Given a channel $Y=X+Z$ and a random variable $Z$ such that $$ Z=\begin{cases} 0, &w.p. \frac{1}{10}\\ Z^* &w.p. \frac{9}{10} \end{cases} $$
such that $Z^* \sim \mathcal{N}(0,\sigma^2)$
What is the capacity of the channel with power constraint $E[X^2] \le P$?
Since, $Z$ has a chance to send the message $X$ noiselessly, the capacity is infinite. But I am having difficulty explicitly calculating it.
Here is what I have tried:
We have,
$$ f_z(z) = \frac{1}{10}\delta(z) + \frac{9}{10}f_{z^*}(z) $$
Assuming $Z$ is independent of $X$, the mutual information is
$$ I(X;Y) = h(Y) - h(Y|X)\\ =h(Y) - h(Z) $$
Given that a Gaussian distribution maximizes entropy for a given variance,
$$ E[Z^2] = \int z^2 f_z(z) dz = \frac{9}{10}E[Z^{*2}] = \frac{9}{10} \sigma^2\\ E[Y^2] = E[X^2] + E[Z^2]\\ \le P+\frac{9}{10}\sigma^2 $$
Then,
$$ I(X;Y) \le \frac{1}{2}\text{log}\bigg(1+\frac{P}{\frac{9}{10}\sigma^2}\bigg) $$
I think I may have done something wrong, since it should be that $C=\infty$.
u r assuming $Z$ to be gaussian in the derivation but $Z$ is a mixed distribution. I am saying this because in ur final capacity upper bound u r assuming $h(Z)=\log(\frac{9 \sigma^2}{10})$ which is not true.
TO exactly calculate: Let $U$ be a random variable which is $0$ is $Z=0$ or else $1$. \begin{equation} = C = I(X;Y) = h(Y)-h(Z) \\ = h(Y|U=0) 1/10 + h(Y|U=1) 9/10 + H(U)-H(U|Y) - ((h(Z|U=0) 1/10 + h(Z|U=1) 9/10)+H(U))\\ = (h(Y|U=0)-h(Z|U=0)) 1/10 + H(U)-H(U|Y) + (h(Y|U=1)-h(Z|U=1)) 9/10 -H(U)\\ \geq (h(X)-h(Z|U=0)) 1/10 + \log \left(1+\frac{P}{\sigma^2} \right) 9/10-H(U|Y)\\ \end{equation} Here $h(Z|U=0) = -\infty$ (define it like this). Hence u get the answer.
Edited: We used $h(Y) = h(Y|U) + H(U) - H(U|Y)$, $h(Z,U) = H(U)+h(Z|U) = h(Z) + H(U|Z) = h(Z)$