The formulation of the Ito form of stochastic differential equations has a Gaussian white noise process (dW) in the diffusion term. $$ dX = F(t,X)dt + G(t,X)dW $$ If the system has a noise input which is known to be nonGaussian, say Weibull, how do I construct the equation. Should I multiply dW with some transformation function?
1 Answer
Firstly, SDE is just an informal way of writing a definition of process $X$, while more formal way is to use so called stochastic integration: $$ X(t)=\int_0^t F(s, X)ds + \int_0^tG(s, X)dW, $$ where the latter term is a limit of $$ \sum_i G(t_i, X_i)(W(t_{i+1})W(t_i)). $$ Let’s state for the sake of simplicity that $G$ depends solely on $t$. Then the terms of the sum are normally distributed and therefore, due to the property of normal distribution the limit is Gaussian too. Moreover, normal distribution is an infinitely divisible distribution, which means that if we have two independent normally distributed random variables, then the sum of them will be also normally distributed. There are different distributions that obey this law (e.g. Stable or Poisson). They form the class of Lévy processes. However, Weibull is not one of them.
Now let us define another nonGaussian process, say $Y(t)$ and replace $W(t)$ with it. Will the sum preserve the distribution now? The answer lies in the definition of $Y(t)$ and how you assume the sum $$ \sum_i Y(t_{i+1})Y(t_i) $$ is distributed in the limit. Once you get it, you could write the SDE.

$\begingroup$ Thank you very much. If instead of finding out a Y(t) and attempting to write the SDE, if I go directly to the EulerMaruyama scheme and write dW in each step as y*sqrt(dt) where y is sampled from a Weibull distribution, am I making a fundamental error? $\endgroup$ Commented Nov 2, 2017 at 11:12

$\begingroup$ Yes. Euler scheme is an approximation, which converges to the desired stochastic process. In your case sum of $y \sqrt{dt}$ converges to normal distribution regardless if $y$ is sampled from Weibull or another distribution with finite variance. $\endgroup$ Commented Nov 2, 2017 at 12:13