= Solution
Let the input law again be $(s,t,u)$. The second output symbol has <probability> $1/2$ regardless of the input, and the two remaining output probabilities <sum> to $1/2$. Hence
$$
H(Y)\leq h_2(1/2)+\frac12\cdot1=\frac32,
$$
with equality when the first and third output symbols each have <probability> $1/4$. The row entropies are $1,3/2,1$, so
$$
H(Y\mid X)=1+\frac t2,
\qquad I(X;Y)\leq\frac12-\frac t2\leq\frac12.
$$
Take $t=0$ and $s=u=1/2$. The output is $(1/4,1/2,1/4)$ and both bounds are attained. Therefore
$$
\boxed{p_X=(1/2,0,1/2),\qquad C=1/2\text{ bit per use}.}
$$
The equality conditions also show uniqueness of the optimum input law. The middle row is the average of the outer rows and carries extra <conditional entropy> without introducing a new output law; <removing a redundant input symbol from a channel> explains why it need not be used.
\Image[/past-exam-of-the-communication-theory-course-of-the-university-of-oxford/2010-channel-capacities.png]
{title=Mutual information along symmetric input distributions for the three channels}
{description=The first two channels attain log2(3/2) at uniform input, while the third attains one half bit by omitting its middle input.}
Back to article page
