1}$ $\displaystyle=\frac{1}{\epsilon}{\bf
Q}_{\underline{\tilde{Z}}_{k}}^{-1/2}\left[{\bf
I}-\frac{\underline{\tilde{y}}_{k}\underline{\tilde{y}}_{k}^{\dagger}}{\epsilon+\|\underline{\tilde{y}}\|^{2}}\right]{\bf
Q}_{\underline{\tilde{Z}}_{k}}^{-1/2}.$ (445)

We further compute approximations are for small $\epsilon$:

$\displaystyle\underline{y}_{k}^{\dagger}\left({\bf
Q}_{\underline{Y}}^{(k)}\right)^{-1}\underline{y}_{k}=\frac{\|\underline{\tilde{y}}_{k}\|^{2}}{\epsilon+\|\underline{\tilde{y}}_{k}\|^{2}}\approx
1$ (446) $\displaystyle{\bf H}_{k}=\left(\underline{y}_{k}{\rm
E}\left[{\left.\underline{\bar{X}}^{\dagger}\right|\mathcal{E}_{k}}\right]+\epsilon^{1/2}{\rm
E}\left[{\left.\underline{\tilde{Z}}_{k}\underline{\bar{X}}^{\dagger}\right|\mathcal{E}_{k}}\right]\right)\left({\bf
Q}_{\underline{\bar{X}}}^{(k)}\right)^{-1}$
$\displaystyle\quad\;\;\approx\underline{y}_{k}{\rm
E}\left[{\left.\underline{\bar{X}}^{\dagger}\right|\mathcal{E}_{k}}\right]\left({\bf
Q}_{\underline{\bar{X}}}^{(k)}\right)^{-1}.$ (447)

We can now treat the limit of large $K$ for which $\epsilon$ approaches zero,
i.e., we choose a different auxiliary model for each
$\underline{Y}=\underline{y}$. Applying the Woodbury and Sylvester identities
(31)-(32) several times, (131) becomes

$\displaystyle I_{1}(A;\underline{Y})=\int_{\mathbb{C}^{N}}p(\underline{y})$
$\displaystyle\left[\log\det\left({\bf I}+\left({\bf
Q}_{\underline{\bar{X}}}^{(\underline{y})}-\underline{E}_{\underline{y}}\,\underline{E}_{\underline{y}}^{\dagger}\right)^{-1}{\bf
Q}_{\underline{\bar{X}}}\left({\bf
Q}_{\underline{\bar{X}}}^{(\underline{y})}\right)^{-1}\underline{E}_{\underline{y}}\,\underline{E}_{\underline{y}}^{\dagger}\right)\right.$
$\displaystyle\left.-{\rm tr}\left({\left({\bf
Q}_{\underline{\bar{X}}}^{(\underline{y})}\left({\bf
D}_{\underline{\bar{X}}}^{(\underline{y})}\right)^{-1}{\bf
Q}_{\underline{\bar{X}}}^{(\underline{y})}-\underline{E}_{\underline{y}}\,\underline{E}_{\underline{y}}^{\dagger}\right)^{-1}\underline{E}_{\underline{y}}\,\underline{E}_{\underline{y}}^{\dagger}}\right)\right]d\underline{y}$
(448)

where

$\displaystyle\underline{E}_{\underline{y}}$ $\displaystyle={\rm
E}\left[{\underline{\bar{X}}|\underline{Y}=\underline{y}}\right]$
$\displaystyle{\bf Q}_{\underline{\bar{X}}}^{(\underline{y})}$
$\displaystyle={\rm
E}\left[{\left.\underline{\bar{X}}\,\underline{\bar{X}}^{\dagger}\right|\underline{Y}=\underline{y}}\right]$
$\displaystyle{\bf D}_{\underline{\bar{X}}}^{(\underline{y})}$
$\displaystyle={\bf Q}_{\underline{\bar{X}}}-{\bf
Q}_{\underline{\bar{X}}}^{(\underline{y})}.$

