Let X and Y be independent random variables with probability density functions
$$f_X(x) = e^{-x} , x>0$$
$$f_Y(y) = 2e^{-2y} , y>0$$
Derive the PDF of $Z_1 = X + Y$
other cases: $Z =min(X,Y)$ , $Z =1/Y^2 $ , $Z =e^{-2y} $
Just considering the 1st part, I understand to go from the fact that
$P(X + Y<= z)$ then $$\int_{0}^{z} f_{xy}(z-y,y) dx $$ since they are independent I integrate $f_x(z-y) f_y (y)$ wrt y. I come to $-e^{-z}$
This is as much as I have managed to pick up, but still very much unsure.
What would I need to look out for in the other cases as for $min(X,Y)$ id have no idea of how to start.
Best Answer
It appears that \begin{align*} P(X+Y<z)&=\int_0^\infty\int_0^{-y+z} f_X(x)f_Y(y)\,dxdy\tag 1\\ &=\int_0^\infty\int_0^{-y+z} e^{-x}\cdot2e^{-2y}\,dxdy\\ &=\int_0^\infty 2e^{-2y}\left(1-e^{-(-y+z)}\right)\,dy\\ &=\int_0^\infty 2e^{-2y}\,dy-2e^{-z}\int_0^\infty e^{-y}\,dy\\ &=1-2e^{-z} \end{align*} where $(1)$ is true by independence.
As for $Z = \min\{X,Y\}$, I would go after the cdf \begin{align*} P(Z\leq z) &= 1-P(Z>z)\\ &=1-P(X> z, Y> z)\\ &=1-P(X> z)P(Y> z)\tag2\\ \end{align*} where $(2)$ is true by independence.