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<Paper uid="W05-0502">
  <Title>Simulating Language Change in the Presence of Non-Idealized Syntax</Title>
  <Section position="6" start_page="15" end_page="15" type="evalu">
    <SectionTitle>
5 Results
</SectionTitle>
    <Paragraph position="0"> In all of the following results, the bound on aj and bj is N = 5000, the sigmoid slope is k = 2, the probability that an agent is replaced when selected is pr = 0.001, and there are 40 agents in the population configured in a loop where each agent talks to its two neighbors. See Figure 1 for a key to the notation used in the figures.</Paragraph>
    <Paragraph position="1"> First, let us consider the base line LEARN-ALWAYS algorithm. Typical sample populations, such as the one shown in Figure 2, tend to be globally and locally incoherent, with neighboring agents favoring completely different grammars. The results are even worse without the biased speech algorithm.</Paragraph>
    <Paragraph position="2"> A sample run using the PARAMETER-CRUCIAL learning algorithm is shown in Figure 3. This population is quite coherent, with neighbors generally favoring similar grammars, and most speakers using non-V2 languages. Remember that the picture represents the internal data of each agent, and that their speech is biased to be more regular than their experience. There is a region of SVO+V2 spanning the second row, and a region of SVO+pro-drop on the fourth row with some SVO+V2+pro-drop speakers. Another sample dominated by V2 with larger regions of SVO+V2+pro-drop is shown in Figure 4.</Paragraph>
    <Paragraph position="3"> A third sample dominated by non-pro-drop speakers is shown in Figure 5. The MCFTP algorithm starts with a population of all Amax and one of Amin and returns a sample that is a possible future of both; hence, both V2 and pro-drop may be lost and gained under this simulation.</Paragraph>
    <Paragraph position="4"> In addition to sampling from the stationary distribution pi of a Markov chain, MCFTP estimates the chain's mixing time, which is how large t must be for the distribution of Xt to be e-close to pi (in total variation distance). The mixing time is roughly how long the chain must run before it &amp;quot;forgets&amp;quot; its initial state. Since this Markov chain is not quite monotonic, the following should be considered a heuristic back-of-the-napkin calculation for the order of magnitude of the time it takes for a linguistic environment to forget its initial state. Figures 3 and 4 require 29 and 30 doubling steps in MCFTP, which indicates a mixing time of around 228 steps of the Markov chain. Each agent has a probability pr of dying and being replaced if it is selected. Therefore, the probability of an agent living to age m is (1[?]pr)mpr, with a mean of (1[?]pr)/pr. For pr = 0.001, this gives an average life span of 999 listening interactions. Each agent is selected to listen or be replaced with probability 1/40, so the average lifespan is approximately 40,000 steps of the Markov chain, which is between 215 and 216. Hence, the mixing time is on the order of 228[?]16 = 4096 times the lifespan of an individual agent. In real life, taking a lifespan to be 40 years, that corresponds to at least 160,000 years. Furthermore, this is an underestimate, because true human language is far more complex and should have an even longer mixing time. Thus, this simulation suggests that the linguistic transitions we observe in real life taking place over a few decades are essentially transient behavior.</Paragraph>
  </Section>
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