一様分布

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一様分布とは

一様分布 (英：uniform distribution) とは、すべての事象が起こる確率が等しい連続型かつ離散型な確率分布のこと。

\[ X\sim\mathrm{Unif}(a,b) \]

確率密度関数

\[ \def\arraystretch{1.5} f_X(x;a,b) = \begin{cases} \dfrac{1}{b-a} &\text{if } a\le x\le b \\ 0 &\text{otherewise} \vphantom{\dfrac 0 0} \\ \end{cases} \]

確率密度関数のグラフ：

# Python 3
from scipy.stats import uniform
import matplotlib.pyplot as plt

cases = [
    (1, 3), # a, b
    (2, 5),
]

plt.figure()
for a,b in cases:
    x = range(a,b+1)
    dist = uniform.pdf(x,a,b-a)
    plt.plot(x, dist, label="$a={},b={}$".format(a,b))

plt.xlim(0)
plt.ylim(0,1)
    
plt.title("PDF of uniform distribution")
plt.xlabel("$x$")
plt.ylabel("$f_X(x;a,b)$")
plt.legend()
plt.show()

累積分布関数

\[ F_X(x;a,b) = \begin{cases} 0 &\text{if } x\lt a \vphantom{\dfrac 0 0} \\ \dfrac{x-a}{b-a} &\text{if } a\le x\le b \\ 1 &\text{if } b\lt x \vphantom{\dfrac 0 0} \\ \end{cases} \]

累積分布関数の導出：

$x\lt a$ の場合、

\[ \begin{aligned} F(x) &= \int_{-\infty}^{x\lt a} f(u)~du \\ &= 0 \vphantom{\int} \\ \end{aligned} \tag{1} \]

$a\le x\le b$ の場合、

\[ \begin{aligned} F(x) &= \int_{-\infty}^{a\le x\le b} f(u)~du \\ &= \int_{-\infty}^a f(u)~du + \int_a^{x\le b} f(u)~du \\ &= \int_a^{x\le b} \frac{1}{b-a}~du \quad \because (1) \\ &= \frac{x-a}{b-a} \end{aligned} \tag{2} \]

$b\lt x$ の場合、

\[ \begin{aligned} F(x) &= \int_{-\infty}^{b\lt x}f(u)~du \\ &= \int_{-\infty}^a f(u)~du + \int_a^b f(u)~du + \int_b^{b\lt x}f(u)~du \\ &= \left.\frac{t-a}{b-a}\right|_{t=b} + \int_b^{b\lt x}f(u)~du \quad \because (1)(2) \\ &= 1 \vphantom{\int} \\ \end{aligned} \tag{3} \]

$(1)(2)(3)$ により、

\[ \therefore F_X(x;a,b) = \begin{cases} 0 &\text{if } x\lt a \vphantom{\dfrac 0 0} \\ \dfrac{x-a}{b-a} &\text{if } a\le x\le b \\ 1 &\text{if } b\lt x \vphantom{\dfrac 0 0} \\ \end{cases} \]

累積分布関数のグラフ：

# Python 3
from scipy.stats import uniform
import matplotlib.pyplot as plt

cases = [
    (1, 3), # a, b
    (2, 5),
]

plt.figure()
for a,b in cases:
    x = range(a,b+1)
    dist = uniform.cdf(x,a,b-a)
    plt.plot(x, dist, label="$a={},b={}$".format(a,b))

plt.xlim(0)
    
plt.title("CDF of uniform distribution")
plt.xlabel("$x$")
plt.ylabel("$F_X(x;a,b)$")
plt.legend()
plt.show()

期待値

\[ E[X] = \frac{a+b}{2} \quad (X\sim\mathrm U(a,b)) \]

期待値の導出：

\[ \begin{aligned} E[X] &= \int_{-\infty}^\infty uf_X(u;a,b)~\mathrm du \cr &= \int_{-\infty}^a uf_X(u;a,b)~\mathrm du + \int_{a}^b uf_X(u;a,b)~\mathrm du + \int_{b}^\infty uf_X(u;a,b)~\mathrm du \cr &= \int_{a}^b uf_X(u;a,b)~\mathrm du \cr &= \int_{a}^b u\frac{1}{b-a}~\mathrm du \cr &= \frac{1}{b-a}\int_a^b u~\mathrm du \cr &= \frac{1}{b-a}\left[\frac{1}{2}u^2\right]_a^b \cr &= \frac{1}{b-a}\left(\frac{b^2}{2}-\frac{a^2}{2}\right) \cr &= \frac{a+b}{2} \cr \cr \therefore E[X] &= \frac{a+b}{2} \end{aligned} \]

分散

\[ V[X] = \frac{(b-a)^2}{12} \quad (X\sim\mathrm U(a,b)) \]

分散の導出：

\[ \begin{aligned} V[X] &= E[X^2] - E[X]^2 \cr &= \int_{-\infty}^\infty u^2 f_X(u;a,b)~\mathrm du - \left(\frac{a+b}{2}\right)^2 \cr &= \int_{a}^b u^2 f_X(u;a,b)~\mathrm du - \left(\frac{a+b}{2}\right)^2 \cr &= \frac{1}{b-a}\int_a^b u^2~\mathrm du - \left(\frac{a+b}{2}\right)^2 \cr &= \frac{1}{b-a}\left[\frac{1}{3}u^3\right]_a^b - \left(\frac{a+b}{2}\right)^2 \cr &= \frac{1}{b-a}\left(\frac{b^3}{3}-\frac{a^3}{3}\right) - \frac{(a+b)^2}{4} \cr &= \frac{a^2+ab+b^2}{3} - \frac{a^2+2ab+b^2}{4} \cr &= \frac{(b-a)^2}{12} \cr \cr \therefore V[X] &= \frac{(b-a)^2}{12} \end{aligned} \]