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In this answer the mean-square limit of a succession of stochastic processes $Q_n$ is calculated in this simple way:

  • Check whether the limit of their mean value is finite or not: $$\tag{1}\displaystyle \lim_{n\to\infty}\operatorname{ E}[Q_n]=:Q\lt\infty$$
  • Make sure that the variance vanishes in the limit: $$\tag{2}\displaystyle \lim_{n\to\infty}\operatorname{Var}[Q_n]=0. $$

If both $(1)$ and $(2)$ are realized, then $Q$ is shown to be the mean square limit of $Q_n$. Is this a general strategy? According to what theorem?

UPDATE. As @Adam pointed out, this strategy provides sufficient but not necessary conditions. As an example, consider the integral

$$ Q:=\int_{t_0}^t\mathrm{d}W_s$$ this is by definition the mean-square limit of $$ Q_n=\sum_{i=1}^n\bigg(W(t_i)-W(t_{i-1})\bigg)=W(t)-W(t_0) $$ where $t_n=t$, and $t_i-t_{i-1}=(t-t_0)/n$. But then, according to the definition of wiener process, $W(t)-W(t_0)\sim\mathcal{N}(0, t-t_0)$, so $$ \mathrm{E}[Q_n]=0,\\ \mathrm{Var}[Q_n]=t-t_0\nrightarrow 0 \quad \mathrm{as} \quad n\to\infty $$

and yet, $W(t)-W(t_0)$ is supposed to be the answer.

ric.san
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1 Answers1

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Yes. For equation (1) to make sense, the limit $Q$ must be a constant; I'll write it as $q$ to avoid confusion. We then have

$$\begin{align*} \mathbb E[(Q_n-q)^2] &= \mathbb E[(Q_n - \mathbb E Q_n + \mathbb E Q_n - q)^2]\\ &=\mathbb E[(Q_n - \mathbb E Q_n)^2 + 2(Q_n - \mathbb E Q_n)(\mathbb E Q_n - q) + (\mathbb E Q_n - q)^2]\\ &=\mathbb V \mathrm{ar}[Q_n] + (\mathbb E Q_n - q)^2\\ &\to 0; \end{align*}$$ thus the $Q_n$ have mean-square limit $q$.

Adam
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  • Why does $Q$ have to be constant? Shouldn't $\displaystyle Q(t)<\infty\ \forall,t$ be enough? – ric.san Mar 06 '23 at 09:03
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    I'm saying it's a constant for fixed $t$; it's not random. @ric.san – Adam Mar 06 '23 at 10:50
  • Isn't there a way to make sense of $(1)$ if $Q$ is a random variable (just depending upon $t$ but not $n$)? – ric.san Mar 07 '23 at 13:13
  • Well, $Q$ is a limit of expectations, so as written it's non-random by definition. If you mean something else, you'd have to say what. @ric.san – Adam Mar 07 '23 at 13:30
  • Ok, but then I don't know how to make sense of the example I added some hours ago – ric.san Mar 07 '23 at 13:36
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    I'm not sure what you're asking here? Yes, it's possible to converge in mean square without satisfying (1) and (2); the conditions are sufficient but not necessary. @ric.san – Adam Mar 07 '23 at 13:47
  • Aren't there necessary and sufficient conditions? Otherwise, how can one find such a limit? By guessing? – ric.san Mar 07 '23 at 13:48
  • There are lots of ways to show mean square convergence; the approach usually depends on the problem. @ric.san – Adam Mar 07 '23 at 13:55
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    I see... thanks for your patience. My original problem was actually to find a criterium to evaluate $$\int_{t_0}^t\mathrm{d}W_s \qquad \mathrm{and} \qquad\int_{t_0}^tW_s,\mathrm{d}W_s$$ but I think that's another question. – ric.san Mar 07 '23 at 14:01