I came across a problem where the marginal distribution of a random variable $Y$, $f(y) = c/y^2$ and $f(x|y) = 1/y$.

Can I simply multiply these two to get $f(x,y)$ the joint distribution of $X$ and $Y$, which in this case will be $c/y^3$. And then integrate it over all $Y$ to find the marginal distribution of $X$.


1 Answer 1


In effect, yes, $f(x,y)=f(x|y)f(y)$, which is essentially just elementary probability rules.

See the relevant section of Wikipedia's article on Conditional probability distribution.

However, you need to take proper care about the limits of $X|Y=y$ (and $Y$ for that matter), since that will define where $(X,Y)$ lives.

Were ranges given?

Then, yes, once you have the joint you could compute a marginal for $X$, but again, you must take care about the limits. Not dealing properly with them is perhaps one of the biggest sources of errors for beginners.

  • $\begingroup$ So even for continuous random variables, probability density function f(x|y) can be multiplied by f(y) to get f(x,y)? $\endgroup$ Commented Oct 3, 2014 at 1:44
  • $\begingroup$ See the link in the updated answer. $\endgroup$
    – Glen_b
    Commented Oct 3, 2014 at 1:47
  • $\begingroup$ Okay. Thanks! I checked the article on wikipedia. The conditional distribution of X given Y is Uniform on [0,y] and the marginal distribution of y exists only if y is greater than 2. So while calculating the joint, the only thing to keep in mind would be that it is valid for y > 2. Right? $\endgroup$ Commented Oct 3, 2014 at 1:52
  • 3
    $\begingroup$ No, you also need to worry about the fact that $0\leq x\leq y$. This is important. The phrasing "marginal exists only if $y>2$" seems odd to me. Was a lower limit of $y$ given that was not 2? $\endgroup$
    – Glen_b
    Commented Oct 3, 2014 at 1:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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