Mean and Variance of Brownian Motion Integrated Over a Time Interval
Summary
The document considers the ordinary time integral of Brownian motion over an interval starting after time zero. It uses the representation of that integral as an Itô integral with a deterministic time-dependent integrand, then applies conditional expectation and Itô isometry to derive its moments. Given the Brownian value known at the interval’s start, the conditional mean is that starting value multiplied by the interval length.
The variance follows from the squared Itô integrand and the conditional mean. The response emphasizes conditioning on the information available at the start time: the unconditional mean need not be zero when the initial Brownian value is random and nonzero. Its final variance expression mixes conditional and unconditional notation, so interpretation depends on whether variance is conditional on the starting information or taken over the full probability space. The explanation does not expand the integral or fully distinguish those two variance concepts.
Key ideas
- For an interval beginning at a positive time, condition on the Brownian value known at its start.
- The conditional mean of the time integral is the starting value multiplied by the interval length.
- The Itô isometry computes the second moment from the squared integrand.
- Variance requires subtracting the square of the appropriate mean, with conditional and unconditional versions distinguished.
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# Integral of Brownian motion w.r.t. time and integral not starting at zero
# Integral of Brownian motion w.r.t. time and integral not starting at zero
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 by Jan Stuller (score 2, accepted)
https://quant.stackexchange.com/a/63739
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:
$$Var\left(\int_{u=s}^{u=t}(t-u)dW_u\right)=\mathbb{E}\left[\left(\int_{u=s}^{u=t}(t-u)dW_u\right)^2\right]-\mathbb{E}\left[\int_{u=s}^{u=t}(t-u)dW_u\right]^2=\\=\int_{u=s}^{u=t}(t-u)^2du-(k(t-s))^2$$Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.