# Multivariate Random Variables

$f(x,y) = {2\over 5}(2x+3y) \quad for\quad 0<x,y<1$

and we want to know the distribution of $2X+3Y$

I did it in a very lousy way which is let $U=2X+3Y ,\; V=X$

Then have $\;f_{U,V}(u,v)$ and then integer by V to get $\;f_U(u)$

So I tried $f_{U,V}(u,v) = f_{X,Y}(v,{u-2v\over 3})*{2\over5}={2u\over 15}$

$\int_0^\infty{2u\over 15}dv=$ then I think i might be wrong here....

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Probably you mean that the joint density function of $X$ and $Y$ is $\frac{2}{5}(2x+3y)$ in the unit square, and $0$ elsewhere. – André Nicolas Oct 23 '12 at 6:16
I think your approach is a good one, not lousy. :) – Patrick Li Oct 23 '12 at 12:35

For the record, the limits of integration must be dealt with carefully since the end result is $$f_U(u)=\tfrac1{15}(u^2\mathbf 1_{0\lt u\lt2}+2u\mathbf 1_{2\lt u\lt3}+u(5-u)\mathbf 1_{3\lt u\lt 5}).$$

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That's the part I am confused, since my $f(u,v)={2\over 15}u \;$ but i am not sure why we need to integrate by parts? – user1489975 Oct 23 '12 at 14:37
@user1489975: Who said we need to integrate by parts? – joriki Oct 23 '12 at 15:04
Oh, sorry I mean I don't know why we need to split u into 3 parts. – user1489975 Oct 23 '12 at 15:14
user: You might want to add to your post the detailed computations you made, so far. This would help to spot the mistakes you made, if you made some. – Did Oct 23 '12 at 18:44
Re what you added, obviously the last integral cannot be on (0,+oo). Let me suggest to add to the densities the limits of their arguments, as indicator functions. – Did Oct 24 '12 at 5:19

Can you calculate the $F(a):=P(2X+3Y\le a)$ for a fixed $a$? It goes like: \begin{align} F(a)=P(2X+3Y\le a) &= P\left(3Y\le a-2X\right) = \\ &=\int_0^{1}\int_0^{\frac{a-2x}3}f(x,y)dydx \end{align} Then differentiate it.

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Why integrate-then-differentiate since the hypothesis and the conclusion both use densities? – Did Oct 23 '12 at 12:53