Let be $f(x,y) = \frac{1}{n!}(x-y)^n e^{-x} (0<y<x)$ a joint density function. I want to find the joint density function of $(U,V)$ where U = X and V = $e^{X-Y}$.
Now, I already found that $X$ ~ $\Gamma(n+1,1)$ and $Y$ ~ Exp(1).
I also found the density function of $V = e^{X-Y}$. However I do not think I should multiply them together because they are not independent.
Is there another way to find this joint density without finding the marginals? To me it seems it could be possible because X is "not changing" in the next density. Is that true?
Any idea?
Best Answer
Generally, joint density cannot be obtained by examing its marginal densities. We should consider the joint density of $(X,Y)$ as a whole, and use this to get that of $(U,V)$. To this end, we can use the transformation of variables rule, which is saying that $$ f(u,v) = f(x(u,v), y(u,v))\left|\det \frac{\partial (x,y)}{\partial (u,v)}\right|. $$ In the given situation, we have a $1$-$1$ mapping $(X,Y)\leftrightarrow (U,V)$ and $$ f(x,y)=\frac{1}{n!}(x-y)^ne^{-x}1_{x>y>0},\\ X=U,\quad Y=U-\log V. $$ (Here, $1_{(x,y)\in A}$ denotes the indicator function, which takes $1$ if the given condition is satisfied and $0$ otherwise.) Now, we have $$ \frac{\partial (x,y)}{\partial (u,v)}=\left[\begin{array}{cc}1&0\\1& -\frac{1}{v}\end{array}\right],\quad \left|\det \left[\begin{array}{cc}1&0\\ 1& -\frac{1}{v}\end{array}\right]\right|=\frac{1}{v}. $$ By the transformation rule, $$\begin{eqnarray} f(u,v) &=& \frac{1}{n!}\left(\log v\right)^n e^{-u}1_{u>\log v, v>1}\left|\det \frac{\partial (x,y)}{\partial (u,v)}\right|\\&=& \frac{1}{n!v}\left(\log v\right)^n e^{-u}1_{u>\log v>0}. \end{eqnarray}$$ This gives the joint density function of $(U,V)$.