The question is : Let 0<q<p, suppose U follows the gamma distribution with parameters p, V follows the beta distribution with parameters q, p-q, and U and V are independent of each other. Proof: UV follow the Gamma distribution with parameters p.
I tried to make W = UV, then find the distribution function of W, and then write the density function, but I couldn't get the result.
Which method should I use to proof this? Please give me some tips. Thank you so much.
I think there is still a typo in the problem statement: the parameter for $UV$ should be $q$, not $p$.
You have $$f_{U,V}(u, v) \, du \, dv = \frac{\lambda^p}{\Gamma(p)} u^{p-1} e^{-\lambda u} \frac{\Gamma(p)}{\Gamma(q)\Gamma(p-q)} v^{q-1} (1-v)^{p-q-1} \, du \, dv$$
Let $X=UV$ and $Y=U(1-V)$. The Jacobian of the transformation $u(x,y) =x+y$ and $v(x,y) = \frac{x}{x+y}$ is $\left|\det \begin{bmatrix} 1 & 1\\\frac{y}{(x+y)^2} & -\frac{x}{(x+y)^2}\end{bmatrix}\right| = \frac{1}{x+y}$. So, \begin{align} f_{U,V}(u,v) \, du \, dv &= f_{U,V}\left(x+y, \frac{x}{x+y}\right) \frac{1}{x+y} \, dx \, dy \\ &= \frac{\lambda^p}{\Gamma(p)} (x+y)^{p-1} e^{-\lambda x+y} \frac{\Gamma(p)}{\Gamma(q)\Gamma(p-q)} \left(\frac{x}{x+y}\right)^{q-1} \left(1-\frac{x}{x+y}\right)^{p-q-1} \frac{1}{x+y} \, dx \, dy \\ &= \frac{\lambda^q}{\Gamma(q)} x^{q-1} e^{-\lambda x} \cdot \frac{\lambda^{p-q}}{\Gamma(p-q)} y^{p-q-1} e^{-\lambda y} \, dx \, dy. \end{align} We see from the factorization of this joint density that $X=UV \sim \text{Gamma}(q, \lambda)$ and $Y=U(1-V) \sim \text{Gamma}(p-q, \lambda)$ are independent.