# 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 $$X_{t} \sim N\left(0, \int_{0}^{t} h^{2}(s) d s\right)$$.

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.

• Very nice...+1. Apr 26 at 8:24