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Zaare is offline
Apr7-05, 06:43 AM
P: 54
Quote Quote by Zone Ranger
let X_1 =1 on [0,1]
let X_2=1 on [0,1/2] , 0 otherwise
let x_3=1 on [1/2,1] , 0 ow
let x_4=1 on[0,1/3] ,0 ow
let x_5=1 on [1/3,2/3] , 0 ow
Taking this example, if I understand correctly, X_n does not converge almost surely (2) since it does not converge to a single value but rather keeps dividing the interval [0,1] into halfs, and "jumps back and forth" within the interval.
X_n does not converge in mean square to X (3) because of the same reason, namely the expected value of [tex](X_n-X)^2[/tex] kepps changing as [tex]{n \to \infty }[/tex].
However, I don't know how to show that X_n converges in probability.