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I have a joint PDF that has gone through some transformations of

$f(x,y) = 12x\displaystyle\frac{1-y}{y^3}$,$0, $0

It definitely is a valid PDF as it has a double integration along its support that equals 1

I am trying to find the marginal PDF of $X$.

However integrating along y gives a definite integral that does not converge:

$\displaystyle\int_{0}^1 12x\displaystyle\frac{1-y}{x^3}dy$

Any ideas how else I can find that marginal PDF of $X$?

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    What are the limits on $x$ and $y$? Or in other words, what is the support?2012-05-24
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    Sorry I somehow cut that bit out. 02012-05-24
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    Then, contrarily to what you declare in the post, if f is the density of a joint PDF, the integral over y should converge. Or rather, it should for almost every x. (Please check your post, at the moment the two formulas for f are not compatible.)2012-05-24
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    Draw a picture! Then integrate, $y=\sqrt{x}$ to $y=1$. No problem.2012-05-24
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    Didier you can try to check, but this shows that it is a PDF http://bit.ly/KuqAK8 and this shows it does not converge http://bit.ly/KlhM8C Andre - Wouldn't I then be losing x values that are less than root x and greater than zero?2012-05-24
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    @Seraphya: In the second calculation, you gave the program the wrong integral, and it calculated correctly. I would rather you found the answer partly by yourself, so will wait a while before answering, in the hope you can push things through. The region in which the joint density lives is the first quadrant region that is below the line $y=1$ and **above** the curve $x=y^2$. So when we "integrate $y$ out" the variable $y$ runs from $y=\sqrt{x}$ to $y=1$, **not** from $y=0$ to $y=1$. Remember the $0\lt x\lt y^2$, which in particular says that $y \gt\sqrt{x}$.2012-05-24
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    @Seraphya: After you find the answer, send a message. I did the calculation, so can check whether you are right.2012-05-24
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    I see I should redefine the support as 02012-05-25
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    @Seraphya: Yes, I also had $6+6x-12x^{1/2}$ over the same interval. If you like, you can write up the answer reasonably carefully, and post it as the solution, and, after waiting a little while, accept it.2012-05-25
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    @AndreNicholas I get 6 + 6x -12 (x)^1/2 and my density g(x) integrates to 1 whereas 6+6x-12 integrates to -3. Mine does look like Seraphyra's. Maybe something caused the square root of x to come up blank in Andre's post.2012-05-25

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