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If I have an inequality, e.g.: $\mathbb{E}|X(t+δ)-X(t)|²≤(aδ+bδ²)K$ say $a,b$ and $K$ are just constants, $X$,an arbitrary stochastic processes, then if I want to evaluate the limit $lim_{δ→0}$: $\lim_{δ→0}(aδ+bδ²)K=0$ Is it correct to say then: $\lim_{δ→0}\mathbb{E}|X(t+δ)-X(t)|²=0$ i.e. convert the inequality to an equality like this. I was told it was incorrect to do this because the difference of two stochastic processes (say any arbitrary stochastic process) could be negative, but i don't see how this is possible since it is the magnitude squared. Then the only possible solution would be zero

(I've corected the Y, to X(t))and yes euclidean metric.

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    In principe it can of course. But what is the relation betewen $\Delta$ and $Z$? I can not see it. If there is no relation defined between them, the only result is $E(Z^2)=0$ which is incorrect.2012-09-05

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Squeeze/Sandwich Theorem:

Given functions $g(x) \leq f(x) \leq h(x)$, and that $\lim_{x\to a}g(x)$ = $\lim_{x\to a}h(x) = L$. Then, $\lim_{x\to a}f(x) = L$.

In your case, you have $f(δ) =\mathbb{E}|X-Y|²$, $g(δ) = 0$ and $h(δ) = (aδ+bδ²)K$. Looks legit.

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    $0 \leq \mathbb{E}|X(t+δ) - X(t)|²$ no?2012-09-05