Gaussianity and Covariance of Deterministic Brownian Integrals
Summary
The note considers an Itô integral of a deterministic function against Brownian motion and explains why it has zero mean, a covariance determined by the overlap of the integration intervals, and Gaussian distributions. The mean follows from the martingale property, while the covariance is obtained using the Itô isometry over the shared interval from zero to the smaller time.
For Gaussianity, the explanation approximates the deterministic integrand with simple functions. Their integrals are sums of Brownian increments with constant coefficients, hence are normally distributed; the Itô integral is their mean-square limit. The note contrasts this with a general stochastic integrand, whose integral need not be Gaussian. The question mentions a desired moment-generating-function and Itô-lemma approach, but the answer instead uses basic Itô integral properties and approximation. The displayed covariance notation in the question appears to contain an indexing typo; the intended integrand is evaluated over the integration variable.
Key ideas
- An Itô integral with a square-integrable integrand is a martingale and has constant mean, which is zero when it starts at zero.
- The covariance at two times is determined by the integrand's squared magnitude over their shared time interval.
- A deterministic integrand can be approximated by simple functions whose Brownian integrals are Gaussian sums.
- A stochastic integrand does not in general produce a Gaussian integral.
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Full text
# Integration of a deterministic function w.r.t. a Brownian motion
# Integration of a deterministic function w.r.t. a Brownian motion
Help me solve this problem:
Let $W_t$ be a Brownian motion and suppose $X_t = \int_{0}^{t}\delta _{s}dW_{s}$ where $\delta _{s}$ is a deterministic function. Then show that $X_t$ is a Gaussian process with mean, $m(t) = 0$ and covariance function $\rho (s,t)=\int_{0}^{min(s,t)}\delta _{s}^{2} ds$.
Edit: I am looking for a specific approach which utilizes Ito's Lemma and moment generating functions.
## Answer by Kevin (score 4)
https://quant.stackexchange.com/a/53214
These are key properties of the Ito integral. Let $(X_t)$ be a cadlag, adapted process with $\int_0^t \mathbb{E}[X_s^2]\mathrm{d}s<\infty$. Then, the Ito integral $$I_t=\int_0^t X_s\mathrm{d}B_s$$ is well-defined, a martingale and satisfies the Ito isometry.
- The martingale property tells you that $I_t$ has a constant mean and since $I_0=0$, we obtain $\mathbb{E}[I_t]=0$ for all $t$. This holds if $X_s$ is a stochastic process (not necessarily a deterministic function).
- Question (ii) follows from the zero mean property: \begin{align*} \mathbb{C}\mathrm{ov}(I_t,I_s)&=\mathbb{E}[I_tI_s] \\ &= \mathbb{E}\left[\int_0^t X_u\mathrm{d}B_u \int_0^s X_\tau\mathrm{d}B_\tau\right] \\ &= \mathbb{E}\left[\int_0^t \int_0^s X_uX_\tau \mathrm{d}B_u \mathrm{d}B_\tau \right] \\ &= \mathbb{E}\left[\int_0^{t\wedge s} X_u^2 \mathrm{d}u \right]. \end{align*} This again holds if $X_s$ is a stochastic process.
- For this point, we need that $X_s$ is a deterministic function. Recall that the Ito integral $I_t$ is defined as a limit (in mean squared sense): You take simple processes $X_s^n$ to approximate the integrand $X_s$ and define the integral of simple processes with respect to Brownian motion, i.e. \begin{align*} I_t^n=\int_0^t X_s^n\mathrm{d}B_s := \sum_{i=1}^n C_{i-1}(B_{s_i}-B_{s_{i-1}})+C_n(B_t-B_{s_n}). \end{align*} If $X_s$ is deterministic, then the $C_i$ are constants and $I_t^n$ is a sum of normally distributed random variables (the increments of Brownian motions). Thus, the Ito integral $I_t$ is just the limit of normally distributed sums and thus, Gaussian itself. If $X_s$ is any stochasic process, then the $C_i$ are random variables and $I_t^n$ is not necessarily normally distributed.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.