I'm trying to determine the channel capacity of the following DMC
Converting this into a PTM, I get the following PTM = [[1,0,0], [1/3,1/3,1/3], [0,0,1]]
I can see this is not symmetric or weakly symmetric. I'm not sure if it is decomposable or not, but I'm guessing that it is not? In that case I want to make H(Y) uniform so that it gets maximized.
$min H(Y|X) = p_{0}H(1, 0, 0) + p_{1}H(1/3, 1/3, 1/3) + p_{2}H(0, 0, 1)$ $= p_{0}0 + p_{1} + p_{2}0$
$max H(Y) = H(p_{0}1 + p_{1}1/3 + p_{2}0, p_{0}0 + p_{1}1/3 + p_{2}0, p_{0}0 + p_{1}1/3 + p_{2}1)$
$p_{0} + \frac{p_{1}}{3} = 1/3, \frac{p_{1}}{3} = 1/3, \frac{p_{1}}{3} + p_{2} = 1/3$
$p_{1} = 1, p_{2} = 0, p_{0} = 0$
$min H(Y|X) = 0H(1, 0, 0) + 1H(1/3, 1/3, 1/3) + 0H(0, 0, 1) = 1$
$C = H(Y) - H(Y|X) = 1 - 1 = 0$ bits
However in this case I'm getting 0 bits and this doesn't seem right. What am I doing wrong?
You can't minimize/maximize each term in $I(X;Y) = H(Y)- H(Y|X) $ separately. What you really have is
$$ H(Y|X) = p_1 \log 3$$
$$ H(Y) = H\left(p_0+\frac13 p_1,\frac13 p_1, p_2+\frac13 p_1\right)$$
Now, for a fixed $p_1$, it's easy to see that $I(X;Y)$ is maximized by $p_0 = p_2 = \frac12 (1-p_1)$. Hence we need to find teh value of $p_1$ that maximizes
$$I(X;Y)=H(Y)- H(Y|X)=H\left(\frac12 - \frac16 p_1,\frac13 p_1, \frac12 - \frac16 p_1\right)-p_1 \log3$$
Or $$I(X;Y) = H(a,1-2a,a)- (1-2a) 3 \log3$$
with $a = \frac12 - \frac16 p_1$. To maximize this, We could find the derivative of this wrt $a$ and compute the root numerically.
I get: $C \approx 1.0265$ bits, with $p_1 \approx 0.055$
BTW: this channel can be seen a (non-binary) Z-channel.