If $\underline{\bar{X}},\underline{Y}$ are jointly CSCG then using (24)-(25)
we have

$\displaystyle\underline{E}_{\underline{y}}={\rm
E}\left[{\underline{\bar{X}}\,\underline{Y}^{\dagger}}\right]{\bf
Q}_{\underline{Y}}^{-1}\cdot\underline{y}$ (449) $\displaystyle{\bf
Q}_{\underline{\bar{X}}}^{(\underline{y})}-\underline{E}_{\underline{y}}\,\underline{E}_{\underline{y}}^{\dagger}={\bf
Q}_{\underline{\bar{X}}}-{\rm
E}\left[{\underline{\bar{X}}\,\underline{Y}^{\dagger}}\right]{\bf
Q}_{\underline{Y}}^{-1}{\rm
E}\left[{\underline{\bar{X}}\,\underline{Y}^{\dagger}}\right]^{\dagger}.$
(450)

For example, if $\underline{Y}={\bf H}\underline{X}+\underline{Z}$ where ${\bf
H},\underline{X},\underline{Z}$ are mutually independent and ${\rm
E}\left[{\underline{Z}}\right]={\bf 0}$ then we have (cf. (429))

$\displaystyle\underline{E}_{\underline{y}}$
$\displaystyle=\int_{\mathbb{C}^{N\times M}}p({\bf h}|\underline{y})\,{\rm
E}\left[{\underline{\bar{X}}|\underline{Y}=\underline{y},{\bf H}={\bf
h}}\right]\,d{\bf h}$ $\displaystyle=\int_{\mathbb{C}^{N\times M}}p({\bf
h}|\underline{y})\,{\bf Q}_{\underline{\bar{X}}}\,{\bf h}^{\dagger}\left({\bf
I}+{\bf h}{\bf Q}_{\underline{X}}{\bf
h}^{\dagger}\right)^{-1}\underline{y}\,d{\bf h}$ (451) $\displaystyle={\rm
E}\left[{\left.{\bf Q}_{\underline{\bar{X}}}{\bf H}^{\dagger}\left({\bf
I}+{\bf H}{\bf Q}_{\underline{X}}{\bf
H}^{\dagger}\right)^{-1}\right|\underline{Y}=\underline{y}}\right]\cdot\underline{y}$
(452)

where we have applied (449) with conditioning on the event ${\bf H}={\bf h}$.
Similarly, we apply a conditional version of (450) and the step (451) to
compute (cf. (430))

$\displaystyle{\bf Q}_{\underline{\bar{X}}}^{(\underline{y})}$
$\displaystyle=\int_{\mathbb{C}^{N\times M}}p({\bf h}|\underline{y})\,{\rm
E}\left[{\left.\underline{\bar{X}}\,\underline{\bar{X}}^{\dagger}\right|\underline{Y}=\underline{y},{\bf
H}={\bf h}}\right]\,d{\bf h}$ $\displaystyle=\int_{\mathbb{C}^{N\times
M}}p({\bf h}|\underline{y})\left({\bf
Q}_{\underline{\bar{X}}}^{(\underline{y},{\bf
h})}+\underline{E}_{\underline{y},{\bf h}}\,\underline{E}_{\underline{y},{\bf
h}}^{\dagger}\right)\,d{\bf h}$ $\displaystyle={\rm E}\left[{\left.{\bf
Q}_{\underline{\bar{X}}}^{(\underline{y},{\bf
H})}+\underline{E}_{\underline{y},{\bf H}}\,\underline{E}_{\underline{y},{\bf
H}}^{\dagger}\right|\underline{Y}=\underline{y}}\right]$ (453)

