If we have a Random variable X that is $$2e^{-2x}$$ and Y that is an exponential distribution with X as its parameter $\lambda$, then how do we go around solving this question? I thought that we should find its CDF first by using $$P[Xe^{-XY} < Y] $$ but then I run into the issue of how to get X on one side. I took the log $$P[Log(X)*-XY < log(Y)] $$ and I am lost on how we can continue next. Help would be appreciated. Thanks!
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What you wrote does not make sense.
You are given that
$$X\sim\text{Exp}(2)$$
$$(Y|X=x)\sim\text{Exp}(x)$$
Assuming that you are requested to find $f_Y(y)$ you have to use the following
$$\begin{align} f_Y(y) & =\int f_{XY}(x,y)dx\\ & = \int_0^{\infty}2x e^{-x(2+y)}dx\\ & =\frac{2}{(2+y)^2} \underbrace{\int_0^{\infty}(2+y)^2x e^{-x(2+y)}dx}_{=1}\\ & =\frac{2}{(2+y)^2}\mathbb{1}_{[0;\infty)}(y) \end{align} $$
The integral in the brackets is 1 as it is the integral of a $\text{Gamma}[2;(2+y)]$
tommik
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Now you surely know that
$$f(x,y)=f(x)f(y|x)$$
do the multiplication and you get the requested result.
– tommik Nov 18 '21 at 16:05