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

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.2012-10-23
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    I think your approach is a good one, not lousy. :)2012-10-23

2 Answers 2

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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?2012-10-23
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    @user1489975: Who said we need to integrate by parts?2012-10-23
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    Oh, sorry I mean I don't know why we need to split u into 3 parts.2012-10-23
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    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.2012-10-23
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    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.2012-10-24
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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?2012-10-23