Joint pdf of $f(x,y)=e^{-(x+y)}$

Question

I am asked to find the joint pdf of $\ f(x,y)=e^{-(x+y)} $ where $x,y$ are between $0$ and $\infty$ or $0$ otherwise.

I can see its an exponential distribution, which means its continuous so I started a double integral like this:

$\displaystyle\int_{0}^{\infty} \int_{0}^{\infty}e^{-x}e^{-y} dy dx$

I tried substituting in $u=e^{-x}$ and $du= -e^{-x} $

$v=e^{-y}$ and $\int v= \int e^{-y}dy=-e^{-y} $

My result after substituting y for 0 and $\infty$ is $\int_{0}^{\infty} e^{-x}-e^{-x}dx$ = $-e^{-x}-e^{-x}$ and with limits applied I get $2$. I can see that the answer should be EXP(1) since the exponential pdf is $\lambda e^{-\lambda x} $ so for my example $\ e^{-(x+y)} $ it must be $1$. (EXP(1) is the answer in the back of the text book for the marginal pdfs). Am I on the right track with my integration please or am I doing all of this wrong? Thanks.

Answer 1

Try the fact that $$\int_0^\infty \int_0^\infty e^{(-x-y)}dxdy=\int_0^\infty\left(\int_{-\infty}^{-y} e^{u}(du)\right)dy=\int_0^\infty e^{-y}dy=1$$

here letting $u=-x-y$ and $du=-dx$ (why?). But I am not sure if this is what you need to find for your problem in general, as prior commenters have pointed out.

Answer 2

For the marginal pdfs, note that you split up $e^{-(x+y)}$ into the product of a function of $x$ and a function of $y$. Or rather, note that it was possible to do this. That says something about whether or not the two variables are independent. What does it say? What then can we say about the marginal pdfs?

As for the joint cdf, I think there's some sort of, uh, formula the books generally give us, but I'm no good at remembering formulas. What if, instead of integrating

$$\int_0^\infty \int_0^\infty e^{-x} e^{-y} \, dy \, dx$$

you integrated

$$\int_0^a \int_0^b e^{-x} e^{-y} \, dy \, dx$$

? That would get you $F(a, b)$. Let me know if you have any questions.