# Assume that $X$ and $Y$ have joint probability density function $f_{X,Y}$. Calculate the joint probability density function of $U = XY$ and $V = X/Y$

Assume that $$X$$ and $$Y$$ have the following joint probability density function $$f_{X,Y}(x,y) = \begin{cases} \displaystyle\frac{1}{x^{2}y^{2}} & \text{if}\,\,x\geq 1\,\,\text{and}\,\,y\geq 1\\\\ \,\,\,\,\,0 & \text{otherwise} \end{cases}$$

(a) Calculate the joint probability density function of $$U = XY$$ and $$V = X/Y$$.

(b) What are the marginal density functions?

MY SOLUTION

(a) To begin with, notice that $$u \geq 1$$ and $$v > 0$$. Moreover, we have $$X = \sqrt{UV}$$ and $$Y = \sqrt{U/V}$$. From whence we conclude that \begin{align*} f_{U,V}(u,v) = f_{X,Y}(\sqrt{uv},\sqrt{uv^{-1}})|\det J(u,v)| \end{align*}

where $$J(u,v)$$ is given by \begin{align*} \begin{vmatrix} \displaystyle\frac{\partial x}{\partial u} & \displaystyle\frac{\partial x}{\partial v} \\ \displaystyle\frac{\partial y}{\partial u} & \displaystyle\frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} \displaystyle\frac{v}{2\sqrt{uv}} & \displaystyle\frac{u}{2\sqrt{uv}} \\ \displaystyle\frac{1}{2\sqrt{uv}} & \displaystyle-\frac{\sqrt{u}}{2\sqrt{v^{3}}} \end{vmatrix} = -\frac{1}{4v} - \frac{1}{4v} = -\frac{1}{2v} \end{align*}

Therefore we have \begin{align*} f_{U,V}(u,v) = \frac{1}{2u^{2}v} \end{align*}

(b) Once you have the joint probability density function, you can determine the marginal distributions through the formulas $$\begin{cases} f_{U}(u) = \displaystyle\int_{-\infty}^{+\infty}f_{U,V}(u,v)\mathrm{d}v\\\\ f_{V}(v) = \displaystyle\int_{-\infty}^{+\infty}f_{U,V}(u,v)\mathrm{d}u \end{cases}$$

My question is: where things went wrong? I cannot find out where are the miscalculations of $$f_{U,V}$$ since its integral does not converge to $$1$$. Any help is appreciated. Thanks in advance!

• Part (a) does not include the support for the function, therefore how can you know where to integrate in part (b). – Graham Kemp Feb 1 at 0:53
• Perhaps I have misunderstood your question, but I have pointed out that $(u,v)\in[1,+\infty)\times(0,+\infty)$. Is this it you were talking about? – user1337 Feb 1 at 0:55
• That's not the support. As hypermova notes, you have to be very careful with the transformation. – Graham Kemp Feb 1 at 1:12

Consider $$U=XY, V=X/Y$$, gives us that $$X^2/U=V$$ and $$Y^2V=U$$
Then as $$1\leq X, 1\leq Y$$, we have $$\{(U,V):1\leq U ~,~ 1/U\leq V\leq U\}$$ as the support.
• Just integrate \begin{align}f_U(u) &= \mathbf 1_{1\leq u}\dfrac 1{2u^2} \int_{1/u}^u \dfrac 1v\mathsf d v\\ f_V(v) &=\mathbf 1_{0<v< 1}\dfrac 1{2v}\int_{1/v}^\infty\dfrac1{u^2}\mathsf du+\mathbf 1_{1\leq v}\dfrac 1{2v}\int_{v}^\infty\dfrac1{u^2}\mathsf du \end{align} – Graham Kemp Feb 1 at 3:34
The problem lies in that the domain $$\left(x,y\right)\in\left[1,\infty\right)^2$$ and $$\left(u,v\right)\in\left[1,\infty\right)\times\left(0,\infty\right)$$ are not equivalent under your transformation. Instead, you may check that $$\left\{\left(xy,x/y\right):\left(x,y\right)\in\left[1,\infty\right)^2\right\}=\left\{y\le x\right\}\cap\left\{y\ge 1/x\right\}.$$