Good morning,
recently I had to solve the two-dimensional SDE, and the solution process I found was
$$\left\{\begin{array}{rcl}\xi^1_t&=&\xi^1_0+\int_0^t dw(s),\\ \xi_2^t&=&\xi_0^2+\int_0^t(\xi_s^1)^2ds\end{array}\right.,$$
where $w(t)$ is a standard one-dimensional brownian motion. Then $\xi_t^1$ has a normal distribution with mean $\xi_0^1$ and variance $t$. What kind of process is $\xi_t^2$? More generally what can I say on the whole process $(\xi_t^1,\xi_t^2)$ (as a Joint process)?
I didn't find any reference in the literature unfortunately so that's why I'm asking. Thank you for all your kind replies.
This is only a partial answer, but I hope it could be useful. Let us set $$\xi^1_t=N(\xi^1_0,t)=X$$ Reminding that the sum of $k $ squared standard normal distributions corresponds to a chi-squared distribution $\chi_k^2$ (where $k $ denotes the degrees of freedom), for a single non-standard normal distribution $Y=N (\mu, \sigma^2) $ we have
$$ \left (\frac {Y-\mu}{\sigma} \right)^2=\chi_1^2$$
Applying this to $X$, we get
$$\left (\frac {X-\xi_0^1}{\sqrt {t}} \right)^2=\chi_1^2$$
and then
$$(X-\xi_0^1)^2=t \, \chi_1^2$$
$$X^2 = 2X \xi_0^1 -(\xi_0^1)^2+t \, \chi_1^2$$
Note that in the RHS of the last equation, the first term is our initial normal distribution $X $ scaled by a factor $2 \xi_0^1$, so that its expected value is $2 (\xi_0^1)^2$. The second term is a constant, and the third term is a chi-square distribution with one degree of freedom (which by definition has an expected value of $1$) scaled by a factor $t$. So, due to linearity of expectation, we get
$$E (X^2)=(\xi_0^1)^2+t$$
We can now rewrite the expression giving $\xi_2^t$ as
$$\xi_2^t=\xi_0^2 +2 \xi_0^1 \int_0^t X \\ - \xi_0^1 \int_0^t ds + t \int_0^t \chi_1^2 $$
and grouping the constant terms into $J $
$$ \xi_2^t= 2 \xi_0^1 \int_0^t X + t \int_0^t \chi_1^2 +J $$
Therefore the distribution of $\xi_2^t$ is given by the sum of a constant term, a scaled integral of the initial normal distribution $X=\xi^1_t$, and a scaled integral of the chi-square distribution corresponding to the square of the standardized $X$. The integral of $X$ can be expressed in terms of the so-called error function, considering that the CDF $F (x ) $ of a generic normal distribution with mean $\mu $ and variance $\sigma^2$ is
$$F (x)=\frac {1}{2} \left[ 1+erf \left( \frac {x-\mu}{\sigma \sqrt {2}} \right) \right] $$
The integral of $\chi_1^2$ can be estimated by reminding that the CDF $C_r(x) $ of a generic chi-square distribution $\chi_r^2$ with $r $ degrees of freedom is
$$C_r (x)=P \left( \frac {1}{2} r, \frac {1}{2} x \right) $$
where $P (m,n) $ is a regularized gamma function.