计算机视觉考试题库 计算机视觉答案

文章目录

• 习题解答

• 3.1
• 3.2
• 3.3
• 3.4
• 3.5
• 3.6
• 3.7
• 3.8
• 3.9
• 3.10
• 3.11
• 3.12

习题解答

$\color{red}{答案仅供参考}$

3.1

$E[x]=1\times \lambda+0\times (1-\lambda)=\lambda$
$由上一章习题2.10可得：$
$E[(x-\mu)^2]=E[x^2]-E[x]E[x]$
$又E[x]=\lambda =\mu，所以$
$\begin{aligned} E[(x-E[x])^2] & =E[(x-\mu)^2] \\ & =E[x^2]-E[x]E[x]\\ & =1\times \lambda-\lambda^2\\ & =\lambda(1-\lambda) \end{aligned}$
$得证$

3.2

$Pr(\lambda)=\frac{\Gamma[\alpha+\beta]}{\Gamma[\alpha]\Gamma[\beta]}\lambda^{\alpha-1}(1-\lambda)^{\beta-1}$
$对\lambda进行求导$
$\begin{aligned} Pr'(\lambda) & =\frac{\Gamma[\alpha+\beta]}{\Gamma[\alpha]\Gamma[\beta]}[(\alpha-1)\lambda^{\alpha-2}(1-\lambda)^{\beta-1}+\lambda^{\alpha-1}(\beta-1)(1-\lambda)^{\beta-2}*(-1)] \end{aligned}$

$\lambda=\frac{\alpha-1}{(\alpha-1)+(\beta-1)}$
$代入可得$
$\text {mode}=\frac{\Gamma[\alpha+\beta]}{\Gamma[\alpha]\Gamma[\beta]}\left(\frac{\alpha-1}{(\alpha-1)+(\beta-1)}\right)^{\alpha-1}\left(\frac{\beta-1}{(\alpha-1)+(\beta-1)} \right)^{\beta-1}$

3.3

$联列方程组$
$\begin{dcases} & \mu=\frac{\alpha}{\alpha+\beta}\\ & \sigma^2=\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)} \end{dcases}$
$从上式得到\beta=\frac{\alpha}{\mu}-\alpha，代入下式得到$
$\begin{dcases} & \alpha=\frac{\mu^2(1-\mu)}{\sigma^2}-\mu \\ & \beta=\frac{\mu(1-\mu^2)}{\sigma^2}+\mu-1 \end{dcases}$

3.4

$已知贝塔分布$
$\begin{aligned} Pr(\lambda) & =\frac{\Gamma[\alpha+\beta]}{\Gamma[\alpha]\Gamma[\beta]}\lambda^{\alpha-1}(1-\lambda)^{\beta-1} \\ & =\text{exp}[log[\Gamma[\alpha+\beta]]+(\alpha-1)log[\lambda]+(\beta-1)log[1-\lambda]-log[\Gamma[\alpha]]-log[\Gamma[\beta]]] \\ &=\text{exp}[(\alpha-1)log[\lambda]+(\beta-1)log[1-\lambda]-[log[\Gamma[\alpha]]+log[\Gamma[\beta]]-log[\Gamma[\alpha+\beta]]]] \end{aligned}$
$已知指数族的形式为$
$Pr(x|\bm\theta)=a[x]\text{exp}[\bm b[\bm\theta]^{\text{T}}\bm c[x]-d[\bm\theta]]$
$对比可得$
$\begin{dcases} a[x]& =1 \\ \bm b[\bm\theta] &= \begin{bmatrix} \alpha-1 \\ \beta-1 \end{bmatrix} \\ & \;\\ \bm c[x]&=\begin{bmatrix} log[x] \\ log[1-x] \end{bmatrix} \\ d[\bm\theta]& =log[\Gamma[\alpha]]+log[\Gamma[\beta]]-log[\Gamma[\alpha+\beta]] \end{dcases}$

3.5

$\begin{aligned} \Gamma[z+1] & =\displaystyle\int_0^\infty t^{z}\text e^{-t}\text dt\\ & =-\int_0^\infty t^z\text d(\text e^{-t}) \\ & =-\left. t^z\text e^{-t} \right|_0^\infty+\int_0^\infty \text e^{-t}\text d(t^z)\\ & =0+z\int\text e^{-t}t^{z-1} \text dt\\ & =z\Gamma[z] \end{aligned}$
$得证$

