Integral of Brownian motion w.r.t. time and integral not starting at zero

Question

I'm new to stochastic calculus and try to calculate (1) mean and (2) variance of $$\int_s^t W_u du$$ where $W_u$ is a Brownian motion. I already found this helpful answer, where it was shown that $\int_0^t W_u du \sim \mathcal{N}(0, \frac{1}{3}t^3)$ Using the same logic I can show that $$\int_s^t W_u du = \int_s^t (t-u) dW_u $$ Can I follow that $$\mathbb{E}\biggl[\int_s^t W_u du \biggl] = \mathbb{E}\biggl[\int_s^t (t-u) dW_u \biggl] = 0 \text{ ?}$$ and if the mean is zero $$Var\biggl[\int_s^t W_u du \biggl] = Var\biggl[\int_s^t (t-u) dW_u \biggl] = \mathbb{E}\biggl[\biggl(\int_s^t (t-u) dW_u \biggl)^2\biggl] = \mathbb{E}\biggl[\int_s^t (t-u)^2 du \biggl] \text{ ?}$$ And if so, why is this true?

Many thanks in advance!

Answer 1

If $s>0$, and the integral runs from $u=s$, then the integral only makes sense if we condition on what we know as of time $s$: we can write $W(s)=k$, where $k$ is some constant known at time $s$, i.e. the value of the Brownian motion $W_u$ known at time $u=s$ (can be zero, but doesn't need to be).

Then, we have:

$$\mathbb{E}\left[\int_{u=s}^{u=t}W_udu|\mathcal{F}_s\right]=\int_{u=s}^{u=t}\mathbb{E}[W_u|\mathcal{F}_s]du=k(t-s)$$

Above, $\mathcal{F}_s$ is the sigma-algebra as of time $s$, i.e. "the information known as of time $s$".

The variance can be computed using Ito Isometry which you rightly state.

For any adapted process $X_t$, Ito Isometry states that:

$$\mathbb{E}\left[\int_{u=s}^{u=t}X_udW_u\right]^2=\int_{u=s}^{u=t}\mathbb{E}[X_u^2]du$$

If you need the proof of Ito Isometry, it's just about writing out the Ito Integral from first principles as sum of Brownian increments and using Ito's lemma. Let me know if you need the proof.

So basically the variance will be:

$$Var\left(\int_{u=s}^{u=t}(t-u)dW_u\right)=\mathbb{E}\left[\left(\int_{u=s}^{u=t}(t-u)dW_u\right)^2\right]-\mathbb{E}\left[\int_{u=s}^{u=t}(t-u)dW_u\right]^2=\\=\int_{u=s}^{u=t}(t-u)^2du-(k(t-s))^2$$