卡方分布、t分布、F分布的期望与方差的计算

2021-03-12 本文已影响0人 Cache_wood

卡方分布

$设X_1,X_2,……X_n独立同分布，且服从公共分布N(0,1),则Y =X_1^2+X_2^2+……X_n^2服从自由度为n的卡方分布.$

卡方分布的期望和方差

$Var(X_i) = E(X_i^2)-E(X_i)^2 \\ E(X_i^2) = Var(X_i) + E(X_i)^2 = 1+0^2 = 1\\ E(\chi^2) = nE(X_i^2) = n\\ Var(X_i^2) = E(X_i^4)-E(X_i^2)^2 = 3-1^2 = 2\\ Var(\chi_i^2) = nVar(X_i^2) =2n$

t分布

$随机变量X_1,X_2独立，且X_1\sim N(0,1),X_2\sim\chi_i^2(n), t= \frac{X_1}{\sqrt{X_2/n}}\sim t(n)$

t分布的期望

$E(T) = E(\frac{X_1}{\sqrt{X_2/n}})=nE(\frac{X_1}{\sqrt{X_2}})\\ 由于X_1,X_2彼此独立，所以E(\frac{X_1}{\sqrt{X_2}}) = E(X_1)E(\frac{1}{\sqrt{X_2}})\\ E(X_1) = 0,即E(T)=0\\$

t分布的方差

$Var(T)= E(T^2)-E(T)^2 = E(T^2)\\ E(T^2) = E(\frac{X_1^2}{X_2/n})=nE(\frac{X_1^2}{X_2})\\ 由于X_1,X_2彼此独立，所以E(\frac{X_1^2}{X_2}) = E(X_1^2)E(\frac{1}{X_2})\\ 由于X_1\sim N(0,1),所以E(X_1^2) = 1\\ 由于X_2\sim\chi^2(n),所以E(\frac{1}{X_2}) = \int_0^{\infty}\frac{1}{x}p(x)dx\\ =\int_0^{\infty}\frac{1}{x}\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}e^{-\frac{x}{2}}dx\\ =\int_0^{\infty}\frac{1}{x}\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}e^{-\frac{x}{2}}dx\\ =\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\int_0^{\infty}x^{\frac{n}{2}-2}e^{-\frac{x}{2}}dx\\ =\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\frac{{\Gamma(\frac{n}{2}-1)}}{(\frac{1}{2})^{\frac{n}{2}-1}}\\ =\frac{1}{n-2}\\ 所以E(T^2) = nE(X_1^2)E(\frac{1}{X_2})=\frac{n}{n-2}$

F分布

$随机变量X_1,X_2独立，且X_1~\chi^2(m),X_2~\chi^2(n),z则F=\frac{X_1/m}{X_2/n}\sim F(m,n)$

F分布的期望

$E(F) = E(\frac{X_1/m}{X_2/n})=\frac{n}{m}E(\frac{X_1}{X_2})\\ 因为X_1\sim\chi^2(m),X_2\sim\chi^2(n),X_1,X_2彼此独立\\ 所以E(\frac{X_1}{X_2})=E(X_1)E(\frac{1}{X_2})\\ E(X_1) = m,E(\frac{1}{X_2})=\frac{1}{n-2}\\ 所以E(F) = \frac{n}{m}E(\frac{X_1}{X_2})=\frac{n}{n-2}$

F分布的方差

$E(F^2) = \frac{n^2}{m^2}E(\frac{X_1^2}{X_2^2})\\ 由于X_1\sim\chi^2(m),所以E(X_1^2) = \int_0^{\infty}{x^2}p(x)dx\\ =\int_0^{\infty}{x^2}\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}e^{-\frac{x}{2}}dx\\ =\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\int_0^{\infty}x^{\frac{n}{2}+1}e^{-\frac{x}{2}}dx\\ =\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\frac{{\Gamma(\frac{n}{2}+2)}}{(\frac{1}{2})^{\frac{n}{2}+2}}\\ =m(m+2)\\ 由于X_2\sim\chi^2(n),所以E(\frac{1}{X_2^2}) = \int_0^{\infty}\frac{1}{x^2}p(x)dx\\ =\int_0^{\infty}\frac{1}{x^2}\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}x^{\frac{n}{2}-1}e^{-\frac{x}{2}}dx\\ =\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\int_0^{\infty}x^{\frac{n}{2}-3}e^{-\frac{x}{2}}dx\\ =\frac{(\frac{1}{2})^{\frac{n}{2}}}{\Gamma(\frac{n}{2})}\frac{{\Gamma(\frac{n}{2}-2)}}{(\frac{1}{2})^{\frac{n}{2}-2}}\\ =\frac{1}{(n-2)(n-4)}\\ 所以E(F^2) = \frac{n^2}{m^2}E(X_1^2)E(\frac{1}{X_2^2})=\frac{n^2(m+2)}{m(n-2)(n-4)}\\ 所以Var(F) = E(F^2)-E(F)^2=\frac{2n^2(m+n-2)}{m(n-2)^2(n-4)}$