Distribution of a vector with stochastic components. I have to prove that the following vector has normal distribution:
$$
\left(  W_t, \int_0^t W_s \, ds\right)
$$
where $W_t$ is a Brownian motion.
I tried just by a forward computation on the characteristic function using the Ito derivative. In order to make an easier computation let
$$
\Gamma_t=exp\left( i \xi W_t + i \eta \int_0^t W_s \, ds \right)
$$
and
$$
Y_t = \int_0^t W_s \, ds
$$
We have:
$$
d \Gamma_t = i\xi \Gamma_t dW_t+i \eta\Gamma_t dY_t  + \frac{1}{2}\left( -|\xi|^2\Gamma_t \, dt\, - i\eta \Gamma_t \, d\langle Y_t\rangle -2\xi \eta \Gamma_t d \langle Y_t, W_t\rangle\right)
$$
Now, since we have that
$$
dY_t = W_t dt
$$
it follows that
$$
d \langle Y_t \rangle = 0
$$
and
(this is a standard formal computation)
$$
d \langle Y_t ,W_t\rangle= dY_t dW_t = W_t dt dW_t = 0
$$
Passing to the average:
\begin{align}
\mathbb{E} \left[ \Gamma_t \right] &=1 +i \eta \int_0^t \mathbb{E} \left[\Gamma_s dY_s \right]-\int_0^t\frac{|\xi|^2}{2} \mathbb{E} \left[ \Gamma_s \right] \, ds \\
&= 1 + i \eta \int_0^t \mathbb{E} \left[ \Gamma_s W_s  \right] ds -\int_0^t\frac{|\xi|^2}{2} \mathbb{E} \left[ \Gamma_s \right] \, ds 
\end{align}
At this point I'm trying to get an ODE for $\mathbb{E}[\Gamma_t]$ but I'm a little bit stuck
 A: Recall the following statement:

Let $X: \Omega \to \mathbb{R}^n$ be a Gaussian random vector and $A \in \mathbb{R}^{k \times n}$ a (deterministic) matrix. Then $Y := A \cdot X$ is Gaussian.

Since $(W_t)_{t \geq 0}$ is a Brownian motion, we know that $X:=(W_{t_0},W_{t_1},\ldots,W_{t_n})$ is Gaussian for any $t_0,\ldots,t_n$. If we choose $t_j := \frac{j}{n} t$, $n \in \mathbb{N}$, and define
$$A := \begin{pmatrix} n^{-1} & \ldots & n^{-1} & 0 \\ 0 & \ldots & 0 & 1 \end{pmatrix} \in \mathbb{R}^{2 \times (n+1)}$$
then, by the above statement, the vector
$$A \cdot X = \left( \frac{1}{n} \sum_{j=0}^{n-1} W_{t_j}, W_{t_n} \right) = \left( \sum_{j=0}^{n-1} W_{t_j} (t_{j+1}-t_j), W_t \right)$$
is Gaussian. Consequently, we find that
$$\left( \int_0^t W_s \, ds, W_t \right) = \lim_{n \to \infty} \left( \sum_{j=0}^{n-1} W_{t_j} (t_{j+1}-t_j), W_t \right)$$
is Gaussian as a pointwise limit of Gaussian random variables (see e.g. this question).
A: At the end I think I've found the solution. Of course the problem is the non linearity of the term:
$$
\mathbb{E}[\Gamma_t W_t] 
$$
In this case one can use this trick:
$$
\partial_\xi \Gamma_t = i \Gamma_t W_t.
$$
This comes in particoular from the exponential structure of the charateristic function so that it seems like to be a quite usefull trick for the future.
However let me continue the computation:
\begin{align}
\mathbb{E}[\Gamma_t] & = 1+i \eta\int_0^t \frac{1}{i}\partial_\xi\mathbb{E}[\Gamma_s] \, ds -\frac{|\xi|^2}{2}\int_0^t\mathbb{E}[\Gamma_s] \, ds \\
& = 1+\int^t_0 \eta \partial_\xi\mathbb{E}[\Gamma_t]- \frac{|\xi|^2}{2}\mathbb{E}[\Gamma_t] \, ds
\end{align}
This is a good formulation for us since it means that the left side of the equation is a solution for the following PDE:
$$
\partial_t V(t, \xi , \eta)=\eta \partial_\xi V(t,\xi,\eta)-\frac{|\xi|^2}{2}V(t,\xi,\eta)
$$
with initial condition $V(0,\xi,\eta)=1$.
Now we are at "home" since we can use all our PDE-weapon to win. For example by choosing the solution of the form:
$$
V=\exp(P(t,\xi,\eta)).
$$
where
$$
P(t,\xi,\eta)=a t^A \xi^2+b t^B \eta^2+c t^C \eta \xi.
$$
Now the idea is just make a computation and choose suitable parameters. By appling the differentials operators we get the relation:
$$
a A t^{A -1} \xi^2 + b B t^{B-1} \eta^2 + c C t^{C-1} \eta \xi = 2 a A t^A \xi \eta + c t^C \eta^2-\frac{\xi^2}{2}
$$
This equation, as equivalence between two polynomes in the $\xi$ and $\eta$ variable, leads us to the following system for the coefficients:
$$
\left\{
\begin{array}{rcl}
A a t^{A-1} & = & -\frac{1}{2} \\
Bbt^{B-1} & = & c t^C \\
2 a A t^A & = & cCt^{C-1}
\end{array}
\right.
$$
Ok...now this gives us (finally)
$$
\left\{
\begin{array}{rcl}
A & = & 1 \\
a & = & -\frac{1}{2} \\
B & = & 3 \\
b & = & -\frac{1}{6} \\
C & = & 2\\
c & = & -\frac{1}{2}.
\end{array}
\right.
$$
