pnorm(-4,0,1)
3.16712418331199e-05
중간고사
1.
\(X_1, \dots , X_n\)이 \(N(\mu, \sigma^2)\)( 단, \(\mu\)는 그 값이 알려져 있음)로부터의 랜덤표본이다.
\(\sigma^2\)의 추정량으로 \(T_1(X) = \frac{1}{n}\sum^n_{i=1}(X_i - \mu)^2\)과 \(T_2(X) = \frac{1}{n-1}\sum^n_{i=1}(X_i - \bar{X})^2\)을 고려하고 있다. 다음에 답하시오.
\(\star\)
\(\frac{n T_1(X)}{\sigma^2} \sim \chi^2 (n)\), \(\frac{(n-1) T_2(X))}{\sigma^2} \sim \chi^2 (n-1)\)
\(E(\frac{n T_1(X)}{\sigma^2}) = n\)
\(\therefore E(T_1(X)) = \sigma^2\)
\(E(\frac{(n-1) T_2(X)}{\sigma^2}) = n-1\)
\(\therefore E(T_2(X)) = \sigma^2\)
\(V(\frac{n T_1(X)}{\sigma^2}) = \frac{n^2}{\sigma^4}V(T_1(X)) = 2n\)
\(\therefore V(T_1(X)) = \frac{2\sigma^4}{n}\)
\(V(\frac{(n-1) T_2(X)}{\sigma^2}) = \frac{(n-1)^2}{\sigma^4}V(T_2(X)) = 2(n-1)\)
\(\therefore V(T_2(X))= \frac{2\sigma^4}{n-1}\)
(a)
\(T_1(X)\)과 \(T_2(X)\)의 기댓값을 각각 구하시오.
answer
- \(E(T_1(X))\)
\(= E(\frac{1}{n}\sum^n_{i=1}(X_i - \mu)^2) = \frac{E((X-\mu)^2)}{n} = \frac{n\sigma^2}{n} = \sigma^2\)
- \(E(T_2(X))\)
\(= E(\frac{1}{n-1}\sum^n_{i=1}(X_i - \bar{X})^2)\)
\(\star\)
\(\sum(X-\mu)^2 = \sum(X - \bar{X} + \bar{X} - \mu)^2\)
\(= \sum((X-\bar{X})^2 + (\bar{X} - \mu)^2 + 2(X-\bar{X})(\bar{X}-\mu))\)
\(= \sum(X - \bar{X})^2.+ n(\bar{X} - \mu)^2\)
\(\star\)
\(= E(\frac{\sum(X-\mu)^2 - n(\bar{X}-\mu)}{n-1}^2) = \frac{1}{n-1}[n\sigma^2 - n\frac{\sigma^2}{n}] = \frac{(n-1)\sigma^2}{n-1} = \sigma^2\)
(b)
\(T_1(X)\)과 \(T_2(X)\)의 분산을 각각 구하시오.
answer
- \(Var(T_1(X))\)
\(= Var(\frac{\sigma^2}{n} \frac{1}{\sigma^2}\sum^n_{i=1}(X_i - \mu)^2)\)
\(= \frac{\sigma^4}{n^2}2n\)
\(= (\frac{2\sigma^4}{n})\)
\(\star\)
\(V = \sum(\frac{X_i - \mu}{\sigma})^2 \sim chi^2(n)\)
\(\star\)
- \(Var(T_2(X))\)
\(= Var(\frac{1}{n-1}\sum^n_{i=1}(X_i - \bar{X})^2)\)
\(\star\)
\(S_n^2 = \sum^n_{i=1}\frac{(X_i - \bar{X}_n)^2}{n-1}\)
\(\frac{(n-1)S_n^2}{\sigma^2} = \frac{\sum^n_{i=1}(X_i - \bar{X})^2}{\sigma^2} \sim \chi^2_{n-1}\)
\(\star\)
\(= Var(\frac{\sigma^2}{n-1} \times \frac{(n-1)S^2}{\sigma^2})\)
\(\frac{\sigma^4}{(n-1)^2} \times 2(n-1)\)
\(= \frac{2\sigma^4}{n-1}\)
2.
