For a random variable $X\sim \text{Exp}(\lambda)$ ($\mathbb{E}[X] = \frac{1}{\lambda}$) I feel intuitively that $\mathbb{E}[X|X > x]$ should equal $x + \mathbb{E}[X]$ since by the memoryless property the distribution of $X|X > x$ is the same as that of $X$ but shifted to the right by $x$.

However, I’m struggling to use the memoryless property to give a concrete proof. Any help is much appreciated.

Thanks.

**Answer**

$\ldots$ by the memoryless property the distribution of $X|X > x$ is the same as that of $X$ but shifted to the right by $x$.

Let $f_X(t)$ denote the probability density function (pdf) of $X$. Then, the mathematical

formulation for what you correctly

state $-$ namely,

the *conditional* pdf of $X$ given that $\{X > x\}$ is the same as that of

$X$ but shifted to the right by $x$ $-$ is that $f_{X \mid X > x}(t) = f_X(t-x)$.

Hence, $E[X\mid X > x]$, the *expected value* of $X$ given that $\{X > x\}$ is

$$\begin{align}

E[X\mid X > x] &= \int_{-\infty}^\infty t f_{X \mid X > x}(t)\,\mathrm dt\\

&= \int_{-\infty}^\infty t f_X(t-x)\,\mathrm dt\\

&= \int_{-\infty}^\infty (x+u) f_X(u)\,\mathrm du

&\scriptstyle{\text{on substituting}~u = t-x}\\

&= x + E[X].

\end{align}$$

Note that we have not explicitly used the density of $X$ in the calculation,

and don’t even need to integrate *explicitly* if we simply remember that

(i) the area under a pdf is $1$ and (ii) the definition of expected value of a continuous random variable in terms of its pdf.

**Attribution***Source : Link , Question Author : mchen , Answer Author : Dilip Sarwate*