0

Let $X$ and $Y$ be random variable with joint pdf:

$$f(x,y)= \begin{cases} \frac{1}{4} e^{-\frac{1}{2}(x+y)} & x \geq 0, y \geq 0 \\ 0 & \text{ otherwise} \end{cases} $$

$U= \frac{1}{2}(X-Y)$, $V=Y$.

I know that $f(u,v)= \frac{\partial^2}{dudv}F(u,v)$ and therefore $$F(u,v)= P(\frac{1}{2}(X-Y)\leq u, Y\leq v) = \iint \frac{1}{4}e^{-\frac{1}{2}(x+y)}dydx$$ with the first integral ($dy$) from $x-2u$ to $v$ and the second integral from - infinity to $2u+v$. This is where I'm stuck. I'm having a hard time finding the partial of $\frac{\partial^2}{dudv}$. I need to prove this equals:

$$f(u,v)= \frac{1}{2}e^{-u-v}$$ if $(u,v)$ exists in $A$ ($A$ is a subset of $(u,v)$) otherwise $f(u,v) = 0$.

Tom
  • 9,978
fred
  • 61
  • 1
  • 6

1 Answers1

1

No need to reinvent the wheel here, hopefully you know multivariable calculus and how to change variables. Note that $(U,V)$ is a linear transformation of $(X,Y)$, namely $$ \left[\begin{array}{c} U\\ V \end{array}\right] = \left[\begin{array}{cc} 1/2 & -1/2\\ 0&1 \end{array}\right] \left[\begin{array}{c} X\\ Y \end{array}\right]. $$ Hence $$f_{(U,V)}(u,v) = f_{(X,Y)}(x,y)|\det\frac{d(X,Y)}{d(U,V)}| = f_{(X,Y)}(2u+v,v)\cdot 2 $$ $$ =1_{2u+v\geq 0,v\geq 0}\frac{1}{4}e^{-\frac12(2u+v + v)} \cdot 2 = \frac12e^{-u-v}1_{2u+v\geq 0,v\geq 0} $$ as desired.

nullUser
  • 27,877