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I need some hints for the following question:

Suppose $X,X_1,X_2, \cdots \in L^2(\Omega)$ are random variables that converge in mean square. Show that $Var[X_n] \rightarrow Var[X]$.

Convergence in mean square implies that as $n \rightarrow \infty$ we have that $\mathbb{E}[(X_n-X)^2] \rightarrow 0$.

I tried to use the definition of variance $Var[X]=\mathbb{E}[X^2]-\mathbb{E}[X]^2$ and trying to prove that $|Var[X_n]-Var[X]|\leq \mathbb{E}[(X_n-X)^2]$ but I don't get any result.

Albanian_EAGLE
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1 Answers1

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Using the Cauchy-Schwarz inequality, we have that $$ |\operatorname E(X_n-X)|\le\bigl(\operatorname E|X_n-X|^2\bigr)^{1/2}\to0 $$ and $\operatorname EX_n\to\operatorname EX$. We also have that $\operatorname EX_n^2\to\operatorname EX^2$ using the continuity of the norm $(\operatorname E|X|^2)^{1/2}$ of $L_2(\Omega)$. Finally, $$ \operatorname{Var}X_n=\operatorname EX_n^2-(\operatorname EX_n)^2\to\operatorname EX^2-(\operatorname EX)^2=\operatorname{Var}X. $$

Phira
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Cm7F7Bb
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  • I dont understand the part where you say that it follows from the continuity of the norm. Can you elaborate a little more on that? I tried to prove this part myself using $\mathbb{E}[X_n^2]+ \mathbb{E}[X^2]\rightarrow 2\mathbb{E}[X_nX]$ which follows from mean square convergence. Knowing that $\mathbb{E}[X_n^2]+ \mathbb{E}[X^2]\geq 2\mathbb{E}[X_nX]$ it means that we should have the equality case, which implies that $\mathbb{E}[X_n^2] \rightarrow \mathbb{E}[X^2]$. Is this correct? – Albanian_EAGLE Mar 07 '14 at 15:49
  • @Albanian_EAGLE I'm afraid I don't understand your notation $\operatorname EX_n^2+\operatorname EX^2\to2\operatorname E[X_nX]$ since we have sequences on both sides. What does $x_n\to y_n$ mean? Why should we have the equality case? If we do have the equality case, then $\operatorname E|X_n-X|^2=0$ and $X_n=X$ almost surely for each $n\ge1$. – Cm7F7Bb Mar 07 '14 at 23:22
  • @Albanian_EAGLE See this answer for the continuity of the norm: http://math.stackexchange.com/questions/703579/convergence-of-random-variables-in-mean/703619#703619 – Cm7F7Bb Mar 07 '14 at 23:47