For independent random variables X ∼ Exp(1) and Y ∼ Exp(2), find the density of the random variable Z = X + Y .
My work:
For any exponential distribution with parameter $\lambda$ the function is
$f(x) = \lambda e^{-\lambda x}$
$f_x(x) = e^{-x}$
$f_y(x) = 2e^{-2y}$
Therefore the joint density function is: $f_{xy}(x) = 2e^{-x-2y} , x,y\ge 0$ and $0$ otherwise
After this I do not know how to $f_{X+Y}$
Best Answer
You can use convolution and get $$f_Z(z) = \int_0^z f_X(x)f_{Y}(-x+z)|1|\,dx.$$
I suspect that you might rather compute the cdf of $Z$ $$P(Z\leq z) = P(X+Y\leq z).$$