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Re: Are those variances equal?
Posted:
Jul 1, 2006 11:40 AM


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." > > > > Allright, i take that as a "definitely maybe". I'll look into > what condition i have and hopefully something will popup.
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.



