Moran Process: worked solutions

Solutions to the example questions on the Moran Process page.

Question 1 [25 marks]

(a) The two random choices. [3]

At each step one individual is selected to reproduce, with probability proportional to its fitness, and one individual is selected uniformly at random to die. A copy of the reproducing individual replaces the one that dies.

(b) Fitnesses. [6]

With \(v_H = 2\) Hawks and \(v_D = 2\) Doves,

\[ \begin{aligned} f_H &= (2 - 1)A_{HH} + 2 A_{HD} = 1 \cdot 0 + 2 \cdot 3 = 6, \\ f_D &= (2 - 1)A_{DD} + 2 A_{DH} = 1 \cdot 2 + 2 \cdot 1 = 4. \end{aligned} \]

(c) Probability the number of Hawks increases. [7]

The number of Hawks rises by one when a Hawk reproduces and a Dove dies:

\[ \frac{2 f_H}{2 f_H + 2 f_D} \cdot \frac{v_D}{N} = \frac{12}{12 + 8} \cdot \frac{2}{4} = \frac{3}{5} \cdot \frac{1}{2} = \frac{3}{10}. \]

(d) Fixation and absorbing states. [5]

The fixation probability of a type is the probability that the population eventually becomes composed entirely of that type. The process has two absorbing states, all Hawks and all Doves: once only one type remains there is no other type to copy, so the state can no longer change.

(e) Direction of drift. [4]

The number of Hawks falls by one when a Dove reproduces and a Hawk dies:

\[ \frac{2 f_D}{2 f_H + 2 f_D} \cdot \frac{v_H}{N} = \frac{8}{20} \cdot \frac{2}{4} = \frac{2}{5} \cdot \frac{1}{2} = \frac{1}{5}. \]

Hence \(\gamma = \dfrac{P(\text{decrease})}{P(\text{increase})} = \dfrac{1/5}{3/10} = \dfrac{2}{3} < 1\). Since \(\gamma < 1\) the process is more likely to gain a Hawk than to lose one from this state, so it tends to drift towards more Hawks.

Question 2 [25 marks]

(a) Definitions. (Bookwork.) [4]

The Moran process assumes a constant population of \(N\) individuals of \(m\) types with a fitness function. At each step one individual is selected to reproduce with probability proportional to its fitness, one individual is selected uniformly at random to be removed, and a copy of the reproducing individual replaces it. This repeats until a single type remains.
The fixation probability of a type is the probability that the population eventually becomes composed entirely of individuals of that type.
Neutral drift is the case where the mutant has the same fitness as the residents (\(r = 1\), equivalently selection intensity \(w = 0\)); a single mutant then fixes with probability \(1/N\).
The relative fitness \(r\) of a mutant is its fitness divided by the resident fitness.

The formula quoted in the question follows from the general fixation result. Because the mutant has constant relative fitness, with \(f_M = r\) and \(f_R = 1\) every ratio is the same:

\[ \gamma_k = \frac{f_R}{f_M} = \frac{1}{r} = r^{-1} \qquad \text{for all } k. \]

The products in the general formula are therefore powers of \(r^{-1}\), so for a single mutant (\(i = 1\)),

\[ \rho = \frac{1}{1 + \sum_{k=1}^{N-1} (r^{-1})^k} = \frac{1}{\sum_{k=0}^{N-1} r^{-k}}. \]

Summing the geometric series, valid for \(r \neq 1\), gives

\[ \rho = \frac{1 - r^{-1}}{1 - r^{-N}}. \]

(b)(i) Neutral drift. [3]

When \(r = 1\) all individuals are equally likely ancestors, so \(\rho = 1/N\).

(b)(ii) \(N = 4\), \(r = 2\). [3]

\[ \rho = \frac{1 - \tfrac{1}{2}}{1 - \tfrac{1}{16}} = \frac{\tfrac{1}{2}}{\tfrac{15}{16}} = \frac{8}{15} \approx 0.533. \]

(b)(iii) \(N = 4\), \(r = \tfrac{1}{2}\). [3]

\[ \rho = \frac{1 - 2}{1 - 16} = \frac{-1}{-15} = \frac{1}{15} \approx 0.067. \]

(b)(iv) Comparison. [5]

The neutral value is \(1/4 = 0.25\). An advantageous mutant (\(r = 2\)) fixes more often (\(\tfrac{8}{15} \approx 0.53\)), while a disadvantageous one (\(r = \tfrac{1}{2}\)) fixes far less often (\(\tfrac{1}{15} \approx 0.07\)); so selection raises or lowers fixation relative to drift. Even so, the strongly advantageous mutant fixes only about half the time: starting as a single individual it is very likely to be lost by chance in the first few steps, before its fitness advantage can act. Selection improves a rare mutant's odds but is far from guaranteeing that it takes over.

