作者idphobia5566 (idphobia5566)
看板Math
标题[分析] converge in probability
时间Sun Feb 13 22:08:02 2011
iid
X1,...,Xn ~ UNIF(a,b) , a<b , let X(n) be the largest order statistic.
(a)What does exp[X(n)] converge in probability? Show your work.
(b)Find the limiting distribution of exp[-n(b-X(n))/(b-a)] . Show your work.
最大顺序统计量之pdf: f_x:n(x) = n(x-a)^n-1/(b-a)^n , a<x<b
想法是
let y = exp(x) , 找出y=exp[X(n)]的pdf，然後用mgf法取极限来求
但这样不知道怎麽积分...因为式子很难看
f(y)= n{[ln(y)-a]^(n-1)}/[y*(b-a)^n] , exp(a) < y < exp(b)
M_y(t)=E[exp(ty)] 超级难积分
请问有没有更好的方法可以做，谢谢
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1F：→ yhliu :P[X(n)≦z] = [(z-a)/(b-a)]^n when a<z<b 02/13 23:39
2F：→ yhliu :故 X(n) converges in probability to b. 02/13 23:40
3F：→ idphobia5566:是观察说当n到无穷大时，z=b才会机率是1吗 02/13 23:49