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Suppose that $X_1$ and $X_2$ are independently uniformly distributed on the interval (0,1). Find the joint and marginal distributions of $U=X_1X_2$ and $V=X_1/X_2$.

I think that $f_U(u) = \int^1_0f_{X_1}(x_1)f_{X_2}(u/x_1){1\over{|x_1|}} dx_1$

I'm not sure if that is right. If it is right, would the marginal distribution just be $\int^1_0 X_1X_2 dx_1$ and $\int^1_0 X_1X_2 dx_2$?

Thanks for your help.

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    $\begingroup$ HINT : $$ X_1=\sqrt{UV}\quad\text{ and }\quad X_2=\sqrt{\frac{U}{V}}, $$ then you find the Jacobian. $\endgroup$ – Tunk-Fey Apr 12 '14 at 15:04
  • $\begingroup$ Thanks, I found the Jacobian to be $-{\sqrt {U/V}}\over 2\sqrt{UV}$ is that right? $\endgroup$ – user123965 Apr 13 '14 at 10:33
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    $\begingroup$ Yep, it's almost correct but you can simplify that into $\dfrac{1}{2V}$ and Jacobian is always positive. Now, you obtain the regions of $U$ and $V$ based on the regions of $X_1$ and $X_2$. $\endgroup$ – Tunk-Fey Apr 13 '14 at 11:29
  • $\begingroup$ $0<UV<1$ and $0<U/V<1$, so $0<U<V$? $\endgroup$ – user123965 Apr 13 '14 at 17:40
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    $\begingroup$ Yep! That's correct. Now, try to plot them and integrate it carefully. Here is the plot I made using WolframAlpha. $\endgroup$ – Tunk-Fey Apr 13 '14 at 17:55
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The joint PDF of $U$ and $V$ is $$ f_{U,V}(u,v)=f_{X_1,X_2}(x_1,x_2)\cdot|J|=\frac1{2v}. $$ Now for the regions. You have $0\le x_1\le1$, this region is corresponding to $$0\le \sqrt{uv}\le1\;\Rightarrow\;0\le uv\le 1\;\Rightarrow\;0\le v\le \frac1u.$$ It's due to $X_1=\sqrt{UV}$. You also have $0\le x_2\le1$, this region is corresponding to $$0\le \sqrt{\dfrac uv}\le1\;\Rightarrow\;0\le \dfrac uv\le1\;\Rightarrow\;0\le u\le v.$$ It's due to your transformation $X_2=\sqrt{\dfrac UV}$. If you plot those regions (check the plot here), you can obtain the marginal PDFs of $U$ and $V$ as follow $$ \begin{align} f_U(u)&=\int_{v=u}^{\frac1u}f_{U,V}(u,v)\ dv\ ;\quad\text{the region is bounded by $v=u$ and $v=\frac1u$ if you see it from $u$-axis}\\ &=\int_{v=u}^{\frac1u}\frac1{2v}\ dv\\ &=\left.\frac 12\ln v\right|_{v=u}^{\frac1u}\\ &=\frac 12\ln \left(\frac1u\right)-\frac 12\ln u\\ &=-\ln u\;;\quad\text{ for } 0\le u\le1. \end{align} $$ and $$ \begin{align} f_V(v)&=\int_{u=0}^{v}f_{U,V}(u,v)\ du\ ;\quad\text{the region is bounded by $u=0$ and $u=v$ if you see it from $v$-axis}\\ &=\int_{u=0}^{v}\frac1{2v}\ du\\ &=\left.\frac u{2v}\right|_{u=0}^{v}\\ &=\frac12\;;\quad\text{ for } 0\le v\le1, \end{align} $$ and $$ \begin{align} f_V(v)&=\int_{u=0}^{\frac1v}f_{U,V}(u,v)\ du\ ;\quad\text{the region is bounded by $u=0$ and $u=\frac1v$ if you see it from $v$-axis}\\ &=\int_{u=0}^{\frac1v}\frac1{2v}\ du\\ &=\left.\frac u{2v}\right|_{u=0}^{\frac1v}\\ &=\frac 1{2v^2}\;;\quad\text{ for } 1\le v\le\infty. \end{align} $$ It can also be written as $$ f_V(v)= \left\{ \begin{array}{l l} \frac12&\;;\quad\text{ for } 0\le v\le1,\\\\ \frac 1{2v^2}&\;;\quad\text{ for } 1\le v\le\infty. \end{array} \right. $$

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