# 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.

$$\ldots$$ 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.