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 '17 at 21:01
  • $\begingroup$ Technically it's a joint distribution with $\gamma$ being a second random variable, so it varies while $\alpha$ remains constant. $\endgroup$ – John Shannon Apr 19 '17 at 21:56
  • $\begingroup$ Try integrating the pdf instead of the cdf with the (now) correct form of $f(\gamma)$. $\endgroup$ – HStamper Apr 19 '17 at 23:52

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.