The marginal distribution of $X$ is $U(0, 1)$. The conditional distribution of $Y$, given $X = x$, is $U(0, e^x)$.

Find $f(y)$, the marginal PDF of $Y$ and give its support.

This should be a simple problem, but every time I attempt it I keep getting the wrong pdf for $Y$. I end up with:

$$f(y) = 1 - \frac1e$$ on $$0 < y < \frac e{e-1}.$$

But this doesn't seem to be quite right. I'm not sure where I'm going wrong.

For the joint PDF, I obtained: $$f(x, y) = e^{-x}$$ on $$0<x<1\ and \ 0 < y < e^x $$

  • $\begingroup$ Perhaps show some details of your work so that the mistake can be detected. For example, what is the joint PDF of $(X,Y)$ and what is the support of $(X,Y)$? $\endgroup$ – Just_to_Answer Nov 5 '17 at 1:46
  • $\begingroup$ Thanks for editing with the intermediate work. Now picturing that support, when you are looking for the marginal of $Y$, one would be going across horizontally for some $y$. That is, integration would be $dx$ running from $x=\ln y$ to $x=1$. $\endgroup$ – Just_to_Answer Nov 5 '17 at 2:11

You're answer is fine. I'm not sure why it wouldn't make sense. I'm assuming you integrated across all x values (0 to 1) of $f$$($$y$$|$$x$$)$$f$$($$x$$)$ to get your answer. Normalising gives the range you specify.

  • $\begingroup$ -1. Can you provide the details confirming the OP's $pdf$ and the support for the marginal of $Y$ are correct? $\endgroup$ – Just_to_Answer Nov 5 '17 at 2:43
  • $\begingroup$ $f(y|x) = e^{-x}$, $f(x) = 1$. $\endgroup$ – stuart stevenson Nov 5 '17 at 2:59
  • $\begingroup$ That is the easy part of writing down the given information. Then how specifically do you get the marginal PDF of $Y$? What are the limits of integration? What is the support of $Y$? $\endgroup$ – Just_to_Answer Nov 5 '17 at 3:44

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