Modeling the Decay of a Pack of Cannibalistic HyenasA population of p 0 hyenas has run out of food in their ecosystem, and so sadly they have resorted to eating each other.Hyenas need to consume one meal a day, and so exactly once per day, any given hyena will kill another hyena. The time at which this happens is random, meaning each hyena's mealtime is uniformly distributed throughout a set of 24 hours.Assuming that the last hyena will get hungry and die trying to eat himself, after how long will the population of hyenas become extinct?I'm can think of two different ways to answer this (neither of which I know how to solve). The first is simpler and less accurate (and will therefore merit less glory).

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Modeling the Decay of a Pack of Cannibalistic HyenasA population of       p    0   hyenas has run out of food in their ecosystem, and so sadly they have resorted to eating each other.Hyenas need to consume one meal a day, and so exactly once per day, any given hyena will kill another hyena. The time at which this happens is random, meaning each hyena&#039;s mealtime is uniformly distributed throughout a set of 24 hours.Assuming that the last hyena will get hungry and die trying to eat himself, after how long will the population of hyenas become extinct?I&#039;m can think of two different ways to answer this (neither of which I know how to solve). The first is simpler and less accurate (and will therefore merit less glory).

relatatt9 · Accepted Answer

Step 1If the consumption time is random, each hyena cannot consume meals exactly every day. The easy resolution is to say that if p is the current population, we have             d      p              d      t        =  −  p  ,  p  (  0  )  =      p    0  , where time is measured in days. The solution to this is   p  =      p    0    exp  ⁡  (  −  t  ) which ignores the discrete nature of coyotes.Step 2Another solution is that each coyote has a time of day when it consumes another, and that time of day is the same every day for as long as that coyote is alive. The difference will be small as long as the population is large, but in this case you have at least one death per day, so there is no long tail.

Hugo Stokes · Accepted Answer

Step 1Let n be the number of hyenas at the beginning of a day. It suffices to show that the fraction of hyenas surviving to the next day is asymptotic to some constant.Arrange the hyenas from left to right by their planned meal times. Now construct the directed graph on the hyenas where an edge points from hyena A to hyena B if A eats B. Observe that every valid graph is equally probable.What does this graph look like? A component consists of a path of leftward edges terminating in a single rightward edge. The number of surviving hyenas is the number of components in the graph.Let the number of such graphs be f(n). We can count the number of such graphs with an EGF. We will find that   f  (  n  +  1  )      /    f  (  n  )  ∼  α      n    d   for some constants   α and d.Step 2Now we can calculate the expected number of components of size k, since we can count the number of ways to choose a component of size k and then fill in the rest of the graph in   f  (  n  −  k  )  ∼      α    d        n          k      d        f  (  n  ) ways. Summing over k will yield the expected number of components, which should be asymptotically cn for some constant c.To prove the desired concentration result, we can use Chebyshev&#039;s inequality. Calculating the variance of the number of components is equivalent to counting the expected number of pairs of components.

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