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## Homework Statement

The joint probability density function of the random variable (X, Y) is given by:

[tex] f(x,y) = \frac{2x}{y^2} \text{where} \; 0 \leq x\leq 1 \; \text{and} \; y\geq 1 [/tex]

and 0 elsewhere.

Find the probability density function of the folowing random variable:

U=X+Y

## Homework Equations

I'm not sure of this equation, (have I mentioned my book should be burned?) but I think ultimately I'm looking for

[tex] \int_{-\infty }^{\infty } f(u-v,v) \mathrm{d} v [/tex]

## The Attempt at a Solution

[tex] U=X+Y [/tex]

[tex] x= u-v [/tex]

[tex] y=v [/tex]

the jacobian of the transformation is 1, so we can leave out further mention of it in equations.

My integral is [tex] \int_a^b \frac{2u-v}{v^2} \mathrm{d} v [/tex]

I'm not sure about the bounds. Since v=y, should the bounds on v equal the bounds on y? That is, from 1 to infinity?

I already tried that and got a non-convergent integral (the sad story of my day today) :(

[tex] \int_1^{\infty } \frac{2u-v}{v^2} \mathrm{d} v = [/tex]

[tex] \int_1^{\infty } \frac{2u}{v^2} \mathrm{d} v - \int_1^{\infty } \frac{1}{v} \mathrm{d} v = [/tex]

[tex] 2u \int_1^{\infty } v^{-2} \mathrm{d} v - \int_1^{\infty } \frac{1}{v} \mathrm{d} v = [/tex]

[tex] -2u(\frac{1}{v} |_1^\infty ) - \mathrm{ln} v |_1^{\infty } [/tex]

And since [itex] \mathrm{ln} (\infty ) = \infty [/itex], part of my problem says [itex] 0-\infty [/itex] Which is not good...

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