I have the following problem:
Suppose that $Y \sim \text{Poi}(1)$. Let $y \ge 0$ be an integer. Given that $Y = y$, suppose that $X$ has the discrete uniform law on $\{ 0, 1, \dots, y \}$.
(a) Show that $E[X \vert Y] = \dfrac{1}{2}Y$, and use this to find $E[X]$.
The provided solution is as follows:
We are given that $P(X = x \vert Y = y) = (y + 1)^{-1}$ for $x \in \{ 0, 1, \dots, y \}$, and $P(X = x \vert Y = y) = 0$ otherwise.
(a) From the conditional PMF, we can compute the conditional expectation of $X$ given $Y = y$:
$$\begin{align} E[X \vert Y = y] &= \sum_{x = 0}^y xP(X = x \vert Y = y) \\ &= (y + 1)^{-1} \sum_{x = 0}^y x \\ &= (y + 1)^{-1} \left[ \dfrac{y(y + 1)}{2} \right] \\ &= \dfrac{y}{2}. \end{align}$$
Therefore, $E[X \vert Y] = \dfrac{Y}{2}$. Now, since $Y \sim \text{Poi}(1)$, we know that $E[Y] = 1$. We now compute $E[X]$ using the law of total expectation:
$$\begin{align} E[X] &= E[E[X \vert Y]] \\ &= E \left[ \dfrac{Y}{2} \right] \\ &= \dfrac{1}{2} E[Y] \\ &= \left( \dfrac{1}{2} \right) (1) \\ &= \dfrac{1}{2}. \end{align}$$
I don't understand where the conclusion that $P(X = x \vert Y = y) = (y + 1)^{-1}$ at the immediate beginning came from. I would greatly appreciate it if someone would please take the time to clarify this.
Casually, for a moment we could forget that $y$ is a variable and pretend that it now serves as a given constant, so that the sample space of $X$ has exactly $y+1$ points 'deterministically'. Thus, the pmf follows from there.