Given the joint pdf of random variables $X_1$ and $X_2$, I'm trying to find the pdf of $W=w_1X_1 + w_2X_2$ I'm given $f_{X_1,X_2}(x_1,x_2)=e^{(-x_1-x_2)}$ where $x_1>0$ and $x_2>0$.
Also, $w_1$ and $w_2$ are both constants greater than zero.
I'm trying to first determine the cdf of $W$ for which I have
$$F_W(w)=P(W \leq w) = P(w_1X_1 + w_2X_2 \leq w)=P(X_1 \leq \frac{w-w_2X_2}{w_1})$$
I think my problem is choosing the limits of integration. I tried $0 < X_1 < \frac{w-w_2X_2}{w_1}$ and $0<X_2<\infty$ which seems to make the integral go to infinity. I'm trying to play with some options but any idea I have doesn't make sense. Can anyone point me in the right direction? 
 A: First let us compute the marginal distribution 
$$f_{X_1}(x_1) = \int_0^{+\infty} f_{X_1X_2}(x_1,x_2)dx_2 = e^{-x_1}\int_0^{+\infty} e^{-x_2}dx_2= e^{-x_1}  $$
$$f_{X_2}(x_2) = \int_0^{+\infty} f_{X_1X_2}(x_1,x_2)dx_1 = e^{-x_2}\int_0^{+\infty} e^{-x_1}dx_1= e^{-x_2}  $$
Let $Y_2=w_2X_2$ then $$f_{Y_2}(y)= f_{X_2}(\frac{y_2}{w_2}) =  e^{-\frac{y_2}{w_2}} $$
$Y_1=w_1X_1$ then $$f_{Y_1}(y)= f_{X_1}(\frac{y_1}{w_1}) =  e^{-\frac{y_1}{w_1}} $$
The pdf of $$W=Y_1+Y_2$$ is then by taking convolution 
$$f_W(w) = \int f_{Y_1}(y_1) f_{Y_2}(w-y_1) dy_1= \int e^{-\frac{y_1}{w_1}}e^{-\frac{w-y_1}{w_2}}dy_1$$
A: You've already found that $\Pr(W\le w) = \Pr\left( X_1\le \dfrac{w-w_2 X_2}{w_1} \right)$.  Then you can say
\begin{align}
\Pr\left( X_1\le \dfrac{w-w_2 X_2}{w_1} \right) & = \operatorname{E}\left( \Pr\left( X_1\le\frac{w-w_2 X_2}{w_1} \mid X_2\right) \right) \\[8pt]
& = \operatorname{E} \left( \left.\begin{cases} 1 - e^{-(w-w_2 X_2)/w_1} & \text{if }w-w_2 X_2 > 0, \\ 0 & \text{if }w-w_2X_2\le 0, \end{cases}\right\} \right) \\[8pt]
& = \int_0^{w/w_2} (1 - e^{-(w-w_2 x_2)/w_1}) e^{-x_2}\, dx_2.
\end{align}
Then the integral can be simplified.
