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Is the limit of a $L^2$-convergent sequence of random variables unique up to a.e.? In other words, if $X$ and $Y$ are both limits, will $X=Y$ a.e.? If yes, is Ito integral, which is defined as $L^2$ limit of a sequence of Ito integrals of simple processes, defined only up to a.e.?

Conversely, if the sequence converges to a random variable $X$, and $Y$ is another random variable same as $X$ a.e., will $Y$ also be the limit of the $L^2$-convergent sequence?

Similar questions for a sequence of random variables that converges in probability.

Thanks in advance!

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1 Answer 1

Hint: $\|X-Y\|_2\leqslant\|X_n-X\|_2+\|X_n-Y\|_2$. And some random variables $X$ and $Y$ such that $\|X-Y\|_2=0$ are such that...

Application: If $X_n\to X$ in $L^2$ and $X_n\to Y$ in $L^2$, then $\|X_n-X\|_2+\|X_n-Y\|_2\to0$ hence $\|X-Y\|_2=0$. For every positive $u$, $$ (X-Y)^2\geqslant u^2\cdot[|X-Y|\geqslant u], $$ hence $$ \|X-Y\|_2^2\geqslant u^2\cdot\mathrm P(|X-Y|\geqslant u). $$ If $\|X-Y\|_2=0$, this shows that $\mathrm P(|X-Y|\geqslant u)=0$ for every positive $u$. That is, $\mathrm P(|X-Y|\gt0)=0$, which is equivalent to $X=Y$ almost surely.

One can adapt this proof to the convergence in probability. For every positive $u$, $$ [|X-Y|\geqslant2u]\subseteq[|X_n-X|\geqslant u]\cup[|X_n-Y|\geqslant u], $$ hence $$ \mathrm P(|X-Y|\geqslant2u)\leqslant\mathrm P(|X_n-X|\geqslant u)+\mathrm P(|X_n-Y|\geqslant u). $$ Maybe you can continue...

Edit Recall that the limits almost sure, in $L^p$ and in probability are only defined almost surely, that is, if $X_n\to X$ in either one of these three acceptions and if $X=Y$ almost surely, then $X_n\to Y$ as well. If $X_n\to X$ in $L^2$ for example, use $\|X_n-Y\|_2\leqslant\|X_n-X\|_2+\|X-Y\|_2$ and the fact that $\|X-Y\|_2=0$ if (and only if) $X=Y$ almost surely. Thus $\|X_n-Y\|_2\leqslant\|X_n-X\|_2$ and $\|X_n-X\|_2\to0$, which implies that $\|X_n-Y\|_2\to0$.

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Thank you! I am not sure how the inequality is used to explain? I just modified my questions because I mixed two different questions together. –  Ethan Nov 27 '11 at 20:30
    
Thanks for details! I am thinking about the part to be continued. A different question is that (maybe you haven't seen it): if the sequence converges to a random variable X, and Y is another random variable same as X a.e., will Y also be the limit of the L2-convergent sequence? –  Ethan Nov 27 '11 at 21:29
    
See edit. $ $ $ $ –  Did Nov 27 '11 at 22:07

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