Calculating marginal distribution via integration

Question

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?

Answer 1

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.

Answer 2

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:

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