a vague step in the proof of variance of geometric distribution

Question

I've been reading a probability textbook recently and get stuck by the following steps in a proof of variance of geometric random variable, the idea is to use formula:$Var(X)=E[X^2]−E[X]^2$, $E[X]$ on the right hand side is easy to calculate with a divide and conquer method, so we just need to find $E[X^2]$. \begin{equation}E[X^2]=E[X^2|X=1]P(X=1)+E[X^2|X>1]P(X>1)\end{equation} note that X is the geometric random variable. This first step is just the same divide and conquer method as the one we used to find $E[X]$, not hard to understand, and the first item on the right hand side equals $1*p$. But the author then says $E[X^2|X>1] = E[(X+1)^2]$ and I don't understand why. I found one explanation said: \begin{equation}E[X^2|X>1]=E[(X−1)^2+2(X−1)+1)|X−1>0]\end{equation} and then \begin{equation}E[X^2|X>1]=E[(X−1)^2|X−1>0]+2E[(X−1)|X−1>0]+1\end{equation} then \begin{equation}E[X^2|X>1]=E[X^2]+2E[X]+1\end{equation} but I still can't convince myself the last step above is right. Is this some sort of "change of variable" method? How does it actually work in this case? Someone please explain it in detail, maybe with some step by step proof, thanks in advance.

Answer 1

The explanation you have found is essentially using the memorylessness property of the geometric distribution, specifically $$\Pr[X > x + s \mid X \ge s] = \Pr[X > x].$$

What is happening is that if we define the conditional random variable $$Y = X - 1 \mid X > 1 = \begin{cases} X - 1, & X > 1 \\ \text{undefined}, & X \le 1 \end{cases}$$ where $X \sim \operatorname{Geometric}(p)$, then $Y$ will have the same geometric distribution, namely $Y \sim \operatorname{Geometric}(p)$. This is because for positive integers $y$, $$\Pr[Y = y] = \Pr[X - 1 = y \mid X > 1] = \frac{\Pr[X = y + 1]}{\Pr[X > 1]} = \frac{(1-p)^{(y+1)-1} p}{1 - p} = (1-p)^{y-1} p.$$