# Integrating deterministic function with respect to Brownian motion

I have looked everywhere for a satisfactory answer to this, including Shreve's textbooks, but I can't find one.

If I want to integrate a some deterministic function f(t) with respect to brownian motion, i.e. $$\int_0^tf(s)\mathrm dB_s$$ what is the solution?

I understand how to decompose this into a Ito integral. I also understand that the integral must have a normal distribution, and it is relatively simple to calculate the mean $= 0$ and the variance: $$\mathsf{Var} = \mathsf E (\int_0^tf(s)\mathrm dB_s)^2 = \mathsf E\int_0^tf^2(s)\mathrm ds.$$ so does this mean I can write the answer as $$\int_0^tf(s)\mathrm dB_s = \sqrt{\mathsf{Var}}B_t?$$ Because this would have a normal distribution with the correct mean and variance. It also makes sense to write the solution as N(0, variance).

Is either of these a correct solution to the integral of a generic deterministic function with respect to B.M.?

any help would be greatly appreciated.

Thanks, Paul

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## migrated from mathoverflow.netJul 16 '13 at 20:31

This question came from our site for professional mathematicians.

sorry I couldn't figure out how to format the equations correctly. I am working on tediting them; I just wanted to get this question out there ASAP. – Paul Jul 16 '13 at 14:44
I flagged that this question better fits MSE. Moreover, indeed for any fixed time $t$ the random variable $$I_t := \int_0^t f_s\mathrm dB_s$$ is normally distributed. However, if we consider $(I_t)_{t\geq 0}$ as a process, then there is a certain dependence between random variables $I_{t'}$ and $I_{t''}$ for $t'\neq t''$, which I guess is uniquely determined by covariance (in case $(I_{t'},I_{t''})$ are jointly Gaussian). Note that you can't "decompose" it into the Riemann integral, but you can apply Euler-Maryama scheme to simulate the result. – Ilya Jul 16 '13 at 14:53
Ilya, thanks for your prompt response. I am a little confused at your notation t'/=t''. Could you maybe explain what this would mean if f(t)=exp(t), so that f=f'=f''? – Paul Jul 16 '13 at 15:03
That means $t'$ not equal to $t''$. – Ilya Jul 16 '13 at 15:05
so you mean t'=dt/dB?...Apologies because I have never seen this notation before. could you please point me towards an equation that shows this dependence? – Paul Jul 16 '13 at 15:10