(c) Limit. [4]

For \(r > 1\), \(r^{-N} \to 0\) as \(N \to \infty\), so

\[ \rho = \frac{1 - r^{-1}}{1 - r^{-N}} \to \frac{1 - r^{-1}}{1} = 1 - r^{-1}. \]

(d) Interpretation. [3]

Even in an arbitrarily large population an advantageous mutant fixes with probability only \(1 - 1/r\), bounded away from \(1\): a beneficial mutation is likely to be lost while still rare. For \(r = 2\) this limit is \(\tfrac{1}{2}\).

Question 3 [23 marks]

(a) Fixation formula. (Bookwork.) [4]

\[ \rho_i = \frac{1 + \sum_{k=1}^{i-1} \prod_{j=1}^{k} \gamma_j} {1 + \sum_{k=1}^{N-1} \prod_{j=1}^{k} \gamma_j}, \qquad \gamma_i = \frac{p_{i, i-1}}{p_{i, i+1}}. \]

(b)(i) Fitnesses. [3]

\[ \begin{aligned} f_H(1) &= 0 + 2(3) = 6, & f_D(1) &= 1 + 2 = 3, \\ f_H(2) &= 0 + 3 = 3, & f_D(2) &= 2 + 0 = 2. \end{aligned} \]

(b)(ii) Ratios. [4]

\[ \begin{aligned} \gamma_1 &= \frac{f_D(1)}{f_H(1)} = \frac{3}{6} = \frac{1}{2}, \\ \gamma_2 &= \frac{f_D(2)}{f_H(2)} = \frac{2}{3}. \end{aligned} \]

(b)(iii) Fixation from one Hawk. [4]

\[ \rho_1 = \frac{1}{1 + \gamma_1 + \gamma_1 \gamma_2} = \frac{1}{1 + \tfrac{1}{2} + \tfrac{1}{3}} = \frac{1}{\tfrac{11}{6}} = \frac{6}{11} \approx 0.545. \]

(b)(iv) Fixation from two Hawks. [4]

\[ \rho_2 = \frac{1 + \gamma_1}{1 + \gamma_1 + \gamma_1 \gamma_2} = \frac{1 + \tfrac{1}{2}}{\tfrac{11}{6}} = \frac{\tfrac{3}{2}}{\tfrac{11}{6}} = \frac{9}{11} \approx 0.818. \]

(c) Interpretation. [4]

A single Hawk fixes with probability \(\tfrac{6}{11}\), and starting from two Hawks fixation is more likely still at \(\tfrac{9}{11}\): the more Hawks initially present, the more likely they take over. Applying the criterion, a single Hawk has \(\rho_1 = \tfrac{6}{11} \approx 0.545 > \tfrac{1}{3} = 1/N\), so a single Hawk fixes more often than a neutral mutant would: Hawks are favoured by selection.

Question 4 [25 marks]

(a) The ratio is state-independent. [4]

In state \(i\) the number of mutants rises by one when a mutant is copied and a resident is removed, and falls by one when a resident is copied and a mutant is removed:

\[ \begin{aligned} P_{i \to i+1} &= \frac{i f_M}{i f_M + (N - i) f_R} \cdot \frac{N - i}{N}, \\ P_{i \to i-1} &= \frac{(N - i) f_R}{i f_M + (N - i) f_R} \cdot \frac{i}{N}. \end{aligned} \]

Taking the ratio, the shared denominator and the factor \(i(N - i)/N\) cancel:

\[ \gamma_i = \frac{P_{i \to i-1}}{P_{i \to i+1}} = \frac{(N - i) f_R \cdot i}{i f_M \cdot (N - i)} = \frac{f_R}{f_M} = \frac{1}{r} = r^{-1}, \]

which is independent of \(i\).

(b) Fixation of a single mutant. [6]

With \(\gamma_k = \gamma = r^{-1}\) for all \(k\), the products in the general formula are powers of \(\gamma\), so

\[ \rho_1 = \frac{1}{1 + \sum_{k=1}^{N-1} \gamma^k} = \frac{1}{\sum_{k=0}^{N-1} \gamma^k}. \]

For \(\gamma \neq 1\) the geometric sum gives \(\sum_{k=0}^{N-1}\gamma^k = \dfrac{1 - \gamma^N}{1 - \gamma}\), hence

\[ \rho_1 = \frac{1 - \gamma}{1 - \gamma^N} = \frac{1 - r^{-1}}{1 - r^{-N}}. \]

For \(\gamma = 1\), that is \(r = 1\), the sum is \(N\) and \(\rho_1 = 1/N\).

