Given joint probability density function:
\begin{align}
f_{X,Y}(x,y)=
\begin{cases}
24x(1-y)&0<x<y<1\\
0&\text{otherwise}
\end{cases}.
\end{align}
Given transformation: $M=\dfrac{X+Y}{2}$ and $W=\dfrac{X}{2}$, find the joint p.d.f. of $M$ and $W$.
I try as follows.
Since $0<x<y<1$ The range of transformation is
$$
M=\dfrac{X+Y}{2}\geq \dfrac{X}{2}=W,
$$
so $0\leq M\leq 1$, $M\geq W$, and $0\leq W\leq 1$.
The invers of transformation is
$X=2W$ and $Y=2M-2W$.
The absolute value of Jacobian is
\begin{align}
|J|=
\begin{vmatrix}
\dfrac{dX}{dM}&\dfrac{dX}{dW}\\
\dfrac{dY}{dM}&\dfrac{dY}{dW}
\end{vmatrix}
=
\begin{vmatrix}
0&2\\
2&-2
\end{vmatrix}
=4.
\end{align}
The p.d.f. of transformation:
\begin{align}
g_{M,W}(m,w)&=f_{X,Y}(x,y)|J|\\
&=f_{X,Y}(2w,2m-2w)|J|\\
&=24(2w)(1-2m+2w)4\\
&=192w(1-2m+2w).
\end{align}
Now, we have
\begin{align}
g_{M,W}(m,w)=
\begin{cases}
192w(1-2m+2w)&0\leq M\leq 1, M\geq W,\text{ and }0\leq W\leq 1\\
0&\text{otherwise}
\end{cases}.
\end{align}
Now I want to check my answer with double integrating joint p.d.f.
\begin{align}
\int\limits_{0}^{1}\int\limits_{0}^{m} 192w(1-2m+2w) \,dw\,dm
\end{align}
and the result is $16$.(I use maple)
So we can conclude $g_{M,W}(m,w)$ is not a p.d.f.
Why this is happen? Am I have a mistake with my answer?
Best Answer
You are missing several additional inequalities satsifesd by $m$ and $w$. The given inequalities for $x$ and $y$ are equiavlen to to the following:
$0 \leq m \leq 1$, $0 \leq w \leq \frac 1 2$, $w \leq \frac m 2$, and $m \leq w+\frac 1 2$.
The third inequality comes from $x \leq y$. The last one comes from $y \leq 1$.
[Always make it a point to check if the inequalities you obatined for the new variables are adequate to give the stated inequalities for the original variables].