Maximizing sum of logarithms (Z-channel capacity)

Question

In the context of information theory, I am trying to maximize the following function (mutual information of the Z-channel's input and output) with respect to $p$ in order to derive Z-channel's capacity: $$I(X;Y)=\mathit{H}(ap)-\mathit{H}(1-a)p$$ where $\mathit{H}(x)=-xlog_2(x)-(1-x)log_2(1-x)$, $0<x<1$ is known as the binary entropy function. So the function is: $$I(X;Y)=-aplog_2(ap)-(1-ap)log_2(1-ap)-\mathit{H}(1-a)p$$ Differentiating with respect to $p$ I get: \begin{eqnarray*} \frac{\partial I(X;Y)}{\partial p}&=&-alog_2(ap)-ap\frac{1}{pln2}+alog_2(1-ap)-(1-ap)\frac{a}{ln2(ap-1)}-\mathit{H}(1-a)\\ &=&log_2((ap)^{-a}e^{-a}(1-ap)^{-a}e^aa^a(1-a)^{1-a})\\ &=&log_2((ap)^{-a}(1-ap)^{-a}a^a(1-a)^{1-a}) \end{eqnarray*} Then: \begin{eqnarray*} \frac{\partial I(X;Y)}{\partial p}=0&\Rightarrow &log_2((ap)^{-a}(1-ap)^{-a}a^a(1-a)^{1-a})=0\\ &\Rightarrow &(ap)^{-a}(1-ap)^{-a}a^a(1-a)^{1-a}=1\\ \end{eqnarray*} I can't solve this. Eventually, I am trying to prove that the value of $p$ that maximizes the function is: $$p=\frac{1}{a(1+2^{\mathit{H}(1-a)/a})}$$ More information on the Z-channel can be found here, but I am using different notation regarding the probabilities.

Answer 1

It is easier to separate the process using the chain rule instead of using the full equation for I: $$I=H(ap)-H(1-a)p$$ $$\frac{dI}{dp}=\frac{dH(ap)}{d(ap)}\frac{d(ap)}{dp}-H(1-a)=\frac{dH(ap)}{d(ap)}a-H(1-a)$$ Now we need to find $H'$ $$\frac{dH}{dx}=-log_2(x)-x\frac{1}{x ln2}+log_2(1-x)+(1-x)\frac{1}{(1-x)ln2}=\frac{-ln(x)-1+ln(1-x)+1}{ln2}=log_2(1-x)-log_2(x)$$ Substituting in the previous equation with $x=ap$: $$\frac{dI}{dp}=a(log_2(1-ap)-log_2(ap))-H(1-a)$$ Setting $dI/dp=0$ $$\frac{H(1-a)}{a}=log_2(1-ap)-log_2(ap)$$ $$2^{\frac{H(1-a)}{a}}=2^{log_2(1-ap)-log_2(ap)}=\frac{2^{log_2(1-ap)}}{2^{log_2(ap)}}=\frac{1-ap}{ap}$$ $$ap2^{\frac{H(1-a)}{a}}+ap=1$$ $$ap(2^{\frac{H(1-a)}{a}}+1)=1$$ $$p=\frac{1}{a(2^{\frac{H(1-a)}{a}}+1)}$$