(c) The selection criterion. [10]

Write \(\gamma = r^{-1}\), so \(r > 1\) is equivalent to \(\gamma < 1\). A short calculation gives

\[ \rho_1 - \frac{1}{N} = \frac{1 - \gamma}{1 - \gamma^N} - \frac{1}{N} = \frac{N(1 - \gamma) - (1 - \gamma^N)}{N(1 - \gamma^N)} = \frac{-g(\gamma)}{N(1 - \gamma^N)}, \]

where \(g(\gamma) = 1 - \gamma^N - N(1 - \gamma)\). We have \(g(1) = 0\) and

\[ g'(\gamma) = -N\gamma^{N-1} + N = N\bigl(1 - \gamma^{N-1}\bigr), \]

which is positive for \(\gamma < 1\) and negative for \(\gamma > 1\). So \(g\) has a strict maximum at \(\gamma = 1\), where \(g(1) = 0\); hence \(g(\gamma) < 0\) for every \(\gamma > 0\) with \(\gamma \neq 1\). Therefore \(-g(\gamma) > 0\), and the sign of \(\rho_1 - 1/N\) is the sign of \(1 - \gamma^N\):

if \(\gamma < 1\) then \(1 - \gamma^N > 0\), so \(\rho_1 > 1/N\);
if \(\gamma > 1\) then \(1 - \gamma^N < 0\), so \(\rho_1 < 1/N\).

Thus \(\rho_1 > 1/N\) if and only if \(\gamma < 1\), that is if and only if \(r > 1\).

(d) The large-population limit. [5]

For \(r > 1\) we have \(r^{-N} \to 0\) as \(N \to \infty\), so

\[ \rho_1 \to \frac{1 - r^{-1}}{1} = 1 - \frac{1}{r}. \]

Even in an arbitrarily large population an advantageous mutant fixes with probability only \(1 - 1/r\), bounded away from \(1\). From part (c) selection fixes the direction, a mutant being favoured exactly when \(r > 1\), for every \(N\); but it does not guarantee fixation. While the mutant is rare, drift can remove it before its fitness advantage takes effect, and it is lost with probability at least \(1/r\).

Marking exercises

Marking exercise 1 (Question 1(b) and (c)).

The fitnesses count self-interactions: an individual does not play itself, which is why the question's formula carries the \((v_i - 1)\) term for the own type. The Hawk value is accidentally unaffected, because \(A_{HH} = 0\) makes the self-term worthless: \(f_H = (2 - 1)(0) + 2(3) = 6\) either way. The Dove value is not: \(f_D = (2 - 1)(2) + 2(1) = 4\), not \(6\), as in the solution above. That coincidence is what makes the slip easy to miss, and is exactly why an examiner recomputes both.

Everything after the fitnesses inherits the error: the types are not equally fit, this is not neutral drift, and the probability that the Hawks increase is

\[ \frac{2 f_H}{2 f_H + 2 f_D} \cdot \frac{2}{4} = \frac{12}{20} \cdot \frac{1}{2} = \frac{3}{10}, \]

not \(\tfrac{1}{4}\). A fair mark is [3] of [6] for (b), one fitness of two, and [2] of [7] for (c), where the structure of the product, reproduction probability times death probability, is right.

Marking exercise 2 (Question 3(b)(iii) and (iv)).

The formula is misremembered: the denominator of the fixation probability is \(1 + \sum_{k} \prod_{j \le k} \gamma_j\), a sum of products of consecutive ratios, not a sum of the ratios themselves. The second term must be \(\gamma_1 \gamma_2 = \tfrac{1}{3}\), not \(\gamma_2 = \tfrac{2}{3}\): walking down from two Hawks to zero requires both downward steps, and the product is what compounds them. The correct values, as in the solution above, are

\[ \rho_1 = \frac{1}{1 + \tfrac{1}{2} + \tfrac{1}{3}} = \frac{6}{11} \approx 0.545, \qquad \rho_2 = \frac{1 + \tfrac{1}{2}}{\tfrac{11}{6}} = \frac{9}{11} \approx 0.818. \]

The conclusion that Hawks are favoured survives, but by luck: both reported numbers are wrong, and with other payoffs the mangled formula can land on the wrong side of \(1/N\). A fair mark is [1] of [4] for each part: the \(\gamma\) values are carried in correctly, but the formula being assessed is wrong in both.