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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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up vote 4 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

Maybe not rigorous, but the idea is correct.

$$ X_n\xrightarrow{a.s}X\Leftrightarrow P(\lim_{n\rightarrow\infty}\lvert |X_n-X\rvert |>\epsilon)=0$$

$$\lvert|X_n-X\rvert|+\lvert|X_{n+1}-X\rvert|\geq\lvert|X_n-X-X_{n+1}+X\rvert|=\lvert|X_n-X_{n+1}\rvert|$$$$\Rightarrow\lim_{n\rightarrow\infty}\lvert|X_n-X\rvert|+\lim_{n\rightarrow\infty}\lvert|X_{n+1}-X\rvert|\geq\lim_{n\rightarrow\infty}\lvert|X_n-X_{n+1}\rvert|$$$$\Rightarrow 0=P(\lim_{n\rightarrow\infty}\lvert|X_n-X\rvert|>\epsilon/2)+P(\lim_{n\rightarrow\infty}\lvert|X_{n+1}-X\rvert|>\epsilon/2)\geq P(\lim_{n\rightarrow\infty}\lvert|X_n-X\rvert|>\epsilon/2\lor\lim_{n\rightarrow\infty}\lvert|X_{n+1}-X\rvert|>\epsilon/2)$$$$=P(\lim_{n\rightarrow\infty}\lvert|X_n-X\rvert|+\lim_{n\rightarrow\infty}\lvert|X_{n+1}-X\rvert|>\epsilon)$$$$\geq P(\lim_{n\rightarrow\infty}\lvert|X_n-X_{n+1}\rvert|>\epsilon)\geq0$$

Since $X_n$s are independent, $X_n$ converges to a constant almost surely.

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It looks like an interesting technique but I don't quite follow: a) How exactly did you use independence? b) Why do you conclude that $X$ is a constant? c) Why do you show a.s. convergence if the question is about convergence in probability? Many thanks in advance. – Leo Jul 29 '15 at 23:11

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