Yale STAT 251 - Stochastic integrals

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Preliminary version 4 April 2004Stochastic integralsSuppose Stdenotes the price of a stock at time t ,for0≤ t ≤ 1. Let 0 = t0< t1<...<tn< tn+1= 1be times at which you buy and sell stock: at time tiyou buy H(ti) stocks at a cost of H (ti)S(ti) then yousell the same stocks at time ti+1for H (ti)S(ti+1). Your total profit will beni=0H(ti)iS where iS = S(ti+1) − S(ti).This formula is also valid for purchases of random numbers of shares. In that case, H(ti) should dependonly on information available at time ti.It is tempting o think that if the times between trades get smaller and smaller then we could passto some limit of continuous trading, with total profit being given by some sort of limit of the sums fortrading in discrete time. To formalize this idea, we need to define a stochastic integral10HsdSs.There is a large class of processes for which stochastic integrals can be defined. A complete treatmentusually takes up a large fraction of the graduate course on Stochastic Calculus. However, with enoughhandwaving I can explain the main ideas.It is easiest to start with a deterministic case.In what follows, I have been sloppy about stating regularity conditions. You should not take theassertions to be true precisely as stated. You need to take the Stochastic Calculus course if you want toknow the truth, almost the whole truth, and hardly anything but the truth.1. Functions of bounded variationSuppose f and g are continuous functions defined on the interval [0, 1]. Remember that the variationof f over a grid G :0= t0< t1< ... <tn< tn+1= 1 is defined asV( f, G) =ni=0| f (ti+1) − f (ti)|,and f is said to be of bounded variation if V( f ) = supGV( f, G) is finite.We mght hope that10g(t)df(t) could be obtained as a limit of approximations from grids,I(g, G) =ni=0g(ti)if where if = f (ti+1) − f (ti).In fact, such a limit does exist, in the sense that there is number J such that I( f, G) → J asmesh(G) := maxi|ti+1− ti| tends to zero. Of course, the limit J is then denoted by10gd f .<1> Theorem.Ifgis continous andfis both continuous and of bounded vaiation, then there is a nuberJfor whichI(g, G) → Jasmesh(G) → 0.Proof. It is enough (Why?) to show that for each >0 there exists a grid Gfor which|I(g, G) − I(g, G)|≤ whenever grid G is a refinement of grid G.Continuity of g on a closed interval ensures that for each >0 there exists a δ>such that<2> |g(t) − g(s)|≤ whenever |t − s|≤δ.Choose G:0= t0< t1< ... <tn< tn+1= 1 as any grid with mesh less than δ.Consider to contributions to both I(g, G) and I(g, G) from the interval [ti, ti+1| when G is arefinement of G. Suppose G puts grid points s0= ti< s1<...<sk< sk+1= ti+1in the interval. Thecontribution to I(g, G) from the interval iskj=0g(sj)jf where jf = f (sj+1) − f (sj)The contribution to I(g, G) isg(ti) f (ti+1) − f (ti)= g(ti)kj=0jf.c David Pollard, 2004 Statistics 251/551Preliminary version 4 April 2004The absolute value of the difference between the two contributions is bounded bykj=0|g(ti) − g(sj)||jf |≤kj=0|jf | because |ti− sj|≤δ.Summing over all i, we conclude that|I(g, G) − I(g, G)|≤V( f, G) ≤ V( f )If it disappoints you that the final bound is not , you should repeat the argument with the  in <2>replaced b /V( f ).For the purposes of this handout, there are two important cases where a function f has boundedvariation.(i) If f is an increasing function on [0, 1] then V( f ) = f (1) − f (0), becauseni=0| f (ti+1) − f (ti)|=ni=0 f (ti+1) − f (ti)= f (1) − f (0)for every grid.(ii) If f (t) =10λ(s) ds, with10|λ(s)| ds < ∞ thenni=0| f (ti+1) − f (ti)|≤ni=0|ti+1tiλ(s) ds|≤ni=0ti+1ti|λ(s)| ds =10|λ(s)| ds.In this case, it is not hard to show that10g(s) df(s) =10g(s)λ(s) ds.A similar method of approximation could be used to definet0gdf for each t in [0, 1]. A better wayis to build the dependence on t into the approximation, by definingI(g, G)t=ni=0g(ti) f (ti+1∧ t) − f (ti∧ t).If t equals ti,wehavetj∧ t = tifor all j ≥ i, which ensures that the all summands for j ≥ i vanish. Ifti< t < ti+1,theith summand becomes g(ti) f (t) − f (ti), which is continuous in t . Indeed, the insertionof the ∧t makes I(g, G)ta continuous function of t. The argument from the proof of Theorem <1> stillworks, leading to the conclusion thatt0gdf is a uniform limit of continuous functions, and hence is itselfcontinuous as a function of t.By various approximation arguments, the integral can also be extended to integrands g that are otcontinuous. I won’t discuss this extension, because we will only need continuous integrands.2. Stochastic integral for BV processesSuppose {Xt(ω) :0≤ t ≤ 1} is a stochastic process for which each sample path X (·,ω) is continuousand of bounded variation. If {Ht(ω) :0≤ t ≤ 1} is another stochastic process, then we can define thestochastic integral pathwise. That is,t0H(s,ω)dX(s,ω) is defined using the method described above foreach ω.It often helps to think of the stochastic integral as defining a new stochastic process H • X withcontinuous sample paths:(H • X)(t,ω)=t0Hs(ω) dXs(ω)3. Stochastic integral with respect to Brownian motionWe know that almost all sample paths of a standard Brownian motion {Bt:0≤ t ≤ 1} haveinfinite total variation. We cannot expect the method from Sections 2 to work to define a stochasticintegralt0Hs(ω) dBs(ω).c David Pollard, 2004 Statistics 251/551Preliminary version 4 April 2004For a smaller class of functions, the definition of the stochastic integral is easy. Suppose H is anelementary process, that is, for some grid G :0= t0< t1< ... <tn< tn+1= 1,H(t,ω) =ni=0H(ti,ω)I{ti< t ≤ ti+1},where H (ti,ω) is a random variable that depends only on information up to time ti. Then we defineH • Bt=t0Hs(ω) dBs(ω) =ni=0H(ti,ω) B(ti+1∧ t,ω)− B(ti∧ t,ω)Again the process H • B has continuous sample paths. It also inherits from B the martingale property. Toestablish this property, we need to show, for s < t and W depending only on information in Fs, that<3> E WHti B(ti+1∧ t) − B(ti∧ t)= E WHti B(ti+1∧ s) − B(ti∧ s).If s ≥ ti+1or t ≤ tithe equality is trivial, because B(ti+1∧ t) − B(ti∧ t) = B(ti+1∧ s) − B(ti∧ s) in bothcases. If s ≤ ti< t,


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Yale STAT 251 - Stochastic integrals

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