Let $B_t (t\geq 0)$ be a standard Brownian Motion, where $B_0=0$. I have $d(B_s^2)= 2B_sdB_s$. I want to find the integral $ \int_0^tB_sdB_s$. Therefore, I get $ \int_0^tB_sdB_s= \frac{1}{2}\int_0^td(B_s^2)$. May I know how should I proceed? Is it possible to relate this to the mean of $B_t^2$, i.e. $\mathbb{E}(B_t^2)$?
2026-04-01 09:42:18.1775036538
Stochastic Integral and Standard Brownian motion
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