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)

enter image description here

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?

  • 1
    $\begingroup$ First you obtained the inverse transformation $x = 2w, y = 2m - 2w$ Next, plug it into the original support $ x < y < 1, 0 < x < 1$ So we have $ 2w < 2m - 2w < 1, 0 < 2w < 1$ which implies $2w < m < 1/2 + w, 0 < w < 1/2$ $\endgroup$ – BGM Apr 5 at 5:49

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].


The functional form of the transformed joint density is correct, but the support is incorrect. All you have to do to rectify this is to first note that the support for $(X,Y)$ is bounded by the lines $$X = 0, \quad X = Y, \quad Y = 1.$$ Because the transformation is linear and invertible, the resulting support in $(M, W)$ space is also bounded by three lines. We invert the transformation to get $$X = 2W, \quad Y = 2(M-W),$$ hence $$2W = 0, \quad 2W = 2(M-W), \quad 2(M-W) = 1,$$ or $$W = 0, \quad M = 2W, \quad M = W + 1/2.$$ This then becomes the support $$0 < 2W < M < W + 1/2.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.