I need to find the expected value of this distribution. I think I just need to integrate the function from $0$ to $\infty$. Having trouble doing this integral though.
$$f(x; \mu) = \frac{1}{\mu} \cdot e^{− x/\mu },\ 0 ≤ x < \infty,\ \mu > 0$$
-1
$\begingroup$
$\endgroup$
-
$\begingroup$ What have you tried? Can you write said integral down? As a hint, integration by parts will help you with the integral you need. $\endgroup$ – AlkaKadri Oct 8 '18 at 23:07
1
$\begingroup$
$\endgroup$
The expected value is $$ \int_0^\infty \frac{x}{\mu} e^{-x/\mu}\ dx. $$
By the change of variables $x/\mu = y$, $dx = \mu\ dy$, and integrating by part we get $$ \int_0^\infty \frac{x}{\mu} e^{-x/\mu}\ dx = \mu \int_0^\infty y e^{-y}\ dx = \mu [-ye^{-y}]_0^{\infty} + \mu \int_0^\infty e^{-y}\ dx = \mu [-e^{-y}]_0^\infty = \mu. $$
