표본정규분포

choyunjeong·2024년 12월 25일

1. 표준정규분포

1) 확률밀도함수

XN(μ,σ2)X\sim N(\mu,\sigma^2)를 따르는 정규분포를 Z=(Xμ)/σZ=(X-\mu)/\sigma라는 표준화 변환을 하면 확률밀도함수는 다음과 같고 ZN(0,1)Z\sim N(0,1)로 표기.

f(x)=12πσexp[(xμ)2/2σ2](<x<)fZ(z)=12πexp(z2/2)(<x<)\begin{aligned} f(x)&=\dfrac{1}{\sqrt{2\pi}\sigma}\exp[-(x-\mu)^2/2\sigma^2]\quad (-\infty<x<\infty) \\[10pt] \rightarrow f_Z(z)&=\dfrac{1}{\sqrt{2\pi}}\exp(-z^2/2) \quad (-\infty<x<\infty) \end{aligned}

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2) 기댓값
XN(0, 1)X\sim N(0,\ 1)를 따를 때

E(X)=x12πexp(x22)dx=xϕ(x)dx=0\begin{aligned} E(X) &=\int_{-\infty}^{\infty}x\cdot\dfrac{1}{\sqrt{2\pi}}\exp\left(-\dfrac{x^2}{2}\right) dx \\[10pt] &=\int_{-\infty}^{\infty}x\cdot \phi(x)dx \\[10pt] &=0 \end{aligned}

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3) 분산
XN(0, 1)X\sim N(0,\ 1)를 따를 때

E(X2)=x212πexp(x22)dx=x2ϕ(x)dx=1\begin{aligned} E(X^2) &=\int_{-\infty}^{\infty}x^2\cdot\dfrac{1}{\sqrt{2\pi}}\exp\left(-\dfrac{x^2}{2}\right) dx \\[10pt] &=\int_{-\infty}^{\infty}x^2\cdot\phi(x)dx \\[10pt] &=1 \end{aligned}

Var(X)=10=1\therefore \text{Var}(X)=1-0=1

  • 참고
    z2ϕ(z)dz=[ϕ(z)+ϕ(z)]dz=1ϕ(z)dz=(d2/dz2)ϕ(z)dz=0\int_{-\infty}^{\infty}z^2\phi(z)dz=\int_{-\infty}^{\infty}[\phi''(z)+\phi'(z)]dz=1 \\[10pt] \int_{-\infty}^{\infty}\phi''(z)dz=(d^2/dz^2)\int_{-\infty}^{\infty}\phi(z)dz=0

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4) 적률생성함수

표준정규 확률변수 ZN(0,1)Z\sim N(0,1)의 적률생성함수는 다음과 같다.

MZ(t)=E(etx)=exp(tz)12πexp(z2/2)dz=12πexp{((zt)2/2+t2/2)}dz=exp(t2/2)\begin{aligned} M_Z(t) &=E(e^{tx}) \\[10pt] &=\int_{-\infty}^{\infty}\exp(tz)\dfrac{1}{\sqrt{2\pi}}\exp(-z^2/2) dz\\[15pt] &=\int_{-\infty}^{\infty}\dfrac{1}{\sqrt{2\pi}}\exp\{(-(z-t)^2/2+t^2/2)\} dz\\[15pt] &=\exp(t^2/2) \end{aligned}

따라서 적률생성함수로 활용한 기댓값과 분산은
MX(1)(t)=(d/dt)(exp(t2/2))=texp(t2/2)M^{(1)}_X(t)=(d/dt)\left(\exp(t^2/2)\right)=t\cdot\exp(t^2/2)\\[10pt]

E(X)=MX(1)(0)=0E(X)=M^{(1)}_X(0)=0

MX(2)(t)=(d/dt)(texp(t2/2))=exp(t2/2)(1+t2)M^{(2)}_X(t)=(d/dt)\left(t\cdot\exp(t^2/2)\right) = \exp(t^2/2)\cdot(1+t^2) \\[10pt]

MX(2)(0)=1M^{(2)}_X(0)=1

Var(X)=MX(2)(0){MX(1)(0)}2=1\therefore \text{Var}(X)=M^{(2)}_X(0)-\left\{M^{(1)}_X(0)\right\}^2=1

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