Multivariate Random Variables

Question

$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....

Answer 1

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.

Answer 2

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