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Let $(X_n,n\geqslant 1)$ be a tight sequence of stochastic processes defined on the same probability. Suppose $\lVert X_n\rVert_{L^2}$ converges to $\lVert X\rVert_{L^2}$. Under what conditions do we have $L^2$ convergence?

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closed as unclear what you're asking by R W, Daniel Moskovich, Todd Trimble, Andrey Rekalo, David White Sep 30 '13 at 13:15

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1 Answer 1

In this context, we have $L^2$ convergence if and only if $X_n\to X$ in probability.

Indeed, $L^2$ convergence always implies convergence in probability. Conversely, if $X_n\to X$ in probability and $\mathbb E(X_n^2)\to\mathbb E(X^2)$, then $\{X_n^2,n\geqslant 1\}\cup\{X^2\}$ is a uniformly integrable family and a standard argument allows us to conclude convergence in $L^2$.

Since the $X_n$ are real valued, convergence in probability implies converges in distribution, which in turn implies tightness.

An example where $\{X_n\}$ is tight and $\mathbb E(X_n^2)\to\mathbb E(X^2)$ but we don't have convergence in probability is the following: take $(\xi_i,i\geqslant 1)$ a sequence of independent centered identically distributed random variables, $\mathbb E(\xi_1^2)=1$. Define $X_n:=\frac 1{\sqrt n}\sum_{j=1}^n\xi_j$.

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I'm not sure the question is of research level. –  Davide Giraudo Sep 30 '13 at 9:15

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