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!
$\endgroup$ 32 Answers
$\begingroup$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.
$\endgroup$ 3 $\begingroup$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$$ \left\{\left(xy,x/y\right):\left(x,y\right)\in\left[1,\infty\right)^2\right\}=\left\{y\le x\right\}\cap\left\{y\ge 1/x\right\}. $$
$\endgroup$ 0
