概要
連続確率分布の1つである一様分布 について解説します。
確率密度関数
確率変数 X X X が次のような確率密度関数をもつとき、X X X は区間 [ a , b ] [a, b] [ a , b ] の(連続)一様分布 (uniform distribution) に従うという。
f X ( x ) = { 1 b – a a ≤ x ≤ b 0 その他の場合
f_X(x) =
\begin{cases}
\frac{1}{b – a} & a \le x \le b \\
0 & その他の場合
\end{cases}
f X ( x ) = { b – a 1 0 a ≤ x ≤ b その他の場合
確率密度関数である
∫ − ∞ ∞ 1 b – a d t = 1 b – a [ t ] a b d t = 1
\int_{-\infty}^{\infty} \frac{1}{b – a} dt
= \frac{1}{b – a} [t]^b_a dt
= 1
∫ − ∞ ∞ b – a 1 d t = b – a 1 [ t ] a b d t = 1
累積確率関数
P ( X ≤ x ) = ∫ − ∞ x 1 b – a d t
P(X \le x) = \int_{-\infty}^x \frac{1}{b – a} dt
P ( X ≤ x ) = ∫ − ∞ x b – a 1 d t x < a x < a x < a の場合、f X ( x ) = 0 f_X(x) = 0 f X ( x ) = 0 なので、
∫ − ∞ x 1 b – a d t = 0
\int_{-\infty}^x \frac{1}{b – a} dt = 0
∫ − ∞ x b – a 1 d t = 0 a ≤ x ≤ b a \le x \le b a ≤ x ≤ b の場合、
∫ − ∞ x 1 b – a d t = ∫ a x 1 b – a d t = x – a b – a
\int_{-\infty}^x \frac{1}{b – a} dt
= \int_a^x \frac{1}{b – a} dt
= \frac{x – a}{b – a}
∫ − ∞ x b – a 1 d t = ∫ a x b – a 1 d t = b – a x – a x > b x > b x > b の場合、f X ( x ) = 0 , ( x < a , x > b ) f_X(x) = 0, (x < a, x > b) f X ( x ) = 0 , ( x < a , x > b ) なので、
∫ − ∞ x 1 b – a d t = ∫ a b 1 b – a d t = 1
\int_{-\infty}^x \frac{1}{b – a} dt
= \int_a^b \frac{1}{b – a} dt
= 1
∫ − ∞ x b – a 1 d t = ∫ a b b – a 1 d t = 1 よって、
P ( X ≤ x ) = { 0 x < a x – a b – a a ≤ x ≤ b 1 x > b
P(X \le x) =
\begin{cases}
0 & x < a \\
\frac{x – a}{b – a} & a \le x \le b \\
1 & x > b
\end{cases}
P ( X ≤ x ) = ⎩ ⎨ ⎧ 0 b – a x – a 1 x < a a ≤ x ≤ b x > b
k 次積率
E [ X k ] = ∫ a b x k 1 b – a d x = 1 b – a [ x k + 1 k + 1 ] a b = 1 b – a b k + 1 – a k + 1 k + 1
\begin{aligned}
E[X^k]
&= \int_a^b x^k \frac{1}{b – a} dx \\
&= \frac{1}{b – a} \left[\frac{x^{k + 1}}{k + 1}\right]_a^b \\
&= \frac{1}{b – a} \frac{b^{k + 1} – a^{k + 1}}{k + 1}
\end{aligned}
E [ X k ] = ∫ a b x k b – a 1 d x = b – a 1 [ k + 1 x k + 1 ] a b = b – a 1 k + 1 b k + 1 – a k + 1 ここで、b k – a k = ( b – a ) ∑ i = 0 k – 1 a i b k – 1 – i b^k – a^k = (b – a) \sum_{i = 0}^{k – 1} a^i b^{k – 1 – i} b k – a k = ( b – a ) ∑ i = 0 k –1 a i b k –1– i を利用すると、
1 b – a b k + 1 – a k + 1 k + 1 = 1 k + 1 ( b – a ) ∑ i = 0 k a i b k – i b – a = 1 k + 1 ∑ i = 0 k a i b k – i
\begin{aligned}
\frac{1}{b – a} \frac{b^{k + 1} – a^{k + 1}}{k + 1}
&= \frac{1}{k + 1} \frac{(b – a) \sum_{i = 0}^k a^i b^{k – i}}{b – a} \\
&= \frac{1}{k + 1} \sum_{i = 0}^k a^i b^{k – i}
\end{aligned}
b – a 1 k + 1 b k + 1 – a k + 1 = k + 1 1 b – a ( b – a ) ∑ i = 0 k a i b k – i = k + 1 1 i = 0 ∑ k a i b k – i
期待値
k k k 次積率の結果より、
E [ X ] = a + b 2
\begin{aligned}
E[X] = \frac{a + b}{2}
\end{aligned}
E [ X ] = 2 a + b
分散
k k k 次積率の結果より、
E [ X 2 ] = 1 3 ( a 2 + a b + b 2 )
E[X^2] = \frac{1}{3} (a^2 + ab + b^2)
E [ X 2 ] = 3 1 ( a 2 + ab + b 2 ) V a r [ X ] = E [ X 2 ] – [ E ( X ) ] 2 = 1 3 ( a 2 + a b + b 2 ) – ( a + b ) 2 4 = a 2 – 2 a b – b 2 12 = ( b – a ) 2 12
\begin{aligned}
Var[X]
&= E[X^2] – [E(X)]^2 \\
&= \frac{1}{3} (a^2 + ab + b^2) – \frac{(a + b)^2}{4} \\
&= \frac{a^2 – 2ab – b^2}{12} \\
&= \frac{(b – a)^2}{12}
\end{aligned}
Va r [ X ] = E [ X 2 ] – [ E ( X ) ] 2 = 3 1 ( a 2 + ab + b 2 ) – 4 ( a + b ) 2 = 12 a 2 –2 ab – b 2 = 12 ( b – a ) 2
標準偏差
S t d [ X ] = V a r [ X ] = b – a 2 2
Std[X] = \sqrt{Var[X]} = \frac{b – a}{2 \sqrt{2}}
St d [ X ] = Va r [ X ] = 2 2 b – a
積率母関数
m X ( t ) = E [ e t X ] = 1 b – a ∫ a b e t x d x = 1 b – a [ e t x t ] a b = e t b – e t a t ( b – a )
\begin{aligned}
m_X(t)
&= E[e^{tX}] \\
&= \frac{1}{b – a} \int_a^b e^{tx} dx \\
&= \frac{1}{b – a} \left[\frac{e^{tx}}{t}\right]_a^b \\
&= \frac{e^{tb} – e^{ta}}{t(b – a)}
\end{aligned}
m X ( t ) = E [ e tX ] = b – a 1 ∫ a b e t x d x = b – a 1 [ t e t x ] a b = t ( b – a ) e t b – e t a
scipy.stats の一様分布
scipy.stats.uniform で一様分布に従う確率変数を作成できます。
サンプリング
[-0.17091662 -0.53469003 1.35470988 -1.71785535 1.02110735]
確率密度関数
累積分布関数
統計量
mean 0.0
var 1.3333333333333333
std 1.1547005383792515
コメント