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


1 Answer 1



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.

  • 2
    $\begingroup$ Very nice...+1. $\endgroup$ Commented Apr 26, 2021 at 8:24

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