For a random sequenc $X_n$, if its expectation $E|X_n|=0$, does that mean it converges in mean to $0$? For convergence in mean to $0$ we need $E|X_n-0|\rightarrow0$
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If $X_n$ converges to $0$ in mean, then $EX_n\to 0$. It is however not sufficient. Convergence in mean means $E|X_n - X| \to 0$, where $X$ is the limit. – user251257 Jul 24 '15 at 15:34
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I up-voted this question and its vote total is not zero, so someone down-voted it. Could that person explain what the objections are? – Michael Hardy Jul 24 '15 at 20:18
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Convergence in mean to $0$ requires $E |X_n| \to 0$, which is a much stronger condition. Those absolute values are important!
EDIT: I see you have now changed the statement to $E|X_n| = 0$. Then that does imply convergence in mean to $0$. But you should be aware that $E|X_n| = 0$ is equivalent to: $X_n = 0$ with probability $1$.
Robert Israel
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what I am confused is not the absolute value part, it is the constant part. In other words, if a sequence is a constant, 0,0,0,0...... does that mean its limit is 0? – martingale Jul 24 '15 at 15:38
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@martingale Yes. Do not confuse limit of a sequence with limit of random variables though – Ant Jul 24 '15 at 15:39
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That $X_n$ converges in mean to $X$ as $n\to\infty$ means that $$ \operatorname{E}|X_n-X|\to0\text{ as }n\to\infty. $$
So convergence in mean to $0$ means that $$ \operatorname{E}|X_n-0|\to0\text{ as }n\to\infty. $$
The fact that $\operatorname{E}X_n=0$ does not at all imply that $\operatorname{E}|X_n|=0$.
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what if I have the absolute value part? Does $E|X_n|=0$ imply limit is zero? – martingale Jul 24 '15 at 15:46
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If $\operatorname{E}|X_n|=0$ for every $n$ then $\lim\limits_{n\to\infty}\operatorname{E} |X_n| = \lim\limits_{n\to\infty} 0 = 0$. ${}\qquad{}$ – Michael Hardy Jul 24 '15 at 15:59