I am working through a book right now in which there is a short introduction to Brownian motion and Wiener processes. I assume it is not treated nearly as rigorous as in mathematics but still more of a mathematical question. I am stuck at some point in the derivation of a pdf.

We introduce a Wiener process as \begin{equation} w(t, w_0 \vert \eta) = w_0 + \int_{t_0}^t\eta(\tau) d \tau \,, \quad w(t=t_0) = w_0 \,, \end{equation} where $\eta(t)$ is some white noise function with the following properties: \begin{gather} \langle \eta(t_1) ...\eta(t_{2n+1}) \rangle_\eta = 0 \\ \langle \eta(t_1) ...\eta(t_{2n}) \rangle_\eta = \sigma^n \sum_{i_1,...,i_{2n} \in (1,...,2n)} \delta(t_{i_1} - t_{i_2})...\delta(t_{i_{2n-1}} - t_{i_{2n}}) \,, \end{gather} where we sum over all permutations of the set of indices, $\sigma$ is a diffusion coefficient and $\langle \cdot \rangle_\eta$ is the average over all noise functions.

Now we want to define a pdf for this process by \begin{equation} p(w,t \vert w_0, t_0) = \langle \delta(w - w(t,w_0 \vert \eta)) \rangle_\eta \,. \end{equation} Using \begin{equation} \delta(x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{iqx} dq \end{equation} we get \begin{equation} p(w,t \vert w_0, t_0) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{iq(w-w_0)} \left\langle \exp \left( iq \int_{t_0}^t \eta(\tau) d\tau \right) \right\rangle_\eta dq \,. \end{equation} In the next step we want to calculate the average over $\eta$ of the exponential. Here the author of the book presents the equation \begin{equation} \left\langle \exp \left( iq \int_{t_0}^t \eta(\tau) d\tau \right) \right\rangle_\eta = \exp \left( -\frac{q^2}{2} \int_{t_0}^t\int_{t_0}^t \langle \eta(\tau) \eta(\tau') \rangle_\eta d\tau d\tau' \right) \,. \end{equation} I struggle to find a derivation of this equality as it seems very non-trivial to me. I already tried a series expansion approach, but this only leads to the exponent of the right side under further assumptions, not the exponential function. Does someone have an idea? I would be thankful for any suggestions!


1 Answer 1


Given the normally distributed random variable $Z\sim {N}(\mu,\sigma^2)$, it is possible to show that its moment-generating function is

$$\langle \exp({tZ})\rangle=\exp({\mu t + \sigma^2 t^2/2})=\exp \left(t{\langle Z\rangle + t^2\,\text{Var}(Z) /2} \right) $$

If the mean value is zero, then $\text{Var}(Z)=\langle Z^2\rangle$ and $\langle \exp(tZ)\rangle = \exp(t^2{\langle Z^2\rangle /2})$.

Now, use this fact with $t=iq$ (call this analytic continuation of the Gaussian integral if you like) and $Z = \int_a^b \eta(s) ds$ is the integral of white noise. Therefore, $Z$ is the Brownian motion (aka Wiener process) and is normally distributed, so we can use the result above to get:

$$\left\langle \exp\left({i q \int_a^b \eta }\right) \right\rangle = \exp\left({-\frac{q^2}{2} \left\langle \left(\int_a^b \eta\right)^2 \right\rangle } \right) $$

Finally, realise that

$$\left\langle \left(\int_a^b \eta\right)^2 \right\rangle = \left\langle \int_a^b ds \int_a^b df \,\eta(f) \, \eta(s) \right\rangle = \int_a^b ds \int_a^b df \,\left\langle \eta(f) \, \eta(s) \right\rangle $$
In the end, it is just an application of the moment-generating function of a Gaussian random variable.

  • 1
    $\begingroup$ Thanks alot. Im still learning about probability theory, so this connection was too far away for me to make yet. This was really helpful! $\endgroup$
    – Pascal S.
    Commented Jul 14, 2023 at 17:33

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