Using Itô Isometry to Calculate Integrated Brownian Motion Moments
Summary
The document works through three expectations involving the Itô integral of time against a Wiener process: its mean, its second moment, and its covariance with the terminal Brownian value. It treats the integrand as deterministic and uses the Gaussian distribution of such an integral to establish that its mean is zero, provided the squared integrand is integrable.
Itô’s isometry gives the second moment as the integral of the squared integrand, yielding T cubed over three for an integrand equal to time. The generalized isometry gives the cross expectation with the terminal value as the integral of time, yielding T squared over two. These results apply to the stated Brownian setting and deterministic integrand; the answer also corrects a notation slip in the original question, where the terminal value should be used.
Key ideas
- An Itô integral with a deterministic square-integrable integrand has zero mean.
- Itô’s isometry computes its second moment from the integral of the squared integrand.
- The generalized isometry computes cross moments between two Brownian integrals.
- For the time integrand, the second moment is T cubed over three.
- The cross expectation with the terminal Brownian value is T squared over two.
Tags
Full text
# Integration over function of Wiener process
# Integration over function of Wiener process
I wish to calculate below 3 expectations of a typical Wiener process -
- $E \left[ \int\limits_{0}^{T} tdW_t \right]$
- $E \left[ \left( \int\limits_{0}^{T} tdW_t \right)^2 \right]$
- $E \left[W_T \int\limits_{0}^{T} tdW_t \right]$
How should approach them?
## Answer by Kevin (score 6, accepted)
https://quant.stackexchange.com/a/57068
- Question 1) The Itô integral of a deterministic function is Gaussian, see here or here, i.e. $$\int_0^T f(u)\mathrm{d}W_u \sim N\left( 0,\int_0^T f(u)^2\mathrm{d}u\right).$$ The answer is thus zero. We of course need to require that $\int_0^T f(u)^2\mathrm{d}u<\infty$.
- Question 2) The simple version of Itô's isometry reads as $$\mathbb{E}\left[\left(\int_0^T X_u\mathrm{d}W_u\right)^2\right]=\mathbb{E}\left[\int_0^TX_u^2\mathrm{d}u\right].$$ Setting $X_u=u$, the answer is to question two is thus $\int_0^T u^2\mathrm{d}u=\frac{1}{3}T^3$.
- Question 3) Itô's isometry generalises to $$\mathbb{E}\left[\left(\int_0^T X_u\mathrm{d}W_u\right)\left(\int_0^T Y_u\mathrm{d}W_u\right)\right]=\mathbb{E}\left[\int_0^TX_uY_u\mathrm{d}u\right].$$ Thus,
$$\mathbb{E}\left[W_T\int_0^T u\mathrm{d}W_u\right]=\mathbb{E}\left[\left(\int_0^T 1\mathrm{d}W_u\right)\left(\int_0^T u\mathrm{d}W_u\right)\right]=\mathbb{E}\left[\int_0^T u\mathrm{d}u\right]=\frac{1}{2}T^2.$$
(Note: There is a typo in your question, the first Brownian motion should be $W_T$ and not $W_t$.)
## Answer by StackG (score 3)
https://quant.stackexchange.com/a/57067
For this type of problem you need to use the Ito isometry
- The first one is 0 due to symmetry of $W_t$ around 0
- A really similar problem is solved with working in this post (I've copied the algebra below): http://www.quantopia.net/interview-questions-vii-integrated-brownian-motion/
- Looks like stochastic integration by parts might help here (also used in the post above)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.