Brownian motion from two gaussian noise processes 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.
Question:


*

*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?

*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?

 A: 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:

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), $\mathrm{Var}\left(\int_0^tu_\tau\mathrm{d}W_\tau\right) = \frac{1}{2\alpha^2}(e^{2\alpha t} -1) - \frac{t}{\alpha}$
Appendix
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)

