Following "Introduction to Mathematical Statistics" of Hogg et al. (Exercise 11.2.2) I am trying to calculate the posterior distribution of $\theta$ with the following information given:
Let $X_1,X_2,...,X_{10}$ be a random sample of size $n=10$ from a gamma distribution with $\alpha = 3$ and $\beta=\frac{1}{\theta}$. Suppose we believe that $\theta$ has a gamma distribution with $a=10$ and $b=2$.
I tried to calculate the posterior in general form first and then substitute the variables with the concrete values above. For the Prior I have:
$$ f_{\Theta}(\theta)=\frac{b^a}{\Gamma(a)} \cdot \theta^{a-1} \cdot e^{-b\theta} $$ And for the likelihood:
$$ f_{X\mid \Theta}(x\mid\theta)=\frac{\beta^\alpha}{\Gamma(\alpha)} \cdot x^{\alpha-1} \cdot e^{-\beta x}$$ Therefore for the posterior the following should be true:
$$ f_{\Theta \mid X_1,X_2,...,X_n}(\theta\mid x_1,x_2,...,x_n)\propto \prod_{I=1}^{n}\frac{\beta^\alpha}{\Gamma(\alpha)} \cdot x_i^{\alpha-1} \cdot e^{-\beta x_i} \cdot \frac{b^a}{\Gamma(a)} \cdot \theta^{a-1} \cdot e^{-b\theta}$$ Now removing all terms independent of $\theta$ my result is the following: $$f_{\Theta \mid X_1,X_2,...,X_n}(\theta\mid x_1,x_2,...,x_n)\propto \beta^{n\alpha} \cdot e^{-\beta\sum_{I=1}^{n}x_i}\cdot\theta^{a-1}\cdot e^{-b\theta}$$
Inserting $\beta=\frac{1}{\theta}$ and aggregating: $$f_{\Theta \mid X_1,X_2,...,X_n}(\theta\mid x_1,x_2,...,x_n)\propto \theta^{a-n\alpha-1} \cdot e^{-\frac{\sum_{I=1}^{n}x_i}{\theta}-b\theta}$$
Finally I am struggling to relate this to any common distribution. Also taking the following hint from the exercise into account: "Can the posterior distribution be related to a chi-square distribution?" Can anybody help?


1 Answer 1


I think it is only a matter of parametrizing the gamma density. If you consider $\beta,b$ as the scale parameter, Gamma density is the following


Thus the likelihood is

$$p(\mathbf{x}|\beta)\propto\theta^{30}\cdot e^{-\theta \Sigma_i X_i}$$

The prior is

$$\pi(\theta)\propto \theta^9\cdot e^{-\theta/2}$$

Thus the posterior is

$$\pi(\theta|\mathbf{x})\propto\theta^{39}\cdot e^{-\theta(1/2+\Sigma_i X_i)}$$

Now we immediately recognize that the posterior is a $Gamma\Big[40;\frac{2}{2 \Sigma_i X_i+1}\Big]$

Obviously it is related with a $\chi_{(80)}^2$ which can be obtained by a simple linear transformation of the Posterior


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