My question is based on this post. In the question, $X$ and $Y$ are independent random variables that follow a poisson distribution, and $Z = X + Y$. This is the math in question: \begin{align} P(Y=y\mid X+Y=z) &= \frac{P(Y=y \text{ and }X+Y=z)}{P(X+Y=z)} \\[1ex] &= \frac{P(Y=y \text{ and }Y=z-x)}{P(X+Y=z)} \\[1ex] &= \frac{P(Y=y)\times P(Y=z-x)}{P(X+Y=z)} && \text{because, independence} \end{align}
I have two questions:
(1) $P(Y=y \text{ and }X+Y=z)$ in the numerator was changed to $P(Y=y \text{ and }Y=z-x)$. I'm not sure how one can change $X$ to $x$, since I would imagine $P(Y=y \text{ and }Y=z-X)$ is the right step, since $X$ is a random variable, not a constant like $x$. Is that a typo?
(2) If what I said above is correct, are we claiming independence because of the memoryless nature of the Poisson distribution?
I think it should be
\begin{align} P(Y=y\mid X+Y=z) &= \frac{P(Y=y \text{ and }X+Y=z)}{P(X+Y=z)} \\[1ex] &= \frac{P(Y=y \text{ and }X=z-y)}{P(X+Y=z)} \\[1ex] &= \frac{P(Y=y)\times P(X=z-y)}{P(X+Y=z)} && \text{because, independence} \end{align}
They are independent because we are told that $X$ and $Y$ are independent.