I am working through a physics 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 maybe someone can help. 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 indecies, $\sigma$ 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. Does someone have an idea how to go from a single to the double integral in the exponent and get rid of the imaginary unit? Also how is it possible to get the averaging over $\eta$ iside the exponential? My first guess was to approach it like the Gaussian integral by squaring and taking the root in the exponent but that did not work out. Thanks for any suggestions!


1 Answer 1


The result is not an identity for arbitrary processes, but it is exact in the case of your delta-correlated $\eta$. Expand in a formal power series in $q$

$$\tag{1} \left< \exp\left({iq\int^t dt'\eta(t')}\right)\right> =1 +iq \int\limits^tdt_1 \left< \eta(t_1) \right> +\frac{(iq)^2}{2!}\int\limits^tdt_1\int\limits^t dt_2 \left<\eta(t_1)\eta(t_2)\right>+\cdots $$

On the RHS of (1) all terms with an odd number of $\eta$ vanish, a generic even term is

$$\tag{2} T_{2n}=\frac{(iq)^{2n}}{(2n)!}\int dt_1 \dots dt_{2n} \left<\eta(t_1)\dots\eta(t_{2n}) \right> $$

I'll use the notation

$$\tag{3} G_{ij}=\left<\eta(t_i)\eta(t_j)\right> $$

So that your third expression reads

$$\tag{4} \left<\eta(t_1)\dots\eta(t_{2n}) \right> = \sum G_{i_1j_1} \dots G_{i_n j_n} $$

Using (4) we ask: how many terms has the sum over distinct permutations for some fixed $n$? There are $(2n)!$ ways to arrange the $t_i$, which we must discount for nondistinct arrangements: the order of the $G$s doesn't matter (which are $n!$ ways), and within each $G$ the order of the $t$s don't matter ($2^n$ times). Consequently there are $\frac{(2n!)}{2^n n!}$ terms in the expansion of (2), which are identical when integrated over so that

$$\tag{5} T_{2n}=\frac{(2n)!}{2^n n!} \frac{(iq)^{2n}}{(2n)!} \left[\int dt_1 \int dt_2 \ G_{12} \right]^n = \frac{1}{n!} \left[ -\frac{q^2}{2} \int\limits^tdt_1\int\limits^t dt_2 \left<\eta(t_1)\eta(t_2)\right> \right]^n $$

and the sum over $n$ in (5) is again an exponential which yields your result

$$\tag{6} \left< \exp\left({iq\int^t dt'\eta(t')}\right)\right>=\sum\limits_{n=0}^\infty T_{2n} = \exp \left( -\frac{q^2}{2} \int\limits^tdt_1\int\limits^t dt_2 \left<\eta(t_1)\eta(t_2)\right>\right) $$

Note for an arbitrary process there would appear on the RHS an infinite sum in the exponential (see cumulant expansion).

  • $\begingroup$ Awesome solution! I tried proving it using power series expansion as in (1) and (2) but failed to go further. I then used an identity for the moment generating function for normal distributed random variables as in en.wikipedia.org/wiki/… which got me the solution too. But I like your proof and in-depth explaination more. Thank you very much! $\endgroup$
    – Pascal S.
    Commented Aug 7, 2023 at 12:40
  • $\begingroup$ @PascalS. no problem, glad it's useful! I considered answering using a cumulant generating function $W=\ln \left< e^{\int \eta} \right>$ but this would have required knowing a priori that the cumulants of order 3 and greater vanish for your process, so instead decided this would be more complete :) $\endgroup$
    – Sal
    Commented Aug 7, 2023 at 17:35

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