Let $X$~$\Gamma(\alpha,\lambda)$ and $Y$~ $\Gamma(\beta,\lambda)$, where $\Gamma(r,\lambda)$ denotes the Gamma distribution with density $f_{r,\lambda}(x)=\lambda e^{-\lambda x}\frac{(\lambda x)^{r-1}}{\Gamma(r)}\mathbb{1}_{[o,\infty)}(x)$ for $\lambda >0, r>0$. Denoting by $f_X$ the density of $X$ and $f_Y$ the density of $Y$, assume that the density of $(X,Y)$ is $f_X\cdot f_Y$.Define $Z=X+Y$ and $U=\frac X{X+Y}$
- Show that the density of $(Z,U)$ can be written as the product of two densities.
- Compute the density of $U$
What I did so far:
For 1: We have $f_X(x)=\lambda e^{-\lambda x}\frac{(\lambda x)^{\alpha -1}}{\Gamma(\alpha)}\mathbb{1}_{[o,\infty)}(x)$ and similarly $f_Y(y)=\lambda e^{-\lambda y}\frac{(\lambda y)^{\beta -1}}{\Gamma(\beta)}\mathbb{1}_{[o,\infty)}(y)$. Hence $f_{(X,Y)}=\lambda^2e^{-\lambda(x+y)}\frac{(\lambda x)^{\alpha -1}(\lambda y)^{\beta -1}}{\Gamma(\alpha)\Gamma(\beta)}\mathbb{1}_{[o,\infty)}(x)\mathbb{1}_{[o,\infty)}(y)=\lambda^{\alpha+\beta}e^{-\lambda(x+y)}\frac{x^{\alpha-1}y^{\beta-1}}{\Gamma(\alpha)\Gamma(\beta)}\mathbb{1}_{[o,\infty)}(x)\mathbb{1}_{[o,\infty)}(y)$
Then form the application $g(x,y)=(x+y,\frac x{x+y})=(z,u)$ we find its inverse to be $g^{-1}(z,u)=(zu,z(1-u))$ such that the Jacobian matrix is $J_{g^{-1}}(z,u)= \begin{pmatrix}u&1-u \\z&-z \end{pmatrix} \Rightarrow |\det(J_{g^{-1}}(z,u))|=|-z|=z$
So we have $f_{(Z,U)}(z,u)=f_{(X,Y)}(zu,z(1-u))|\det(J_{g^{-1}}(z,u))|=\frac{\lambda^{\alpha+\beta}}{\Gamma(\alpha)\Gamma(\beta)}e^{-\lambda z}z^{\alpha+\beta-1}u^{\alpha-1}(1-u)^{\beta-1}\mathbb{1}_{[o,\infty)}(z)\mathbb{1}_{[o,\infty)}(u)$
How can I say that this is a product of two densities? I don't know if the request is ambiguos or maybe do I miss something?
For 2: $f_U(u)=\frac{\lambda^{\alpha+\beta}}{\Gamma(\alpha)\Gamma(\beta)}u^{\alpha-1}(1-u)^{\beta-1}\int_0^\infty e^{-\lambda z}z^{\alpha+\beta-1} dz$
Since the integral using just partial integration seems to be infinite to compute, I used a calculator which returned that$\int_0^\infty e^{-\lambda z}z^{\alpha+\beta-1} dz=-\lambda^{-\alpha-\beta}\Gamma(\alpha+\beta,\lambda z)$.
Now this can be true, but we've never seen this "incomplete gamma function", so I was a litlle concerned about this result, is there a "smoother" way to compute it?
Best Answer
Divide and multiply by the term $\Gamma(\alpha+\beta)$ in your completed density $f(Z, U)$ to get
$$f(Z, U)=\frac{\lambda^{\alpha+\beta}}{\Gamma(\alpha+\beta)}z^{\alpha+\beta-1}e^{-\lambda z}\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}u^{\alpha-1}(1-u)^{\beta-1}, z>0, 0<u<1$$
As you can see this has been factored into the product of a $\text{Gamma}(\alpha+\beta,\lambda)$ density and a $\text{Beta}(\alpha,\beta)$ density.
Thus $U\sim\text{Beta}(\alpha,\beta)$ with density shown.
At this point we have answered both problems.
You could also integrate out the $Z$ to find the density of $U$ (not necessary as they are independent), in which case notice that $$\int_0^\infty z^{\alpha+\beta-1}e^{-\lambda z}=\frac{\Gamma(\alpha+\beta)}{\lambda^{\alpha+\beta}}$$
and we would end up with the density of the beta distribution (but again, was not necessary here as the joint density is factorizable into a function of $U$ and a function of $Z$)