Let $X_n$, $n\in \mathbb{N}$ be a sequence of random variables which converges in probability to $X$, i.e. $X_n \stackrel{Prob}{\longrightarrow} X$. Furthermore it is known that $X \in L^2$. Does this imply $X_n \stackrel{L^2}{\longrightarrow} X$?
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See also http://math.stackexchange.com/questions/66029/lp-and-lq-space-inclusion – Willie Wong Jun 17 '13 at 10:43
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No. For example, consider $X_n=\sqrt n\cdot\chi_{[0,1/n]}$ in $L_2([0,1])$. $(X_n)$ converges to the zero function in probability and $\Vert X_n\Vert_{L_2}=1$ for each $n$.
David Mitra
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Thanks for the counterexample. I know uniform integrability is sufficient to conclude. Do you know any other assumption such that the L^2 convergence holds? – Harry Jun 17 '13 at 10:53
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@Harry If you have pointwise convergence and if everything is in $L_2$, then $\Vert X_n \Vert_{L_2}\rightarrow\Vert X\Vert_{L_2} $ implies $\Vert X_n-X\Vert_{L_2}\rightarrow0$. – David Mitra Jun 17 '13 at 10:59
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Take a sequence $X_n$ that converges in probability to zero (the zero function) but such that no $X_n$ is square integrable. The zero function is in $L_2$ but the sequence does not converge in $L_2$ to zero.
Rabee Tourky
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Thanks! I know uniform integrability is sufficient to conclude. Do you know some other assumption such that the L^2 convergence holds? – Harry Jun 17 '13 at 10:52