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What does this expression mean - $\lim_{n\rightarrow\infty} E|X_n-X|=0$? $X_n$ is a sequence of random variables and $X$ is a random variable. What does this expression imply? Can I say that the sequence $X_n$ converges to $X$ in probability and almost sure convergence?

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This is called convergence in the mean or convergence in the $L^{1}$-norm. In general, if \begin{eqnarray} \lim_{n \to \infty} \mathbb{E}(| X_{n} - X|^{p}) = 0, \end{eqnarray} then $X_{n}$ is said to converge to $X$ in the $L^{p}$-norm (provided that $\mathbb{E}(|X_{n}|^{p})$ is finite for all $n \geq 1$). Analytically, there are nice implications of such convergence. For example, convergence in an $L^{p}$-norm implies convergence in an $L^{q}$-norm if $p \geq q$. (See

Markov's inequality states \begin{eqnarray} \mathbb{P}(|X_{n} - X| > \epsilon) \leq \epsilon^{-p} \, \mathbb{E}(|X_{n} - X|^{p}). \end{eqnarray} Thus, $L^{p}$-norm convergence implies convergence in probability.

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Wiki page mentions convergence in mean implies convergence in probability. Why is that? – user957 Nov 4 '10 at 7:39

When $\lim _{n \to \infty } E|X_n - X| = 0$, we say that $X_n$ converges in mean to $X$. It is very well known that this implies that $X_n$ converges in probability to $X$, but not that $X_n$ converges almost surely to $X$. Consider a sequence $(X_n)$ of independent random variables such that $P(X_n = 0) = 1 - 1/n$ and $P(X_n = 1) = 1/n$. Then $E|X_n|=1/n$, and hence $X_n$ converges in mean to $0$. However, since $\sum\nolimits_{n = 1}^\infty {P(X_n = 1)} = \infty $ and the $X_n$ are independent, almost surely the sequence $(X_n)$ contains infinitely many $1$'s; in particular $X_n$ does not converge to $0$ in the almost sure sense.

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