PRML 5章演習問題　5.33～5.37

5.33

関節の位置
$\displaystyle{ x_{0}=L1\cos(\theta_{1})\\ y_{0}=L1\sin(\theta_{1}) }$
関節の位置から見た先端の位置
$\displaystyle{ x_{10}=L2\cos(\theta_{2}-(\pi-\theta_{1}))=-L2\cos(\theta_{1}+\theta_{2}))\\ y_{10}=L2\sin(\theta_{2}-(\pi-\theta_{1}))=-L2\sin(\theta_{1}+\theta_{2})) }$
先端の位置
$\displaystyle{ x_{1}=L1\cos(\theta_{1})-L2\cos(\theta_{1}+\theta_{2}))\\ x_{2}=L1\sin(\theta_{1})-L2\sin(\theta_{1}+\theta_{2})) }$

5.34

$E_{n}=-\ln \sum_{k}\pi_{k} N_{nk}$
より
$\displaystyle{ \frac{\partial E_{n}}{\partial a_{m}^{\pi}}= \frac{\partial E_{n}}{\partial p(t|x_{n})}\frac{\partial p(t|x_{n})}{\partial a_{m}^{\pi}}= -\frac{1}{p(t|x_{n})}\sum_{k}\frac{\partial \pi_{k}}{\partial a_{m}^{\pi}}N_{nk}\\ \hspace{15pt} =-\frac{1}{p(t|x_{n})}\sum_{k} \pi_{k}(I_{km}-\pi_{m})N_{nk} =\frac{\pi_{m} \sum_{k} \pi_{k}N_{nk}-\pi_{m}N_{nm}}{p(t|x_{n})}\\ \hspace{15pt} =\pi_{m}-\gamma_{nm} }$

5.35

$\displaystyle{ \frac{\partial E_{n}}{\partial a_{ml}^{\mu}}= \frac{\partial E_{n}}{\partial p(t|x_{n})}\frac{\partial p(t|x_{n})}{\partial a_{ml}^{\mu}}= -\frac{1}{p(t|x_{n})}\sum_{k}\pi_{k}\frac{\partial N_{nk}}{\partial a_{ml}^{\mu}}\\ \hspace{15pt}= -\frac{\pi_{m} N_{nm}}{p(t|x_{n})}\frac{t_{nk}-\mu_{ml}}{\sigma_{m}^{2}}\\ \hspace{15pt}= \gamma_{nm} \frac{\mu_{ml}-t_{nk}}{\sigma_{m}^{2}} }$

式変形に
$\displaystyle{ \frac{\partial }{\partial \mu_{ml}}\exp \left(-\frac{1}{2\sigma^{2}}||t-\mu_{m}||^{2} \right)= \exp \left(-\frac{1}{2\sigma^{2}}||t-\mu_{m}||^{2} \right)\frac{t-\mu_{m}}{\sigma^{2}}(-1_{l})=\\ \hspace{15pt} \exp \left(-\frac{1}{2\sigma^{2}}||t-\mu_{m}||^{2} \right)\frac{-t_{l}+\mu_{ml}}{\sigma^{2}},\,\,\,1_{l}:l要素のみ1、他0のベクトル }$
を用いた。

5.36

$\displaystyle{ \frac{\partial E_{n}}{\partial a_{m}^{\sigma}}= \frac{\partial E_{n}}{\partial p(t|x_{n})}\frac{\partial p(t|x_{n})}{\partial a_{m}^{\sigma}}= \frac{\partial E_{n}}{\partial p(t|x_{n})}\frac{\partial p(t|x_{n})}{\partial \sigma_{m}} \frac{\partial \sigma_{m}}{\partial a_{m}^{\sigma}}\\ \hspace{15pt}= -\frac{1}{p(t|x_{n})}\sum_{k}\pi_{k}\frac{\partial N_{nk}}{\partial \sigma_{m}}\sigma_{m}= -\frac{\pi_{m} N_{nm}}{p(t|x_{n})} \left( \frac{-L}{\sigma_{m}}+\frac{||t_{n}-\mu_{m}||^{2}}{\sigma_{m}^{3}} \right) \sigma_{m}\\ \hspace{15pt}= \gamma_{nm} \left(L-\frac{||t_{n}-\mu_{m}||^{2}}{\sigma_{m}^{3}}\right) }$

式変形に
$\displaystyle{ \frac{\partial }{\partial \sigma} \left\{\frac{1}{(2\pi)^{L/2}}\frac{1}{|\sigma^{2}I|}\exp \left(-\frac{1}{2\sigma^{2}}||t-\mu||^{2} \right)\right\}\\= \frac{1}{(2\pi)^{L/2}}\frac{-L}{\sigma}\frac{1}{|\sigma^{2}I|}\exp \left(-\frac{1}{2\sigma^{2}}||t-\mu||^{2} \right)+ \frac{1}{(2\pi)^{L/2}} \frac{1}{|\sigma^{2}I|} \exp \left(-\frac{1}{2\sigma^{2}} ||t-\mu||^{2} \right) \left(\frac{||t-\mu||^{2}}{\sigma^{3}} \right)\\= \frac{1}{(2\pi)^{L/2}}\frac{1}{|\sigma^{2}I|}\exp \left(-\frac{1}{2\sigma^{2}}||t-\mu||^{2} \right) \left(\frac{-L}{\sigma}+\frac{||t-\mu||^{2}}{\sigma^{3}} \right) }$
を用いた。

5.37

(5.158)

$\displaystyle{ E[t|x]=\int t p(t|x)dt=\sum_{k} \pi_{k} \int t N_{k} dt=\sum\pi_{k}\mu_{k} }$

(5.160)

(2.62)より
$\displaystyle{ \int t^{T}t N_{k}dt = Tr[\int tt^{T} N_{k}dt ] = Tr[\mu_{k}\mu_{k}^T+\sigma_{k}^{2}I]= \mu_{k}^{T}\mu_{k}+L\sigma_{k}^{2} }$
であることに注意すると

$\displaystyle{ E[||t-E[t|x]||^{2}|x]= \int \{t^{T}t-E[t|x]^{T}t-t^{T}E[t|x]+E[t|x]^{T}E[t|x]\} \sum_{k}\pi_{k}N_{k}dt\\ \hspace{15pt}= \sum_{k} \pi_{k} \{\int t^{T}tN_{k}dt-2E[t|x]^{T}\int tN_{k}dt +E[t|x]^{T}E[t|x]\}\\ \hspace{15pt}= \sum_{k} \pi_{k} \{ \mu_{k}^{T}\mu_{k}+L\sigma_{k}^{2} -2E[t|x]^{T}\mu_{k} +E[t|x]^{T}E[t|x] \}\\ \hspace{15pt}= \sum_{k} \pi_{k} \{ L\sigma_{k}^{2}+||\mu_{k}-E[t|x]||^2 \}\\ \hspace{15pt}= \sum_{k} \pi_{k} \{ L\sigma_{k}^{2}+||\mu_{k}-\sum_{l}\pi_{l}\mu_{l}||^2 \} }$
(5.160)はLが抜けてるような

trsing’s diary

勉強、読んだ本、仕事で調べたこととかのメモ。