UA MATH564 概率论 QE练习题3
UA MATH564 概率論 QE練習(xí)題3
- 第一題
- 第二題
- 第三題
這是2015年1月的1-3題。
第一題
Part a
Obvioulsy, EU=EW=0EU = EW = 0EU=EW=0,
VarU=Var(Y?Z)=1+1?2γ=2?2γVarW=Var(Y+Z)=1+1+2γ=2+2γVar U = Var(Y - Z) = 1 + 1 - 2\gamma = 2 - 2\gamma \\ Var W = Var(Y + Z) = 1 + 1 + 2\gamma = 2+ 2 \gammaVarU=Var(Y?Z)=1+1?2γ=2?2γVarW=Var(Y+Z)=1+1+2γ=2+2γ
So U~N(0,2?2γ),W~N(0,2+2γ)U \sim N(0,2-2\gamma),W \sim N(0,2+2\gamma)U~N(0,2?2γ),W~N(0,2+2γ).
Part b
Cov(U,W)=Cov(Y?Z,Y+Z)=Var(Y)?Var(Z)+Cov(Y,Z)?Cov(Y,Z)=0Cov(U,W) = Cov(Y-Z,Y+Z) = Var(Y) - Var(Z) + Cov(Y,Z) - Cov(Y,Z) = 0Cov(U,W)=Cov(Y?Z,Y+Z)=Var(Y)?Var(Z)+Cov(Y,Z)?Cov(Y,Z)=0
Notice they are bivariate normal, so they are independent.
Part c
參考UA MATH564 概率論VI 數(shù)理統(tǒng)計(jì)基礎(chǔ)2 多元正態(tài)分布中正態(tài)分布的條件分布:Y=(Y1′,Y2′)′,μ=(μ1′,μ2′)′,Y1,μ1∈Rr×1,Y2,μ2∈R(m?r)×1Y = (Y_1',Y_2')',\mu = (\mu_1',\mu_2')',Y_1,\mu_1 \in \mathbb{R}^{r \times 1},Y_2,\mu_2 \in \mathbb{R}^{(m-r) \times 1}Y=(Y1′?,Y2′?)′,μ=(μ1′?,μ2′?)′,Y1?,μ1?∈Rr×1,Y2?,μ2?∈R(m?r)×1,AA′=[V11V12V21V22]AA' = \left[ \begin{matrix} V_{11} & V_{12} \\ V_{21} & V_{22} \end{matrix} \right]AA′=[V11?V21??V12?V22??],V11∈Rr×r,V22∈R(m?r)×(m?r),V12∈Rr×(m?r),V21∈R(m?r)×rV_{11} \in \mathbb{R}^{r \times r},V_{22} \in \mathbb{R}^{(m-r)\times (m-r)},V_{12} \in \mathbb{R}^{r \times (m-r)},V_{21} \in \mathbb{R}^{(m-r) \times r}V11?∈Rr×r,V22?∈R(m?r)×(m?r),V12?∈Rr×(m?r),V21?∈R(m?r)×r,則
E[Y1∣Y2]=μ1+V12V22?1(Y22?μ2)Var(Y1∣Y2)=V11,2=V11?V12V22?1V11E[Y_1|Y_2] = \mu_1 + V_{12}V_{22}^{-1}(Y_{22} - \mu_2) \\ Var(Y_1|Y_2) = V_{11,2} = V_{11} - V_{12}V_{22}^{-1}V_{11}E[Y1?∣Y2?]=μ1?+V12?V22?1?(Y22??μ2?)Var(Y1?∣Y2?)=V11,2?=V11??V12?V22?1?V11?
Now compute
Cov(X,W)=Cov(X,Y+Z)=Cov(X,Y)=ρCov(X,W) = Cov(X,Y+Z) = Cov(X,Y) = \rhoCov(X,W)=Cov(X,Y+Z)=Cov(X,Y)=ρ
So
E[X∣W]=ρ2+2γWVar(X∣W)=1?ρ2+2γE[X|W] = \frac{\rho}{2+2\gamma}W \\ Var(X|W) = 1-\frac{\rho}{2+2\gamma}E[X∣W]=2+2γρ?WVar(X∣W)=1?2+2γρ?
第二題
參考UA MATH564 概率論I 求離散型隨機(jī)變量的分布1例3。
第三題
Part a
By LLN, 1n∑i=1nXi?1=1/Hn→pE(1/X1)\frac{1}{n}\sum_{i=1}^n X_i^{-1} = 1/H_n \to_p E(1/X_1)n1?∑i=1n?Xi?1?=1/Hn?→p?E(1/X1?),
E(1/X1)=∫121xdx=ln?2E(1/X_1) = \int_1^2 \frac{1}{x}dx = \ln 2 E(1/X1?)=∫12?x1?dx=ln2
By property of convergence in probability, Hn→p1/ln?2=cH_n \to _p 1/\ln 2= cHn?→p?1/ln2=c
Part b
參考UA MATH564 概率論V 中心極限定理中的Delta方法,
Zn=g(Xˉ)?g(μ)[g′(μ)]2σ2n→DN(0,1)Z_n = \frac{g(\bar{X})-g(\mu)}{\sqrt{[g'(\mu)]^2 \frac{\sigma^2}{n}}} \to_D N(0,1)Zn?=[g′(μ)]2nσ2??g(Xˉ)?g(μ)?→D?N(0,1)
Let Hn=11n∑i=1nXi?1,g(μ)=cH_n = \frac{1}{\frac{1}{n}\sum_{i=1}^n X_i^{-1}},g(\mu) = cHn?=n1?∑i=1n?Xi?1?1?,g(μ)=c,
σ2=E(1/X12)?(ln?2)2=∫121x2dx?(ln?2)2=12?(ln?2)2\sigma^2 = E(1/X_1^2) - (\ln 2)^2 = \int_1^2 \frac{1}{x^2}dx -(\ln 2)^2 = \frac{1}{2} - (\ln 2)^2σ2=E(1/X12?)?(ln2)2=∫12?x21?dx?(ln2)2=21??(ln2)2
[g′(μ)]2σ2n=1μ4σ2n=1n12?(ln?2)2(ln?2)4\sqrt{[g'(\mu)]^2 \frac{\sigma^2}{n}} = \sqrt{\frac{1}{\mu^4} \frac{\sigma^2}{n}} = \frac{1}{\sqrt{n}} \sqrt{\frac{\frac{1}{2} - (\ln 2)^2}{(\ln 2)^4}}[g′(μ)]2nσ2??=μ41?nσ2??=n?1?(ln2)421??(ln2)2??
So
n(Hn?1/ln?2)~dN(0,12?(ln?2)2(ln?2)4)\sqrt{n}(H_n - 1/\ln 2) \sim _d N(0,\frac{\frac{1}{2} - (\ln 2)^2}{(\ln 2)^4})n?(Hn??1/ln2)~d?N(0,(ln2)421??(ln2)2?)
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