where

$\displaystyle{\bf Q}_{\underline{\bar{X}}}^{(\underline{y},{\bf h})}$
$\displaystyle={\bf Q}_{\underline{\bar{X}}}-{\bf Q}_{\underline{\bar{X}}}{\bf
h}^{\dagger}\left({\bf I}+{\bf h}{\bf Q}_{\underline{X}}{\bf
h}^{\dagger}\right)^{-1}{\bf h}{\bf Q}_{\underline{\bar{X}}}$
$\displaystyle\underline{E}_{\underline{y},{\bf h}}$ $\displaystyle={\rm
E}\left[{\underline{\bar{X}}|\underline{Y}=\underline{y},{\bf H}={\bf
h}}\right]$ $\displaystyle={\bf Q}_{\underline{\bar{X}}}\,{\bf
h}^{\dagger}\left({\bf I}+{\bf h}{\bf Q}_{\underline{X}}{\bf
h}^{\dagger}\right)^{-1}\underline{y}.$

Appendix F

Proof of Lemma 4

Let $\bar{P}={\rm E}\left[{|\bar{X}|^{2}}\right]$ and write

$\displaystyle\bar{X}=\sqrt{\bar{P}}\,\bar{U},\quad
X(s_{T})=\sqrt{P(s_{T})}\,U(s_{T}).$ (454)

Since the $U(s_{T})$ are CSCG we have

$\displaystyle
U(s_{T}^{\prime})=\rho(s_{T}^{\prime},s_{T})\,U(s_{T})+Z(s_{T}^{\prime})$
(455)

where $\rho(s_{T}^{\prime},s_{T})={\rm
E}\left[{U(s_{T}^{\prime})\,U(s_{T})^{\text{*}}}\right]$ and

$\displaystyle
Z(s_{T}^{\prime})\sim\mathcal{CN}(0,1-|\rho(s_{T}^{\prime},s_{T})|^{2})$ (456)

is independent of $U(s_{T})$. As in (111), define

$\displaystyle\bar{X}$
$\displaystyle=\sum_{s_{T}^{\prime}}w(s_{T}^{\prime})X(s_{T}^{\prime})$
$\displaystyle=\sum_{s_{T}}w(s_{T}^{\prime})\sqrt{P(s_{T}^{\prime})}\left[U(s_{T})\rho(s_{T}^{\prime},s_{T})+Z(s_{T}^{\prime})\right]$
$\displaystyle=\sqrt{\bar{P}}\,\bar{\rho}(s_{T})\,U(s_{T})+\sum_{s_{T}^{\prime}}w(s_{T}^{\prime})\sqrt{P(s_{T}^{\prime})}Z(s_{T}^{\prime})$
(457)

where, assuming that $\bar{P}>0$, we have

$\displaystyle\bar{\rho}(s_{T})={\rm
E}\left[{\bar{U}\,U(s_{T})^{\text{*}}}\right]=\sum_{s_{T}^{\prime}}w(s_{T}^{\prime})\sqrt{\frac{P(s_{T}^{\prime})}{\bar{P}}}\,\rho(s_{T}^{\prime},s_{T}).$
(458)

Observe that $\sqrt{\bar{P}}\,\bar{\rho}(s_{T})\,U(s_{T})$ is the LMMSE
estimate of $\bar{X}$ given $U(s_{T})$.

Using Lemma 2, we have the auxiliary variables

$\displaystyle\tilde{h}=\frac{{\rm
E}\left[{Y\bar{X}^{\text{*}}}\right]}{\bar{P}},\quad\tilde{\sigma}^{2}={\rm
E}\left[{|Y|^{2}}\right]-|\tilde{h}|^{2}\bar{P}$ (459)

and the GMI

$\displaystyle I_{1}(A;Y)=\log\left(\frac{{\rm E}\left[{|Y|^{2}}\right]}{{\rm
E}\left[{|Y|^{2}}\right]-|\tilde{h}|^{2}\bar{P}}\right).$ (460)

