Given that $B_t$ is the standard Brownian motion, I need to contrast the mean and variance of the stochastic integral $\int\limits_{0}^t B_s dB_s = \frac{1}{2}(B_t^2 - t)$ with the non-stochastic integral $\int\limits_0^t B_s ds$. So I obtain zero means for both of the integrals. I obtain $Var(\int\limits_{0}^t B_s dB_s) = \frac{1}{2}t^2$ while $Var(\int\limits_{0}^t B_s ds) = \frac{1}{3}t^3$. I am wondering what is the implication of this result? How should I interpret the observation that the stochastic integral of the Brownian motion has lower variance compared with the non-stochastic integral?


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.