conditional probability p(x|y) if y = x + e

Question

Assuming x, y, $\epsilon$ are variables, and $y = x + \epsilon$, then:

$$p(x=s|y=t) = p(t-s) = p(\epsilon) $$

Is that exist? It seems true but still don't know how to prove it formally. Can someone help me with the proof? Thx!

Answer 1

That notational abbreviation is atrocious and should be discouraged whenever you encounter it.

It causes more confusion than its savings on typesetting is worth.

Hopefully one of these notations make the derivation clearer.

$$\begin{align}&p_{x\mid y}(s\mid t)&\qquad&=\mathsf P(x=s\mid y=t)\\=~&p_{y-\epsilon\mid y}(s\mid t)&&=\mathsf P(y-\epsilon=s\mid y=t)\\=~&p_{\epsilon\mid y}(s{-}t\mid t)&&=\mathsf P(\epsilon =s-t\mid y=t)\\=~&p_\epsilon(s-t)&&=\mathsf P(\epsilon=s-t)\end{align}$$

First we change the variable, substituting $x\gets y-\epsilon$.   Then note $\{\epsilon=s-y\}=\{\epsilon=s-t\}$ under the condition that $y=t$.   Finally $y$ and $\epsilon$ are independent.