# Combining Coefficients in a Fokker-Planck equation

Question: What is the most appropriate way to combine two diffusion processes in a way that "averages" their behavior?

Details

A diffusion process can be parameterized by two functions $(\mu,\sigma)$. This corresponds to an Ito process: $$dX_t = \mu(X_t)dt + \sigma(X_t)dW_t$$ as well as a Fokker-Planck equation: $$\partial_t p(x,t) = -\sum_i\partial_i[\mu(x) p(x,t)] +\frac{1}{2}\sum_i\sum_j \partial_{ij}[D_{ij}(x) p(x,t)]$$ where the diffusion tensor is $D(x)=\sigma(x)\,\sigma(x)^T$, which also appears in the equation for the infinitesimal generator. Furthermore, it is common to consider $D$ as the contravariant (inverse) metric tensor of a Riemannian manifold, i.e. $D=g^{-1}$.

Ignoring $\mu$, notice there are three reasonable ways to define the variance of the process: $\sigma$, $D$, or $g$.

My question is this: I have two diffusion processes, one with $\sigma$, $D$, and $g$, and another with $\hat{\sigma}$, $\hat{D}$, and $\hat{g}$. I want to create a new process with behaviour similar to the average between these two processes. So then my question is which variance measure should I combine?

I.e. which should I take: (1) $\tilde{\sigma}=\sigma+\hat{\sigma}$, (2) $\tilde{g}=g+\hat{g}$, or (3) $\tilde{D}=D+\hat{D}$?

The application is not a physical one (in the experimental sense), but I would be happy with a physical reason. Indeed, I think any of the three would produce some kind of combined/average process, but it's not clear which one is the most sensible or natural.

My initial thoughts: my most immediate thought is (1), because it is the most "direct" to the process. However, if we consider two metrics, it is sensible to add them (i.e. the sum of two metric tensors is still a metric), so perhaps (2) is reasonable as well.

• 1.The average of diffusion processes is not a diffusion process. 2. What do you mean by "similar"? – zhoraster Aug 14 '17 at 19:41
• @zhoraster 1. As the answerer below suggests, if written as $Y_t = (X_t + \hat{X}_t)/2=\int \frac{1}{2}(\mu+\hat{\mu})dt + \int \frac{1}{2}(\sigma+\hat{\sigma})dt$ (with component-wise integration), it seems to be, maybe with some reasonable assumptions on $\mu$ and $\sigma$. Please enlighten me :) 2. Apologies for the vagueness; please feel free to define one. Indeed, my question could be interpreted as asking for a similarity measure, since that would let me choose an option (1-3) sensibly. – user3658307 Aug 14 '17 at 20:13
• Under the integral you have $\mu(X) +\hat \mu(\hat X)$. There is no way to make $(X+\hat X) /2$ in the argument. For this reason, the answer below is some nonsense. – zhoraster Aug 15 '17 at 9:23
• @zhoraster I see what you mean. Thanks for your comment. But what if we define the new process as: $$Y_t = \int \frac{1}{2}(\mu(Y_t) + \hat{\mu}(Y_t)) dt + \int \frac{1}{2}(\sigma(Y_t) + \hat{\sigma}(Y_t)) dW_t$$ instead? – user3658307 Aug 15 '17 at 13:44
• Then it's not a diffusion process. – zhoraster Aug 15 '17 at 14:01

Diffusion process is linear, so as an option one can consider the arithmetic mean of random variables: $$\tilde{X_t}=\frac{X_t+\hat{X_t}}2.$$ For independent processes, $n=1$ and constant diffusion coefficients it corresponds to $$\tilde{D}=\frac{D+\hat{D}}4.$$
• Thanks. So is your suggestion $(X_t+\hat{X}_t)/2$, which is the same as $(dX_t+d\hat{X}_t)/2$, implying $\tilde{\sigma}=(\sigma + \hat{\sigma})/2$, i.e. option (1)? – user3658307 Aug 11 '17 at 16:16
• @user3658307 For one dimensional case $D=\sigma^2$. – Andrew Aug 11 '17 at 17:23