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Hi guys,how can I prove that the in the limit the quadratic variation of of a Wiener process is bounded and a.s. =t?Here's is what I thought:1. The normal W is unbounded since its trajectories are fractal, thus infinite.2. If we look at very very small steps the the quadratic variation is always smaller: So in a way, if the process is "infinite" and the quadratic variation is smaller and the quadratic variation cannot be infinite.3. To prove that the quadratic variation is bounded (and =t) I need to find the expectation of the quadratic variation to be =t and the variance to be =0. How do I do this?

For expectation, exchange sum and expectation, use the fact that increments are normal, thus you'll get SUM var(Wtj-Wtj-1)=SUM(t_tj-t_tj-1)=[telescope rule]=t-0=t.How to deal with variance - I don't know.

Nobody?

here come my two cents to deal with variance.following magnolija's notation, we need to compute var(SUM (Wtj-Wtj-1)^2) = SUM(var(Wtj-Wtj-1)^2) (because of the independent increments) <= SUM 3*(tj-tj-1)^2 (because V(x^2) = E(x^4) - (E(x^2))^2 <= E(x^4) = 3*sigma^4, for x \sim Normal(0, sigma)) <= 3*max_i (ti-ti-1) * SUM (ti-ti-1) --> 0 (because of the maximand converges to zero and final term equals to t due to telescope rule). given that exepcation converges to t and the variance vanishes, we apply the chebyshev to conclude that [W]=t, p-a.s.

mjy, I think it's not p-a.s., it's the limit in probability, no? e.g. L_2(P)-lim => lim-in-P, a.s. is a different kind of limit

Zerdr0n, thanks a lot for pointing that out. yes, the sum converges in probability, which is a weaker convergence concept than almost sure convergence. ** added on May 7th 2008 **if the sum converges in probability, then one might find a subsequence of the original sequence of Riemannian subdivision, which converges almost surely. ************************

QuoteOriginally posted by: mjyhere come my two cents to deal with variance.following magnolija's notation, we need to compute var(SUM (Wtj-Wtj-1)^2) = SUM(var(Wtj-Wtj-1)^2) (because of the independent increments) <= SUM 3*(tj-tj-1)^2 (because V(x^2) = E(x^4) - (E(x^2))^2 <= E(x^4) = 3*sigma^4, for x \sim Normal(0, sigma)) <= 3*max_i (ti-ti-1) * SUM (ti-ti-1) --> 0 (because of the maximand converges to zero and final term equals to t due to telescope rule). given that exepcation converges to t and the variance vanishes, we apply the chebyshev to conclude that [W]=t, p-a.s.Thanks mjy! There is one thing I don't understand yet.So we have then we use the fact thatnow I think you wrote but isn't ?In other words, why is there a multiplication in here: "3*max_i (ti-ti-1) * SUM (ti-ti-1)" and why is it not a minus?Thanks for your help!

what i wrote is the following:SUM(var(Wtj-Wtj-1)^2) = SUM(E[(Wtj-Wtj-1)^4] - E[(Wtj-Wtj-1)^2]^2) <= SUM(E[(Wtj-Wtj-1)^4]) = 3 SUM (tj-tj-1)^2 <= 3*max_j (tj-tj-1) * SUM (ti-ti-1) --> 0.the first equality follows from the definition of the variance. then the weak inequality (<=) follows from the fact that E[(Wtj-Wtj-1)^2]^2 is nonnegative for all j. the second equality is due to the fourth moment of a gaussian r.v. as I indicated in the bracket below. the final weak inequality holdes because max_j (tj-tj-1) >= tj-tj-1, for all j.

QuoteOriginally posted by: mjywhat i wrote is the following:SUM(var(Wtj-Wtj-1)^2) = SUM(E[(Wtj-Wtj-1)^4] - E[(Wtj-Wtj-1)^2]^2) <= SUM(E[(Wtj-Wtj-1)^4]) = 3 SUM (tj-tj-1)^2 <= 3*max_j (tj-tj-1) * SUM (ti-ti-1) --> 0.the first equality follows from the definition of the variance. then the weak inequality (<=) follows from the fact that E[(Wtj-Wtj-1)^2]^2 is nonnegative for all j. the second equality is due to the fourth moment of a gaussian r.v. as I indicated in the bracket below. the final weak inequality holdes because max_j (tj-tj-1) >= tj-tj-1, for all j.Man! I'm such an idiot. O_o I thought your '<=' was not an inequality but some sort of an arrow, and so obviously nothing made sense to me... Haha.Well, thanks again. One more question: What's the 'telescope rule'? I googled it and couldn't really find anything, but I'd like to read some proof and I'm certain it should be online somewhere. Maybe under an alternative name?

QuoteOriginally posted by: CrashedMintWell, thanks again. One more question: What's the 'telescope rule'? I googled it and couldn't really find anything, but I'd like to read some proof and I'm certain it should be online somewhere. Maybe under an alternative name?In continuous time telescope rule is presented by INTEGRAL(0:T) dZ(t)=Z(T)-Z(0). This also holds in discrete case as SUM(i=0:T) delta Z(i) = Z(T)-Z(0). I hope the idea is clear: you have a sum of differences in sequential time periods and when you write everything explicitly, all terms vanishes, just the last minus first are left.

QuoteOriginally posted by: magnolijaQuoteOriginally posted by: CrashedMintWell, thanks again. One more question: What's the 'telescope rule'? I googled it and couldn't really find anything, but I'd like to read some proof and I'm certain it should be online somewhere. Maybe under an alternative name?In continuous time telescope rule is presented by INTEGRAL(0:T) dZ(t)=Z(T)-Z(0). This also holds in discrete case as SUM(i=0:T) delta Z(i) = Z(T)-Z(0). I hope the idea is clear: you have a sum of differences in sequential time periods and when you write everything explicitly, all terms vanishes, just the last minus first are left.So, to put this into layman terms:1. dW is just a really tiny bit of a Wiener process. If I "sum up" all these really tiny bits starting from W(0)=0 then I will just end at W(t). 2. dt is a really tiny period of time. If I sum up all these itty bitty changes from t=0 until t=t then I end up at... t.Correct, right?

yes, you can say so. the following might also be helpful. http://mathworld.wolfram.com/TelescopingSum.html