习题解答

$答案仅供参考$

4.1

$$\begin{array}{rl}\hat{\mu },{\hat{\sigma }}^2& =\underset{\mu ,{\sigma }^2}{\mathrm{argmax}}\left[\sum _{i=1}^I\text{log}\left[{\text{Norm}}_{x_i}[\mu ,{\sigma }^2]\right]\right]\\ & =\underset{\mu ,{\sigma }^2}{\mathrm{argmax}}\left[-0.5I\text{log}[2\pi ]-0.5I\text{log}{\sigma }^2-0.5\sum _{i=1}^I\frac{(x_i-\mu {)}^2}{{\sigma }^2}\right]\end{array}$$
$求似然对数L对{\sigma }^2的微分，并令结果为0$
$$\begin{array}{rl}\frac{\partial L}{\partial {\sigma }^2}& =-0.5I\frac{1}{{\sigma }^2}+0.5\sum _{i=1}^I\frac{(x_i-\mu {)}^2}{{\sigma }^4}=0\end{array}$$
$整理得到$
$${\hat{\sigma }}^2=\sum _{i=1}^I\frac{(x_i-\hat{\mu }{)}^2}{I}$$
$得证$

4.2

$$\begin{array}{rl}\hat{\mu },{\hat{\sigma }}^2& =\underset{\mu ,{\sigma }^2}{\mathrm{argmax}}\left[\sum _{i=1}^I\text{log}\left[{\text{Norm}}_{x_i}[\mu ,{\sigma }^2]\right]+\text{log}[{\text{NormInvGam}}_{\mu ,{\sigma }^2}[\alpha ,\beta ,\gamma ,\delta ]]\right]\\ & =\underset{\mu ,{\sigma }^2}{\mathrm{argmax}}\left[-0.5I\text{log}[2\pi ]-0.5I\text{log}{\sigma }^2-0.5\sum _{i=1}^I\frac{(x_i-\mu {)}^2}{{\sigma }^2}+\text{log}\left[\frac{\sqrt{\gamma }{\beta }^{\alpha }}{\sqrt{2\pi }\Gamma [\alpha ]}\right]-(\alpha +1.5)\text{log}[{\sigma }^2]-\frac{2\beta +\gamma (\delta -\mu {)}^2}{2{\sigma }^2}\right]\end{array}$$
$求似然对数L对\mu 的微分，并令结果为0$
$$\begin{array}{rl}\frac{\partial L}{\partial \mu }& =\sum _{i=1}^I\frac{x_i-\mu }{{\sigma }^2}+\frac{\gamma (\delta -\mu )}{{\sigma }^2}\\ & =\frac{{\sum }_{i=1}^I{x}_i-I\mu +\gamma \delta -\gamma \mu }{{\sigma }^2}\\ & =0\end{array}$$
$整理得到$
$$\hat{\mu }=\frac{{\sum }_{i=1}^I{x}_i+\gamma \delta }{I+\gamma }$$
$同理求似然对数L对{\sigma }^2的微分，并令结果为0$
$$\begin{array}{rl}\frac{\partial L}{\partial {\sigma }^2}& =-\frac{I}{2{\sigma }^2}+\frac{\sum (x_i-\mu {)}^2}{2{\sigma }^4}-\frac{2\alpha +3}{2{\sigma }^2}+\frac{2\beta +\gamma (\delta -\mu {)}^2}{2{\sigma }^4}\\ & =\frac{\sum (x_i-\mu {)}^2+2\beta +\gamma (\delta -\mu {)}^2}{2{\sigma }^4}-\frac{I+3+2\alpha }{2{\sigma }^2}\\ & =0\end{array}$$
$整理得到$
$$\widehat{{\sigma }^2}=\frac{{\sum }_{i=1}^I(x_i-\mu {)}^2+2\beta +\gamma (\delta -\mu {)}^2}{I+3+2\alpha }$$

4.3

$已知$
$$L=\sum _{k=1}^6{N}_k\text{log}[{\lambda }_k]+\nu \left(\sum _{k=1}^6{\lambda }_k-1\right)$$
$求似然对数L对{\lambda }_k的微分，并令结果为0$
$$\begin{array}{rl}\frac{\partial L}{\partial {\lambda }_k}& =\frac{N_k}{{\lambda }_k}+\nu \\ & =0\end{array}$$
$整理得到$
$$\widehat{{\lambda }_k}=\frac{N_k}{-\nu }$$
$又因为$
$$\sum _{k=1}^6{\lambda }_k=1$$
$所以$
$$-\nu =\sum _{m=1}^6{N}_m$$
$综上$
$$\widehat{{\lambda }_k}=\frac{N_k}{{\sum }_{m=1}^6{N}_m}$$
$得证$

