# What is the posterior distribution of θ? Is the Gamma a conjugate prior for an exponential likelihood?

A manufacturer is interested in the time to failure of his batteries.

Suppose the time to failure of the batteries has an exponential distribution:

$$p(x│\theta)=\theta\exp(-\theta x)$$

Note that the mean of this population is $$1/\theta.$$

The manufacturer is interested in estimating $$\theta.$$ Suppose that $$n$$ batteries are randomly selected and let their failure times be $$x_1, x_2,\ldots, x_n.$$

Suppose we use a gamma prior for $$\theta$$ $$p(\theta│\alpha,\beta)=\beta^\alpha/(\Gamma(\alpha)) \theta^{(\alpha-1) }\exp(-\beta\theta)$$

a) What is the posterior distribution of $$\theta$$? Is the Gamma a conjugate prior for an exponential likelihood?

b) Suppose an expert believes that the mean time to failure is $$120$$ hours. Furthermore, he claims that he is $$95\%$$ sure that the mean time to failure is between $$100$$ and $$150$$ hours. What would be a suitable choice for $$\alpha$$ and $$\beta$$ in the prior?

Gamma distribution is indeed a conjugate prior for an exponential likelihood, because they share the same kernel for the parameter $$\theta$$, i.e.

The Gamma prior for $$\theta$$ has

$$p(\theta;a,b)\propto \theta^{a-1}e^{-b\theta}$$

whereas the exponential likelihood for all the independent identically distributed observations $$x_{1}, x_{2},..., x_{n}$$ is

$$l(\theta;x_{1}, x_{2},..., x_{n}) = \theta e^{-\theta x_{1}}\theta e^{-\theta x_{2}}...\theta e^{-\theta x_{n}} = \theta^{n} e^{-\theta \sum_{i=1}^{n}x_{i}}$$

You can observe that both prior and likelihood have the same kernel, i.e. $$\theta$$ in a power multiplied by exponential of $$\theta$$, this makes it possible to multiply $$p(\theta;a,b)$$ with $$l(\theta ;x_{1}, x_{2},..., x_{n})$$ and get a standard output.

Hence, the posterior will be

$$p(\theta|x_{1}, x_{2},..., x_{n}) \propto l(\theta;x_{1}, x_{2},...,x_{n}) p(\theta; a,b) = \theta^{n} e^{-\theta\sum_{i=1}^{n}}\theta^{a-1}e^{-b\theta}=\theta^{n+a-1}e^{-\theta(b+\sum_{i=1}^{n}x_{i})}$$

This kernel might remind you the kernel of a Gamma distribution, so $$\theta\sim Gamma(a+n,b+\sum_{i=1}^{n}x_{i})$$

Now if you know that apriori the mean of $$\theta$$ is 120, you can translate it as $$\mathbb{E}[\theta]_{p(\theta)} = \frac{a}{b}=120$$ which is the expected value of $$\theta$$ under the prior $$p(\theta;a,b)$$.

So, you have one equation $$a=120b \ \ \ (*)$$

Then you also, have the information that $$\mathbb{P}(100\leq \theta \leq 150) = 0.95$$

$$\Rrightarrow \mathbb{P}(\theta \leq 150)-\mathbb{P}(\theta \leq 100) = 0.95$$

$$\Rrightarrow \frac{\gamma(a,b 150)}{\Gamma(a)} -\frac{\gamma(a,b 100)}{\Gamma(a)}=0.95 \Rightarrow \frac{\gamma(a,50b)}{\Gamma(a)} = 0.95 \ \ \ (**)$$

Now plugging in the $$(*)$$ into $$(**)$$ you get

$$\frac{\gamma(120b,50b)}{\Gamma(120b)} = 0.95 \ \ (***)$$ which depends only on $$b$$.

So, what you can do is define a grid of values for $$b$$ and calculate $$(***)$$ with the use of $$R$$ language or any other language, and check if this equality holds $$\frac{\gamma(120b,50b)}{\Gamma(120b)} = 0.95$$ and denote as $$b^{*}$$ the $$b$$ for which you found that the equality holds, then calculate as $$a^{*}=120b^{*}$$. And those $$a^{*},b^{*}$$ are the parameters that reflect your prior beliefs.