Long-term survival of a species requires diversity: a case of Tribbles

We will demonstrate that the long-term survival of the species requires diversity. Here, “diversity” means the difference in personal preferences and, eventually, their divergence from rationality on the individual level. It might be related to or completely unrelated to physical parameters such as shape, size, social background, or skin/eye colour.

A logical yet striking conclusion is that if every species member is entirely rational (given their experience and current circumstances), the species will eventually die off with 100% certainty. An alternative approach that we tend to follow naturally – it ensures our survival as a species – is the probability matching. A simple example of the probability matching is as follows: suppose there are two possible outcomes, one that occurs with the probability of 95% and the other one that occurs with the probability of 5%; then we optimise for the first outcome with the frequency of 95% and the second outcome with the frequency of 5%.

The ideas below come from Professor Andrew Lo‘s book Adaptive Markets, which I warmly recommend. Although its subject is financial markets theory, this book discusses many subjects. One of them is a simple mathematical model that shows that the long-term survival of a species requires diversity (see pages 190-205).

The Model: The Valley vs. Mountains

Only valley tribbles will survive
Only valley tribbles will survive

In the model presented in the book, a tribble is a simple creature that lives only for one season, produces 3 offspring, and then dies of old age, unless it dies of a natural disaster earlier.

Any tribble can choose where to live: in the valley or the mountains. Once they make their choice, they cannot change their mind.

Each season, a natural disaster strikes. If the weather is sunny (the more likely scenario), the drought in the mountains will kill all the tribbles that live there. Those who live in the valley will survive. However, if it rains, the valley will be flooded. All tribbles who have chosen to nest in the valley will drown. Those who are up in the mountains will survive.

Individual Survival vs. Species Survival: Quick Conclusions

Only mountain tribbles will survive
Only mountain tribbles will survive

Obviously, on the individual level, it makes sense to live in the valley because the sunny weather is more likely, and thus, the chances of survival of those who live there are greater.

However, if all tribbles are rational (on the individual level), the first rainy season will wipe out the whole population.

Therefore, individual irrationality (the “unhealthy obsession” with the mountains?) of a part of the population seems to be a prerequisite for species survival. In other words, the eccentricity of mountain tribbles appears to be an adaptation feature on the level of the population as a whole.

The Optimal Number of Irrational Tribbles

Naive Approach: Maximising The Expected Number

We need to define what “optimal” means; this is not trivial.

If we wish to maximise the expected number of tribbles, then it is easy to show that there should be 0 irrational tribbles. However, there is a substantial probability that the rain kills all tribbles. Thus, this is a perilous strategy.

Let’s put some numbers on this claim. Let

  • s be the probability that it is sunny; this implies that the probability of rain is 1-s.
  • v be the proportion of tribbles who choose the valley; it means that 1-v out of all tribbles will choose the mountains.
  • k be the number of children (k for kids) each tribble has.
  • G be the number of generations that we set as the horizon for our study.
  • N be the number of tribbles at the beginning of our study.

So, if there are 0 irrational tribbles (v=1), the population will grow by the factor $k^G \text{ with probability } s^G$ and will be $0 \text{ with probability } 1-s^G$
If each tribble has 3 children, the chances of rain are 5%, and we consider a 100-generation horizon,

  • with the probability of 99.4%, the tribbles will become extinct;
  • with the probability of 0.6%, there will be approximately 515377520732011000000000000000000000000000000000 times more tribbles than there were initially.

So, even though this strategy maximises the expected number of tribbles, it is too risky to be accepted because of the high probability of tribbles’ extinction.

Thus, we must balance the expected population size and the risk. The “optimal strategy” is, in fact, a subjective concept; its choice depends on the decision maker(s) ‘s risk appetite.

Approach From The Book: Maximising The Expected Growth Rate

Let’s consider optimising the expected logarithmic growth rate of population growth, as suggested in the book. If it is sunny, the number of tribbles in the next generation will grow by the factor of $k v$

… and if it rains, the number of tribbles in the next generation will grow by the factor of $k (1-v)$

Therefore, the expected logarithmical rate of population growth per generation will be $log(k) + s \ log(v) + (1 – s) log(1-v)$

Solving its first derivative w.r.t. v yields that the maximal rate of growth is reached when $v = s$

The implication is that if the chances of sunny weather are 95% (s = 95%), 95% of the tribbles should nest in the valley (v = 95%) to maximise the expected population growth rate. As the book mentions, this strategy is called probability matching.

Reasons For Maximising The Expected Growth Rate

There is no immediate reason to optimise the expected population growth rate: unlike the expected population size, it has no “physical” meaning. However, we expect the risk to be much lower since the strategy is much less extreme than the “always choose the best” strategy we have seen earlier. Also, it is very convenient from a numerical/computational point of view.

