Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

find the marginal distributions of $$ f(x,y) = 2xe^{-y}, \quad 0 < x,\quad x^2 < y $$

I have drawn the support, with $y = x^2$. Not sure how to proceed (tried it on wolfram advanced calculus app and says it doesn't converge)

share|cite|improve this question
There are two different marginal distributions that you can compute. Does Wolfram claim that both integrals are diverge? If only one diverges, which one? – Dilip Sarwate Apr 16 '14 at 22:06
up vote 0 down vote accepted

The marginal in $x$ is $$ \int_{x^2}^\infty 2xe^{-y} dy = 2x \left. (- e^{-y}) \right|_{x^2}^\infty = 2xe^{-x^2}, \quad x > 0. $$ The marginal in $y$ is $$ \int_0^{\sqrt{y}} 2xe^{-y} dx = e^{-y} \left. (x^2) \right|^{\sqrt{y}}_0 = ye^{-y}, \quad y > 0. $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.