3.6

书上关于共轭分布讲的比较简单，较为详细的可以参见https://www.jianshu.com/p/bb7bce40a15a

$需要证明先验分布和后验分布有相同的形式，即证明$
$Pr(\mu)*Pr(x|\mu)\propto Pr(\mu)$
$\begin{aligned} Pr(x|\mu)*Pr(\mu) & =\frac{1}{2\pi \sigma_p}\text{exp}\left[-0.5\left[(x-\mu)^2+\frac{}{}\frac{(\mu-\mu_p)^2}{\sigma_p^2} \right] \right] \\ & =\frac{1}{2\pi \sigma_p}\text{exp}\left[-0.5\left[x^2-2\mu x+\mu^2+\frac{\mu^2}{\sigma_p^2}-\frac{2\mu\mu_p}{\sigma_p^2}+\frac{\mu_p^2}{\sigma_p^2}\right] \right] \\ & =\frac{1}{2\pi \sigma_p}\text{exp}\left[-0.5\left[x^2+\frac{\mu_p^2}{\sigma_p^2}\right] -0.5\left[\mu^2(1+\frac{1}{\sigma_p^2})-2\mu(x+\frac{\mu_p}{\sigma_p^2})\right] \right] \\ &(因为提取出来的前两项与\mu无关，可以将其作为常量)\\ & =\kappa_1\text{exp}\left[-0.5\left( \frac{\sigma_p^2+1}{\sigma_p^2}\mu^2-2\frac{\sigma_p^2x+\mu_p}{\sigma_p^2}\mu\right) \right]\\ &(这里使用配方法)\\ & =\kappa_1\text{exp}\left[-0.5\left[ \frac{\left(\mu-\frac{\sigma_p^2x+\mu_p}{\sigma_p^2+1}\right)^2}{\frac{\sigma_p^2}{\sigma_p^2+1}}-常量\right] \right]\\ & =\kappa_2\text{exp}\left[-0.5 \frac{\left(\mu-\frac{\sigma_p^2x+\mu_p}{\sigma_p^2+1}\right)^2}{\frac{\sigma_p^2}{\sigma_p^2+1}}\right]\\ & \propto Pr(\mu) \end{aligned}$
$得证$

3.7

$和3.4相似，已知正态分布$
$\begin{aligned} Pr(x) & =\frac{1}{\sqrt{2\pi\sigma^2}}\text{exp}\left[ -0.5\frac{(x-\mu)^2}{\sigma^2}\right]\\ & =\text{exp}\left[ -0.5(\frac{x^2}{\sigma^2}-\frac{2\mu x}{\sigma^2}+\frac{\mu^2}{\sigma^2})-log\left[\sqrt{2\pi\sigma^2} \right]\right] \end{aligned}$
$已知指数族的形式为$
$Pr(x|\bm\theta)=a[x]\text{exp}[\bm b[\bm\theta]^{\text{T}}\bm c[x]-d[\bm\theta]]$
$对比可得$
$\begin{dcases} a[x]& =1 \\ \bm b[\bm\theta] &= \begin{bmatrix} 1/\sigma^2\\ \mu/\sigma^2 \end{bmatrix} \\ & \;\\ \bm c[x]&=\begin{bmatrix} -0.5x^2\\ x \end{bmatrix} \\ d[\bm\theta]& =\frac{\mu^2}{2\sigma^2}+log\left[ \sqrt{2\pi\sigma^2}\right] \end{dcases}$

3.8

$已知$
$Pr(\mu,\sigma^2)=\frac{\sqrt{\gamma}}{\sigma\sqrt{2\pi}}\frac{\beta^\alpha}{\Gamma[\alpha]}\left(\frac{1}{\sigma^2} \right)^{\alpha+1}\text{exp}\left[-\frac{2\beta+\gamma(\delta-\mu)^2}{2\sigma^2}\right]$
$对\mu进行求导$
$\begin{aligned} \frac{\partial Pr(\mu,\sigma^2)}{\partial \mu}=\text constant\cdot \text e[\cdots]\left( -\frac{1}{2\sigma^2}\right)(2\gamma)(\delta-\mu)(-1) \end{aligned}$
$令\frac{\partial Pr(\mu,\sigma^2)}{\partial \mu}=0，解得$
$\mu=\delta$

