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Re: Are those variances equal?
Posted:
Jul 1, 2006 11:40 AM
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Konrad Viltersten wrote:
>>>Suppose you start off with this expression.
>>>V[W_t / t^2]
>>>and you wish to show that it tends towards 0 as t->oo.
>>>
>>>What we tried is this rephrasal.
>>>(1/t) * V[W_t / t]
>>>Does it hold?
>>>
>>>And then, is it possible to use the fact that
>>>lim t->oo (W_t / t) = 0 (a.s.)
>>>and do a rewriting as follows?
>>>
>>>lim t->oo (V[W_t / t]) = V[lim t->oo (W_t / t)]
>>
>>1. V(a W_t) = a^2 V(W_t) for any scalar a.
>
>
> True, but how does it work if we have a process, let's
> say a Wiener process like {W_t/t}_t>=0. Is it still OK
> to regard the t as a scalar? I'd say so, because it's a
> deterministic value but i'm a little unsure...
Yes, it makes no difference. The important thing is that t does not
depend upon w (the variable denoting the probability space).
>>2. Convergence a.s. cannot be converted to convergence in variance
>>without some kind of "dominated convergence."
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>
>
> Allright, i take that as a "definitely maybe". I'll look into
> what condition i have and hopefully something will pop-up.
It is quite often possible to make these things work, but it isn't
automatic or necessarily straightforward. But, for example, if W_t is a
Weiner process, you even have an explicit formula.
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