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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?

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Sunhwa
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    $X|X>1\overset{d}= X+1$, hence $E=E(X+1)^2=E(X^2)+2E(X)+1=\sigma^2+(\mu+1)^2$. – A.S. Jan 25 '16 at 01:20
  • I see, this is clear thank you @A.S.! if you had posted this as an answer I would have accepted it first(: – Sunhwa Jan 25 '16 at 04:07

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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).

V. Vancak
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  • Thank you @user230329 , I can tell what you did here, but because of what A.S. also posted .. there were two things I was misunderstanding~ the $X^2$ as a different r.v. as you said, and $E[X^2|X>1] = E[(X+1)^2]$ – Sunhwa Jan 25 '16 at 04:12