I have the following stochastic process represented by $$S_t = 1+\int_0^t exp\left(\sigma B_2 - \sigma^2\frac{s}{2}\right)dB_s,$$ where $B_t$ is a standard Brownian motion.

I would like to compute its expectations. I thought I could do it in two ways: $(1)$ using the fact that expectation of Ito integral is $0$ and $(2)$ using direct computations. But when I do this, I get two different answers.

Clearly, I must be doing something wrong, but I wonder if someone could point the problem to me.

(Expectations of Ito integral) Since the integrand is square integrable for a given t, one can simply use the fact that Ito integral with respect to Brownian motion will be 0. Thus,


(Direct computations) $$\mathbb{E}[S_t]=1+ \mathbb{E}\left[\int_0^t exp\left(\sigma B_s - \sigma^2\frac{s}{2}\right)dB_s\right]$$ Since integrand is non-negative, I thought I could apply Tonelli's Theorem to change the order of integration. But then $$\mathbb{E}[S_t] = 1+ \int_0^t \mathbb{E} \left[exp\left(\sigma B_s - \sigma^2\frac{s}{2}\right)\right]dB_s \\ = 1+ \int_0^t \mathbb{E} \left[exp\left(\sigma B_s \right)\right] exp\left(-\sigma^2\frac{s}{2}\right)dB_s \\ = 1+ \int_0^t exp\left(\sigma^2 \frac{s}{2} \right) exp\left(-\sigma^2\frac{s}{2}\right)dB_s \\ =1 + \int_0^t 1 dB_s \\ =1 + B(s) \neq 1$$

So it seems that Tonelli's (and Fubini's) theorem do not apply here. Is this because the integration is wrt to Brownian motion? Is there a way to compute this expectation directly?


You cannot write $E\int_0^{t} Y(s)dB_s$ as $\int_0^{t} (EY(s))dB_s$. The left side is a number and the right side is a random variable.

  • $\begingroup$ I see, that explains it. Thank you! $\endgroup$ – Mdoc Mar 27 at 6:09

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