# Reasoning about change of beginning index in summation for probability problem

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.

• You do not have $y=x$. $y$ is an index, and its values effectively start from $y=x$ because the terms in the sum corresponding to $y<x$ are zero. Commented May 1, 2020 at 18:44
• @GReyes Oh, yes, you're right. I seem to have confused myself. Thank you for the clarification. Commented May 1, 2020 at 18:46

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*}

• I don't see how your answer is even addressing my question. We already know that $P(X = x \vert Y = y) = (y + 1)^{-1}$; it's the summation that is of concern. Commented May 1, 2020 at 19:49
• @ThePointer Does the modification to the answer helps? Or I'm not following your question. Correct me If I'm wrong, but you asked why the summation starts at $y=x$. As far as I understand, my answer is justifying, with the aid of the indicator function, why that is the case.
– RLC
Commented May 1, 2020 at 20:04
• Oh, yes, my apologies; it wasn't clear to me what you were saying before the edit. Thank you for the clarification. Commented May 1, 2020 at 20:07
• One last thing: Do you know how the author concluded that $\sum\limits_{z = x + 1}^\infty P(Y = x + 1) = P(Y \ge x + 1)$ at the end? It isn't evident to me. Commented May 1, 2020 at 20:22
• @ThePointer Yes. I edited the answer. Note that there is a typo in your formulation: it should be $P(Y=z)$ in the summation.
– RLC
Commented May 1, 2020 at 21:21