二項分布(binomial distribution)の積率母凾数(moment-generating function)と期待値(expected value)・分散(variance)

二項分布(binomial distribution)の積率母凾数(moment-generating function)と期待値(expected value)・分散(variance)

二項分布

$$ \begin{eqnarray} X&\sim&B(n, p) \\f(X=x)&=& \begin{cases} _n\mathrm{C}_x\;p^x(1-p)^{n-x} & x \in \left\{0,1,2, \dotsc ,n\right\} \\0 & x \notin \left\{0,1,2, \dotsc ,n\right\} \end{cases}\;\cdots\;確率密度凾数 \end{eqnarray} $$

積率母凾数

$$ \begin{eqnarray} M_X(t)&=&\mathrm{E}\left[e^{tx}\right] \\&=&\sum_{k=0}^n e^{tx}\;_n\mathrm{C}_x\;p^x\left(1-p\right)^{n-x} \\&=&\sum_{k=0}^n\;_n\mathrm{C}_x\;\left( e^{t} p \right)^x \left(1-p\right)^{n-x} \\&=&\;_n\mathrm{C}_0\;\left( e^{t} p \right)^0 \left(1-p\right)^{n-0} +\;_n\mathrm{C}_1\;\left( e^{t} p \right)^1 \left(1-p\right)^{n-1} +\;_n\mathrm{C}_2\;\left( e^{t} p \right)^2 \left(1-p\right)^{n-2} +\cdots +\;_n\mathrm{C}_n\;\left( e^{t} p \right)^n \left(1-p\right)^{n-n} \\&=&\left\{e^{t} p + \left(1-p\right) \right\}^n \\&&\;\cdots\;(A+B)^D=\;_D\mathrm{C}_0\;A^0B^{D-0}+\;_D\mathrm{C}_1\;A^1B^{D-1}+\cdots+\;_D\mathrm{C}_D\;A^DB^{D-D}=\sum_{k=0}^D \;_D\mathrm{C}_k\;A^kB^{D-k}\;(二項関係) \\&=&\left(e^{t}p - p + 1\right)^n \end{eqnarray} $$

(原点周りの)一次モーメント = 期待値

$$ \begin{eqnarray} \mathrm{E}\left[X\right]&=&M^{(1)}_X(t) \\&=&\left.\frac{\mathrm{d} M_X(t)}{\mathrm{d}t}\right|_{t=0} \\&=&\left.\frac{\mathrm{d}}{\mathrm{d}t} \left(e^{t}p - p + 1\right)^n \right|_{t=0} \\&=&\left.\frac{\mathrm{d}}{\mathrm{d}u} u^n \frac{\mathrm{d}u}{\mathrm{d}t} \right|_{t=0} \;\cdots\;u=e^{t}p - p + 1,\frac{\mathrm{d}u}{\mathrm{d}t}=\frac{\mathrm{d}}{\mathrm{d}t}\left(e^{t}p - p + 1\right)=e^{t}p \\&=&\left.nu^{n-1} e^{t}p \right|_{t=0} \\&=&\left.n\left( e^{t}p - p + 1 \right)^{n-1} e^{t}p \right|_{t=0} \\&=&\left.npe^{t} \left( e^{t}p - p + 1 \right)^{n-1} \right|_{t=0} \\&=&npe^{0} \left( e^{0}p - p + 1 \right)^{n-1} \\&=&np\cdot1 \left( 1\cdot p - p + 1 \right)^{n-1} \\&=&np \left(1\right)^{n-1} \\&=&np\;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/blog-post_30.html}{Xの期待値の定義から求めた二項分布の期待値}と同じ \end{eqnarray} $$

