確率分布まとめ | 期待値・分散・モーメント母関数の導出、定義や覚えておきたい性質を解説

\begin{eqnarray}
E[X] &=& \sum_{k=0}^nkP(X=k) \\
&=& \sum_{k=0}^nk\binom{n}{k}p^k(1−p)^{n−k} \\
&=& \sum_{k=0}^nk\frac{n!}{k!(n-k)!}p^k(1−p)^{n−k} \\
&=& np\sum_{k=1}^n\frac{(n-1)!}{(k-1)!(n-k)!}p^{k-1}(1−p)^{n−k} \\
&=& np\sum_{k=0}^{n-1}\frac{(n-1)!}{(k)!(n-1-k)!}p^{k}(1−p)^{n−1-k} \\
&=& np(p+1-p)^{n-1} \quad\because 二項定理\\
&=& np
\end{eqnarray}

\begin{eqnarray}
E[X^2] &=& \sum_{k=0}^nk^2P(X=k) \\
&=& \sum_{k=1}^nk(k-1)\binom{n}{k}p^k(1−p)^{n−k}+\sum_{k=1}^nk\binom{n}{k}p^k(1−p)^{n−k} \\
&=& n(n-1)\sum_{k=2}^n\binom{n-2}{k-2}p^k(1−p)^{n−k}+E[X] \\
&=& n(n-1)p^2\sum_{k=2}^{n}\binom{n-2}{k-2}p^{k-2}(1−p)^{n−k}+np \\
&=& n(n-1)p^2\sum_{k=0}^{n-2}\binom{n-2}{k}p^{k}(1−p)^{n−k-2}+np \\
&=& n(n-1)p^2(p+1-p)^{n-2}+np \quad\because 二項定理 \\
&=& n(n-1)p2+np
\end{eqnarray}
従って、\(V[X]=E[X^2]-E[X]^2=n(n-1)p^2+np-n^2p^2=np-np^2=np(1-p)\)

\begin{eqnarray}
M_X(t) &=& E[e^{tk}] \\
&=& \sum_{k=0}^ne^{tk}P(X=k) \\
&=& \sum_{k=0}^ne^{tk}\binom{n}{k}p^k(1−p)^{n−k} \\
&=& \sum_{k=0}^n\binom{n}{k}(pe^{t})^k(1-p)^{n-k} \\
&=& (pe^t+1-p)^n \quad\because 二項定理 \\
\end{eqnarray}

\[P(X=k)=\binom{n}{k}p^k(1−p)^{n−k}\]
\(p=\lambda /n\) とおくと、
\[\binom{n}{k}(\frac{\lambda}{n})^k(1-\frac{\lambda}{n})^{n-k}\]
\(n\to\infty\)とすると、
\[\frac{\lambda^k}{k!}​e^{-\lambda}\]
となる。

\begin{eqnarray}
E[X] &=& \int_\infty^\infty xf(x)dx \\
&=& \frac{1}{\sqrt{2\pi\sigma^2}}\int_\infty^\infty x\exp(-\frac{(x-\mu)^2}{2\sigma^2})dx \\
&=& \frac{1}{\sqrt{2\pi\sigma^2}}\int_\infty^\infty (x-\mu+\mu)\exp(-\frac{(x-\mu)^2}{2\sigma^2})dx \\
&=& \frac{1}{\sqrt{2\pi\sigma^2}}\int_\infty^\infty (x-\mu)\exp(-\frac{(x-\mu)^2}{2\sigma^2})dx+\frac{\mu}{\sqrt{2\pi\sigma^2}}\int_\infty^\infty\exp(-\frac{(x-\mu)^2}{2\sigma^2})dx \\
&=& \frac{1}{\sqrt{2\pi\sigma^2}}\int_\infty^\infty (x-\mu)\exp(-\frac{(x-\mu)^2}{2\sigma^2})dx+\mu
\end{eqnarray}

第１項について、\(y=\frac{x-\mu}{\sigma}\) で置換すると、

\begin{eqnarray}
E[X] &=& \frac{1}{\sqrt{2\pi\sigma^2}}\int_\infty^\infty\sigma y\exp(-y^2/2)\sigma dy+\mu \\
&=& \frac{\sigma}{\sqrt{2\pi}}\int_\infty^\infty y\exp(-y^2/2)dy+\mu \\
&=& \mu
\end{eqnarray}

従って、

\[E[X]=\mu\]

\begin{eqnarray}
E[X^2] &=& \int_\infty^\infty x^2f(x)dx \\
&=& \int_\infty^\infty x^2\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2})dx \\
&=& \int_\infty^\infty \{(x-\mu)^2+2\mu x-\mu^2)\}\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2})dx \\
&=& \int_\infty^\infty (x-\mu)^2\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2})dx+2\mu\int_\infty^\infty x\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2})dx-\mu^2\int_\infty^\infty \frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2})dx \\
&=& \int_\infty^\infty (x-\mu)^2\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2})dx+2\mu^2-\mu^2
&=& \int_\infty^\infty (x-\mu)^2\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2})dx+\mu^2
\end{eqnarray}

第１項について、\(y=\frac{x-\mu}{\sigma}\) で置換すると、

\begin{eqnarray}
E[X^2] &=& \int_\infty^\infty\sigma^2 y^2\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-y^2/2)\sigma dy+\mu^2 \\
&=& \frac{\sigma^2}{\sqrt{2\pi}}\int_\infty^\infty y^2\exp(-y^2/2)dy+\mu^2 \\
&=& \sigma^2+\mu^2
\end{eqnarray}

\(V[X]=E[X^2]-(E[X])^2\) （参考）より、

\[V[X]=\sigma^2+\mu^2-\mu^2=\sigma^2\]

\begin{eqnarray}
M_X(t) &=& E[e^{tX}] \\
&=& \int_\infty^\infty e^{tx}f(x)dx \\
&=& \int_\infty^\infty e^{tx}\frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2})dx \\
&=& \int_\infty^\infty \frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{(x-\mu)^2}{2\sigma^2}+tx)dx \\
&=& e^{\mu t+\frac{\sigma^2t^2}{2}}\int_\infty^\infty \frac{1}{\sqrt{2\pi\sigma^2}}\exp(-\frac{\{x-(\mu+\sigma^2t)\}^2}{2\sigma^2})dx \\
&=& e^{\mu t+\frac{\sigma^2t^2}{2}}
\end{eqnarray}