UA MATH564 概率论 依概率收敛的一个例题
UA MATH564 概率論 依概率收斂的一個例題
Part (a)
Let Yn~U(?1/n,1/n)Y_n \sim U(-1/n,1/n)Yn?~U(?1/n,1/n), Zn~U(n,n+1)Z_n \sim U(n,n+1)Zn?~U(n,n+1), Yn,ZnY_n,Z_nYn?,Zn? are independent and Xn=n?1nYn+1nZnX_n = \frac{n-1}{n}Y_n + \frac{1}{n}Z_nXn?=nn?1?Yn?+n1?Zn?.
EXn=n?1nEYn+1nEZn=2n+12n→1,asn→∞EX_n = \frac{n-1}{n}EY_n + \frac{1}{n}EZ_n = \frac{2n+1}{2n}\to 1,\ as\ n \to \inftyEXn?=nn?1?EYn?+n1?EZn?=2n2n+1?→1,?as?n→∞
Part (b)
Var(Xn)=(n?1)2n2Var(Yn)+1n2Var(Zn)=(n?1)23n4+112n2→0,asn→∞Var(X_n) = \frac{(n-1)^2}{n^2}Var(Y_n) + \frac{1}{n^2}Var(Z_n) = \frac{(n-1)^2}{3n^4} + \frac{1}{12n^2} \to 0,\ as\ n \to \inftyVar(Xn?)=n2(n?1)2?Var(Yn?)+n21?Var(Zn?)=3n4(n?1)2?+12n21?→0,?as?n→∞
Part (c)
By part (a) and part (b), XnX_nXn? converges to 1 in mean square. By property of convergence in probability Xn→p1X_n \to_p 1Xn?→p?1.
總結
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