4.4

$已知$
$$\begin{array}{rl}{\hat{\lambda }}_{1\cdots 6}& =\underset{{\lambda }_{1\cdots 6}}{\mathrm{argmax}}\left[\prod _{i=1}^{I}Pr(x_i|{\lambda }_{1\cdots 6})Pr({\lambda }_{1\cdots 6})\right]\\ & =\underset{{\lambda }_{1\cdots 6}}{\mathrm{argmax}}\left[\prod _{i=1}^I{\text{Cat}}_{x_i}[{\lambda }_{1\cdots 6}]{\text{Dir}}_{{\lambda }_{1\cdots 6}}[{\alpha }_{1\cdots 6}]\right]\\ & =\underset{{\lambda }_{1\cdots 6}}{\mathrm{argmax}}\left[\prod _{k=1}^6{\lambda }_k^{N_k}\cdot (与{\lambda }_k无关的量)\cdot \prod _{k=1}^6{\lambda }_k^{{\alpha }_k-1}\right]\\ & =\underset{{\lambda }_{1\cdots 6}}{\mathrm{argmax}}\left[\prod _{k=1}^6{\lambda }_k^{N_k+{\alpha }_k-1}\right]\end{array}$$
$通过拉格朗日因子增强约束，似然对数为$
$$L=\sum _{k=1}^6(N_k+{\alpha }_k-1)\text{log}[{\lambda }_k]+\nu (\sum _{k=1}^6{\lambda }_k-1)$$
$求似然对数L对{\lambda }_k的微分，并令结果为0$
$$\begin{array}{rl}\frac{\partial L}{\partial {\lambda }_k}& =\frac{N_k+{\alpha }_k-1}{{\lambda }_k}+\nu \\ & =0\end{array}$$
$整理得到$
$$\widehat{{\lambda }_k}=\frac{N_k+{\alpha }_k-1}{-\nu }$$
$又因为$
$$\sum _{k=1}^6{\lambda }_k=1$$
$所以$
$$-\nu =\sum _{m=1}^6(N_m+{\alpha }_m-1)$$
$综上$
$$\widehat{{\lambda }_k}=\frac{N_k+{\alpha }_k-1}{{\sum }_{m=1}^6(N_m+{\alpha }_m-1)}$$
$得证$

4.5

$(i)$
$$\begin{array}{rl}Pr(x_{1\cdots I})& =\int \prod _{i=1}^{I}Pr(x_i|\theta )Pr(\theta )\text{d}\theta \\ & =\iint \prod _{i=1}^I{\text{Norm}}_{x_i}[\mu ,{\sigma }^2]\cdot {\text{NormInvGam}}_{\mu ,{\sigma }^2}[\alpha ,\beta .\gamma ,\delta ]\text{d}\mu \text{d}{\sigma }^2\\ & =\iint \kappa \cdot {\text{NormInvGam}}_{\mu ,{\sigma }^2}[\tilde{\alpha },\tilde{\beta },\tilde{\gamma },\tilde{\delta }]\text{d}\mu \text{d}{\sigma }^2\\ & =\kappa =balabala\end{array}$$

$(ii)$
$$\begin{array}{rl}Pr(x_{1\cdots I})& =\int \prod _{i=1}^{I}Pr(x_i|\theta )Pr(\theta )\text{d}\theta \\ & =\int \prod _{i=1}^I{\text{Cat}}_{x_i}[{\lambda }_{1\cdots I}]\cdot {\text{Dir}}_{{\lambda }_{1\cdots I}}[{\alpha }_{1\cdots I}]\text{d}{\lambda }_{1\cdots I}\\ & =\int \kappa \cdot {\text{Dir}}_{{\lambda }_{1\cdots I}}[{\tilde{\alpha }}_{1\cdots I}]\text{d}{\lambda }_{1\cdots I}\\ & =\kappa =balabala\end{array}$$

