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.

  • 1
    $\begingroup$ What is the point of this parameterization? Isn't it redundant to have $\gamma$ and $\alpha$? Also, equation 2 shouldn't have $x$, I don't think. $\endgroup$
    – HStamper
    Apr 19, 2017 at 21:01
  • $\begingroup$ Technically it's a joint distribution with $\gamma$ being a second random variable, so it varies while $\alpha$ remains constant. $\endgroup$ Apr 19, 2017 at 21:56
  • $\begingroup$ Try integrating the pdf instead of the cdf with the (now) correct form of $f(\gamma)$. $\endgroup$
    – HStamper
    Apr 19, 2017 at 23:52

1 Answer 1


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: $$ g(x)=\frac{\beta}{\alpha}\left(\frac{x}{\alpha}\right)^{\beta-1} \left(1+\theta\left(\frac{x}{\alpha}\right)^\beta\right)^{-1/\theta-1} $$


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