Let $X_{n}(\omega)$ be a sequence of r.v. such that $|X_{n}(\omega)|\leq 1$ and $\underset{n\rightarrow \infty}{\lim}X_{n}(\omega)=C$ a.s., where $C\in [0,1]$. By using dominated convergence we have $\underset{n\rightarrow \infty}{\lim}E[X_{n}(\omega)]=C$. What can we say about $\underset{n\rightarrow \infty}{\lim}E[(X_{n}(\omega)-E[X_{n}(\omega)])^2]$? Also what further condition do we need in order to have $\underset{n\rightarrow \infty}{\lim}E[(X_{n}(\omega)-E[X_{n}(\omega)])^2]=0$?
Almost sure convergence and Variance
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probability-theory
convergence
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3If $X_n \to C$ and $E[X_n] \to C$, then $(X_n - E[X_n])^2 \to \ldots$ – 2017-01-11
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0I thought that from almost sure convergence I could not say much about L2 convergence. Is there any problem with interchange of limit and expectation? – 2017-01-11
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0The boundedness of $X_n$ makes $L^2$ just as easy as $L^1$. – 2017-01-11
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0ok thank you Prof. Robert Israel – 2017-01-12