# Probability density function (pdf) of a Wiener process

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 $$$$w(t, w_0 \vert \eta) = w_0 + \int_{t_0}^t\eta(\tau) d \tau \,, \quad w(t=t_0) = w_0 \,,$$$$ where $$\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 $$$$p(w,t \vert w_0, t_0) = \langle \delta(w - w(t,w_0 \vert \eta)) \rangle_\eta \,.$$$$ Using $$$$\delta(x) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{iqx} dq$$$$ we get $$$$p(w,t \vert w_0, t_0) = \frac{1}{2\pi} \int_{-\infty}^\infty e^{iq(w-w_0)} \left\langle \exp \left( iq \int_{t_0}^t \eta(\tau) d\tau \right) \right\rangle_\eta dq \,.$$$$ In the next step we want to calculate the average over $$\eta$$ of the exponential. Here the author of the book presents the equation $$$$\left\langle \exp \left( iq \int_{t_0}^t \eta(\tau) d\tau \right) \right\rangle_\eta = \exp \left( -\frac{q^2}{2} \int_{t_0}^t\int_{t_0}^t \langle \eta(\tau) \eta(\tau') \rangle_\eta d\tau d\tau' \right) \,.$$$$ 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!

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.

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