Levyの反転定理

\(P(\cdot)\)を確率測度とし,その特性関数を\( \varphi _X (t) \)とすると 以下が成立する.

\[ P(a < X < b) + \frac{1}{2}(P(X=a)+P(X=b)) = \lim_{T \to \infty} \frac{1}{2 \pi} \int ^{T} _{-T} \frac{e^{-ita}-e^{-itb}}{it} \varphi _X(t) dt \]

  • 証明
    \[ \begin{align} F(T) &= \int ^{T} _{-T} \frac{e^{-ita}-e^{-itb}}{it} \varphi _X(t) dt \\ &= \int ^{T} _{-T} \frac{e^{-ita}-e^{-itb}}{it} \int ^{\infty} _{-\infty} e^{itx} dF_X(x) dt \\ &= \int ^{\infty} _{-\infty} \int ^{T} _{-T} \frac{e^{-ita}-e^{-itb}}{it} e^{itx} dt dF_X(x) \: (Fubiniの定理より) \\ &= \int ^{\infty} _{-\infty} \int ^{T} _{-T} \frac{e^{-it(a-x)}-e^{-it(b-x)}}{it} dt dF_X(x) \\ &= \int ^{\infty} _{-\infty} \int ^{T} _{-T} \frac{\cos(t(a-x))-i\sin(t(a-x))-\cos(t(b-x))+i\sin(t(b-x))}{it} dt dF_X(x) \\ &= \int ^{\infty} _{-\infty}\left(\int ^{T} _{-T} \frac{\cos(t(a-x))-\cos(t(b-x))}{it} dt + \int ^{T} _{-T} \frac{i\sin(t(b-x))-i\sin(t(a-x))}{it} dt \right) dF_X(x) \end{align} \] \(\frac{\cos(t(a-x))-\cos(t(b-x))}{it}\)は奇関数なので,積分範囲が対称だと0になる. \[ \begin{align} &= \int ^{\infty} _{-\infty}\left(\int ^{T} _{-T} \frac{\sin(t(b-x))-\sin(t(a-x))}{t} dt \right) dF_X(x) \\ &= \int ^{\infty} _{-\infty}\left(\int ^{T} _{-T} \frac{\sin(t(b-x))}{t} dt - \int ^{T} _{-T} \frac{\sin(t(a-x))}{t} dt \right) dF_X(x) \end{align} \]

\(F(T)\)に\(\frac{1}{2\pi}\)を乗して,\(T \to \infty\)の極限を取る.

\[ \begin{align} \lim_{T \to \infty} \frac{1}{2 \pi} F(T) &= \lim_{T \to \infty} \frac{1}{2 \pi} \int ^{\infty} _{-\infty}\left(\int ^{T} _{-T} \frac{\sin(t(b-x))}{t} dt - \int ^{T} _{-T} \frac{\sin(t(a-x))}{t} dt \right) dF_X(x) \\ &= \frac{1}{2 \pi} \int ^{\infty} _{-\infty} \left(\lim _{T \to \infty} \int ^{T} _{-T} \frac{\sin(t(b-x))}{t} dt - \lim _{T \to \infty} \int ^{T} _{-T} \frac{\sin(t(a-x))}{t} dt \right) dF_X(x) \end{align} \] \(\frac{\sin(t(a-x))}{t}\)は\(t\)の関数と見ると,偶関数なので, \[ \begin{align} f_a(x) &= \lim _{T \to \infty} \int ^{T} _{-T} \frac{\sin(t(a-x))}{t} dt = \lim _{T \to \infty} 2 \int ^{T} _{0} \frac{\sin(t(a-x))}{t} dt \\ &= \left\{ \begin{array}{ll} \pi & (x < a) \\ 0 & (x = a) \\ -\pi & (x > a) \\ \end{array} \right. \end{align} \]

から, \[ \begin{align} \frac{1}{2 \pi} \int ^{\infty} _{-\infty} \left(\lim _{T \to \infty} \int ^{T} _{-T} \frac{\sin(t(b-x))}{t} dt - \lim _{T \to \infty} \int ^{T} _{-T} \frac{\sin(t(a-x))}{t} dt \right) dF_X(x) &= \frac{1}{2 \pi} \int ^{\infty} _{-\infty} (f_b(x) - f_a(x)) dF_X(x) \end{align} \]

\(f_a(x)\)は指示関数で書き直すと,\(f_a(x) = \pi (I _{x < a}(x) - I _{x > a}(x)) \) \[ \begin{align} \frac{1}{2 \pi} \int ^{\infty} _{-\infty} (f_b(x) - f_a(x)) dF_X(x) &= \frac{1}{2 \pi} \int ^{\infty} _{-\infty} (\pi (I _{x < b}(x) - I _{x > b}(x)) - \pi (I _{x < a}(x) - I _{x > a}(x))) dF_X(x) \\ &= \frac{1}{2} \int ^{\infty} _{-\infty} (I _{x > b}(x) - I _{x < b}(x) - I _{x < a}(x) + I _{x > a}(x)) dF_X(x) \\ &= \frac{1}{2} \int ^{\infty} _{-\infty} (2 I _{a < x < b}(x) + I _{x > a \land x = b}(x) + I _{x < b \land x = a}(x)) dF_X(x) \\ &= P(a < x < b) + \frac{1}{2}(P(X=a)+P(X=b)) \end{align} \]

証明終わり.

分布関数と特性関数の1対1対応

2つの分布関数\(F_{X_1},F_{X_2}\)とその特性関数を\(\varphi_{X_1},\varphi_{X_2}\)とすると, \[ F_{X_1}(x) = F_{X_2}(x),\;\forall x \in \overline{\mathbb{R}} \Leftrightarrow \varphi_{X_1}(t)=\varphi_{X_2}(t),\;\forall t \in \overline{\mathbb{R}} \]

  • 証明

  1. \(F_{X_1}(x) = F_{X_2}(x),\;\forall x \in \overline{\mathbb{R}} \Rightarrow \varphi_{X_1}(t)=\varphi_{X_2}(t),\;\forall t \in \overline{\mathbb{R}}\)
    特性関数の定義より自明.

  2. \(F_{X_1}(x) = F_{X_2}(x),\;\forall x \in \overline{\mathbb{R}} \Leftarrow \varphi_{X_1}(t)=\varphi_{X_2}(t),\;\forall t \in \overline{\mathbb{R}}\)
    Levyの反転定理で,\(a,b\)が連続点ならば,\(P(a) = 0, P(b) = 0\)なので, \[ \begin{align} P(a < X < b) &= F_X(b) - F_X(a) \\ &= \lim_{T \to \infty} \frac{1}{2 \pi} \int ^{T} _{-T} \frac{e^{-ita}-e^{-itb}}{it} \varphi _X(t) dt \\ \end{align} \]

    \(\varphi_{X_1}(t)=\varphi_{X_2}(t)\)ならば\( F_{X_1}(b) - F_{X_1}(a) = F_{X_2}(b) - F_{X_2}(a) \)が上式から得られ, 分布関数の定義より,\(a \to -\infty\)とすると,\( F_{X_1}(b) = F_{X_2}(b) \)となる.改めて, \(\varphi_{X_1}(t)=\varphi_{X_2}(t)\)ならば,\( F_{X_1}(x) = F_{X_2}(x) \)