Deriving the Product Expectation for Itô Integrals
Summary
The document asks how to compute the expected product of two Itô integrals driven by the same Brownian motion. Under suitable adaptedness, independence, and integrability conditions, the stated identity reduces this expectation to the time integral of the expected product of the two integrands. This is a cross-integral counterpart to the familiar Itô isometry.
The answer sketches a derivation by approximating each stochastic integral with sums over time intervals. Products from distinct, non-overlapping Brownian increments have expectation zero because the later increment is independent of the information available before it and has mean zero. Terms using the same interval contribute the expected product of the integrands multiplied by the interval length, since the increment’s second moment equals that length. Passing from these sums to the integral requires conditions that justify limits and exchange of expectation with summation. The question’s stated future-increment independence assumption is unusual; standard formulations generally use adapted integrands and appropriate square-integrability rather than requiring that exact condition.
Key ideas
- The expected product of two Itô integrals driven by the same Brownian motion is governed by the expected product of their integrands.
- In a partition-based argument, non-overlapping increments contribute zero by independence and zero conditional mean.
- Matching increments contribute through their variance, which equals the time-step length.
- The sketch relies on assumptions that justify the limiting argument and the existence of the integrals.
Tags
Full text
# Simplifying the expectation of the product of two stochastic integrals
# Simplifying the expectation of the product of two stochastic integrals
Let $f(t, \omega), g(t, \omega)$ be functions that are independent of the increments of the Brownian motion $w(t, \omega)$ in the future. That is, $f(t, \omega), g(t, \omega)$ are independent of $w(t + s, \omega) - w(t, \omega)$ for all $s > 0$. I want to show that for any $0 \leq \tau \leq t \leq T$,
$$E\left(\int_\tau^t f(s,\omega) dw(s,\omega) \int_{\tau}^t g(s,\omega)dw(s,\omega)\right) = \int_{\tau}^{t} E[f(s,\omega)g(s,\omega)] ds$$
This is somewhat like a linearity property in some sense. I'm not really sure how to show it. If it helps, I know how to show the following:
$$E\left(\int_\tau^t f(s,\omega) dw(s,\omega)\right) = 0,$$
but I don't quite see how to apply that here (it doesn't seem related). I was thinking this might have to do with Fubini's theorem (since the expectation is really just an integral), but I'm not so sure.
Any help is appreciated. Maybe there is some clever way to apply Ito's Lemma here.
## Answer by ir7 (score 3)
https://quant.stackexchange.com/a/63315
Providing only a sketch here, using Ito integral definition (and commuting limit, summations and expectation), the result boils down to studying the expectation term:
$$ E\left[ f_{t_{i-1}} (W_{t_i}-W_{t_{i-1}}) \cdot g_{t_{j-1}} (W_{t_j} - W_{t_{j-1}}) \right]. $$
If the intervals don't overlap, $i\not= j$, and say $t_i \leq t_{j-1}$, then $f_{t_{i-1}} g_{t_{j-1}}(W_{t_i}-W_{t_{i-1}})$ is independent of $W_{t_j}-W_{t_{j-1}}$, so the expectation term above is equal to
$$ E\left[ f_{t_{i-1}} g_{t_{j-1}} (W_{t_i}-W_{t_{i-1}})\right] \cdot E\left[ (W_{t_j} - W_{t_{j-1}}) \right] = 0.$$
If the intervals overlap, $i=j$, then $f_{t_{i-1}} g_{t_{i-1}}$ is independent of $W_{t_i}-W_{t_{i-1}}$, so the expectation term above is equal to
$$ E\left[ f_{t_{i-1}} g_{t_{i-1}} \right] E\left[ (W_{t_i} - W_{t_{i-1}})^2 \right] = E\left[ f_{t_{i-1}} g_{t_{j-1}} (t_{i} - t_{i-1})\right]. $$
Have also a look at Ito isometry (and its textbook proofs).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.