Here is my problem. Consider four independent exponential distributions $X^A_1$, $X^B_1$, $X^A_2$, $X^B_2$ where $X^A_1$ and $X^B_1$ are $\exp(\lambda_1)$ and $X^A_2$ and $X^B_2$ are $\exp(\lambda_2)$.
There is another random variable $\mu$ where $\mu=\mu^G$ when $X_1=X^A_1+X^B_1 < X_2=X^A_2+X^B_2$ and $\mu=\mu^B$ otherwise.
In this setup, I'd like to calculate $E[e^{-rX_1}\mu]$.
The approach I've taken is use
$E[e^{-rX}\mu]=E[e^{-rX}\mu^G|X_1<X_2]P(X_1<X_2)+E[e^{-rX}\mu^B|X_1>X_2]P(X_1>X_2)$,
and since $X_1$ and $X_2$ follow gamma distribution with (2,$\lambda_1$) and (2,$\lambda_2$), respectively, I calculated the density function $f_Y(y)$ where $Y=X_1-X_2$. And it is easy to show
$$f_{{X_1},Y}(x_1,y)=f_{X_1}(x_1)f_{X_2}(x_1-y),$$
and from this point, I obtained the conditional density $f_{X_1}(x_1|Y=y)$ and tried to calculated the conditional expectation.
But, I ended up with having a complicated form in the integrand when calculating the conditional expectation $E[e^{-rX}\mu^G|X_1<X_2]$, and probably I could proceed further, but I'd like to ask you if there is an easier way to get $E[e^{-rX}\mu]$ without calculating all the density functions. Thank you very much!