[1]
\(\xi\)
\(\displaystyle \xi = \int^\infty_{-\infty} x(pf_1(x) + (1-p)f_2(x))dx\)
\(\displaystyle = p\mu_1 + (1-p)\mu_2\)
\(\tau^2\)
\(\displaystyle E[X^2] \)\(\displaystyle= \int^\infty_{-\infty} x^2(pf_1(x) + (1-p)f_2(x))dx\)
\(\displaystyle = p(\mu_1^2 + \sigma^2) + (1-p)(\mu_2^2 + \sigma^2)\)
\(\displaystyle \tau^2 = E[X^2] – \xi^2\)
\(\displaystyle = p(\mu_1^2 + \sigma^2) + (1-p)(\mu_2^2 + \sigma^2) \)\(- (p\mu_1 + (1-p)\mu_2)^2\)
\(\displaystyle = \sigma^2 \)\(+p(1-p)(\mu_1-\mu_2)^2\)
これは\(p=1/2\)の時最大値をとる。
[2]
\(p\)
\(\displaystyle E[\gamma(X)] = \int \frac{pf_1(x)}{g(x)}g(x)dx\)
\(\displaystyle = \int pf_1(x)dx = p\)
\(\mu_1\)
\(\displaystyle \frac{1}{p}E[\gamma(X)X] = \frac{1}{p} \int \frac{pf_1(x)}{g(x)}xg(x)dx\)
\(\displaystyle = \int x f_1(x)dx = \mu_1 \)
\(\mu_2\)
\(\displaystyle \frac{1}{1-p}E[\{1-\gamma(X)\}X] \)\(\displaystyle= \frac{1}{1-p} \int \left(1-\frac{pf_1(x)}{g(x)}\right)xg(x)dx\)
\(\displaystyle =\frac{1}{1-p} \int \frac{(1-p)f_2(x)}{g(x)}xg(x)dx\)
\(\displaystyle = \int x f_2(x)dx = \mu_2 \)
[3]
\(\displaystyle \gamma(x)= \frac{pf_1(x)}{g(x)} \)
\(\displaystyle = \frac{pf_1(x)}{pf_1(x)+(1-p)f_2(x)} \)
\(\displaystyle = \frac{1}{1+\frac{(1-p)f_2(x)}{pf_1(x)}} \)
\(\displaystyle = \frac{1}{1+\frac{(1-p)\frac{1}{\sqrt{2\pi\sigma^2}}\exp \left(-\frac{(x-\mu_2)^2}{2\sigma^2}\right)}{p\frac{1}{\sqrt{2\pi\sigma^2}}\exp \left(-\frac{(x-\mu_1)^2}{2\sigma^2}\right)}} \)
\(\displaystyle = \frac{1}{1+\frac{1-p}{p}\exp \left(-\frac{(\mu_1-\mu_2)(x-(\mu_1+\mu_2)/2)}{\sigma^2} \right)} \)
[4]
公式の解答を参照。
ただし、初期値の設定では、\(p=0.5\)\(, \mu_1=\mu_2 = \bar{x}\)のように設定すると学習が進まないため、\(p=0.5\)\(, \mu_1= (重量の重い順で下位半分の平均),\)\(\mu_2 = (重量の重い順で上位半分の平均)\)のように設定するとよさそう。
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