$对\sigma^2进行求导$
$\begin{aligned} \frac{\partial Pr(\mu,\sigma^2)}{\partial \sigma^2}=balabala \end{aligned}$
$令\frac{\partial Pr(\mu,\sigma^2)}{\partial \sigma^2}=0，代入\mu的值，解得$
$\sigma^2=\frac{2\beta}{2\alpha+3}$
$综上$
$\begin{dcases} \mu&=\delta \\ \sigma^2&=\frac{2\beta}{2\alpha+3} \end{dcases}$
$代入可得$
$\text {mode}=balabala(自己代进去就可以)$

3.9

$\begin{aligned} &\prod_{i=1}^I \text{Bern}_{x_i}[\lambda]\cdot\text{Beta}_\lambda[\alpha,\beta] \\ =&\prod_{i=1}^I\lambda^{x_i}(1-\lambda)^{1-x_i}\frac{\Gamma[\alpha+\beta]}{\Gamma[\alpha]\Gamma[\beta]}\lambda^{\alpha-1}(1-\lambda)^{\beta-1}\\ =& \kappa\cdot \frac{\Gamma[\alpha+\beta+I]}{\Gamma\left [\alpha+\sum x_i\right]\Gamma[\beta+\sum(1-x_i)]}\lambda^{\alpha+\sum x_i-1}(1-\lambda)^{\beta+\sum(1-x_i)-1}\\ =& \kappa\cdot\text{Beta}_\lambda[\widetilde{\alpha},\widetilde{\beta}] \end{aligned}$
$得证$

3.10

$\begin{aligned} &\prod_{i=1}^I\text{Cat}_{x_i}[\lambda_{1\cdots K}]\cdot \text{Dir}_{\lambda_{1\cdots K}}[\alpha_{1\cdots K}] \\ =& \left[\prod_{i=1}^{I}\prod_{j=1}^{K}\lambda_j^{x_{ij}} \right]\frac{\Gamma\left[\displaystyle\sum_{j=1}^{K}\alpha_j\right]}{\displaystyle\prod_{j=1}^K\Gamma[\alpha_j]}\prod_{j=1}^K\lambda_j^{\alpha_j-1}\\ &(x_{ij}是x_i 取到j号结果的次数) \\ =& \left[\prod_{j=1}^{K}\lambda_j^{\sum_{i=1}^I x_{ij}} \right] \frac{\Gamma\left[\displaystyle\sum_{j=1}^{K}\alpha_j\right]}{\displaystyle\prod_{j=1}^K\Gamma[\alpha_j]} \prod_{j=1}^K\lambda_j^{\alpha_j-1} \\ =& \frac{\Gamma\left[\displaystyle\sum_{j=1}^{K}\alpha_j\right]}{\displaystyle\prod_{j=1}^K\Gamma[\alpha_j]} \prod_{j=1}^K\lambda_j^{\alpha_j-1+N_j} \\ &(设\displaystyle\sum_{i=1}^I x_{ij}=N_j，即取I次总共取到j号结果的总次数) \\ =& \frac{\Gamma\left[\displaystyle\sum_{j=1}^{K}\alpha_j\right]}{\displaystyle\prod_{j=1}^K\Gamma[\alpha_j]} \frac{\displaystyle\prod_{j=1}^K\Gamma[\alpha_j+N_j]}{\Gamma\left[\displaystyle\sum_{j=1}^{K}(\alpha_j+N_j)\right]} \frac{\Gamma\left[\displaystyle\sum_{j=1}^{K}(\alpha_j+N_j)\right]}{\displaystyle\prod_{j=1}^K\Gamma[\alpha_j+N_j]} \prod_{j=1}^K\lambda_j^{\alpha_j-1+N_j} \\ &（\because \displaystyle\sum_{j=1}^{K}N_j=I，即取I次总共取到所有结果的总次数，因为取了I次，所以等于I）\\ =& \frac{\Gamma\left[\displaystyle\sum_{j=1}^{K}\alpha_j\right]}{\displaystyle\prod_{j=1}^K\Gamma[\alpha_j]} \frac{\displaystyle\prod_{j=1}^K\Gamma[\alpha_j+N_j]}{\Gamma\left[I+\displaystyle\sum_{j=1}^{K}\alpha_j\right]} \cdot \text{Dir}_{\lambda_{1\cdots K}}\left[\widetilde{\alpha_{1\cdots K}} \right] \\ & \\ =& \widetilde{\kappa} \cdot \text{Dir}_{\lambda_{1\cdots K}}\left[\widetilde{\alpha_{1\cdots K}} \right] \end{aligned}$
$得证$

