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Let $X_n$ be a sequence of independent random variable which converges in probability to $X$. Prove $X$ is a constant.

Can someone give me a hint how I should go about proving this? I tried proving this by contradiction by saying $X$ taking 2 different values, but this can still still happen because $X$ is only almost surely constant.

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1 Answer 1

up vote 2 down vote accepted

Hint 1: Passing to a subsequence, we can assume $X_n \to X$ almost surely.

Hint 2: Do you know the Kolmogorov zero-one law? $X$ is a tail random variable.

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On the subsequence it converges almost surely. Consider any interval [a,b) of R, where [a,b) are dyadic fractions. The probability X lies in one of the interval is either 0 or 1, by Kolmogorov Zero-One Law. Then we take the intersection of these interval to show X is almost surely constant. –  Lost1 Oct 27 '12 at 18:35
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