If the $P(s_{T})$ are fixed then so is ${\rm E}\left[{|Y|^{2}}\right]$ because
$U(s_{T})$ is CSCG and independent of $Z$ given $S_{T}=s_{T}$. The GMI (460)
is thus maximized by maximizing $|\tilde{h}|^{2}\bar{P}$. We compute

$\displaystyle|\tilde{h}|^{2}\bar{P}=\left|\sum_{s_{T}}P_{S_{T}}(s_{T})\frac{{\rm
E}\left[{\left.Y\bar{X}^{\text{*}}\right|S_{T}=s_{T}}\right]}{\sqrt{\bar{P}}}\right|^{2}$
$\displaystyle\overset{(a)}{=}\left|\sum_{s_{T}}P_{S_{T}}(s_{T}){\rm
E}\left[{\left.Y\,U(s_{T})^{\text{*}}\right|S_{T}=s_{T}}\right]\bar{\rho}(s_{T})^{\text{*}}\right|^{2}$
(461) $\displaystyle\leq\left(\sum_{s_{T}}P_{S_{T}}(s_{T})\left|{\rm
E}\left[{\left.Y\,U(s_{T})^{\text{*}}\right|S_{T}=s_{T}}\right]\right|\right)^{2}$
(462)

where step $(a)$ follows because we have the Markov chain
$A-[U(S_{T}),S_{T}]-Y$ which implies that $Y$ and the $Z(s_{T}^{\prime})$ in
(457) are independent give $S_{T}=s_{T}$.

Equality holds in (462) if the summands in (461) all have the same phase and
$|\bar{\rho}(s_{T})|=1$ for all $s_{T}$. But this is possible by choosing
$X(s_{T})$ as given in (134) so that $U(s_{T})=e^{j\phi(s_{T})}\,U$, and the
receiver weights as

$\displaystyle
w(\tilde{s}_{T})=\sqrt{\frac{\bar{P}}{P(\tilde{s}_{T})}}\,e^{-j\phi(\tilde{s}_{T})}$
(463)

for one $\tilde{s}_{T}\in\mathcal{S}_{T}$ with $P(\tilde{s}_{T})>0$, and
$w(s_{T})=0$ otherwise. We then have $\bar{X}=\sqrt{\bar{P}}\,U$ and

$\displaystyle\rho(s_{T}^{\prime},s_{T})$
$\displaystyle=e^{j(\phi(s_{T}^{\prime})-\phi(s_{T}))}$ (464)
$\displaystyle\bar{\rho}(s_{T})$ $\displaystyle=e^{-j\phi(s_{T})}$ (465)

and the resulting maximal $I_{1}(A;Y)$ is given by (132)-(133).

######  Remark 75.

The full correlation permits many choices for the $w(s_{T})$ and hence these
weights do not seem central to the design. However, including weights can be
useful if the codebook is not designed for the CSIR. For example, suppose $A$
has independent entries $X(s_{T})$ for which we compute

$\displaystyle\bar{\rho}(s_{T})$
$\displaystyle=\frac{w(s_{T})\sqrt{P(s_{T})}}{\sqrt{\sum_{s_{T}^{\prime}}|w(s_{T}^{\prime})|^{2}P(s_{T}^{\prime})}}$
(466)

and thus (461) becomes

$\displaystyle\frac{\left|\sum_{s_{T}}P_{S_{T}}(s_{T}){\rm
E}\left[{\left.YX(s_{T})^{\text{*}}\right|S_{T}=s_{T}}\right]\,w(s_{T})^{\text{*}}\right|^{2}}{\sum_{s_{T}}|w(s_{T})|^{2}P(s_{T})}.$
(467)

Using Bergström’s inequality (or the Cauchy-Schwarz inequality), the
expression (467) is maximized by

