Why Combining Independent Brownian Integrals Preserves Brownian Covariance
Summary
The document considers a process formed from two independent Brownian motions with deterministic, time-varying weights. The weights are chosen so their squared values sum to one. Although the questioner’s covariance expansion introduces cross terms, independence means the two Brownian integrals have zero covariance with each other; those terms should not be added. Applying the Itô isometry separately to the integrals gives a total variance equal to elapsed time.
For an increment over an interval, each integral is normally distributed, and the two are independent. Their sum is therefore normal with mean zero and variance equal to the interval’s length. The answer establishes this increment distribution, but it does not provide a full proof of the stated covariance between process values at different times. The result relies on independent Brownian motions, deterministic integrands, and the chosen normalization of the weights; it is a stochastic-calculus explanation rather than a trading strategy or empirical market study.
Key ideas
- The two Brownian integrals have zero cross-covariance because their driving motions are independent.
- The Itô isometry gives each integral’s variance by integrating the square of its deterministic weight.
- The squared weights sum to one, so the combined increment variance equals the interval length.
- Independent normal increments remain normal when summed.
Tags
Full text
# Ito isometry and the covariance of an Ito process
# Ito isometry and the covariance of an Ito process
Let $(B_t)_{t \geq 0}$ et $(W_t)_{t \geq 0}$ be two independent Brownian motions and let $f: \mathbb{R} \rightarrow \mathbb{R}$ a deterministic function of time. We define the following process: \begin{equation} X_t = \int_0^t \frac{1}{\sqrt{1 + f(u)^2}} dB_u + \int_0^t \frac{f(u)}{\sqrt{1 + f(u)^2}} dW_u. \end{equation}
We want to prove that
- $\forall s,t \geq 0 \; X_{t+s} - X_t \sim N(0,s)$;
- $\forall s,t \geq 0 \; E(X_t) = 0 \text{ and } Cov(X_t, X_s) = min \{ s,t \}$.
Regarding (2), the Ito integrals are each of null expectation, hence $E(X_t) = 0$ must be true. Now, without loss of generality, assume $s < t$
\begin{align} COV(X_t, X_s) = \int_0^t \frac{1}{\sqrt{1 + f(u)^2}} \frac{1}{\sqrt{1 + f(u)^2}} \mathbb{I}(u \leq s) du \; + \\ \int_0^t \frac{f(u)}{\sqrt{1 + f(u)^2}} \frac{f(u)}{\sqrt{1 + f(u)^2}} \mathbb{I}(u \leq s) du + \\ \int_0^t \frac{2}{\sqrt{1 + f(u)^2}} \frac{f(u)}{\sqrt{1 + f(u)^2}} \mathbb{I}(u \leq s) du \\ COV(X_t, X_s) = \int_0^s \frac{1 + f(u)^2 + 2f(u)}{1 + f(u)^2} du = \int_0^s du + 2 \int_0^s \frac{f(u)}{1 + f(u)^2}du \end{align} Clearly, I did something wrong, but I don't see where. It looks like I applied the Ito isometrie correctly here. Anyone sees the problem?
## Answer by Idonknow (score 1)
https://quant.stackexchange.com/a/50146
Recall that for any deterministic function $g,$ Ito's integral follows a normal distribution: $$\int_0^t g(u) dW_u \sim N\left(0,\int_0^t g^2(u) du\right).$$ Therefore, since $$X_{t+s} - X_t = \int_t^{t+s} \frac{1}{\sqrt{1 + f(u)^2}} dB_u + \int_t^{t+s} \frac{f(u)}{\sqrt{1 + f(u)^2}} dW_u,$$ each integral follows a normal distribution and they are independent, so $X_{t+s}-X_t$ follows a normal distribution with mean $0$ and variance $$\mathbb{E}\left( \int_t^{t+s} \frac{1}{\sqrt{1 + f(u)^2}} dB_u \right)^2 + \mathbb{E}\left( \int_t^{t+s} \frac{f(u)}{\sqrt{1 + f(u)^2}} dW_u \right)^2 = \mathbb{E}\left( \int_t^{t+s}du \right) = s.$$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.