Can also see this note from Statistics Theory on Bayes estimation.

1 Bayesian Inference

${\displaystyle X}$: observed data. ${\displaystyle \mathrm{\Theta }}$: unknown parameter(s). All continuous random variables.
Law of Total Probability:$$f_X(x)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{f}_{x|\mathrm{\Theta }=\theta }(x)f_{\mathrm{\Theta }}(\theta )\mathrm{d}\theta .$$
Bayes Rule for Continuous RV:$$f_{\mathrm{\Theta }|X=x}(\theta )=\frac{f_{X|\mathrm{\Theta }=\theta }(x)\cdot f_{\mathrm{\Theta }}(\theta )}{{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{f}_{X|\mathrm{\Theta }=\theta }(x)f_{\mathrm{\Theta }}(\theta )\mathrm{d}\theta }.$$
If ${\displaystyle X}$ or ${\displaystyle \mathrm{\Theta }}$ is discrete, use p.m.f instead of p.d.f.

For ${\displaystyle d}$ dimensions, ${\displaystyle \overrightarrow{X}=(X_1,\cdots ,X_d),\overrightarrow{\theta }=({\theta }_1,\cdots ,{\theta }_d)}$, ${\displaystyle \sum _{i=1}^d{\theta }_i=1}$. ${\displaystyle (\overrightarrow{X}|\overrightarrow{\mathrm{\Theta }}=\overrightarrow{\theta })\sim \mathrm{Multinomial}(n,{\theta }_1,\cdots ,{\theta }_d)}$.$$\mathbb{P}(\overrightarrow{X}=\overrightarrow{x}|\overrightarrow{\mathrm{\Theta }}=\overrightarrow{\theta })=(\genfrac{}{}{0ex}{}{n}{x_1,\cdots ,x_d}){\theta }_1^{x_1}\cdots {\theta }_d^{x_d}\mathbf{1}\{\sum _{i=1}^n{x}_i=n\}\prod _{i=1}^d\{x_i\in \{0,\cdots ,n\}\}.$$

where$$\begin{array}{rl}& {\mu }_n=\left(\frac{{\sigma }^2}{n{\sigma }_0^2+{\sigma }^2}\right){\mu }_0+\left(\frac{n{\sigma }_0^2}{n{\sigma }_0^2+{\sigma }^2}\right)\stackrel{\frac{1}{n}\sum _{i=1}^n{X}_i}{\overbrace{{\mu }_{\mathrm{ML}}}},\\ & \frac{1}{{\sigma }_n^2}=\frac{1}{{\sigma }_0^2}+\frac{n}{{\sigma }^2}.\end{array}$$
(Precision = prior precision + data precision)

1. ${\displaystyle {\mu }_n\to {\mu }_{\mathrm{ML}}}$ as ${\displaystyle n\to \mathrm{\infty }}$.
2. Precisions are additive.
3. Precision gets large as sample size gets large.
4. For a finite ${\displaystyle n}$, if ${\displaystyle {\sigma }_0^2\to \mathrm{\infty }}$, then ${\displaystyle {\mu }_n\to {\mu }_{\mathrm{ML}}}$ and ${\displaystyle {\sigma }_n^2\to \frac{{\sigma }^2}{n}}$.

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2. ${\displaystyle \mu }$ known, ${\displaystyle {\sigma }^2}$ unknown.
   Put a prior on precision ${\displaystyle \mathrm{\Lambda }=\frac{1}{{\sigma }^2}}$. Then$$f_{\overrightarrow{X}|\mathrm{\Lambda }=\lambda }(\overrightarrow{x})={\left(\frac{\lambda }{2\pi }\right)}^{\frac{n}{2}}\exp \{-\frac{\lambda }{2}\sum _{i=1}^n(x_i-\mu {)}^2\}.$$Conjugate prior ${\displaystyle \mathrm{\Lambda }\sim \mathrm{Gamma}({\alpha }_0,{\beta }_0)}$:$$\begin{array}{rl}{f}_{\mathrm{\Lambda }}(\lambda )& =\frac{{\beta }_0^{{\alpha }_0}}{\mathrm{\Gamma }({\alpha }_0)}{\lambda }^{{\alpha }_0-1}{\mathrm{e}}^{-{\beta }_0\lambda }.\\ f_{\mathrm{\Lambda }|\overrightarrow{X}=\overrightarrow{x}}(\lambda )& \propto f_{\overrightarrow{X}|\mathrm{\Lambda }=\lambda }(\overrightarrow{x})f_{\mathrm{\Lambda }}(\lambda )\\ & \propto {\lambda }^{{\alpha }_0+\frac{n}{2}-1}\exp \{-\lambda [{\beta }_0+\frac{1}{2}\sum _{i=1}^n(x_i-\mu {)}^2]\},\end{array}$$ ${\displaystyle {\sigma }_{\mathrm{ML}}^2=\frac{1}{n}\sum _{i=1}^n(x_i-\mu {)}^2}$. Then ${\displaystyle (\mathrm{\Lambda }|\overrightarrow{X}=\overrightarrow{x})\sim \mathrm{Gamma}({\alpha }_n,{\beta }_n)}$. ${\displaystyle {\alpha }_n={\alpha }_0+\frac{n}{2},{\beta }_n={\beta }_0+\frac{n}{2}{\sigma }_{\mathrm{ML}}^2}$.

