# Special case of channel garbling/degradation

Let $$P(X,Y)$$ be a finite state distribution. Let’s build the Markov chain $$X – Y – Y^1$$ where $$Y^1$$ is a cover of $$Y$$ featuring only non-singleton subsets of $$Y$$ with probability $$P(X, Y, Y^1)$$ defined as
$$begin{equation} p(x, y, y^1) := begin{cases} frac{p(x,y^1)}{r_y} & text{if } y in y^1 \ 0 & text{otherwise} end{cases} end{equation}$$

I am finding that $$P(Y^1 mid Y)$$ is a degradation of $$P(X mid Y)$$, i.e. that
$$p(y^1 mid y) = sum_x p(y^1 mid x) cdot p(x mid y) quad text{for all} quad y in Y quad text{and} quad y^1 in Y^1$$
This result is confusing me, since we are not dealing with a chain $$Y – X – Y^1$$. The proof that I have drafted is below. Is the proof correct, or am I missing something?

Proof

From the relationship
$$frac{p(x, y)}{p(y)} = frac{p(x, y mid y^1)}{p(y mid y^1)},$$
we have
$$begin{equation} p(y^1 mid y) = frac{p(x,y,y^1)}{p(x, y)} = begin{cases} frac{1}{r^1_y} & text{if } y in y^1 \ 0 & text{otherwise} end{cases} end{equation}$$
Furthermore we have
$$sum_{x} p(y^1 mid x) cdot p(x mid y) = sum_{x} frac{p(x,y^1) cdot p(x, y)}{p(x) cdot p(y)} = begin{cases} sum_{x} frac{p(x) cdot p(x, y)}{r^1_y cdot p(x) cdot p(y)} = frac{1}{r^1_y} & text{if } y in y^1 \ 0 & text{otherwise} end{cases}$$