\(POI(\lambda)\) 분포로부터 랜덤표본 \(X_1,X_2,\dots, X_n\)을 얻었다.
(a)
\(\sqrt{n}(\bar{X}^2_n - \lambda^2)\)의 극한 분포를 구하시오.
answer
Delta Method
\(\sqrt{n}(\bar{X}^2_n - \mu^2) \xrightarrow[]{d} N(0,4\mu^2\sigma^2)\) 단, \(\mu \neq 0\)
\(\mu = \lambda\), \(\sigma^2 = \lambda\)
\(\sqrt{n}(\bar{X}^2_n - \lambda^2) \xrightarrow[]{d} N(0,4\lambda^4)\)
(b)
\(E(\sum^n_{i=1}a_i X_i) = \lambda\)가 되기 위한 \(a_i\)의 조건은 무엇인가?
answer
\(E(\sum^n_{i=1}a_iX_i) = \lambda\)
\(= E(a_1X_1 + a_2X_2 + \dots a_nX_n) = a_1E(X_1) + a_2E(X_2) + \dots + a_nE(X_n)\)
\(= a_1\lambda + a_2\lambda + \dots + a_n\lambda = \sum^n_{i=1}a_i \lambda\)
\(\therefore \sum^n_{i=1} a_i= 1\)이어야 한다.
(c)
\(\sum^n_{i=1}a_iX_i\)의 분산을 구하시오.
answer
\(Var(\sum^n_{i=1}a_iX_i) = Var(a_1X_1 + a_2X_2 + \dots + a_n X_n)\)
\(= a_1^2\lambda + a_2^2\lambda + \dots + a_n^2\lambda = \sum^n_{i=1} a_i^2 \lambda\)
3.
확률변수 \(X\)의 확률밀도함수가 \(f_X(x) = 3e^{-3x}I(x>0)\)일 떄, \(Y = \frac{1}{X}\)의 확률밀도함수를 구하라.
answer
\(f_Y(y) = -\frac{3}{y^2} exp(-\frac{3}{y})I(y>0)\)
\(g(X) = \frac{1}{X}\), \(g^{-1} (Y) = \frac{1}{y}\)
\(= f_Y(y) = f_X(g^{-1}(y))|\frac{dg^{-1}(y)}{dg}|\)
\(= f_X(\frac{1}{y})(\frac{1}{y^2})\)
\(= 3e^{-\frac{3}{y}}(\frac{1}{y^2})I(y>0)\)
\(= \frac{3}{y^2}e^{-\frac{3}{y}}I(y>0)\)
4.
\(X_1,\dots,X_{16}\)이 \(N(5,25)\)로부터의 랜덤표본 이라고 하자. 이때 \(P(0<\bar{X}_{16} <6,20<S^2_{16}<30)\)을 계산하시오.(R함수 이용)
answer
\(\bar{X}_{16} \sim N(5,\frac{25}{16})\)
\(\frac{\bar{X}_{16} - 5}{\sqrt{25/16}} \sim N(0,1)\)
\(\frac{15 S^2_{16}}{25} \sim \chi^2_{15}\)
\(P(0 < \bar{X}_{16} < 6) = P(\frac{0-5}{\sqrt{25/16}} < \frac{\bar{X}_{16} - 5}{\sqrt{25/16}} < \frac{6-5}{\sqrt{25/16}}) = P(-4 < Z < 0.8)\)
pnorm(0.8,0,1)
0.788144601416603
pnorm(0.8,0,1) - pnorm(-4,0,1)
0.78811293017477
\(P(20 < S^2_{16} < 30) = P(\frac{15 \times 20}{25} < \frac{15 S^2_{16}}{25} < \frac{15 \times 30}{25})\)
pchisq(18,15)
0.737334439327678
pchisq(12,15)
0.320970942909585
pchisq(18,15) - pchisq(12,15)
0.416363496418093
\(P(0<\bar{X}_{16} <6,20<S^2_{16}<30)\)
(pnorm(4,0,1) - pnorm(-4,0,1)) * (pchisq(18,15) - pchisq(12,15))
0.416337122920122
5.