(原点周りの)二次モーメント

$$ \begin{eqnarray} \mathrm{E}\left[X^2\right]&=&M^{(2)}_X(t) \\&=&\left.\frac{\mathrm{d}^2 M_X(t)}{\mathrm{d}t^2}\right|_{t=0} \\&=&\left.\frac{\mathrm{d}^2}{\mathrm{d}t^2} \left(e^{t}p - p + 1\right)^n \right|_{t=0} \\&=&\left.\frac{\mathrm{d}}{\mathrm{d}t} npe^{t} \left(e^{t}p - p + 1\right)^{n-1} \right|_{t=0} \\&=&\left.np\frac{\mathrm{d}}{\mathrm{d}t} e^{t} \left(e^{t}p - p + 1\right)^{n-1} \right|_{t=0} \;\cdots\;\frac{\mathrm{d}}{\mathrm{d}x}cf(x)=c\frac{\mathrm{d}}{\mathrm{d}x}f(x)\;(c:定数) \\&=&\left.np\frac{\mathrm{d}}{\mathrm{d}t} uv \right|_{t=0} \;\cdots\;u=e^t,\;v= \left(e^{t}p - p + 1\right)^{n-1} \\&=&\left.np\left\{\left(\frac{\mathrm{d}}{\mathrm{d}t}u\right)v+u\left(\frac{\mathrm{d}}{\mathrm{d}t}v\right) \right\}\right|_{t=0} \\&=&\left.np\left[\left(e^{t}\right)v+u\left\{\left(n-1\right)\left(e^{t}p - p + 1\right)^{n-2}pe^t\right\} \right]\right|_{t=0} \;\cdots\;\frac{\mathrm{d}u}{\mathrm{d}t}=e^{t},\frac{\mathrm{d}v}{\mathrm{d}t}=\left(n-1\right)\left(e^{t}p - p + 1\right)^{n-2}pe^t \\&=&\left.np\left[\left(e^{t}\right)\left(e^{t}p - p + 1\right)^{n-1}+e^t\left\{\left(n-1\right)\left(e^{t}p - p + 1\right)^{n-2}pe^t\right\} \right]\right|_{t=0} \;\cdots\;u=e^t,\;v= \left(e^{t}p - p + 1\right)^{n-1} \\&=&\left.npe^{t}\left(e^{t}p - p + 1\right)^{n-1}+n\left(n-1\right)p^2e^{2t}\left(e^{t}p - p + 1\right)^{n-2} \right|_{t=0} \\&=&npe^{0}\left(e^{0}p - p + 1\right)^{n-1}+n\left(n-1\right)p^2e^{2\cdot0}\left(e^{0}p - p + 1\right)^{n-2} \\&=&np\cdot1\cdot\left(1\cdot p - p + 1\right)^{n-1}+n\left(n-1\right)p^2\cdot1\cdot\left(1\cdot p - p + 1\right)^{n-2} \\&=&np\left(1\right)^{n-1}+n\left(n-1\right)p^2\left(1\right)^{n-2} \\&=&np+n\left(n-1\right)p^2 \end{eqnarray} $$

二次の中心(化)モーメント / 母平均周りの二次モーメント = 分散

$$ \begin{eqnarray} \mathrm{V}\left[X\right]&=&\mathrm{E}\left[\left(X-\mathrm{E}\left[X\right]\right)^2\right] \\&=&\mathrm{E}\left[X^2\right]-\mathrm{E}\left[X\right]^2 \;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/discrete-random-variable-variance.html}{\mathrm{E}\left[\left(X-\mathrm{E}\left[X\right]\right)^2\right]=\mathrm{E}\left[X^2\right]-\left[X\right]^2} \\&=&M^{(2)}_X(t) -\left(M^{(1)}_X(t)\right)^2 \\&=&np+n\left(n-1\right)p^2 - (np)^2 \\&=&np+n^2p^2-np^2 - n^2p^2 \\&=&np-np^2 \\&=&np(1-p)\;\cdots\;\href{https://shikitenkai.blogspot.com/2019/06/blog-post_75.html}{X(X-1)の期待値を利用した二項分布の分散}と同じ \end{eqnarray} $$