2

Let $X$ and $Y$ be random variables s.t. $X \sim {\Gamma}(r,1)$ and $Y \sim {\Gamma}(s,1)$.

First, I took the Jacobian of $X=UV$ and $Y=U-UV$, which I found was $-U$. To show that $U$ and $V$ are both independent, here's what I did:

$$f_{U,V} (u,v)= \frac{(uv)^{r-1} e^{-uv}}{\Gamma(r)} \times (-u) \times\frac{(u-uv)^{s-1} e^{-(u-uv)}} {\Gamma(s)}$$

A hint I was given was to change this into a gamma function, in the form of $B(\alpha, \beta)=\Gamma(\alpha)\Gamma(\beta)/\Gamma(\alpha+\beta)$... but I'm not so sure this is right because I'm not seeing how this can be done above. Can I get some help?

stochasticboy321
  • 9,003
cambelot
  • 2,411

2 Answers2

5

Note that the Jacobian is $+ u$:

$\left| \begin{array}{cc} \partial_v x & \partial_u x \\ \partial_v y & \partial_u y \end{array}\right| = \left| \begin{array}{cc} u & v \\ -u & 1-v \end{array}\right| = u$

Also note the domains where the density is defined:

$\{uv \ge 0\} \cap \{u - uv \ge 0\} = \{u \ge 0\} \cap \{0 \le v \le 1\}$ .

Rearranging things in the pdf you provided above, and dropping the minus sign because of the argument above.

\begin{align} f_{UV}(u,v) &= \frac{1}{\Gamma(r)\Gamma(s)} u^{r+s-1} e^{-u} v^{r-1} (1-v)^{s-1} \\ &= \frac{v^{r-1}(1-v)^{s-1}}{B(r,s)} \times \frac{u^{r+s-1}e^{-u}}{\Gamma(r+s)} \\ &= g(v) h(u) \end{align}

for $u \ge 0, v \in [0,1]$, and $0$ otherwise.

Both of the terms in the product above are probability densities: $g$ is the density of a Beta distribution, and $h$ is yet another Gamma.

Thus, integrating, away, we'll find that

$f_U (u) = h(u)$

and

$f_V(v) = g(v)$,

leading to $f_{UV}(u,v) = f_U(u) f_V(v)$, and establishing independence.

stochasticboy321
  • 9,003
3

First thing is that in $f_{U,V}$ we take absolute value of the jacobian, so instead of $-u$, it should be $u$, i.e.$$f_{U,V} (u,v)= \frac{(uv)^{r-1} e^{-uv}}{\Gamma(r)} \times u \times\frac{(u-uv)^{s-1} e^{-(u-uv)}} {\Gamma(s)}$$ $U$ varies over $(0,\infty)$, where $V$ varies over $(0,1)$. Then you can find the marginals $f_U$ and $f_V$ by $f_U(u)=\int{f_{U,V}(u,v)dv}$, and $f_V(v)=\int{f_{U,V}(u,v)du}$. They are $$f_U(u) = \dfrac{u^{r+s-1}e^{-u}}{\Gamma(r)\Gamma(s)}\int_{0}^{1}v^{r-1}(1-v)^{s-1}dv=\dfrac{u^{r+s-1}e^{-u}}{\Gamma(r)\Gamma(s)} B(r,s)=\dfrac{u^{r+s-1}e^{-u}}{\Gamma(r+s)}$$ for $u>0$. And, we can similarly find, $$f_V(v) = \dfrac{v^{r-1}(1-v)^{s-1}}{\Gamma(r)\Gamma(s)}\int_{0}^{\infty}u^{r+s-1}e^{-u}du=\dfrac{v^{r-1}(1-v)^{s-1}}{\Gamma(r)\Gamma(s)}\Gamma(r+s)=\dfrac{v^{r-1}(1-v)^{s-1}}{B(r,s)}$$ for $0<v<1$ .Then, note that $f_{U,V}(u,v)=f_U(u)f_V(v)$, so they are independent. Also note that $U \sim Gamma(r+s)$, and $V \sim Beta(r,s)$.

Kaustav
  • 61