# Two parameter model

### Background knowledge about Distribution

$$
\eta \sim \Gamma^{-1}(\alpha,\beta)  \\
p(\eta|\alpha,\beta)=\frac{\beta^\alpha}{\Gamma(\alpha)}(\frac{1}{\eta})^{\alpha+1}e^{-\frac{\beta}{\eta}} \\
Mean: ; \frac{\beta}{\alpha-1}, Var: ; \frac{\beta^2}{(\alpha-1)^2(\alpha-2)}
$$

$$
\theta \sim \chi^2(\nu)\equiv \Gamma(\nu/2,1/2)\ \theta \sim \chi^{-2}(\nu,\tau^2) \equiv \Gamma^{-1}(\nu/2,\nu\tau^2/2)
$$

## One parameter

### Normal model with known mean

$$
y|\sigma^2 \sim N(\mu, \sigma^2) \\
\sigma^2 \sim \chi^{-2}(\nu\_0, \sigma^2\_0) \\
\sigma^2|y \sim \chi^{-2}(\nu\_n,\sigma^2\_n)
$$

$$
\nu\_n=\nu\_0+n \\
\sigma^2\_n = \frac{\nu\_0\sigma^2\_0+ns(y)}{\nu\_0+n} \\
f(y|\sigma^2)=h(y)c(\sigma^2)exp(\sigma^2 s(y))
$$

## Two parameter

### Normal data with a conjugate prior

$$
p(y|\mu,\sigma^2) \propto \sigma^{-n}exp(-\frac{1}{2\sigma^2}\Sigma(y\_i-\mu)^2) \\

p(\mu, \sigma^2)=p(\mu|\sigma^2)p(\sigma^2) \propto \sigma^{-1}(\sigma^2)^{-(\frac{\nu\_0}{2}+1)}exp\[-\frac{1}{2\sigma^2}(\nu\_0\sigma^2\_0+\kappa\_0(\mu\_0-\mu)^2)] \ Ninv\chi^2(\mu\_0,\sigma^2\_0/\kappa\_0;\nu\_0,\sigma^2\_0)
\\
\mu|\sigma^2 \sim N(\mu\_0, \sigma^2/\kappa\_0) \\
\sigma^2 \sim \chi^{-2}(\nu\_0,\sigma^2\_0)
$$

$$
p(\mu,\sigma^2|y) \sim Ninv\chi^2(\mu\_n,\frac{\sigma^2\_n}{\kappa\_n};\nu\_n,\sigma^2\_n) \\
\mu\_n=\frac{\kappa\_0}{\kappa\_0+n}\mu\_0+\frac{n}{\kappa\_0+n}\bar{y} \\
\kappa\_n=\kappa\_0+n \\
\nu\_n=\nu\_0+n
$$

### Example

![](https://1943863620-files.gitbook.io/~/files/v0/b/gitbook-legacy-files/o/assets%2F-MZyJM_SVjd9SJ3tuTbA%2F-Ma4q2C8YjFmLGCIdgb-%2F-Ma4uya3pSepvAC78FJx%2Fimage.png?alt=media\&token=806a72ca-68f6-441b-b88e-0c2a6060515a)

(a) Give your posterior distribution for $$\theta$$

$$
y|\theta \sim N(\theta, 20^2) \\
\theta \sim N(180, 40^2) \\
\theta|y \sim N(\mu, \tau\_n^2) \\
\tau\_n^2=\frac{1}{1/\tau\_0^2 + n/\sigma^2}=\frac{1}{1/1600+n/400}=\frac{1600}{1+4n} \\
\mu\_n=\tau\_n^2(\frac{1}{\tau^2\_0}\mu\_0+\frac{n}{\sigma^2}\bar{y})=\frac{180+600n}{1+4n}
$$

(b) Give a posterior predictive distribution for $$\tilde{y}$$

$$
E(\tilde{y}|y)=E(E(\tilde{y}|\mu)|y)=E(\mu|y)=\mu\_n \\
V(\tilde{y}|y)=E(V(\tilde{y}|\mu)|y)+V(E(\tilde{y}|\mu)|y)=\sigma^2 + \tau^2\_n
$$

$$
\tilde{y}|y \sim N(\mu\_n, 400+\tau^2\_n)
$$

(c) 95% posterior interval for $$\theta$$ and posterior predictive interval for $$\tilde{y}$$(n=10)

95% posterior interval for $$\theta|y$$

$$
(\mu\_n-1.96\frac{\tau\_n}{\sqrt{10}},\mu+1.96\frac{\tau\_n}{\sqrt{10}}) \\
\because \theta|y \sim N(\mu\_n,\tau^2\_n)
$$

In the same way, we can also get posterior predictive interval with the distribution in (b)

(d) Do the same for n=100