# Posterior Distribution of a Prior Variable

Let $X_{1},\dots,X_{n}$ be a random sample from an exponential distribution with density $f(x;\theta)=\theta e^{-\theta x}$, $x>0$ (having mean $1/\theta$). Assume a prior density for $\theta$ which is also exponential with mean $1/\beta$, where $\beta$ is known. Prove that the posterior distribution of $\beta$ is a gamma distribution.

--In general, the posterior distribution would be $\pi(\beta\mid \mathbf{x})=f(\mathbf{x}\mid \beta)\pi(\beta)/m(\mathbf{x})$, where $m(\mathbf{x})$ is the marginal probability of $x$. The part that I am stuck at has to do with being given $f(\mathbf{x}\mid\theta)$, not $f(\mathbf{x}\mid\beta)$. Any help is appreciated!

• If $\beta$ is assumed to be known, then a posteriori it is the same --- it is still known, and has the same value as before. Are you sure it's not the posterior distribution of $\theta$ that you're looking for? – Kirill Jul 13 '13 at 20:53
• It would make more sense to me, but it is the posterior distribution of $\beta$ that I need. It is from a past PhD qualifying exam. – Kirk Fogg Jul 13 '13 at 20:54
• Are you definitely certain it's not a typo? If $\beta$ is known with certainty a priori, it is still known with certainty a posteriori. – Kirill Jul 13 '13 at 20:55
• Because the posterior over $\theta$ is $P(\theta|x) \propto \theta^n e^{-\theta(\beta+\sum_i x_i)}$, and, in fact, is a gamma distribution. – Kirill Jul 13 '13 at 20:58
• Perhaps this is what the question was meant to ask. It seems much more straight forward this way. – Kirk Fogg Jul 13 '13 at 21:03

The likelihood function is $$L(\theta) = \prod_{i=1}^n \left(\theta e^{-\theta x_i} \right) = \theta^n e^{-\theta(x_1+\cdots+x_n)}\text{ for }\theta>0.$$
The prior density is $$f(\theta) = \frac1\beta e^{-\theta/\beta}\text{ for }\theta>0.$$
Multiply the two: $$\text{constant}\cdot \theta^n e^{-\theta\left(x_1+\cdots+x_n+\frac1\beta\right)},$$ where in this case "constant" means not depending on $\theta$.
Possibly you don't even need the normalizing constant. If one parametrizes the family of gamma densities as $$f(t) = \text{constant}\cdot t^{\alpha-1} e^{-\gamma t}\text{ for }t>0,$$ then in this case we have $\alpha = n+1$ and $\gamma=x_1+\cdots+x_n+\frac1\beta$.
By Bayes theorem, the posterior distribution over $\theta$ is given by $$P(\theta\mid x_1,\ldots,x_n) = \frac{P(x_1,\ldots,x_n\mid\theta)P(\theta)}{P(x_1,\ldots,x_n)}.$$ Now, since $x_i$ are all independent, we have $$P(x_1,\ldots,x_n\mid\theta) = P(x_1\mid\theta)\cdots P(x_n\mid\theta) = \theta^n e^{-\theta\sum_i x_i}.$$ Also, the prior is $P(\theta) = \beta e^{-\beta\theta}$. Because $P(x_1,\ldots,x_n)$ is not a function of $\theta$, we can write $$P(\theta\mid x_1,\ldots,x_n) = \text{const}\times\theta^n e^{-\theta\sum_i x_i-\theta\beta},$$ where the constant of proportionality will be fixed by requiring that $\int P(\theta\mid x_1,\ldots,x_n)\,d\theta=1$.