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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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    @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

1 Answers 1

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The form of the probability density was clarified in the comments: $f(x,y) = 12x(1-y)y^{-3}\, [0 (using the Iverson bracket). To prepare integration along $y$, note that $[0 Therefore, $\int f(x,y)\,dy = [0 which evaluates to $[0. This is the required marginal.

(Sanity check: the marginal integrates to $6+6/2-12/(3/2)=1$.)