Stochastic average of a differential equation is not the same as average of its solutions

Question

Assume a static (time-independent) random variable $r$ for which we know its probability distribution $P(r)$. Consider this to be a Gaussian distribution, such that $\langle r \rangle=0$ and $\langle r^2 \rangle=r_0$, etc.

Consider now the following differential equation: $\frac{d}{dt}X(t)=r A X(t)$, where $X=(x_1,x_2)^T$ and $A$ is a time-independent $2\times 2$ matrix.

I am interested in taking the average of the differential equation over the random variable.

If I naively just take the expected value of the previous differential equation, the problem is that I would just get $\frac{d}{dt}X=0$ because $\langle r \rangle=0$. This is not true, as $X$ is a functional of r, $X(r,t)$, so $\langle r X(r,t) \rangle\neq0$.

One possible solution would be to formally solve the differential equation as $X(t)=e^{rAt}X(0)$ and then take the average. I would like to avoid this method, as it is not always aplicable.

What would be the correct way to proceed if I would like to get an averaged differential equation that properly reproduces the results of averaging the solution?

Note: I know about Novikov's theorem which is sometimes used to solve such stochastic differential equations using two-time correlation functions. However, as far as I am aware, it only works if the stochastic variable is dynamical, i.e. r(t) is sampled by the solution of a stochastic differential equation. In my case, my stochastic variable is static, so only the probability distribution is known.

Answer 1

Consider linear evolution equations of the form

$$\tag{1} x'(t)=r A x(t) $$

Where $x$ is an $n$-tuple, $r$ is a scalar random variable and $A$ is an $n\times n$ matrix. The solution of (1) (which we will not need to evaluate) is

$$\tag{2} x(t)=e^{rAt}x(0) $$

You seek a differential equation for the average $\tilde{x}$ such that

$$\tag{3} \tilde{x}(t):=\left< x(t) \right>=\left< e^{rAt}\right>x(0) $$

with initial conditions $\tilde{x}(0)=x(0)$. Differentiating (3) we have

$$\tag{4} \tilde{x}'(t)=K'(t)\tilde{x}(t) $$

Where $e^{K(t)}=\left< e^{rAt}\right>$ is a 'matrix cumulant' generating function. The appropriate mean-field equivalent of (1) is obtained by making the replacement $rA\to K'(t)$. For Gaussian $r$ with mean $\mu$ and variance $\sigma^2$, we can evaluate explicitly

$$\tag{5} K(t)=\mu t A +\frac{\sigma^2 t^2}{2}A^2 $$

Which follows from expanding the exponential and integrating each term against the Gaussian measure. The evolution equations for the averages of (1) are

$$\tag{6} \tilde{x}'(t)=\left[\mu +\sigma^2 A t \right]A \tilde{x}(t) $$

From which we see that, at the level of the differential equation, the 'average' $r$ corresponds to $\tilde{r}(t)=\mu+\sigma^2A t$.

Notes:

1. In the case when $A=A(t)$, equation (2) needs to replaced by the ordered exponential $x(t)=\mathcal{T} [e^{r \int_0^t dt' A(t')}]x(0)$, and the analogues of (4) through (6) will become more complicated.

2. If $r=r(t)$ is a $\delta$ correlated process, a result similar to (6) holds, which can be obtained from the result here.

3. If $r$ is not normally distributed, it is still not necessary to evaluate $\left<e^{rAt}\right>$. Instead, the (scalar) cumulants $\kappa_m$ of $r$ maybe be inserted directly into (1) to obtain (6) via the replacement $rA \to \frac{d}{dt}\sum_m \frac{\kappa_m t^mA^m}{m!}$

Answer 2

You should be careful about how you consider this ODE. A reasonable interpretation is a path-by-path interpretation. That is, for each $\omega \in \Omega$ we can define the differential equation: $$\frac{dX_t}{dt}(\omega ) = r(\omega) A X_t$$ whose solution is the trajectory $$X_t(\omega) = X_0 \exp \left( r(\omega) A t \right)$$ One can then compute the time-average as:

$$\mu_t = E(X_t) = E( X_0 \exp \left( r A t \right))$$

where the expectation is taken relative to the distribution of $r$.

Novikov's criterion gives you conditions under which a certain process is a martingale, but that process has nothing to do with your random ODE.