Conditional expectation of exponential random variable

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

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