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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.

2 Answers 2

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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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    is there an elementary proof (not using Kolmogorov zero-one law) if the $X_n$ as well as $X$ might only take values in $\{a,b\}$ for some integers $ a \not =b$?2018-02-26
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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.2015-07-29