Predictably, a similar problem has arisen in finance. Optimisation of the growth rate in the context of money management is called the Kelly criterion. In the original paper, published by John Larry Kelly Jr. in 1956, the author shows that choosing to optimise the growth rate helps to avoid ruin, but he does not explore the alternatives. As per this discussion, it is claimed that there is no reason why optimising for the growth rate is à priori optimal. The growth rate is just one of many criteria that (implicitly, in this case) include the penalty on risk.

To decide on the optimal strategy, we need measures that factor in both the expected growth and the risk of drastic population reduction/extinction. The science of finance, which deals with similar problems (maximising profit and minimising the risk of default), offers several measures we could study.

Penalty On Risk

Several possible explicit risk measures can define the penalty:

  1. Volatility: roughly speaking, this is the average deviation from the mean;
  2. Expected shortfall: this is the average over α% worst outcomes, α being a user-defined threshold, typically 1% or 0.1%;
  3. Maximum drawdown: it is the maximum loss in-between two peaks;
  4. Lower partial moments, a.k.a. downside deviation: roughly speaking, this is the average return overall returns that are below a certain user-defined threshold.

The first two measures only take the final size of the population as the input, whereas the other two require information about the evolution of the population size. Since we are only interested in the final number of tribbles, the first two measures are more relevant for our study.

The Formulae For The Number of Tribbles

One can show that the number of tribbles can be defined using the binomial distribution (over the sunny day quantity). We can use it to derive several closed-form formulae.

We have a slight technical difficulty because of the discretisation. If there are 50 tribbles, and 5% prefer mountains, then 5% of 50, i.e. 2.5 tribbles, will go to the mountains. Obviously, we can only expect an integer number of tribbles to go to the mountains! So we will do the roundings randomly, with the probability of rounding up or down picked to match the expectations!

With this assumption, the probability of having exactly g rainy reasons is
$ \frac{G!}{g! (G-g)!} (1-s)^g s^{G-g} $
The number of tribbles after g rainy reasons (and (G-g) sunny seasons) is
$ N k^G (1-v)^g v^{G-g} $
The expected number of tribbles is
$ N (k (s v + (1 – s) (1-v)))^G $
The variance of the number of tribbles is
$ N^2 k^{2G} ((s v^2 + (1 – s) (1-v)^2)^G – (s v + (1 – s) (1-v))^{2G})$
The log growth of the number of tribbles is
$ G (\log k + (s \log v + (1 – s) \log (1-v))) $

We can compute the expected shortfall as a sum of a series. The probability that there will be $R$ or more rainy days is
$ \sum_{g=R}^G \frac{G!}{g! (G-g)!} (1-s)^g s^{G-g} $
The expected number of tribbles if there were R or more rainy days is
$ N k^G \sum_{g=R}^G \frac{G!}{g! (G-g)!} ((1-s)(1-v))^g (vs)^{G-g} $
To compute the shortfall, we will first need to find R such that the resulting probability matches the expected shortfall we defined. Then, use this R to compute the expected number of tribbles if there were R or more rainy days – this will be our expected shortfall.

The Optimal Number of Irrational Tribbles: Numerical Model

My simulation results agree with Professor Lo’s statement about the optimality of probability matching for logarithmic growth maximisation.

I have studied the Tribbles’ population dynamics for the horizon of 100 seasons, a range of rain probabilities, and shares of mountain-picking tribbles. I have dropped the number of Tribbles’ children because this parameter only affects the results as a static multiplier, the same for all rain and mountain probabilities.

As the first step, I have plotted the distribution of the sunny days’ quantity and the tribble population size given the number of sunny seasons. Next, I plotted initial measures of strategy success: the expected number of tribbles and the expected number of tribbles over the standard deviation of the number of tribbles. This risk measure was inspired by the Sharpe ratio; the higher the ratio, the better. The conclusions we can draw from these two measures are contradictory: the former would suggest choosing the valley all the time, whereas the other indicates that splitting the choice 50%-50% is the best!

This contradiction is due to the fact that, when we compute the expected number of tribbles, the biggest contribution comes from the valley-almost-always strategies; these are very low-probability strategies (if the probability of rain is not so low) with extremely high values.

Next, we compute the expected number of tribbles excluding the extremely favourable scenarios (this measure is inspired by expected shortfall risk measure. We can observe that, while the probability matching is fully optimal in all settings, it appears to be a “good enough” choice, thus making it a great rule of thumb for situations when a more sophisticated analysis is not available.

Finally, I have plotted the expected growth rate, which indeed peaks when the probabilities of rain and choosing mountains are equal.

The Python code I have used to get the numbers, as well as the results, can be found here.

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