Integrating out a gamma-distributed parameter from a Weibull distribution I'm dealing with a variation of the three-parameter Weibull distribution where the third parameter is randomly distributed over a Gamma distribution. The cdf takes the form:
$$
G(x|\gamma) = 1-\exp\left[-\gamma\left(\frac{x}{\alpha}\right)^\beta \right]
$$
$$
f(\gamma) = \frac{1}{\Gamma(k)\theta^k}\gamma^{k-1}\exp\left(-\frac{\gamma}{\theta}\right)
$$
I've seen it written that it's possible to integrate out the gamma-distributed parameter analytically, but I've had a hard time doing so. I'm assuming it's done by integrating to obtain the cdf:
$$
G(x)= \int_0^{\infty}G(x|\gamma)f(\gamma)d\gamma
$$
My strategy in attempting this was to manipulate the integrals to form Gamma distributions with different parameters, which would then integrate to 1.
\begin{align}
G(x) 
& =\int_0^{\infty}G(x|\gamma)f(\gamma)d\gamma \\
& =\int_0^{\infty}\left(1-\exp\left[-\gamma\left(\frac{x}{\alpha}\right)^\beta 
\right] \right) f(\gamma)d\gamma \\
& =\int_0^{\infty}f(\gamma)d\gamma-\int_0^{\infty}\frac{1}{\Gamma(k)\theta^k}\gamma^{k-1}\exp\left(-\frac{\gamma}{\theta}\right)\exp\left(-\gamma\left(\frac{x}{\alpha}\right)^\beta 
\right)d\gamma \\
& =1-\int_0^{\infty}\frac{1}{\Gamma(k)\theta^k}\gamma^{k-1}\exp\left[-\gamma\left( \left(\frac{x}{\alpha}\right)^\beta+\frac{1}{\theta} \right) \right]d\gamma
\end{align}
The trick I want to use is to define some parameters $k_2$ and $\theta_2$ such that the second integral becomes a Gamma distribution and also integrates to 1. I define them as:
\begin{align}
 k_2 &= k \\
\theta_2 &= \left( \left(\frac{x}{\alpha}\right)^\beta+\frac{1}{\theta} \right)^{-1}
\end{align}
So the integral becomes:
\begin{align}
G(x) &=1-\int_0^{\infty}\frac{1}{\Gamma(k)\theta^k}\gamma^{k-1}\exp\left(-\frac{\gamma}{\theta_2} \right)d\gamma \\
&=1-\frac{{\theta_2}^{k_2}}{\theta^k}\int_0^{\infty}\frac{1}{\Gamma(k_2){\theta_2}^{k_2}}\gamma^{k_2-1}\exp\left(-\frac{\gamma}{\theta_2} \right)d\gamma\\
&=1-\frac{{\theta_2}^{k_2}}{\theta^k} \\
&=1-\theta^{-k}\left( \left(\frac{x}{\alpha}\right)^\beta+\frac{1}{\theta} \right)^{-k} \\
&=1-\left( \theta\left(\frac{x}{\alpha}\right)^\beta+1 \right)^{-k}
\end{align}
This scarcely bears any resemblance to a Weibull cdf, especially since the exponential component was integrated away. I guess there's an error somewhere in the derivation, but I'm stuck here.
 A: Yes, this is actually a variation on the Weibull cdf. In fact, it asymptotically approaches the Weibull cdf as $\gamma$ converges in mean square error to 1. To simplify taking the limits, first normalize the moments of $\gamma$. If $\gamma \sim \Gamma(k,\theta)$, then $E[\gamma] = k\theta$ and $var(\gamma)=k\theta^2$. Setting $k=1/\theta$ normalizes the mean to $1$ and the variance to $\theta$. The cdf is now:
$$ G(x) = 1 - \left(\theta\left(\frac{x}{\alpha}\right)^{\beta}+1\right)^{-\frac{1}{\theta}-1} $$
We want to show the following:
$$\lim_{\theta\rightarrow0}1 - \left(\theta\left(\frac{x}{\alpha}\right)^{\beta}+1\right)^{-\frac{1}{\theta}-1} = 1-\exp\left(-\left(\frac{x}{\alpha}\right)^\beta\right)
$$
Now by definition $e^y=\lim_{n\rightarrow\infty}\left(1+\frac{y}{n}\right)^n$. The desired result follows immediately from substituting $n=-1/\theta$ and $y=(x/\alpha)^\beta$
\begin{align}
  \lim_{\theta\rightarrow0}G(x ;\alpha,\beta,\theta) & =
  \lim_{\theta\rightarrow0}1 - \left(\theta\left(\frac{x}{\alpha}\right)^{\beta}+1\right)^{-\frac{1}{\theta}-1} \\
& = 1-\lim_{\theta\rightarrow0}\left(1+\theta\left(\frac{x}{\alpha}\right)^{\beta}\right)^{-\frac{1}{\theta}-1} \\
& =1-\lim_{-1/\theta\rightarrow\infty} \left(1+\frac{-(x/\alpha)^\beta}{-1/\theta}\right)^{-1/\theta-1} \\
& = 1-\exp\left(-\left(\frac{x}{\alpha}\right)^\beta\right) \\
& = W(x;\alpha,\beta)
\end{align}
Here $W(*;\alpha,\beta)$ is the Weibull distribution with parameters $\alpha$ and $\beta$. You can see that this holds by playing with this graph.
For completeness's sake, the pdf is given by:
$$
g(x)=\frac{\beta}{\alpha}\left(\frac{x}{\alpha}\right)^{\beta-1}
\left(1+\theta\left(\frac{x}{\alpha}\right)^\beta\right)^{-1/\theta-1}
$$
