Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

1 Answer

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 $$ \mathrm E(X)=\iint_{x\geqslant0,x_i\geqslant0}\frac{x(1+x_1+\cdots+x_N)}{c+x(\alpha+\beta(x_1+\cdots+x_N))}\mathrm e^{-x-x_1-\cdots-x_N}\mathrm dx\mathrm dx_1\cdots\mathrm dx_N. $$

share|improve this answer
    
You may want to check your notation again. –  wolfies May 2 '13 at 13:02
    
@wolfies What are you talking about? –  Did May 2 '13 at 14:16
1  
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
    
Notwithstanding the confusion created through your use of internally inconsistent notation, perhaps you could explain what value it contributes to actually solving the problem, other than writing out the mathematical formula for an expectation, and/or expressing a Gamma variate as the sum of n Exponentials?? –  wolfies May 2 '13 at 18:41
show 1 more comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.