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}

$$