Deriving the Gaussian Law of a Deterministic Brownian Integral

Summary

The document derives the distribution of a stochastic integral of a deterministic function against Brownian motion. It applies Itô's lemma to the exponential of the integral multiplied by an imaginary parameter. Taking expectations removes the stochastic-integral term under the usual integrability conditions, leaving an integral equation for the characteristic function.

That equation becomes a first-order differential equation with initial value one. Solving it yields the characteristic function of a centered normal variable whose variance is the integral of the squared deterministic function over time. The normal distribution follows because characteristic functions uniquely determine distributions. The argument is a concise derivation rather than a full statement of technical assumptions: existence and integrability conditions on the function and stochastic integral are not detailed. This result is useful in stochastic modeling, including financial models driven by Brownian noise.

Key ideas

• A deterministic-integrand Brownian integral has mean zero and variance equal to the integral of the squared integrand.
• Itô's lemma applied to an exponential produces an equation for the integral's characteristic function.
• Taking expectations removes the stochastic-integral term when the required integrability conditions hold.
• Solving the resulting differential equation gives the characteristic function of a normal distribution.

Tags

# Ito calculus is Gaussian (using method of characteristic function)


# Ito calculus is Gaussian (using method of characteristic function)












Let $h$ be a deterministic function and define $X_{t}=\int_{0}^{t} h(s) d W_{s} .$ Show that $$\mathbb{E} \exp \left(i u X_{t}\right)=\exp \left(-\frac{u^{2}}{2} \int_{0}^{t} h^{2}(s) d s\right),$$ from which deduce that $X_{t} \sim N\left(0, \int_{0}^{t} h^{2}(s) d s\right)$.

## Answer by ir7 (score 7)

https://quant.stackexchange.com/a/63584

Hints:

First show (using Ito Lemma) that

$$ \exp(iuX_t) = 1 + iu \int_0^t\exp(iuX_s) h(s) d W_s -2^{-1}u^2 \int_0^t\exp(iuX_s) h(s)^2ds$$

Then show (by taking expectations):

$$ E[\exp(iuX_t)] = 1 - 2^{-1}u^2 \int_0^tE[\exp(iuX_s)] h(s)^2ds $$

Finally note that this is an ODE in unknown variable $x(t):=E[\exp(iuX_t)]$, $x(0)=1$:

$$ x'(t) =- 2^{-1}u^2h(t)^2 x(t) $$

and solve it.

For the deduction of the normality of $X_t$, use the fact that two random variables with the same characteristic function are identically distributed.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.