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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!

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  • $\begingroup$ Part (a) does not include the support for the function, therefore how can you know where to integrate in part (b). $\endgroup$ – Graham Kemp Feb 1 at 0:53
  • $\begingroup$ 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? $\endgroup$ – user1337 Feb 1 at 0:55
  • $\begingroup$ That's not the support. As hypermova notes, you have to be very careful with the transformation. $\endgroup$ – Graham Kemp Feb 1 at 1:12
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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.

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  • $\begingroup$ Thank you very much! Could you please include the marginal density functions in your answer? I'd be greatly thankful for that. $\endgroup$ – user1337 Feb 1 at 2:17
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    $\begingroup$ Just integrate $$\begin{align}f_U(u) &= \mathbf 1_{1\leq u}\dfrac 1{2u^2} \int_{1/u}^u \dfrac 1v\mathsf d v\\ f_V(v) &=\mathbf 1_{0<v< 1}\dfrac 1{2v}\int_{1/v}^\infty\dfrac1{u^2}\mathsf du+\mathbf 1_{1\leq v}\dfrac 1{2v}\int_{v}^\infty\dfrac1{u^2}\mathsf du \end{align}$$ $\endgroup$ – Graham Kemp Feb 1 at 3:34
  • $\begingroup$ Thanks again :) $\endgroup$ – user1337 Feb 1 at 4:17
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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 $$ \left\{\left(xy,x/y\right):\left(x,y\right)\in\left[1,\infty\right)^2\right\}=\left\{y\le x\right\}\cap\left\{y\ge 1/x\right\}. $$

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