# Expected Value of function of two random variable

Assume $X_1$ is an Exponential random variable with unit mean ( i.e. $f_{X_1}(x) = e^{-x}$ ) and $X_2$ is an Erlang distribution with shape $N$ and unit rate ( i.e. $f_{X_2}(x) = \frac{x^{N-1}e^{-x}}{(N-1)!}$ ).

I need to compute the expected value of $X=\frac{X_1+X_1X_2}{c+\alpha X_1+\beta X_1X_2}$ where $c$, $\alpha$ and $\beta$ are larger than zero.

I tried to find the expectation by calculating $f_X(x)$ and then $\int_0^\infty xf_X(x)dx$, but I could not calculate integral.

I also tried $E(X)=\int_0^\infty P[X>x] dx$, but again failed calculating $E(X)$.

I am not even sure if $E(X)$ is available in closed form at all or not. I would appreciate if one could help me.

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Are $X_1,X_2$ independent ? –  vanna Jul 30 '12 at 12:30
sorry for incomplete question. yes, $X_1$ and $X_2$ are independent. –  Tural Jul 30 '12 at 13:05
I can get a closed form solution using mathStatica / mma if $c = 0$. If that would help, please let me know and I will post it up. –  wolfies May 2 '13 at 13:00
By definition, $$\mathrm E(X)=\iint_{x\geqslant0,z\geqslant0}\frac{x(1+z)}{c+x(\alpha+\beta z)}\mathrm e^{-x}\frac{z^{N-1}}{(N-1)!}\mathrm e^{-z}\mathrm dx\mathrm dz,$$ which is equivalent to $$\mathrm E(X)=\iint_{x\geqslant0,x_i\geqslant0}\frac{x(1+x_1+\cdots+x_N)}{c+x(\alpha+\beta(x_1+\cdots+x_N))}\mathrm e^{-x-x_1-\cdots-x_N}\mathrm dx\mathrm dx_1\cdots\mathrm dx_N.$$
The OP defines $X$ to be a function of random variables $X_1$ and $X_2$. Your posting expresses $E(X)$ in terms of random variables $X$ and $Z$, which is confusing given that your lower case $x$ on the RHS (which is the OP's Exponential(1) variate) clashes with the very different $X$ on the LHS. –  wolfies May 2 '13 at 16:20
Your posting expresses E(X) in terms of random variables X and Z... No it does not, period. Furthermore, if you think that $x$ "clashes with" $X$, well... this is your problem. (As an aside, let me note that, although I can well understand your desire to promote and advertise a software dear to your heart, you should not feel necessary to invent nonexistent problems in others' answers (neither to downvote these) to promote said software.) –  Did May 2 '13 at 18:22