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If $f(x,y) = \frac{12}{5x (2-x-y)}$ for 0 < x and y < 1, compute the marginal density function $f_Y(y)$

I know that I have to integrate the joint density function with respect to x, but how I figure out the limits of integration? Also, if the bounds were 0 < y < x < 1, what would the limits be then?

Thanks :D

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  • $\begingroup$ you just gave the limits: $x>0, y>1$ $\endgroup$ – Alex Feb 12 '14 at 2:06
  • $\begingroup$ You must have meant $f_Y(y)$ rather than $f_y(Y)$. $\endgroup$ – Michael Hardy Feb 12 '14 at 2:11
  • $\begingroup$ fixed, thanks. But my professor said the limits for the marginal density of y for the first case is from 0 to 1, which I don't understand... $\endgroup$ – girlrockingguna Feb 12 '14 at 2:19
  • $\begingroup$ Your wording "for $0 < x$ and $y < 1$" is ambiguous in that it can be taken to mean that the expression applies for all $x$ and $y$ such that $x$ is positive and $y < 1$. It is most likely that what is meant, and what might even be what your professor actually wrote, is "$0 < x, y < 1$" which is interpreted as shorthand for the more cumbersome $0 < x < 1$ and $0 < y < 1$. So, the limits of integration would be $y=0$ ands $y=1$. $\endgroup$ – Dilip Sarwate Feb 12 '14 at 2:27
  • $\begingroup$ @DilipSarwate: he did write it like "0<x,y<1" So that's shorthand then? What if he had written "0 < x, y < 2". Then would the bounds be 0 < x < 2 and 0 < y < 2? $\endgroup$ – girlrockingguna Feb 12 '14 at 3:10
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You would integrate over the support of X. The first bounds you gave cannot be correct because density functions must be nonnegative which is false for x=y=2. If the second bounds are correct then integrate from y to 1. Check if the resulting density integrates to one.

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  • $\begingroup$ The bounds for the first one are incorrect, sorry, it should be y < 1, my professor said to integrate from 0 to 1 but can you give me a methodology for how you come up with the support of X for both cases? $\endgroup$ – girlrockingguna Feb 12 '14 at 2:17
  • $\begingroup$ Still can't be right for eg x=2 and y=.5. If both are between 0 and 1 that might make sense. The bounds should be given since there is any number of possibilities, i.e. there is no way to figure it out. $\endgroup$ – JPi Feb 12 '14 at 2:24

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