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2.3(標準) 分布関数,密度関数の定義3

Last updated at Posted at 2021-07-08

方針

2.1,2.2でも見た通り確率密度関数での定義を満たすことを示すためには非負性と全確率1を確かめれば良い.

ある確率密度関数fに対し定義域をある定数a以上の領域に限定したい,となったとき,単に「a以上だよ!」とするだけでは全確率1を満たさなくなってしまう.その際,全確率1を満たすための正規化定数は1-F_X(a)となり,これを打ち切り分布という.

答案

関数g(x)は,

\begin{align}
g(x)=
\begin{cases}
\frac{f(x)}{1-F(a)}\ \ \ \ \ \ \mathrm{if}\ a\leq x\\
0\ \ \ \ \ \ \ \ \ \ \ \ \ \ \mathrm{if}\ x<a
\end{cases}
\end{align}

と書けるので,任意のxに対し,g(x)は非負である.また,

\begin{align}
\int_{-\infty}^\infty g(x)dx&= \int_a^\infty\frac{f(x)}{1-F(a)}dx+\int_{-\infty}^a0dx\\
&= \frac{1}{1-F(a)}\{\int_{–\infty}^\infty f(x)dx-\int_{-\infty}^af(x)dx\}\\
&= \frac{1}{1-F(a)}(1-F(a))\\
&= 1
\end{align}

より,g(x)は全確率1を満たす.また,f_X(x)=e^{-x}, 0<xとすると,

\begin{align}
F_X(x)&= \int_0^x e^{-t}dt\\
&= \left[-e^{-t}\right]_0^x\\
&= -e^{-x}+1
\end{align}

なので,a=1で打ち切った分布は

g(x)=\frac{e^{-x}}{1-(e^{-1}+1)}=e^{-(x-1)}

となる.

参考文献

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