# Inequality and limit question

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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Are the processes in some way dependent on $\delta$? Because if they aren't I can't see why the l.h.s. shouldn't be a continuous function of $\delta$ (a constant) and therfore $0$. –  vanguard2k Sep 5 '12 at 11:50
Is $X$ and $Y$ complex? or real? –  Seyhmus Güngören Sep 5 '12 at 12:02
how can the expected value of a squared random variable $Z^2=(X-Y)^2$ be zero?? –  Seyhmus Güngören Sep 5 '12 at 12:10
It could tend to zero without being zero? –  Legendre Sep 5 '12 at 12:12
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. –  Seyhmus Güngören Sep 5 '12 at 12:18
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.
I am assuming $\mathbb{E}|X-Y|²$ refers to the expected value of the square of the euclidean distance of $X$ and $Y$. As far as I know, euclidean distance cannot be negative (property of a metric). –  Legendre Sep 5 '12 at 12:08
$0 \leq \mathbb{E}|X(t+δ) - X(t)|²$ no? –  Legendre Sep 5 '12 at 12:33