4.6

$ToDo$

4.7

$$\begin{array}{rl}\hat{\lambda }& =\underset{\lambda }{\mathrm{argmax}}\left[\sum _{i=1}^I\text{log}\left[{\text{Bern}}_{x_i}[\lambda ]\right]\right]\\ & =\underset{\lambda }{\mathrm{argmax}}\left[\left(\sum _{i=1}^I{x}_i\right)\text{log}[\lambda ]+\left(\sum _{i=1}^{I}1-x_i\right)\text{log}[1-\lambda ]\right]\end{array}$$
$求似然对数L对\lambda 的微分，并令结果为0$
$$\begin{array}{rl}\frac{\partial L}{\partial \lambda }& =\frac{{\sum }_{i=1}^I{x}_i}{\lambda }-\frac{{\sum }_{i=1}^{I}1-x_i}{1-\lambda }\\ & =0\end{array}$$
$整理得到$
$$\hat{\lambda }=\frac{{\sum }_{i=1}^I{x}_i}{I}$$

4.8

$$\begin{array}{rl}\hat{\lambda }& =\underset{\lambda }{\mathrm{argmax}}\left[\text{log}\left[\frac{{\prod }_{i=1}^{I}Pr(x_i|\lambda )Pr(\lambda )}{Pr(x_{1\cdots I})}\right]\right]\\ & =\underset{\lambda }{\mathrm{argmax}}\left[\text{log}\left[\prod _{i=1}^{I}Pr(x_i|\lambda )Pr(\lambda )\right]\right]\\ & =\underset{\lambda }{\mathrm{argmax}}\left[\sum _{i=1}^I\text{log}\left[{\text{Bern}}_{x_i}[\lambda ]\right]+\text{log}[{\text{Beta}}_{\lambda }[\alpha ,\beta ]]\right]\\ & =\underset{\lambda }{\mathrm{argmax}}\left[\left(\sum _{i=1}^I{x}_i\right)\text{log}[\lambda ]+\left(\sum _{i=1}^{I}1-x_i\right)\text{log}[1-\lambda ]+(\alpha -1)\text{log}[\lambda ]+(\beta -1)\text{log}[1-\lambda ]\right]\end{array}$$

$求似然对数L对\lambda 的微分，并令结果为0$
$$\begin{array}{rl}\frac{\partial L}{\partial \lambda }& =\frac{{\sum }_{i=1}^I{x}_i}{\lambda }-\frac{{\sum }_{i=1}^{I}1-x_i}{1-\lambda }+\frac{\alpha -1}{\lambda }-\frac{\beta -1}{1-\lambda }\\ & =0\end{array}$$
$整理得到$
$$\hat{\lambda }=\frac{{\sum }_{i=1}^I{x}_i+\alpha -1}{I+\alpha +\beta -2}$$

4.9

(i)
$$\begin{array}{rl}Pr(\lambda |x_{1\cdots I})& =\frac{Pr(x_{1\cdots I}|\lambda )Pr(\lambda )}{Pr(x_{1\cdots I})}\\ & =\frac{{\prod }_{i=1}^I{\text{Bern}}_{x_i}[\lambda ]\cdot {\text{Beta}}_{\lambda }[\alpha ,\beta ]}{Pr(x_{1\cdots I})}\\ & =\frac{\kappa \cdot {\text{Beta}}_{\lambda }[\tilde{\alpha },\tilde{\beta }]}{Pr(x_{1\cdots I})}\\ & ={\text{Beta}}_{\lambda }[\tilde{\alpha },\tilde{\beta }]\end{array}$$

(ii)
$$\begin{array}{rl}Pr(x^{\ast }|x_{1\cdots I})& =\int Pr(x^{\ast }|\lambda )Pr(\lambda |x_{1\cdots I})\text{d}\lambda \\ & =\int {\text{Bern}}_{x^{\ast }}[\lambda ]\cdot {\text{Beta}}_{\lambda }[\tilde{\alpha },\tilde{\beta }]d\lambda \\ & =\int \kappa (x^{\ast },\tilde{\alpha },\tilde{\beta }){\text{Beta}}_{\lambda }[\breve{\alpha },\breve{\beta }]d\lambda \\ & =\kappa (x^{\ast },\tilde{\alpha },\tilde{\beta })\end{array}$$

4.10

$方法与上面一致，过程略去$
(i)
$$\hat{\lambda }=\frac{\sum x_i}{I}=0$$
$计算Pr(x^{\ast }|\hat{\lambda })$

(ii)
$$\hat{\lambda }=\frac{{\sum }_{i=1}^I{x}_i+\alpha -1}{I+\alpha +\beta -2}=0$$
$计算Pr(x^{\ast }|\hat{\lambda })$
(iii)
$计算Pr(x^{\ast }|x_{1\cdots 4})=\kappa (x^{\ast },\tilde{\alpha },\tilde{\beta })$