Calculating the marginal probability $f(x_1,x_2)=e^{-x_1}e^{-x_2}$ for $0<x_1<x_2<\infty$

Question

How would you calculate the marginal probability of the following pdf. We have the the joint pdf of two variables $X_1,X_2$

$f(x_1,x_2)=2e^{-x_1}e^{-x_2}$ for $0<x_1<x_2<\infty,0 elsewhere$

So first we find the marginal probability of $x_1$ I got $f(x_1)=\int_{x_1}^{\infty}2e^{-x_1}e^{-x_2}dx_2=2e^{-2x_1}$

we then find the proabity of $f(x_2)=\int_{0}^{x_2}2e^{-x_1}e^{-x_2}dx_1=2e^{-x_2}-2e^{-2x_2}$

And I do not think the variables $X_1$ AND $X_2$ are the indepedent because $f(x_1,x_2)\neq f(x_1)f(x_2)$

Answer 1

Your calculated marginal PDFs are correct; there's no independence requirement to calculate them. As you found, the marginal PDF of $X_1$ is $\int_{x_1}^\infty f(x_1,\,x_2)dx_2=2e^{-2x_1}$, whereas the marginal PDF of $X_2$ is $\int_0^{x_2}f(x_1,\,x_2)dx_1=2e^{-x_2}(1-e^{-x_2})$. The product of these, $4e^{-2x_1-x_2}(1-e^{-x_2})$, differs from $f$ because, again as you noted, the variables aren't independent.

Answer 2

As defined $X_1 \lt X_2$, so they aren't independent.