convergence of sequence of random variables 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?
 A: 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 http://en.wikipedia.org/wiki/Convergence_of_random_variables).
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.
A: 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.
