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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.

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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.

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You have obtained:$$\begin{align} f_{X+Y}(z) & = \frac{\operatorname d \mathsf P(X+Y\leq z)}{\operatorname d z} \\[1ex] & = \frac{\operatorname d }{\operatorname d z} \int_0^z \int_0^{z-x} f_X(x)\,f_Y(y)\operatorname d y\operatorname d x \\[1ex] & = \int_0^zf_X(x)\,f_Y(z-x)\operatorname d x \\[1ex] & =\ldots \end{align}$$ Similarly $$\begin{align} f_{\min(X,Y)}(z) & =\frac{\operatorname d 1-\mathsf P(X> z, Y> z)}{\operatorname d z} \\[1ex] & = \frac{\operatorname d \mathsf P(X\leq z)}{\operatorname d z}\cdot\mathsf P(Y> z)+ \mathsf P(X> z)\cdot\frac{\operatorname d \mathsf P(Y\leq z)}{\operatorname d z} \\[1ex] & = f_X(z)\int_z^\infty f_Y(y)\operatorname d y+ f_Y(z)\int_z^\infty f_X(x)\operatorname d x \\[1ex] & =\ldots \\[3ex] f_{1/Y^2}(z) & = \frac{\operatorname d \mathsf P(Y\leq \sqrt{1/z\;})}{\operatorname d z} \\[1ex] & = f_Y\left(\sqrt {1/z\,}\right)\;\left\lvert\frac{\operatorname d \sqrt{1/z\;}}{\operatorname d z}\right\rvert \\[1ex] & =\ldots \\[3ex] f_{e^{-2Y}}(z) & = \frac{\operatorname d \mathsf P(Y\geq -\tfrac 1 2\ln z)}{\operatorname d z} \\[1ex] & =\ldots \end{align}$$

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