# Posterior for Gamma prior and Gamma likelihood with known shape

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?

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

$$f_X(x)=\frac{1}{\Gamma(\alpha)\beta^\alpha}x^{\alpha-1}e^{-x/\beta}$$

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