Conditional probability question about gamma and beta distribution I am now reading a paper which has a time series model given by
$$
y_t|\theta_t\sim N(0,\theta_t^{-1}),\\
\theta_t=e^{r_t}\theta_{t-1}\eta_t,\\
\theta_0|Y_0\sim \text{Gamma}(a_0,b_0),\\
\eta_t\sim \text{Beta}(wa_{t-1},(1-w)a_{t-1}),
$$
where $Y_t$ is the information available at time $t$, and as I understand $r_t$ is deterministic. The Gamma distribution is defined as
$$
f(x;\alpha,\beta)=\frac{x^{\alpha-1}\beta^\alpha\exp(-x\beta)}{\Gamma(\alpha)}.
$$
 The paper says if we combine the transition equation of $\theta_t$ with the prior, then we have that
$$
\theta_1|Y_0\sim \text{Gamma}(wa_0,e^{-r_1}b_0).
$$
Bayes' theorem then delivers
$$
\theta_1|Y_1\sim \text{Gamma}(wa_0+\frac{1}{2},e^{-r_1}b_0+\frac{1}{2}y_1^2).
$$
I am not sure how I can get the distribution of $\theta_1|Y_0$. Thanks for any help!
 A: The statement to prove is:

Let $c\gt0$, $S$ with distribution gamma $(a,b)$, and $U$ with distribution beta $(wa,(1-w)a)$ and independent of $S$.
  Then $cSU$ has distribution gamma $(wa,b/c)$.

First, the distribution of $cX$ is gamma $(v,b/c)$ if and only if the distribution of $X$ is gamma $(v,b)$ hence one can assume without loss of generality that $c=1$. Second, the distribution of $SU$ has density
$$
f(z)=\int_0^1f_U(u)f_S(u^{-1}z)u^{-1}\mathrm du,
$$
where $f_U$ denotes the density of $U$ and $f_S$ the density of $S$. By hypothesis,
$$
f_U(u)\propto u^{wa-1}(1-u)^{(1-w)a-1},\qquad
f_S(s)\propto s^{a-1}\mathrm e^{-bs},
$$
hence, after some compulsory simplifications,
$$
f(z)\propto z^{a-1}\int_0^1u^{-(1-w)a-1}(1-u)^{(1-w)a-1}\mathrm e^{-bz/u}\mathrm du.
$$
The change of variable $u=1/(1+x)$ yields $x$ in $(0,+\infty)$ and $\mathrm du=\mathrm dx/(1+x)^2$ hence
$$
f(z)\propto z^{a-1}\mathrm e^{-bz}\int_0^\infty x^{(1-w)a-1}\mathrm e^{-bzx}\mathrm dx.
$$
Finally, the change of variable $\xi=bzx$ yields $\xi$ in $(0,+\infty)$ and $\mathrm dx=\mathrm d\xi/(bz)$ hence
$$
f(z)\propto z^{a-1}\mathrm e^{-bz}\int_0^\infty \xi^{(1-w)a-1}z^{1-(1-w)a}\mathrm e^{-\xi}z^{-1}\mathrm d\xi\propto z^{wa-1}\mathrm e^{-bz},
$$
which proves the claim.
