UA MATH574 概率论 一个均匀分布的例题2018May/4
UA MATH574 概率论 均匀分布例题
2018May第四题。
Part a
Let Zi=−lnUiZ_i =- \ln U_iZi=−lnUi,
P(Zi≤z)=P(−lnUi≤z)=P(Ui≥e−z)=1−FU(e−z)fZ(z)=e−z,z>0P(Z_i \le z) = P(-\ln U_i \le z) = P(U_i \ge e^{-z}) = 1 - F_U(e^{-z}) \\ f_Z(z) = e^{-z},z>0P(Zi≤z)=P(−lnUi≤z)=P(Ui≥e−z)=1−FU(e−z)fZ(z)=e−z,z>0
This means Zi∼EXP(1)=dΓ(1,1)Z_i \sim EXP(1) =_d \Gamma(1,1)Zi∼EXP(1)=dΓ(1,1). By additivity, ∑i=1nZi∼Γ(n,1)\sum_{i=1}^n Z_i \sim \Gamma(n,1)∑i=1nZi∼Γ(n,1). Note that Vn=e−∑i=1nZiV_n = e^{-\sum_{i=1}^n Z_i}Vn=e−∑i=1nZi,
P(Vn≤v)=P(e−∑i=1nZi≤v)=P(∑i=1nZi≥−lnv)fVn(v)=1v(n−1)ln(1/v)Γ(n)elnv=−n−1Γ(n)lnv,v∈(0,1)P(V_n \le v) = P(e^{-\sum_{i=1}^n Z_i} \le v) = P(\sum_{i=1}^n Z_i \ge -\ln v) \\ f_{V_n}(v) = \frac{1}{v} \frac{(n-1)\ln (1/v)}{\Gamma(n)}e^{\ln v} = -\frac{n-1}{\Gamma(n)}\ln v,v \in (0,1)P(Vn≤v)=P(e−∑i=1nZi≤v)=P(i=1∑nZi≥−lnv)fVn(v)=v1Γ(n)(n−1)ln(1/v)elnv=−Γ(n)n−1lnv,v∈(0,1)
Part b
Notice Ui/SnU_i/S_nUi/Sns should have the same distribution,
E[U1/Sn]=1nE[∑i=1nUi/Sn]=1nE[Sn/Sn]=1nE[U_1/S_n] = \frac{1}{n}E[\sum_{i=1}^n U_i/S_n] = \frac{1}{n}E[S_n/S_n] = \frac{1}{n}E[U1/Sn]=n1E[i=1∑nUi/Sn]=n1E[Sn/Sn]=n1
Part c
Define Y=ln(Vn)−1/Sn=1Sn(−lnVn)=Zˉ/UˉY = \ln (V_n)^{-1/S_n} =\frac{1}{S_n}(-\ln V_n) =\bar{Z}/\bar{U}Y=ln(Vn)−1/Sn=Sn1(−lnVn)=Zˉ/Uˉ. By LLN, Uˉ→p1/2\bar{U} \to_p 1/2Uˉ→p1/2, Zˉ→p1\bar{Z} \to_p 1Zˉ→p1. By property of convergence in probability, Zˉ/Uˉ→p2\bar{Z}/\bar{U} \to_p 2Zˉ/Uˉ→p2 and (Vn)−1/Sn→pe2(V_n)^{-1/S_n} \to_p e^{2}(Vn)−1/Sn→pe2
Part d
1
总结
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