3.11

$\begin{aligned} & \prod_{i=1}^I\text{Norm}_{x_i}\left[\mu,\sigma^2\right] \cdot \text{NormInvGam}_{\mu,\sigma^2}[\alpha,\beta,\gamma,\delta] \\ =& \prod_{i=1}^I \frac{1}{\sqrt{2\pi \sigma^2}} \text{exp}\left[ -0.5\frac{(x_i-\mu)^2}{\sigma^2} \right] \cdot \frac{\sqrt{\gamma}}{\sigma\sqrt{2\pi}} \frac{\beta^{\alpha}}{\Gamma[\alpha]} \left(\frac{1}{\sigma^2} \right)^{\alpha+1} \text{exp}\left[ -\frac{2\beta+\gamma(\delta-\mu)^2}{2\sigma^2} \right] \\ =& \kappa \cdot \frac{\sqrt{\widetilde\gamma}\widetilde\beta^{\widetilde\alpha}}{\Gamma[\widetilde\alpha]} \left(\frac{1}{\sigma}\right)^{2\alpha+2+I} \frac{1}{\sigma\sqrt{2\pi}} \text{exp}\left[ -\frac{\displaystyle\sum_{i=1}^I(x_i-\mu)^2+2\beta+\gamma(\delta-\mu)^2}{2\sigma^2} \right] \\ =& \kappa \cdot \frac{\sqrt{\widetilde\gamma}}{\sigma\sqrt{2\pi}} \frac{\widetilde\beta^{\widetilde\alpha}}{\Gamma[\widetilde\alpha]} \left(\frac{1}{\sigma^2} \right)^{\widetilde\alpha+1} \text{exp}\left[ -\frac{2\widetilde\beta+\widetilde\gamma(\widetilde\delta-\mu)^2}{2\sigma^2} \right] \\ =& \kappa \cdot \text{NormInvGam}_{\mu,\sigma^2}[\widetilde\alpha,\widetilde\beta,\widetilde\gamma,\widetilde\delta] \end{aligned}$
$得证$

3.12

待定
$\begin{aligned} & \prod_{i=1}^I\text{Norm}_{\bm{x_i}}\left[\bm\mu,\bm\Sigma\right] \cdot \text{NorIWis}_{\bm\mu,\bm\Sigma}[\alpha,\bm\Psi,\gamma,\bm\delta] \\ =& \prod_{i=1}^I \frac{1}{(2\pi)^{D/2}|\bm\Sigma|^{1/2}} \text{exp} \left[ -0.5(\bm x-\bm\mu)^\text T\bm\Sigma^{-1}(\bm x-\bm\mu) \right] \cdot \frac{\gamma^{D/2}|\bm\Psi|^{\alpha/2}|\bm\Sigma|^{-(\alpha+D+2)/2}}{2^{\alpha D/2}(2\pi)^{D/2}\Gamma_D[\alpha/2]} \text{exp}\left[ -0.5\left( \text{Tr}[\bm\Psi\bm\Sigma^{-1}]+\gamma(\bm\mu-\bm\delta)^{\text T}\bm\Sigma^{-1}(\bm\mu-\bm\delta) \gamma^{D/2} \right) \right] \\ =& \kappa \cdot \frac{\widetilde{\gamma}^{D/2}|\widetilde{\bm\Psi}|^{\widetilde{\alpha}/2}|\bm\Sigma|^{-(\alpha+D+2+I)/2}}{2^{D(\alpha+I)/2}(2\pi)^{D/2}\Gamma_D[\widetilde{\alpha}/2]} \\ =& \kappa \cdot \frac{\widetilde{\gamma}^{D/2}|\widetilde{\bm\Psi}|^{\widetilde{\alpha}/2}|\bm\Sigma|^{-(\widetilde\alpha+D+2)/2}}{2^{D\widetilde\alpha/2}(2\pi)^{D/2}\Gamma_D[\widetilde{\alpha}/2]} \end{aligned}$