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Say there is a population of mass 1 in which individuals can choose one of two traits (1 and 2). The population share with trait 1 at time $t+1$, $$p_{t+1}=\frac{1}{1+e^{-\beta\left(u_1-u_2+J(2p_t-1)\right)}}$$ for constants $u_1,u_2,\beta,J$, with $u_1>u_2$. I can't calculate them analytically, but it can be shown that for $\beta J>1$, there are at most three stationary points (of which the smallest and largest are stable) and at least one stationary point (which will be in the neighbourhood of $p^*\approx 1$).

Assume $p_{t=0}=0$. Consider two cases defined by two sets of parameters $u_1,u_2,\beta,J$ such that in each, there is one unique stationary point $p^*\approx 1$. How can I work out which system will reach its stationary point faster? In general I would like to know how to calculate the convergence rates analytically, if it's possible, but if not, knowing how to compare two convergent systems and work out the faster one would suffice.

For example, my hunch is that higher $J$ slows the system. For given $\beta,u_1,u_2$, if we take two environments with $J_1$ and $J_2$, respectively, where there exists one stationary point, then the curve of the function with lower $J$ more closely "wraps" around the line $p=p$, which should lead to more transition steps. (I don't know how to explain that better, but hopefully you understand what I mean.)

Even just knowing what terms to search for would be a big help. I've tried searching for literature on convergence rates but given the implicit transition expression, I'm not sure whether my case is applicable to (e.g.) Markov chain theory, etc. As you can tell, this is not my area :) Any search terms would be appreciated.

(Postscript: Ultimately I'm investigating a three-trait system, but I thought this would a good start to get the tools to solve it. However, if your answer also generalises, all the better!)

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  • $\begingroup$ Are you sure, you want to talk about stationary points and not fixed points? $\endgroup$ – Wrzlprmft Aug 1 '17 at 16:55
  • $\begingroup$ My map comes from the expected path of a stochastic process. Individuals share a common payoff matrix but each choice is subject to i.i.d shocks for each individual (distributed $\sim EV_1(0,\beta)$, which might explain the form of the mapping function). So in each period we can't be sure of the exact distribution of shocks across the population, but for large populations we have a pretty good prediction of average group behaviour. Having said that, I'm not entirely sure on the definitions of fixed and stationary, so I could still be wrong :) $\endgroup$ – David Smerdon Aug 3 '17 at 13:06
  • $\begingroup$ $p^*$ is a fixed point if $f(p^*) = f(p)$. It’s a stationary point if $f'(p^*) = 0)$. (Note that some fields may use these terms in a different way.) $\endgroup$ – Wrzlprmft Aug 3 '17 at 13:41
  • $\begingroup$ Ah, I see. Thanks. Then I should have said "three fixed points, of which two are stationary." $\endgroup$ – David Smerdon Aug 4 '17 at 14:46
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Let $f$ be your map, i.e.: $$f(p) := \frac{1}{1+e^{-\beta\left(u_1-u_2+J(2p-1)\right)}}.$$

Suppose your state $p$ is close to the fixed point $p^*$, namely it is $p := p^*+ε$. Then you can estimate how quickly the distance to the fixed point shrinks via the following quantity:

$$ Λ := \left | \frac{f(p) - p^*}{p-p^*}\right | = \left | \frac{f(p^*+ε) - f(p^*)}{ε}\right | ≈ \left | f'(p^*) \right |. $$

With more iterations, the distance behaves like $ε Λ^t$ (where $t$ is the number of iterations). The quantity $λ := \ln(Λ)$ is also known as the local Lyapunov exponent at the point $p^*$. For more details on the exponential behaviour, see this question.

So to compare convergence behaviours close to fixed points, you only need to take a look at the derivative of your map. For multidimensional systems, things are a bit more complicated: There are multiple Lyapunov exponents and the one you need is the largest one, which in turn you can obtain from eigenvalues of the Jacobian of $f$.

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  • $\begingroup$ Thanks; very helpful. Having researched Lyapunov exponents, I'm still a little lost as to how to proceed in my three-dimensional case in terms of predicting which point the system converges to. But as it is a slightly different question to the original post, I've asked it separately here: math.stackexchange.com/questions/2381249/… $\endgroup$ – David Smerdon Aug 3 '17 at 13:30

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