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I'm very sorry if this question sounds very basic, as I just started learning stochastic calculus. I found this link and I would like to clarify whether the following always holds:

$$\mathbb{E}[\int_a^b B_s dB_s] = 0$$

where $a$ and $b$ can be any real number, and $B_s$ is a Standard Brownian Motion. So far, almost all examples I have seen sets $a=0$.

InvadersMustDie
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  • Yes, because the Ito integral always produces a martingale, for sufficiently nice integrands. – Nate Eldredge Oct 03 '19 at 15:23
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    Based on the other link, you may find it helpful to write $\int_a^b B_s,dB_s = \int_0^b B_s,dB_s - \int_0^a B_s,dB_s$ and use linearity of expectation. – Nate Eldredge Oct 03 '19 at 15:24
  • @NateEldredge Ahhh goodness, why didn't I think of that...I really look like a fool for asking this ><...Thank you so much – InvadersMustDie Oct 03 '19 at 15:37

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