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3. Both ${\displaystyle \mu ,{\sigma }^2}$ unknown.$$f_{\overrightarrow{X}|M=\mu ,\mathrm{\Lambda }=\lambda }(\overrightarrow{x})\propto {\left[{\lambda }^{\frac{1}{2}}{\mathrm{e}}^{-\frac{\lambda {\mu }^2}{2}}\right]}^n\exp \{\lambda \mu \sum _{i=1}^n{x}_i-\frac{\lambda }{2}\sum _{i=1}^n{x}_i^2\}.$$ $$f_{M,\mathrm{\Lambda }}(\mu ,\lambda )=\underset{N({\mu }_0,\frac{1}{c\lambda })}{\underbrace{f_{M|\mathrm{\Lambda }=\lambda }(\mu )}}\underset{\mathrm{Gamma}(\alpha ,\beta )}{\underbrace{f_{\mathrm{\Lambda }}(\lambda )}},$$ where$${\mu }_0=\frac{a}{c},\alpha =1+\frac{c}{2},\beta =b-\frac{a^2}{2c},a,b,c>0.$$

2 Model Selection

${\displaystyle \mathrm{\Theta }\sim \mathrm{Bernoulli}\left(\frac{1}{2}\right)}$. Consider Model 0 vs. Model 1.
Prior odds: ${\displaystyle \frac{\mathbb{P}(\mathrm{\Theta }=0)}{\mathbb{P}(\mathrm{\Theta }=1)}}$.
${\displaystyle X_1,\cdots ,X_n|\mathrm{\Theta }=0\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }N(0,1)}$, p.d.f. ${\displaystyle f_0(x)=\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-\frac{x^2}{2}}}$.
${\displaystyle X_1,\cdots ,X_n|\mathrm{\Theta }=1\stackrel{\mathrm{i}.\mathrm{i}.\mathrm{d}}{\sim }\mathrm{Laplace}(0,\sqrt{\frac{\pi }{2}})}$, p.d.f. ${\displaystyle f_1(x)=\frac{1}{\sqrt{2\pi }}{\mathrm{e}}^{-|x|\sqrt{\frac{2}{\pi }}}}$.
Bayes Factor: ${\displaystyle \frac{f_{\overrightarrow{X}|\mathrm{\Theta }=0}(\overrightarrow{x})}{f_{\overrightarrow{X}|\mathrm{\Theta }=1}(\overrightarrow{x})}}$.
Posterior odds: ${\displaystyle \mathrm{BF}\times \text{Prior odds}}$.

Now suppose

Put priors on ${\displaystyle \alpha ,\beta }$ and get RV: ${\displaystyle A,B}$.

For example, $$\begin{array}{r}\log A|\mathrm{\Theta }=0\sim \mathrm{Uniform}(-c,c),\\ \log B|\mathrm{\Theta }=1\sim \mathrm{Uniform}(-c,c).\end{array}$$

${\displaystyle Z=\log A,A=T(Z)={\mathrm{e}}^Z}$. $$\begin{array}{rl}{f}_{A|\mathrm{\Theta }=0}(\alpha )=& f_{\log A|\mathrm{\Theta }=0}(\log \alpha )|\frac{\mathrm{d}}{\mathrm{d}\alpha }\log \alpha |\\ =& \frac{1}{\alpha }{f}_{\log A|\mathrm{\Theta }=0}(\log \alpha )=\frac{\mathbf{1}\{-c<\log \alpha <c\}}{2c\alpha },\\ f_{\overrightarrow{X}|\mathrm{\Theta }=1}(\overrightarrow{x})=& {\int }_0^{\mathrm{\infty }}{f}_{\overrightarrow{X}|\mathrm{\Theta }=1,B=\beta }(\overrightarrow{x})f_{B|\mathrm{\Theta }=1}(\beta )\mathrm{d}\beta .\\ f_{B|\mathrm{\Theta }=1}(\beta )=& \frac{\mathbf{1}\{-c<\log \beta <c\}}{2c\beta },\\ \mathrm{BF}=& \frac{f_{\overrightarrow{X}|\mathrm{\Theta }=0}(\overrightarrow{x})}{f_{\overrightarrow{X}|\mathrm{\Theta }=1}(\overrightarrow{x})}.\end{array}$$
(${\displaystyle c}$ cancels out in the Bayes Factor.)