Capacity of Binary Erasure Channel Consider the the binary erasure channel, with input and output alphabet $\{0,?,1\}$ and channel matrix
\begin{bmatrix}
1-\lambda-\mu &amp; \mu &amp; \lambda\\
0 &amp; 1 &amp; 0\\
\lambda &amp; \mu &amp; 1-\lambda-\mu
\end{bmatrix}
I want to determine its capacity, i.e. maximize $I(X;Y)$ over the probabilities distribution of the input.
I have tried writing and rewriting $I(X;Y)$ in different ways but the expression gets too complicated and I can't find a way to assign values to $p(x)$ so that $I(X;Y)$ is maximized. Any suggestions? Thank you in advance.
By symmetry, we can assume that $\Pr[X=0] = \Pr[X=1] = p$, and so $\Pr[X=?] = 1-2p$. We can then express $I(X;Y)$ as a function of $p,\lambda,\mu$. The capacity of the channel is the maximum of $I(X;Y)$ over all $p \in [0,1/2]$. You can find this maximum using calculus.
In more detail, let us first note that $\Pr[Y = 0] = \Pr[Y = 1] = (1-\mu)p$ and $\Pr[Y=?] = 1-2(1-\mu)p$. Furthermore, $H(Y|X=?) = 0$ and $H(Y|X=0) = H(Y|X=1) = H(1-\lambda-\mu,\lambda,\mu)$. Therefore
$$
I(X;Y) = H((1-\mu)p,(1-\mu)p,1-2(1-\mu)p) - 2pH(1-\lambda-\mu,\lambda,\mu).
$$
The derivative of this with respect to $p$ is
$$
-2(1-\mu)\log((1-\mu)p)+2(1-\mu)\log(1-2(1-\mu)p) - 2H(1-\lambda-\mu,\lambda,\mu) = \\
2(1-\mu) \log \frac{1-2(1-\mu)p}{(1-\mu)p} - 2H(1-\lambda-\mu,\lambda,\mu).
$$
Equating this to zero, we get
$$
\log \frac{1-2(1-\mu)p}{(1-\mu)p} = \frac{H(1-\lambda-\mu,\lambda,\mu)}{1-\mu}
$$
and so
$$
\frac{1-2(1-\mu)p}{(1-\mu)p} = e^{H(1-\lambda-\mu,\lambda,\mu)/(1-\mu)}.
$$
Massaging this, we finally get
$$
p = \frac{1}{(1-\mu)(2 + e^{H(1-\lambda-\mu,\lambda,\mu)/(1-\mu)})}.
$$