4
$\begingroup$

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.

  • 1
    $\xi^2_t$ has a generalized chi-square distribution. There's a lot you can say about it; what exactly do you need? – 2017-01-15
  • 0
    I need to know if it is possible to derive explicitly the joint probability densityof the process $(\xi_t^1,\xi_t^2)$. By the way I already knew that the square of a Gaussian is distributed as a chi-square distribution. So you're telling me that the integral of a chi-square distribution is again a chi-square? wow, that's really interesting – 2017-01-15
  • 1
    1. Explicit joint density? No, I don't think so. More likely only the marginal characteristic function as some infinite product (I'm not sure about the joint one). 2. No, I'm not telling that. You probably didn't see the word "generalized" in my comment. – 2017-01-16
  • 1
    By using the Loeve-Karhunen decomposition or Wiener chaos expansion, $(\xi_t^1,\xi_t^2)$ could be expressed as quadratic functionals of a sequence of I.I.D N(0,1) r.v.s. – 2017-01-22

1 Answers 1

1

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.