Memoryless property of Exponential at $X^2$ > Without computations, find an unconditioned expression for $E[X^2|X>1]$ I have tried following the same steps of this problem: Memoryless property of the exponential distribution By the memory-less property of the exponential. Given $X>1 \Rightarrow X^2 > 1$ and the memory-less property says that $X^2 - 1 > 0$ has this same distribution, then $$ E[X^2|X>1] = E[X^2-1|X^2>1] $$ $$ = E[X^2|X^2>1] - E[1|X^2>1] $$ $$ = E[X^2] - 1 $$ What is wrong with my logic please?

Assuming that $X\sim \exp(\lambda)$ and using $Var(X) = EX^2 - E^2X$ we get $$ E(X^2|X>1) = Var(X|X>1)+E^2(X|X>1) = \frac{1}{\lambda^2} + (1 + \frac{1}{\lambda})^2 $$ where the memory-less property helps to deduce the shift in the expected value and that the variance remains the same.

I guess that your mistake stems from the fact that $X^2$ and $X$ are not the same r.v (different distributions).