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Suppose that we have an IID random sample $\mathbf{x} = (x_1, \dots, x_n)$ from a given distribution with the following PDF:

$$\theta (1 - e^{-x})^{\theta -1}e^{-x}, \, x > 0, \, \theta > 0$$

Also, let's say that we have a non-informative prior distribution such that $\pi(\theta) \propto \frac{1}{\theta^c}$ where $c$ is a given constant. I'd like to calculate the posterior distribution, but to do so, I have to calculate:

$$m(\mathbf{x}) = \int f(\mathbf{x} \mid \theta) \pi(\theta) d\theta = \int_0^{\infty} \left( \prod_{i=1}^n \theta (1-e^{-x_i})^{\theta-1}e^{-x_i} \right)\left(\frac{1}{\theta^c} \right) d \theta$$

I don't see an easy way to integrate the above. The examples I have seen are simple ones such as Poisson with Gamma prior and Binomial with Beta prior. Is there another approach to analytically writing the posterior PDF?

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If you consider the change of variable$$y=\exp\{-x\}\,,$$the pdf of $Y$ is given by $$f(x|\theta)=\theta(1-y)^{\theta-1}\,,\quad 0<y<1$$which means you observe [the log-transform of] a $\text{Beta}(1,\theta)$ distribution. The conjugate distributions in that special setting are of the form $θ^a\,b^θ=θ^a\exp\{-\log(b)\theta\}$, hence are the $$\text{Gamma}(\alpha,\beta)\qquad \alpha,\beta>0$$ distributions, with \begin{align*}\pi(\theta|\mathbf{x})&\propto f(\mathbf{x}|\theta)\pi(\theta)\\ &\propto\theta^n\left(\prod_i(1-y_i)\right)^{\theta}\theta^{\alpha-1}e^{-\beta\theta}\\ &= \theta^{\alpha+n-1}\exp\left(-\left[\beta-\sum_i\log(1-y_i)\right]\theta\right)\end{align*} which corresponds to a $$\text{Gamma}(\alpha+n,\beta-\sum_i\log(1-y_i))$$ distribution. The special case $\alpha=-c+1$, $\beta=0$ leads to a $$\text{Gamma}(1-c+n,-\sum_i\log(1-y_i))$$ posterior distribution.

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  • $\begingroup$ Can you explain in a bit more detail how one might show that $\mathrm{Beta}(1,\theta)$ has $\mathrm{Gamma}(\alpha,\beta)$ conjugate prior? $\endgroup$ – PatternMatching Mar 9 '15 at 22:55
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    $\begingroup$ Since the likelihood is of the form $\theta^a\,b^\theta$, this is the functional form of a Gamma distribution. Multiplying by another density of the form $\theta^c\,d^\theta$ keeps the same structure, hence leads to conjugacy. $\endgroup$ – Xi'an Mar 10 '15 at 9:03
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I was not seeing before that you can move product inside the term of the product being exponentiated by $\theta-1$ and that one should focus on which terms depend on $\theta$. You can then plug into Mathematica and it provides the following:

$$e^{-\sum_i x_i} \int_0^{\infty} \theta^n \left( \prod_{i=1}^n (1 - e^{-x_i}) \right)^{\theta-1} \left( \frac{1}{\theta^c} \right) d \theta = e^{-\sum_i x_i}\frac{\Gamma(n + 1 -c) \left(-\log \left( \prod_{i=1}^n(1-e^{-x_i}\right) \right)^{n-c-1}}{\prod_{i=1}^n (1-e^{-x_i})}$$

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    $\begingroup$ You really shouldn't need Mathematica for that. It's some simple manipulation followed by a very short game of 'spot the density' to notice that $\theta$ is in the form of a gamma density; mutliply and divide by the appropriate constant and cross out the integral that's now 1. $\endgroup$ – Glen_b Mar 8 '15 at 23:02

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