$\displaystyle w(s_{T})=P_{S_{T}}(s_{T})\,\frac{{\rm
E}\left[{\left.YX(s_{T})^{\text{*}}\right|S_{T}=s_{T}}\right]}{P(s_{T})}\cdot
c$ (468)

for some constant $c\neq 0$. The expression (461) is therefore

$\displaystyle\sum_{s_{T}}P_{S_{T}}(s_{T}|h)^{2}\left|{\rm
E}\left[{\left.YU(s_{T})^{\text{*}}\right|S_{T}=s_{T}}\right]\right|^{2}$
(469)

which is generally smaller than ${\rm E}\left[{\,\left|{\rm
E}\left[{\left.YU(S_{T})^{\text{*}}\right|S_{T}}\right]\right|\,}\right]^{2}$
(apply $\sum_{i}a_{i}^{2}\leq(\sum_{i}a_{i})^{2}$ for non-negative $a_{i}$).

######  Remark 76.

The following example shows that more general signaling and more general
$\bar{X}$ can be useful. Consider the channel with two equally-likely states
$\mathcal{S}_{T}=\\{+1,-1\\}$ and $Y=|X|\exp(js_{T}\arg(X))+Z$. We compute

$\displaystyle{\rm E}\left[{Y\,U(+1)^{\text{*}}|S_{T}=+1}\right]=\sqrt{P(1)}$
$\displaystyle{\rm E}\left[{Y\,U(-1)^{\text{*}}|S_{T}=-1}\right]=0$
$\displaystyle{\bar{\rho}}(+1)=\frac{w(1)\sqrt{P(1)}+w(-1)\sqrt{P(-1)}\rho(-1,+1)}{\sqrt{\bar{P}}}$

and one should choose $P(-1)=0$ and $P(1)=2P$ if the power constraint is ${\rm
E}\left[{P(S_{T})}\right]\leq P$. We thus have

$\displaystyle{\rm E}\left[{|Y|^{2}}\right]=P+1,\quad\tilde{P}=\frac{P}{2}$

and therefore (132) gives

$\displaystyle I_{1}(A;Y)=\log\left(1+\frac{P}{2+P}\right).$

However, one can achieve the rate $\log(1+P)$ with other Gaussian $\bar{X}$,
namely linear combinations of both the $X(s_{T})$ and the
$X(s_{T})^{\text{*}}$ in (457). This idea permits circularly asymmetric
$\bar{X}$, also known as _improper_ $\bar{X}$ [94]. Alternatively, the
transmitter can send the complex-conjugate symbols if $S_{T}=-1$.

Appendix G

Proof of Lemma 5

We mimic the steps of Appendix C-D. Consider the SVDs

$\displaystyle{\bf Q}_{\underline{\bar{X}}}$ $\displaystyle={\bf
V}_{\underline{\bar{X}}}\,{\bf\Sigma}_{\underline{\bar{X}}}\,{\bf
V}_{\underline{\bar{X}}}^{\dagger}$ $\displaystyle{\bf
Q}_{\underline{X}(s_{T})}$ $\displaystyle={\bf
V}_{\underline{X}(s_{T})}\,{\bf\Sigma}_{\underline{X}(s_{T})}\,{\bf
V}_{\underline{X}(s_{T})}^{\dagger}.$

Let $\underline{\bar{U}}\sim\mathcal{CN}(0,I)$ and write

$\displaystyle\underline{\bar{X}}$ $\displaystyle={\bf
Q}_{\underline{\bar{X}}}^{1/2}\,\underline{\bar{U}}.$

Since the $\underline{U}(s_{T})$ are CSCG, we have

$\displaystyle\underline{U}(s_{T}^{\prime})={\bf
R}(s_{T}^{\prime},s_{T})\,\underline{U}(s_{T})+\underline{Z}(s_{T}^{\prime})$
(470)

where ${\bf R}(s_{T}^{\prime},s_{T})={\rm
E}\left[{\underline{U}(s_{T}^{\prime})\,\underline{U}(s_{T})^{\dagger}}\right]$
and