확률변수 \(X\)와 \(Y\)의 결함확률밀도함수가 \(f(x,y) = 4I(0<x<1,1<y<2x)\)일 때, 조건부 기댓값 \(E(Y|X=x)\)을 구하시오.
answer
\(E(Y|X=x) = \int^{\infty}_{-\infty} y f_{Y|X=x} (y) dy = \int^{\infty}_{-\infty} y \frac{f(X,y)}{f_X(X)}dy\)
\(\star\)
\(f_X(X) = \int(x,y)dy = \int^{\infty}_{-\infty} 4 I (0<x<1,1<y<2x)dy\)
\(= 4 I (0<x<1)\int_{-\infty<y<\infty,1<y<2x}I(1<y<2x)dy\)
\(= 4 I (0<x<1)\int^{2x}_1 dy\)
\(= 4I(0<x<1)[y]^{2x}_1\)
\(= 4(2x-1)I(0<x<1)\)
\(\star\)
\(= \int^{\infty}_{-\infty} y \frac{4I(0<x<1,1<y<2x)}{4(2x-1)I(0<x<1)} dy\)
\(= \int_{-\infty<y<\infty,1<y<2x} y \frac{I(0<x<1)}{2x-1}dy\)
\(= \frac{I(0<x<1)}{2x-1} \int^{2x}_1 y dy\)
\(= \frac{I(0<x<1)}{2x-1} [\frac{1}{2}y^2]^{2x}_1\)
\(= \frac{I(0<x<1)}{2x-1} (\frac{1}{2}(4x^2-1))\)
\(= \frac{2x+1}{2}I(0<x<1)\)
6.
\(X_1, \dots, X_4\)는 \(Ber(p)\)분포로부터의 랜덤표본이다.
(a)
이 랜덤표본의 결합확률밀도함수를 기술하시오.
answer
\(f(X_1)f(X_2)f(X_3)f(X_4)\)
\(= (1-p)^{1-X_1}p^{X_1}(1-p)^{1-X_2}p^{X_2}(1-p)^{1-X_3}p^{X_3}(1-p)^{1-X_4}p^{X_4}\)
\(= (1-p)^{4-X_1-X_2-X_3-X_4} p^{X_1+X_2+X_3+X_4}\)
(b)
\(X_1(1-X_2)\)의 기댓값을 구하시오.
answer
\(E(X_1(1-X_2))\)
\(= E(X_1 - X_1X_2) = E(X_1) - E(X_1)E(X_2)\)
\(= p - p^2 = p(1-p)\)
(c)
\(E[X_1(1-X_2)|\sum^4_{i=1}X_i = t]\)을 구하시오.
answer
\(\frac{p(1-p)}{4p} = \frac{1-p}{4}\)
(d)
\(E[(\frac{d}{dp}log p^X (1-p)^{1-p})^2]\)을 구하시오.
answer
\(E[(\frac{d}{dp} log p^x (1-p)^{1-X})^2]\)
\(= E((\frac{d}{dp}(X log p - (1-X) log(1-p)))^2)\)
\(= E((\frac{X}{p} - \frac{1-X}{1-p})^2)\)
\(= E((\frac{X-Xp - p + pX}{p(1-p)})^2)\)
\(= E((\frac{X-p}{p(1-p)})^2)\)
\(= \frac{1}{p^2(1-p)^2}(E(X^2)-p^2)\)
\(= \frac{p-p^2+p^2+p^2}{p^2(1-p)^2} = \frac{p(1-p)}{p^2(1-p)^2} = \frac{1}{p(1-p)}\)