Is second moment of exponential also Memory Less?

Question

I know that the exponential is memory less and that means that: $$ E[X\mid X>1]=1+E[X] $$ Now, does the memory less property also hold for the second moment? Specifically, is the following true? $$ E[X^2\mid X\geq1]=1+E[X^2] $$

I don't see why someone downvoted my question ... – Wajahat Apr 02 '15 at 00:00 — Wajahat, Apr 02 '15 at 00:00

score 3 · Accepted Answer · answered Apr 01 '15 at 23:55

3

Actually it should say $$ \operatorname{E}(X^2\mid X\ge 1) = \operatorname{E}((1+X)^2). $$ The conditional distribution of $X-a$ given that $X\ge a$ is the same as the marginal (i.e. unconditional) distribution of $X$. Hence the conditional distribution of $X$ given that $X\ge a$ is the same as the marginal distribution of $a+X$.

answered Apr 01 '15 at 23:55

Michael Hardy

1

Nice, exactly what I was looking for. – Wajahat Apr 01 '15 at 23:57

Is second moment of exponential also Memory Less?

1 Answers1