$\displaystyle\underline{Z}(s_{T}^{\prime})\sim\mathcal{CN}(0,{\bf I}-{\bf
R}(s_{T}^{\prime},s_{T}){\bf R}(s_{T}^{\prime},s_{T})^{\dagger})$ (471)

is independent of $\underline{U}(s_{T})$. As in (111), define

$\displaystyle\underline{\bar{X}}=\sum_{s_{T}^{\prime}}{\bf
W}(s_{T}^{\prime})\,\underline{X}(s_{T}^{\prime})$
$\displaystyle=\sum_{s_{T}^{\prime}}{\bf W}(s_{T}^{\prime})\,{\bf
Q}_{\underline{X}(s_{T}^{\prime})}^{1/2}\left[{\bf
R}(s_{T}^{\prime},s_{T})\,\underline{U}(s_{T})+\underline{Z}(s_{T}^{\prime})\right]$
$\displaystyle={\bf Q}_{\underline{\bar{X}}}^{1/2}\,\bar{\bf
R}(s_{T})\,\underline{U}(s_{T})+\sum_{s_{T}^{\prime}}{\bf
W}(s_{T}^{\prime})\,{\bf
Q}_{\underline{X}(s_{T}^{\prime})}^{1/2}\,\underline{Z}(s_{T}^{\prime})$ (472)

where as in (458), and assuming ${\bf Q}_{\underline{\bar{X}}}\succ{\bf 0}$,
we write

$\displaystyle\bar{\bf R}(s_{T})$ $\displaystyle={\rm
E}\left[{\underline{\bar{U}}\,\underline{U}(s_{T})^{\dagger}}\right]$
$\displaystyle=\sum_{s_{T}^{\prime}}{\bf Q}_{\underline{\bar{X}}}^{-1/2}\,{\bf
W}(s_{T}^{\prime})\,{\bf Q}_{\underline{X}(s_{T}^{\prime})}^{1/2}\,{\bf
R}(s_{T}^{\prime},s_{T}).$ (473)

Observe that the vector ${\bf Q}_{\underline{\bar{X}}}^{1/2}\,\bar{\bf
R}(s_{T})\,\underline{U}(s_{T})$ is the LMMSE estimate of
$\underline{\bar{X}}$ given $\underline{U}(s_{T})$.

Using Lemma 2, we have (see (459))

$\displaystyle\tilde{\bf H}={\rm
E}\left[{\underline{Y}\,\underline{\bar{X}}^{\dagger}}\right]{\bf
Q}_{\underline{\bar{X}}}^{-1},\quad{\bf Q}_{\tilde{\underline{Z}}}={\bf
Q}_{\underline{Y}}-\tilde{\bf H}\,{\bf Q}_{\underline{\bar{X}}}\tilde{\bf
H}^{\dagger}$ (474)

and the GMI (136) that we repeat here for convenience:

$\displaystyle I_{1}(A;\underline{Y})=\log\left(\frac{\det{\bf
Q}_{\underline{Y}}}{\det\left({\bf Q}_{\underline{Y}}-\tilde{\bf H}\,{\bf
Q}_{\underline{\bar{X}}}\tilde{\bf H}^{\dagger}\right)}\right).$ (475)

As in Appendix C-D, if the ${\bf Q}_{\underline{X}(s_{T})}$ are fixed then so
is ${\bf Q}_{\underline{Y}}$ because
$\underline{U}(s_{T})\sim\mathcal{CN}(\underline{0},{\bf I})$ is independent
of $\underline{Z}$ given $S_{T}=s_{T}$. We would like to maximize the GMI
(475). Similar to (461), we have the decomposition

$\displaystyle\tilde{\bf H}{\bf Q}_{\underline{\bar{X}}}\tilde{\bf
H}^{\dagger}=\tilde{\bf D}\,\tilde{\bf D}^{\dagger}$ (476)

where

$\displaystyle\tilde{\bf D}=\sum_{s_{T}}P_{S_{T}}(s_{T}){\rm
E}\left[{\left.\underline{Y}\,\underline{U}(s_{T})^{\dagger}\right|S_{T}=s_{T}}\right]\bar{\bf
R}(s_{T})^{\dagger}.$ (477)

