Find $\mathrm E \eta$ where $\eta = min(\xi_1^3, \xi_2^2)$ is random variable I have a probability problem about random variables.
Need to find $\mathrm E\eta$. $\eta = min(\xi_1^3, \xi_2^2)$. 
$\xi_1$ and $\xi_2$ are independent random variables and have a exponential distribution with parameter 1.
Could you help me please? 
Also, how to solve this kind of problems when I need to find expected value of several random variables? Is there a great way to do it?
 A: Notation: $a \wedge b := \min\{a ,b\}$.
Since $\xi_1,\xi_2$ are independent,
$$
P( \xi_1^3 \wedge \xi_2^2 > t) = P( \xi_1^3 > t, \xi_2^2 > t) = P(\xi_1^3 > t) P(\xi_2^2 > t) = P(\xi_1 > t^{1/3}) P(\xi_2 > t^{1/2})
$$
$$
= e^{-t^{1/3}} e^{-t^{1/2}} = e^{-(t^{1/3}+t^{1/2})}
$$
Then use that $$E\eta = \int_0^\infty P(\eta\geq t)dt =\int_0^\infty e^{-(t^{1/3}+t^{1/2})}dt. $$
You can now use a computer to numerically approximate.
A: There are several ways to proceed. On the plus side, one can derive the exact symbolic answer quite easily with a computer algebra system. On the downside, the solution to the expectation does not appear to have a nice compact form. To illustrate, if $\xi_1$ and $\xi_2$ are independent standard Exponentials, then the joint pdf of $(\xi_1, \xi_2)$ is:

We seek:  $E[min(\xi_1^3, \xi_2^2)]$

All done. To 5 decimal places, the latter is:   0.55248
Notes


*

*The Expect function used above is from the mathStatica package for Mathematica. As disclosure, I should add that I am one of the authors.

*For more detail on the special functions in the output, see: AiryBi andHypergeometricPFQ
