I have the following stochastic process represented by
$$S_t = 1+\int_0^t \exp\left(\sigma B_s – \sigma^2\frac{s}{2}\right)\,dB_s\,,$$
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,
$$\mathbb{E}[S_t]=1$$
(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
$$\mathbb{E}[S_t] = 1+ \int_0^t \mathbb{E} \left[\exp\left(\sigma B_s – \sigma^2\frac{s}{2}\right)\right]\,dB_s \\
= 1+ \int_0^t \mathbb{E} \left[\exp\left(\sigma B_s \right)\right] \exp\left(-\sigma^2\frac{s}{2}\right)\,dB_s \\
= 1+ \int_0^t \exp\left(\sigma^2 \frac{s}{2} \right) \exp\left(-\sigma^2\frac{s}{2}\right)\,dB_s \\
=1 + \int_0^t 1 \,dB_s \\
=1 + B_t \neq 1\,.$$
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?
Best Answer
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.