Maximizing Mutual information of a Fading Channel with Additive Gaussian Noise

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I am considering a problem such that we have a channel $Y=XV+Z$. Assuming $X,V,Z$ all independent

Where $$Z \sim \mathcal{N}(0, N)$$

and $$ V=\begin{cases} \alpha_1, & p\\ \alpha_2, & (1-p) \end{cases} $$

With Power Constraint $E[X^2] \le P$

Here is the approach I have used to attempt to maxmize the mutual information:

\begin{align} I(X;Y) &= h(Y) - h(Y|X)\\ &= h(Y)- h(XV+Z|X)\\ &\le h(Y) - h(XV+Z|X,V)\\ &=h(Y) - p \ h(\alpha_1X+Z|X,V=\alpha_1) - (1-p)\ h(\alpha_2X+Z|X,V=\alpha_2)\\ &= h(Y) - p\ h(Z) - (1-p)\ h(Z)\\ &= h(Y) - h(Z) \end{align}

Then $h(Y)$ is maximized for $Y$ gaussian.

\begin{align} Var(Y) &= Var(XV+Z)\\ &= Var(XV) + Var(Z)\\ &\le \sqrt{Var(X)Var(V)} + E[Z^2] &(\text{Cauchy-Shwarz})\\ &\le \sqrt{PVar(V)} +N \end{align}

Then my upper bound on the mutual information is

\begin{align} I(X;Y) \le \frac{1}{2}log\bigg(1+ \frac{\sqrt{PVar(V)}}{N}\bigg) \end{align}

which I suppose is maximized when $X \sim \mathcal{N}(0,P)$.

Question: Is this the right approach to take to find the maximum?

Typically, I have seen fading channels with deterministic multiplication factors, so I wanted to consider a problem where the multiplicative factor is a random variable.

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extended comment.

Your attempt has the following issues:

  1. the application of the Cauchy inequality is wrong.

  2. $Y$ cannot be Gaussian even with $X$ Gaussian (it is a mixture of Gaussians in that case).

  3. (Most important one) Are you sure conditioning the entropy of an arbitrary distributed real-valued random variable reduces its entropy in this case? "Conditioning reduces entropy" holds for discrete-valued variables, but not always for real-valued variables. The way you employ this inequality suggests you have made some assumption about $X$ distribution (such that the inequality indeed holds). But, doing so, you are effectively restricting yourself to a specific $X$ distribution and you cannot claim you are computing a universal upper bound. You are only computing an upper bound for that specific $X$ distribution.