# Mean of a function of exponential distribution

Let $X$ a random variable exponentially distributed, compute: $$E[ {1}_{\{X>t\}}X]$$ I think that it can be computed as function of a random variable, but my answer was different of answer given. I Think that:

$$E[ {1}_{\{X>t\}}X]=\int_t^\infty x\lambda e^{-\lambda x} dx$$

• Please show us your calculations, then we can tell you where you have made a mistake. – Dominik Dec 5 '16 at 10:02
• What was the result of your calculation, and what was the answer given? – Henry Dec 5 '16 at 10:09
• So far, your equation is correct. How did you evaluate the integral? – Dominik Dec 5 '16 at 10:09
• I found my error, I was computing incorrectly the integral. Thanks. – A P Dec 5 '16 at 10:17

Probably you have figured it out. But to summarize the unanswered question, $$E[ {1}_{\{X>t\}}X]=\int_t^\infty x\lambda e^{-\lambda x} dx$$ Integrate by parts or do as follows:
\begin{align} \int_t^\infty x\lambda e^{-\lambda x} dx&=-\lambda\int_t^\infty \frac{d}{d\lambda}\left( e^{-\lambda x}\right) dx\\ &=-\lambda\frac{d}{d\lambda}\left(\int_t^\infty e^{-\lambda x}dx\right) \\ &=-\lambda\frac{d}{d\lambda}\left(\frac{e^{-\lambda x}}{-\lambda}\biggr\vert^{\infty}_t\right)\\ &=-\lambda\frac{d}{d\lambda}\left(\frac{e^{-\lambda t}}{\lambda} \right)\\ &=-\lambda\frac{-\lambda te^{-\lambda t}-e^{-\lambda t}}{\lambda^2}\\ &=\frac{(1+\lambda t)e^{-\lambda t}}{\lambda} \end{align}