Are Lyapovov exponents related to statistical precision? [closed]

In my understanding, Lyapunov exponents measure the average rate of separation of near-identical chaotic trajectories, while statistical precision, in the context of a predictive model, is a measure of a spread of projections, taking into account the uncertainty of the input data.

This blog post states that:

The precision of a model is the variance of its predictions, without reference to observations.

This is in the context of numerical climate models. These models don't have any explicit variance estimates: a single simulation simply outputs a single value for a given variable at each timestep/grid point. You could calculate an approximation of a Lyapunov exponent by taking rate of change in the variance of an ensemble of simulations with similar initial conditions. But is this actually a measure of model precision?

closed as unclear what you're asking by kjetil b halvorsen, Michael Chernick, user158565, Peter Flom♦Jun 5 at 19:51

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• 1) Precision is independent from accuracy. Most times, precision of an inaccurate result is much more useful than accurate results that are hopelessly imprecise 2) Precision measurements are not one-one with variance, some precise measurements have no second moment, i.e., their variances do not exist (Think Cauchy). Precision has no set best measurement system. 3) Precision can refer to how precise a model is, or how precise the data is, such that is we say precision, we should say of what. 4) One can often withhold data and check modelling precision and accuracy against it. – Carl Jun 4 at 5:41

I am not sure I fully understand the notation that my search results are using to describe Lyapunov exponents, but it reminds me of a Dobrushin coefficient for a Markov chain: $$\delta(K) = \frac{1}{2} \sup_{(x,x')} \left\Vert K(x,\cdot) - K(x',\cdot) \right\Vert_{TV},$$ which can be useful for showing certain chains are ergodic. Note: some books define it without the half in the front, and use a $$\beta$$ instead of a $$\delta$$.
The smaller this number is, the faster the chain forgets its past, or the closer the observations are to iid. This is because you can show that, for two probability measures $$\mu$$ and $$\nu$$, $$\left\Vert \mu K - \nu K \right\Vert_{TV} \le \delta(K)\left\Vert \mu - \nu\right\Vert_{TV}.$$ Because $$0 \le \delta(K) \le 1$$ for any kernel $$K$$, this is saying that applying the kernel to two measures can't make them farther apart, and might even make them closer. This sort of resembles the first equation on here[https://en.wikipedia.org/wiki/Lyapunov_exponent].
You can see this doesn't really have anything to do with precision. This is because $$\mu$$ and $$\nu$$ may have small or large precision, and it doesn't change anything that we're talking about here.
Yeah, we're dealing with long run stability, not dispersion/variance/precision. Meyn and Tweedie talk about some connections between Markov chains and deterministic dynamical systems on page 17. They describe the one that arises "most naturally" is based on the sequence of marginal distributions $$\mu P^k$$, where $$P$$ is some transition kernel, and $$\mu$$ is an initial measure.
On page 141, in the chapter on Feller chains, they define the dynamical system $$(P, \mathcal{M},d_m)$$ to be stable in the sense of Lyapunov if for each measure $$\mu \in \mathcal{M}$$ $$\lim_{\nu \to \mu}\sup_{k \ge 0}d_m\left(\nu P^k, \mu P^k \right) = 0.$$ $$d_m$$ here is some distance between measures. The TV norm I discussed earlier is one such example.