Finding the posterior mean I have been trying to solve the following problem:
Suppose $X_1,...,X_n$ are iid exponential random variables, with density $f(x;\theta) =\theta e^{-\theta x}$ ,and let us suppose that we have a prior on $\theta$ with density $\theta^{\alpha -1}e^{-\beta \theta}$
Now, I found the posterior to be
$$\theta^{n+\alpha-1}e^{-n \theta \bar{x} - \beta \theta}$$
I now want to find the posterior mean, which I have read is given by:
$$\int \theta \theta^{n+\alpha-1}e^{-n \theta \bar{x} - \beta \theta} d\theta= \int\theta^{n+\alpha}e^{-n \theta \bar{x} - \beta \theta} d\theta$$
I have tried to solve this integral, but I don't end up with anything meaningful. I have tried to read some examples online, and can see that the normalising constant plays a role, but I don't see how it all links together.
 A: Suppose that $X_{1},\ldots,X_{n}$ are iid exponential random variables, with density function $f(x;\theta)=\theta e^{-\theta x}$. Then the likelihood function will be
\begin{equation*}
\text{L}(\theta|x)=\prod_{i=1}^{n}f(x_{i};\theta)=\prod_{i=1}^{n}\theta e^{-\theta x_{i}}=\theta^{n} e^{-\theta n\bar{x}}
\end{equation*}
where $n\bar{x}=\sum_{i=1}^{n}x_{i}.$
Now, suppose that we want to use a gamma prior for a quantity $\theta$. So $\theta \sim \text{Ga}(\alpha,\beta).$ Then the prior density function for $\theta$ is
\begin{equation*}
\pi(\theta)=\frac{\beta^{\alpha}}{\Gamma(\alpha)}\theta^{\alpha-1}e^{-\theta \beta}
\end{equation*}
therefore, the posterior density function for $\theta$, $\pi(\theta|x)$ is proportional to the prior density times the likelihood function. We can write $\pi(\theta|x)$ as follows
\begin{equation*}
\pi(\theta|x)=\frac{\pi(\theta)\text{L}(\theta|x)}{\int_{\Theta}\pi(\theta)\text{L}(\theta|x)d\theta}.
\end{equation*}
In order to compute the posterior mean for $\theta$, say $\text{E}(\theta|x)$. We have
\begin{equation*}
\text{E}(\theta|x)=\frac{\int \theta \pi(\theta)\text{L}(\theta|x)d\theta}{\int \pi(\theta)\text{L}(\theta|x)d\theta}.
\end{equation*}
We can calculate all these integrals analytically when the prior distribution is conjugate (if the posterior distribution and the prior belong to the same family of distributions, then the prior is called a conjugate prior) to the likelihood.
Let us first deal with the denominator of the posterior mean $\text{E}(\theta|x)$ [normalising constant]. So,
\begin{align*}
\frac{\beta^{\alpha}}{\Gamma(\alpha)}\int_{0}^{\infty}\theta^{\alpha-1}e^{-\theta \beta} \theta^{n}e^{-n\bar{x}\theta}d\theta&=\frac{\beta^{\alpha}}{\Gamma(\alpha)}\int_{0}^{\infty}\theta^{(\alpha+n)-1}e^{-\theta (\beta+n\bar{x})}d\theta\\
&=\frac{\beta^{\alpha}}{\Gamma(\alpha)}.\frac{\Gamma(\alpha+n)}{(\beta+n\bar{x})^{\alpha+n}}.
\end{align*}
Secondly, the integral in the numerator will be
\begin{align*}
\frac{\beta^{\alpha}}{\Gamma(\alpha)}\int_{0}^{\infty}\theta \theta^{\alpha-1}e^{-\theta \beta} \theta^{n}e^{-n\bar{x}\theta}d\theta&=\frac{\beta^{\alpha}}{\Gamma(\alpha)}\int_{0}^{\infty}\theta^{(\alpha+n+1)-1}e^{-\theta (\beta+n\bar{x})}d\theta\\
&=\frac{\beta^{\alpha}}{\Gamma(\alpha)}.\frac{\Gamma(\alpha+n+1)}{(\beta+n\bar{x})^{\alpha+n+1}}.
\end{align*}
As a result, the posterior mean, $\text{E}(\theta|x)$ is
\begin{equation*}
\text{E}(\theta|x)=\frac{\frac{\beta^{\alpha}}{\Gamma(\alpha)}.\frac{\Gamma(\alpha+n+1)}{(\beta+n\bar{x})^{\alpha+n+1}}}{\frac{\beta^{\alpha}}{\Gamma(\alpha)}.\frac{\Gamma(\alpha+n)}{(\beta+n\bar{x})^{\alpha+n}}}=\frac{\Gamma(\alpha+n+1)}{(\beta+n\bar{x})^{\alpha+n+1}}.\frac{(\beta+n\bar{x})^{\alpha+n}}{\Gamma(\alpha+n)}=\frac{\alpha+n}{\beta+n\bar{x}}.
\end{equation*}
