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:
The law of total probability gives that
$$P(X = x) = \sum_{y = 0}^\infty P(X = x \vert Y = y) P(Y = y).$$
The conditional PMF of $X$ given $Y = y$ is non-zero only when $y \ge x \ge 0$. Therefore, for $x \ge 0$, we have that
$$\begin{align} P(X = x) &= \sum_{y = x}^\infty (y + 1)^{-1} e^{-1} \dfrac{1^y}{y!} \\ &= \sum_{y = x}^\infty e^{-1} \dfrac{1}{(y + 1)!} \\ &= \sum_{z = x + 1}^\infty e^{-1} \dfrac{1}{z!} \\ &= \sum_{z = x + 1}^\infty P(Y = x + 1) \\ &= P(Y \ge x + 1). \end{align}$$
I do not understand why the sum beginning index is switched to $y = x$. I realise that the author reasons that the conditional PMF of $X$ given $Y = y$ is non-zero only when $y \ge x \ge 0$, but I do not see how this allows us to say definitively that we must necessarily have that $y = x$.
I would greatly appreciate it if people would please take the time to explain this.
The mass function of $X$ given $Y=y$ is $$ P(X=x|Y=y)=\dfrac{1}{y+1}\mathbf{1}_{\{0,1,...,y\}}(x),\quad y\geq 0, $$ where $\mathbf{1}_{A}(x)=1$ whenever $x\in A$ and $\mathbf{1}_{A}(x)=0$ whenver $x\notin A$. Therefore, even if the sumation starts at $y=0$, for all values of $y$ such that $x> y$ we have that $\mathbf{1}_{\{0,1,...,y\}}(x)=0$. And when $x\leq y$, $\mathbf{1}_{\{0,1,...,y\}}(x)=1$, so you don't put the indicator:
\begin{align*} P(X=x)&=\sum_{y=0}^{\infty}P(X=x|Y=y)P(Y=y)\\ &=\sum_{y=0}^{\infty}\dfrac{1}{y+1}\mathbf{1}_{\{0,1,...,y\}}(x)P(Y=y)\\ &=\sum_{y=0}^{x-1}\dfrac{1}{y+1}\mathbf{1}_{\{0,1,...,y\}}(x)P(Y=y)\\ &\quad+\sum_{y=x}^{\infty}\dfrac{1}{y+1}\mathbf{1}_{\{0,1,...,y\}}(x)P(Y=y)\\ &=\sum_{y=0}^{x-1}(0)\dfrac{1}{y+1}P(Y=y)+\sum_{y=x}^{\infty}(1)\dfrac{1}{y+1}P(Y=y)\\ &=\sum_{y=x}^{\infty}\dfrac{1}{y+1}P(Y=y) \end{align*}
For the last equality: since $(Y=x+1),(Y=x+2),...$ are disjunt events and $(Y\geq x+1) = (Y=x+1)\cup(Y=x+2)\cup\cdots$, then \begin{align*} \sum_{z=x+1}^{\infty}P(Y=z)=P(\cup_{z=x+1}^{\infty}(Y= z))=P(Y\geq x+1). \end{align*}