3.2. Tail Probabilities and Moments

Remark

Let \(X\) be a (real-valued) random variable. Then as \(n\rightarrow \infty\), \[ \{\omega:|X(\omega)| > n \}\downarrow \phi\implies P(|X|>n)\downarrow 0, \] and because \(P(|X|>t)\) is monotone in \(t\), it follows that \[ \lim_{t\rightarrow \infty}P(|X|>t)=0. \]

자주 쓰인다.

Theorem

If \(X\) is a nonnegative random variable, then \[ E(X)=\int_0^\infty P(X>t)dt = \int_0^\infty P(X\ge t)dt. \]

• 증명: \(X(\omega)=\lambda(0,X(\omega))\)임을 이용한다면, \(A=\{(t,\omega):0<t<X(\omega)\}\)에 대해 \[ E(X)=\int_\Omega X(\omega)dP(\omega)=\int_\Omega \lambda(0,X(\omega))dP(\omega)=\int_\Omega\int_\mathbb{R} I_A(t,\omega) d\lambda(t) dP(\omega) \] 이고 \(A\in \mathcal{R}\times \mathcal{F}\)이다. 때문에 \(I_A(t,\omega)\)는 \(\mathcal{R}\times \mathcal{F}\)-measurable이고 Fubini’s Theorem에 의해 다음을 얻는다: \[\begin{eqnarray*} E(X)&=&\int_\Omega\int_\mathbb{R} I_A(t,\omega) d\lambda(t) dP(\omega)\\ &=&\int_\mathbb{R}\int_\Omega I_A(t,\omega) dP(\omega)d\lambda(t) \\ &=& \int_\mathbb{R}\int_\Omega I_{(0,\infty)}(t)I_{\{X>t\}} dP(\omega)d\lambda(t) \mbox{ }\mbox{ }\mbox{ (두 영역의 교집합을 하면 }0<t<X\mbox{ 기 때문이다.) }\\ &=& \int_\mathbb{R} I_{(0,\infty)}(t) \int_\Omega I_{\{X>t\}} dP(\omega)d\lambda(t)\\ &=& \int_\mathbb{R} P(X>t)d\lambda(t)=\int_\mathbb{R} P(X>t)dt. \end{eqnarray*}\]

• \(P(X\ge t)= P(X>t)\).

Corollary

For any random variable \(X\) and \(0<p<\infty\), \[ E(|X|^p)=\int_0^\infty P(|X|>t^{1/p})dt=\int_0^\infty P(|X|\ge t^{1/p})dt. \]

• 위의 Theorem에 \(|X|^p\) 대입하면 된다.

매우 자주 쓰인다.

Corollary

For a nonnegative random variable \(X\), \[ \sum_{n=1}^{\infty} P(X\ge n)\le E(X)\le \sum_{n=0}^\infty P(X>n), \] and \(E(X)<\infty\) if and only if \(\sum_{n=1}^\infty P(X\ge n)<\infty\).

• 위의 Theorem과 Corollary에서: \[\begin{eqnarray*} E(X)&=&\int_0^\infty P(X\ge t)dt= \int_0^\infty \left(\sum_{n=1}^\infty I_{(n-1,n]}(t)\right)P(X\ge t)dt\\ &\stackrel{\text{MCT}}{=}&\sum_{n=1}^\infty \int_0^\infty I_{(n-1,n]}(t)P(X\ge t)dt\\ &=& \sum_{n=1}^\infty\int_{n-1}^nP(X\ge t)dt\\ &\ge&\sum_{n=1}^\infty\int_{n-1}^nP(X\ge n)dt\\ &=&\sum_{n=1}^\infty P(X\ge n) \int_{n-1}^ndt=\sum_{n=1}^\infty P(X\ge n). \end{eqnarray*}\] Similarly, \[\begin{eqnarray*} E(X)&=&\int_0^\infty P(X> t)dt= \int_0^\infty \left(\sum_{n=0}^\infty I_{(n,n+1]}(t)\right)P(X> t)dt\\ &\stackrel{\text{MCT}}{=}&\sum_{n=0}^\infty \int_0^\infty I_{(n,n+1]}(t)P(X> t)dt\\ &=& \sum_{n=0}^\infty\int_{n}^{n+1}P(X> t)dt\\ &\le&\sum_{n=0}^\infty\int_{n}^{n+1}P(X> n)dt\\ &=&\sum_{n=0}^\infty P(X> n) \int_{n}^{n+1}dt=\sum_{n=0}^\infty P(X> n). \end{eqnarray*}\] Thus, \(\sum_{n=1}^{\infty} P(X\ge n)\le E(X)\le \sum_{n=0}^\infty P(X>n)\).

• Note that \[ \sum_{n=1}^{\infty} P(X\ge n)\le E(X)\le \sum_{n=0}^\infty P(X>n)\le\sum_{n=0}^\infty P(X\ge n)=1+\sum_{n=1}^\infty P(X\ge n)\\ \sum_{n=1}^{\infty} P(X\ge n)\le E(X)\le 1+\sum_{n=1}^\infty P(X\ge n) \] 즉 \(\sum_{n=1}^\infty P(X\ge n)<\infty\)이어야지만 \(E(X)<\infty\)이다(Sandwich로 둘러쌓였기 때문).

Theorem(Tail Rate from pth Moment)

If \(X\in L^p\) for some \(0<p<\infty\), then \[ t^p P(|X|\ge t)\rightarrow 0 \mbox{ }\mbox{ }\mbox{ as }\mbox{ }\mbox{ }t\rightarrow \infty. \]

• 증명: 마코프 부등식(finer version)을 이용한다. \[ \int_{[|X|^p\ge t]}|X|^p \mbox{ }dP \ge \int_{[|X|^p\ge t]}t^p \mbox{ }dP=t^pP(|X|^p\ge t) \mbox{ }\ge 0\mbox{ }\mbox{ }\mbox{ for }t\ge 0\\ \implies \int |X|^p I_{\{|X|^p\ge t\}} \mbox{ }dP \ge t^pP(|X|^p\ge t). \] 여기서 \(|X|^p I_{\{|X|^p\ge t\}}\)에 D.C.T를 적용하기 위해 condition을 확인한다.

1. \(|X|^p I_{\{|X|^p\ge t\}} \le |X|^p\) (\(t\)에 영향을 받지 않는다);

2. \(|X|^p\in L^p\), integrable;

3. \(|X|^p I_{\{|X|^p\ge t\}} \rightarrow 0\) as \(t\rightarrow \infty\);

때문에 \(\int |X|^p I_{\{|X|^p\ge t\}} dP \rightarrow 0\) as \(t\rightarrow \infty\)이고, 따라서 \[ t^pP(|X|^p\ge t)\le \int |X|^p I_{\{|X|^p\ge t\}} \mbox{ }dP \rightarrow 0 \mbox{ }\mbox{ }\mbox{ as }\mbox{ } t\rightarrow \infty\\ \implies t^pP(|X|^p\ge t)\rightarrow 0 \mbox{ }\mbox{ }\mbox{ as }\mbox{ } t\rightarrow \infty. \]

back