As in (461), we have the Markov chain
$A-[\underline{U}(S_{T}),S_{T}]-\underline{Y}$ which implies that
$\underline{Y}$ and the $\underline{Z}(s_{T}^{\prime})$ in (472) are
independent give $S_{T}=s_{T}$. It is natural to expect that the matrix
$\bar{\bf R}(s_{T})$ of correlation coefficients should be “maximized” in some
way. Indeed, the Cauchy-Schwarz inequality gives

$\displaystyle\underline{v}_{1}^{\dagger}\,\bar{\bf
R}(s_{T})\,\underline{v}_{2}={\rm
E}\left[{\underline{v}_{1}^{\dagger}\,\underline{\bar{U}}\cdot\underline{U}(s_{T})^{\dagger}\,\underline{v}_{2}}\right]$
$\displaystyle\leq\sqrt{{\rm
E}\left[{\left|\underline{\bar{U}}^{\dagger}\,\underline{v}_{1}\right|^{2}}\right]}\cdot\sqrt{{\rm
E}\left[{\left|\underline{U}(s_{T})^{\dagger}\,\underline{v}_{2}\right|^{2}}\right]}=\left\|\underline{v}_{1}\right\|\cdot\left\|\underline{v}_{2}\right\|$

for any complex $M$-dimensional vectors $\underline{v}_{1}$ and
$\underline{v}_{2}$. The singular values of ${\bf R}(s_{T})$ are thus at most
1. We will choose the $\underline{U}(s_{T})$ so that the ${\bf R}(s_{T})$ are
unitary matrices and thus all singular values are 1.

Consider the SVD decompositions (138) and a codebook based on scaling and
rotating a common $\underline{U}\sim\mathcal{CN}(\underline{0},{\bf I})$ of
dimension $N$ (see (134)):

$\displaystyle\underline{U}(s_{T})={\bf V}_{T}(s_{T})\,\underline{U}.$ (478)

The receiver chooses $M\times M$ unitary matrices ${\bf V}_{R}(s_{T})$ for all
$s_{T}$ and uses the weighting matrix (cf. (463))

$\displaystyle{\bf W}(\tilde{s}_{T})={\bf Q}_{\underline{\bar{X}}}^{1/2}\,{\bf
V}_{R}(\tilde{s}_{T})\,{\bf V}_{T}(\tilde{s}_{T})^{\dagger}\,{\bf
Q}_{\underline{X}(\tilde{s}_{T})}^{-1/2}$ (479)

for one $\tilde{s}_{T}\in\mathcal{S}_{T}$ with ${\bf
Q}_{\underline{X}(\tilde{s}_{T})}\succ 0$, and ${\bf W}(\tilde{s}_{T})={\bf
0}$ otherwise. These choices give $\underline{\bar{X}}={\bf
Q}_{\underline{\bar{X}}}^{1/2}\,\underline{U}$ and (cf. (464) and (465))

$\displaystyle{\bf R}(s_{T}^{\prime},s_{T})$ $\displaystyle={\bf
V}_{T}(s_{T}^{\prime})\,{\bf V}_{T}(s_{T})^{\dagger}$ (480)
$\displaystyle\bar{\bf R}(s_{T})$ $\displaystyle={\bf V}_{R}(s_{T}){\bf
V}_{T}(s_{T})^{\dagger}.$ (481)

Using (138), (477), and (481) we have

$\displaystyle\tilde{\bf D}=\sum_{s_{T}}P_{S_{T}}(s_{T})\,{\bf
U}_{T}(s_{T})\,{\bf\Sigma}(s_{T})\,{\bf V}_{R}(s_{T})^{\dagger}.$ (482)

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