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Let $B_t$ be Brownian motion, and $X_t$ satisfies the following Ito SDE: $$ d X_t = \mu\, dt + \sigma\, d B_t, $$ and $f$ is a function over $X_T$. I want to calculate $\mathbb{E}[f(X_t)dB_t]$.


It seems that the result is $\mathbb{E}[f(X_t)dB_t] = 0$, but I fail to prove it. I know the following equations hold $$ \mathbb{E}[B_t] = 0, \quad \mathbb{E}[dB_t] = 0. $$ However, since there is an additional term, i.e., $f(X_t)$ in the expectation, $\mathbb{E}[f(X_t)dB_t] = f(X_t)\mathbb{E}[dB_t] = f(X_t) * 0 = 0 $ doesn't hold. Could anyone tell me how to prove it?

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  • $\begingroup$ It seems that this question is related to this answer: Itō Integral has expectation zero $\endgroup$
    – jzin
    Commented Aug 19, 2022 at 4:12
  • $\begingroup$ I think your phrasing is off. What does $dB_t$ even mean outside the context of an SDE? Perhaps you mean $\mathbb{E}[df(X_t)dB_t]$? $\endgroup$
    – Yair Daon
    Commented Aug 19, 2022 at 5:48

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