hi. what is the distribution of the following:

How's the W term being used?

QuoteOriginally posted by: hongjiren2000hi. what is the distribution of the following:I assume f(s) is a deterministic function and W(s) is a BM.You can rewrite your integral as int g(s) dW + parts term. Details to you.If the parts term vanishes, you're in luck. That's because the other is normally distributed, with mean zeroand the variance is given by the Ito isometry, which you can look up if you don''t know it. In that special case, you are done.If the parts term -doesn't- vanish, I'm not sure if your original integral is normal or not.But here's what I would do to find out:Let X(t) = your original integral. Obviously odd moments vanish and and even moments do not. Calculate m2(t) = E[X^2(t)] and m4(t) = E[X(t)^4].How? Let's abbreviate <...> = E[...].Then, you can calculate these moments using <W(s) W(t)> = min[s,t], and <W(t1) W(t2) W(t3) W(t4)> = <W(t1) W(t2)><W(t3) W(t4)> + 2 similar permutations.So, now you have a formula for m2 as a double integral and m4 as a quadruple integral.Next, check if m4(t) = 3 m2(t)^2 ??? If "yes", I would strongly suspect the answer is "your integral is normally distributed with mean zero and variance m2(t)".If "no", a likely answer is "your integral has no simple distribution in general". regards,

a linear combination of normal things will be normal, and an integral is a limit of linear combinations, so the answer will be a normal. It's also clearly symmetric if W_0 =0 so mean is zero. So we just have to get the variance. You could this from a limiting argument.

Yes, I should have seen this from my moment calculation too, as m4(t) = 3 m2(t)^2. In fact, by the general pairwise sum formula for <W(t1) ... W(tn)>, all even moments m_n(t) of the integral can be seen to satisfy: m_n = (n-1)!! [m_2]^(n/2).This provides a second proof of normality. Then, given it's normal, I already gave the variance in my original post: m2(t) = \int(0,t) int(0,t) min[t1,t2] f(t1) f(t2) dt1 dt2

thanks for the help.

If the function f is Riemann-integrable wrt to time from Ito's lemma for the stochastic process H(t,W(t))=F(t)W(t) where dF/dt=f you havedF(t)W(t)=f(t)W(t)dt+f(t)dW(t) <=> f(t)W(t)dt=-f(t)dW(t)+dF(t)W(t) , thus int {f(s)W(s)ds}= int {-f(s)dW(s)}+ F(0)W(0)+F(t)W(t)which is a normal distribution mean zero since W(0)=0 and variance int {f(s)^2ds}+t*F^2. So the problem is to determine F by integrating f. Hope that this helps without moments and limiting stuff.int {} is integral

A mistake seems to be here.First, in Ito's formula for H(t,W(t))=F(t)W(t) with dF/dt=f we have (H(t,x) = F(t)x; H_t(t,x) = f(t)x, H_x(t,x) = F(t), H_xx(t,x) = 0)dF(t)W(t) = f(t)W(t)dt + F(t)dW(t) (last term is correct now),which yieldsint {f(s)W(s)ds} = F(t)W(t) - int {F(s)dW(s)}.We have X = F(t)W(t) ~ N(0, F^2(t)t) and ( note also that F(t)W(t) = int{F(t)dW(s)} )Y = int {F(s)dW(s)} ~ N(0, int{F^2(s)ds} ).Now let find the covariance between X and Y :Cov(X,Y) = Cov( int {F(s)dW(s)}, int{F(t)dW(s)} ) = int{ F(t)F(s)ds }.We finally conclude thatint {f(s)W(s)ds} = X - Y ~ N( 0, int{(F(t)-F(s))^2 ds} )

Thanks i forgot to press caps lock however F(t)W(t) = int{F(t)dW(s)} that you mention is not correct, Ito's lemma shows you that it is not the case, thusdumn i also forgot to take covariance, the proper covariance would be:Cov[int{F(t)dW(s)}, F(t)W(t)]=E[int{F(t)dW(s)}* F(t)W(t)]=F(t)E[int{F(t)dW(s)}* W(t)]=F(t)E[int{F(t)dW(s)}* int{dW(s)}]=F(t)E(int{F(s)ds} (from isometry)and take out the expectation since we have a deterministic function and add this to my original result. Also change the small with the big F at the place indicated by the last guy. I disagree with his covariance

Since W(t) = int{dW(s)} we do have F(t)W(t) = F(t) int{dW(s)} = int{F(t)dW(s)}GreekMartingale, where do you see a mistake here?

The argument in thr F is s and not t thus it should stay inside a ds integral

You seem to miss the point, sorry..When writing F(t) I do mean the argument is t, the upper limit in the integral. So F(t) is just a constant for any expression like int{... ds}.

could you write in Latex so we can both figure out i still dont get your point