# 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.

• Hint: $f_{X|X > a}(x) = f_X(x-a)$ is the mathematical expression corresponding to "shifted to the right by $a$", and so $$E[X\mid X>a] = \int_{-\infty}^\infty xf_{X\mid X>a}(x)\,\mathrm dx= \int_{-\infty}^\infty xf_X(x-a)\,\mathrm dx.$$ Now do a change of variables on the integral on the right. Jan 25, 2013 at 17:19
• Note that $X|X>x$ is a truncated distribution truncated below "$x$".Specially it is shifted exponential distribution and shifted exponential does not have memoryless property.
– A.D
Jan 25, 2013 at 18:53

$$\ldots$$ 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.
For $x>0$, the event $\{X>x\}$ has probability $P\{X>x\}=1-F_X(x)=e^{-\lambda x} > 0$. Hence, $$\newcommand{\E}{\mathbb{E}} \E[X\mid X> x] = \frac{\E[X\,I_{\{X>x\}}]}{P\{X>x\}} \, ,$$ but $$\E[X\,I_{\{X>x\}}] = \int_x^\infty t\,\lambda\,e^{-\lambda t}\,dt = (*)$$ (using Feynman's trick, vindicated by the Dominated Convergence Theorem, because it is fun) $$(*) = -\lambda \int_x^\infty \frac{d}{d\lambda}(e^{-\lambda t}) \,dt = -\lambda \frac{d}{d\lambda} \int_x^\infty e^{-\lambda t} \,dt$$ $$= -\lambda \frac{d}{d\lambda} \left(\frac{1}{\lambda} \int_x^\infty \lambda\,e^{-\lambda t} \,dt\right) = -\lambda\frac{d}{d\lambda}\left(\frac{1}{\lambda}\left(1 - F_X(x)\right)\right)$$ $$= -\lambda\frac{d}{d\lambda}\left(\frac{e^{-\lambda x}}{\lambda}\right) = \left(\frac{1}{\lambda}+x\right)e^{-\lambda x} \, ,$$ which gives the desired result $$\E[X\mid X>x] = \frac{1}{\lambda}+x = \E[X] + x \, .$$
• Although the use of Feynman's trick is interesting, why not just integrate by parts to get $$\int_x^\infty t\lambda e^{-\lambda t}\,dt = -t e^{-\lambda t}\biggr |_x^\infty + \int_x^\infty e^{-\lambda t}\,dt = \left(x + \frac{1}{\lambda}\right)e^{-\lambda x}?$$ Mar 23, 2013 at 14:13