MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

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

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question.If this question can be reworded to fit the rules in the help center, please edit the question.

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 $X_n:=\frac 1{\sqrt n}\sum_{j=1}^n\xi_j$.

share|cite|improve this answer
I'm not sure the question is of research level. – Davide Giraudo Sep 30 '13 at 9:15

Not the answer you're looking for? Browse other questions tagged or ask your own question.