Consider some brownian motion for which we obtained the following solution for the langevin equations

$$ u\left(t\right)=e^{-\alpha t}\int_{0}^{t}e^{\alpha s}\left(\xi\left(s\right)-\xi'\left(s\right)\right)ds $$

Here, $\xi\left(t\right)$ and $\xi'\left(t\right)$ are two independent gaussian white noises with zero mean.


  1. I believe we can compute $\left\langle u\left(t_{1}\right)u\left(t_{2}\right)\right\rangle $ with the usual procedure where we consider $\left\langle \xi\left(t_{1}\right)\xi\left(t_{2}\right)\right\rangle =g\delta\left(t_{2}-t_{1}\right)$ if we consider $\left\langle \xi\left(t_{1}\right)\xi'\left(t_{2}\right)\right\rangle =0$ with the argument that the noise processes are independent; can you confirm if I am correct?
  2. The book I'm reading shows without proving a solution for $\left\langle u\left(t\right)\xi\left(t\right)\right\rangle $ and I am trying to understand how this is computed. I don't understand how the $\xi\left(t\right)$ could go inside the integral of $u\left(t\right)$ for one to be able to use the usal identity that yields the dirac delta. Do you have an idea how this is done? Could you please advise?
  • $\begingroup$ Hello! I am excited to see that you're posting questions and stochastic processes. I am writing to invite you to join me in writing solutions to textbooks in stochastic processes. So far there are two projects, one for the book by Lemons (link), and one for the book by Van-Kampen (link). $\endgroup$
    – DanielSank
    Commented Sep 7, 2018 at 17:17
  • $\begingroup$ "The book I'm reading shows without proving..." What does the book show? The sentence seems incomplete. $\endgroup$
    – DanielSank
    Commented Sep 7, 2018 at 17:18
  • $\begingroup$ Also, I am puzzled by the first equation. It looks like a Laplace transform but it's not clear where it came from or what is the meaning of $u(t)$. Adding some more information might help clarify and invite good answers. $\endgroup$
    – DanielSank
    Commented Sep 7, 2018 at 17:19
  • $\begingroup$ Hi @DanielSank! I tried to reduce my question (and the equation) to focus on the part I don't understand which is dealing with two noise processes and computing that expectation. The book is this, section, pag.153. I'm trying to show equation 4.45 from 4.42, using the solution I obtained for 4.43, not shown in there. $\endgroup$
    – Bella
    Commented Sep 7, 2018 at 17:32
  • 1
    $\begingroup$ To the second question: First you multiply $u(t)$ with $\xi(t)$. Because the latter is independent of $s$ you can pull it inside the integral. If you now take the expectation value and multiply things out, you end up with a $\langle \xi(s) \xi(t) \rangle$ and a $\langle \xi'(s) \xi(t) \rangle$ term. The latter is zero, and the former is $g \delta(s-t)$. Because the integration goes only up to $s=t$ (rather than extending across $t$ on both sides), a physicist might argue that you only get $g/2$ from the integration. $\endgroup$
    – user8153
    Commented Sep 8, 2018 at 17:29

1 Answer 1


So the original SDE is probably (see this answer for derivation) $$ \mathrm{d}u_t = \alpha u_t \mathrm{d}t + \mathrm{d}W_t + \mathrm{d}W'_t $$ from which indeed $$ u_t = e^{\alpha t} \left(\int_0^te^{-\alpha s}\mathrm{d}W_s + \int_0^te^{-\alpha s}\mathrm{d}W'_s\right) = \mathcal{N}\left(0, \frac{1}{\alpha}(e^{2\alpha t} - 1)\right) $$ given the two motions are uncorrelated (the correlated case should also be rather simple by the standard mapping to an uncorrelated pair, i.e. $\mathrm{d}W' = \rho \mathrm{d}W + \sqrt{1-\rho^2} \mathrm{d}Z$, where $\mathrm{d}Z$ is a Brownian motion uncorrelated with $\mathrm{d}W$, $\mathrm{d}W'$, and the rest of the math is the same as above).

And we could verify this easily enough with a bit of ugly Python (see Appendix), plotting the PDFs of analytical vs numerical solutions: PDF of analytical vs (Euler) numerical sol'n to the SDE

So if I interpret your question correctly, you're asking for the Ito integral

$$ \mathbb{E}\left(\int_0^tu_\tau\mathrm{d}W_\tau\right) = \mathbb{E}\left(\int_0^te^{\alpha \tau} \left(\int_0^\tau e^{-\alpha s}\mathrm{d}W_s + \int_0^\tau e^{-\alpha s}\mathrm{d}W'_s\right)\mathrm{d}W_\tau\right) = \mathbb{E}\left(\int_0^te^{\alpha \tau} \left(\int_0^\tau e^{-\alpha s}\mathrm{d}W_s\right)\mathrm{d}W_\tau\right) = \mathbb{E}\left(\int_0^te^{\alpha \tau} e^{-\alpha \tau}\mathrm{d}\tau\right) = t $$

Similarly for variance (though that's not what you asked for), $\mathrm{Var}\left(\int_0^tu_\tau\mathrm{d}W_\tau\right) = \frac{1}{2\alpha^2}(e^{2\alpha t} -1) - \frac{t}{\alpha}$


Analytical vs numerical PDF of the original SDE:

import numpy as np
from   scipy.stats import norm
import matplotlib.pyplot as plt

alpha = -75.
dt = .001
sdt = np.sqrt(dt)
N  = int(.25/dt)
M  = 10000

ud = np.zeros(M)
for m in range(M):
    u = 0
    w1 = np.random.randn(N)*sdt
    w2 = np.random.randn(N)*sdt
    for n in range(N):
        u += alpha*u*dt + w1[n] + w2[n]
    ud[m] = u

yy, xx = np.histogram(ud, bins=64, density=True)
xx = .5*(xx[:-1] + xx[1:])
plt.plot(xx, norm.pdf(xx, 0, np.sqrt(1./alpha*(np.exp(2*alpha*N*dt) - 1))), xx, yy)

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