$E[X^2|X>1] = E[(X+1)^2]$
By the memoryless property, the conditional distribution of $X$, given that $X>1$, is the same as the unconditional distribution of $X+1$. Therefore, the equation stands.
I'm trying to prove $$P(X|X>1) = P(X+1)$$ as given in the statement. However, the equation doesn't come as true according to my calculation. Please let me know where my mistake is. Here is my calculation:
$\begin{align}P(X|X>1) &= \frac{P(X=k)}{P(X>1)} , \quad k>1\\
&= \frac{\lambda e^{-\lambda k}}{e^{-\lambda}}\\
&= \lambda e^{-\lambda k + 1},\quad k>1
\end{align}
$
$\begin{align}P(X+1) &= \lambda e^{-\lambda (x+1)}\\
&= \lambda e^{-\lambda x – \lambda }
\end{align}
$
Best Answer
What we really want to prove is, for $k\ge1$ (since it is trivial when $k<1$), $$P(X=k|X>1) = P(X+1=k)$$ which expands into $$\frac{\lambda e^{-\lambda k}}{e^{-\lambda}}=\lambda e^{-\lambda(k-1)}$$ Simplifying the LHS leads to the RHS, proving the identity: $$\frac{\lambda e^{-\lambda k}}{e^{-\lambda}}=\lambda e^{-\lambda k-\lambda}=e^{-\lambda(k-1)}$$