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This is not a homework question so please be kind not to remove it right away. I am working on some research but have to justify the following argument: Assume $S_t$ is a continuous stochastic process, don't want to make an assumption about distribution, think about something like a smooth function of Brownian motion. I define another process $$Y_t=\frac{1}{t} \int_0^t S_u du$$ Now I am interested in the limit of $Y_t$ as $t$ approaches zero. I would like to know in what sense the argument would hold, the guess is that the limit is $S(0)$. Please suggest a solution or the way to approach this problem.

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How is S defined? That is if really is a continuous function of BM then there is a lot you can say. So, more detail please. –  BSteinhurst Nov 20 '12 at 3:21
    
ok, let's take a particular case, where $S_t$ is according to the following dynamics $dS_t=adt+bdB_t$ where $B_t$ is a Brownian motion, so that $S_t$ has a lognormal distribution. –  Kamil Nov 20 '12 at 3:59
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@ Kamil : Unless mistaken you can think on a path by path basis as long as your process is continuous for every path it is integrable on a fixed time interval with respect to Lebesgue measure "ds", then applying Lebesgue differentiation theorem (wiki) you have that the limit is $S_0$ (it is not a research level question but a good question for math exchange this is why I answer it in comment). Regards. –  The Bridge Nov 20 '12 at 9:51
    
@The Bridge: what if $S_u=u^{1/2}$, don't I have a problem then? Is the statement holds for "any" continuous process $S_u$ regardless how wild it is? –  Kamil Feb 6 '13 at 3:44
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1 Answer

up vote 1 down vote accepted

You can prove it reasoning $\omega$ by $\omega$. I mean, if you are working in a probability space $\Omega$, the continuous path $(S_t)_{t\geqslant 0}$ depends on $\omega\in \Omega$. But for all $\omega$, $t\mapsto S_t(\omega)$ is continuous, so the following convergence holds : $$ Y_t(\omega) \rightarrow S_0(\omega) $$

It is enough to conclude that $Y_t$ converges almost surely to $S_0$ as $t$ goes to zero.

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actually The Bridge had already answered in the same way in the comments. –  Guillaume Nov 20 '12 at 18:36
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@ Guillaume : 1up for honesty ;-) –  The Bridge Nov 20 '12 at 19:24
    
yes, thanks the Bridge and Guillaume, I think the answer now is complete as I was looking for some omega mentioning. Thus, I have the convergence for every single omega and therefore convergence in a.s. sense. –  Kamil Nov 21 '12 at 13:30
    
related to time $0$ question: If I assume $dS_t = adt+bdB_t$ and from above $dY_t = \frac{1}{t}(S_t-Y_t)$ then I can use Ito formula for $u(t,S_t,Y_t)$. I would like that to be a martingale and thus require $u_t+0.5b^2u_{ss}+au_s+\frac{1}{t}(s-y)u_y=0$. The last term is not defined at $t=0$ but from above answer as $t$ approaches $0$ I have $\frac{1}{t}(s-y)=\frac{0}{0}$. Is there anything can be said about the last term in the pde? –  Kamil Nov 22 '12 at 16:14
    
@kamil : something is missing in your sde for $Y$, moreover once modification done can you show that the sde has a solution ? Best regards –  The Bridge Nov 22 '12 at 20:46
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