Conditional expectation of exponential distribution

Question

We were learning about the memoryless property of the exponential distribution, when I came across these seemingly easy problems:

If $X$ ~ $Exp(\lambda)$, find:

• $E[X|X>t]$

For this, I used the tail sum formula. $$P(X>s+t|X>t) = P(X>s)$$ $$\int_0^{\infty}P(X>s+t|X>t)ds = \int_0^{\infty}P(X>s)ds$$

I'm not too sure about this step, because I'm not entirely sure of tail sum expectation working for conditional expectations the way I've written it.

If it is correct, it gives the unsurprising result that $E[X|X>t] = E[X]$

• $E[X^2|X>t]$

This seemed much harder to tackle for me, since I couldn't easily use the memoryless properly as above. I feel like this should also maybe come out cleanly as $E[X^2]$, but I'm not sure how to go about this.

Thanks in advance!

Answer 1

$E(X|X>t)>t$ so we cannot have $E(X|X>t)=EX$ for all $t$. Memoryless property is not useful in evaluating this.

$E(X|X>t)=\frac {\int_t^{\infty} x\lambda e^{-\lambda x} dx} {\int_t^{\infty} \lambda e^{-\lambda x} dx}$ and $E(X^2|X>t)=\frac {\int_t^{\infty} x^2\lambda e^{-\lambda x} dx} {\int_t^{\infty} \lambda e^{-\lambda x} dx}$. Use integration by parts to evaluate the numertors. Both of these numbers depend on $t$.

Answer 2

The memoryless property works in deriving this conditional expectation like so:

$$\begin{align}\mathsf E(X\mid X>t)&=\int_0^\infty (t+s)\,f_X(t+s\mid X>t)\,\mathrm ds\\&=\int_0^\infty (t+s)\,f_X(s)\,\mathrm d s&&\star\\&=t\int_0^\infty f_X(s)\,\mathsf ds+\int_0^\infty s\,f_X(s)\,\mathrm d s\\&=t+\int_0^\infty\int_0^sf_X(s)\,\mathrm d u\,\mathrm d s\\&=t+\int_0^\infty \int_u^\infty f_X(s)\,\mathrm d s\,\mathrm d u\\&=t+\int_0^\infty \mathsf P(X>u)\,\mathrm du\\[2ex]\therefore\quad\mathsf E(X\mid X>t)&=t+\mathsf E(X)\end{align}$$