Duration of courtship effort with memory

Duration of courtship effort with memory

Robert M Seymour

Department of Mathematics

&

Department of Genetics, Evolution and Environment UCL

Peter Sozou LSE

Acknowledgement to

Courtship as extended bargaining

• Courtship between a male and a female is an asymmetric bargaining game extended over time

• Time delay is costly

• Participation involves costs to both male and female energy, predation risk, opportunity cost of time

• Why do they pay these costs?

• Why don’t they mate immediately?

Blue bird of paradise displays to a female by hanging upside down and vocalising for a

prolonged period of time (Frith and Beehler 1998)

Courtship over time

A male signal, e.g. ornamentation, may be costly and can act as an honest signal of the male’s

quality (Zahavi 1975, Grafen 1990)

Great Grey Shrike (Lanius excubitor)

• A raptor-like passerine bird

• Males give prey to females

immediately

before copulation

• Prey are rodents, birds, lizards or large insects

• Females select a mate according to the size of the

prey offered

Tryjanowski, P. & Hromada, M. (2005) Animal Behaviour 69, 529-533

Arthropods : Hanging fly (Bittacus apicalis)

Thornhill, R.(1976) Am. Nat 110, no. 974, 529-548

Human courtship can involve a long sequence of outings, gifts….

And …

The model : male types

There are two types of male:

Good males : high quality - a female wants to mate

- she gets a positive fitness payoff

Bad males: low quality - a female does not want to mate

- she gets a negative fitness payoff

Either type of male wants to mate with a female

- he gets a positive fitness payoff

A female does not have complete information about a male’s type

A priori probability that a random male is good: P

Good male Bad male

Species with facultative paternal care

Male finds female attractive and will stay and help after mating

Male will desert after mating

Species with universal paternal care

Male is in good condition: likely to be a good provider

Male is in poor

condition: likely to be a poor provider Species with

sexual

selection and no paternal

care

Male is in good condition: likely to be of high genetic quality

Male is in poor

condition: likely to be of low genetic quality

The model : game tree per round

M

F

F

quit courtship signal

reject and quit accept

mate solicit new signal

t

game ends

begin next round

One game round - repeated until mate or quit

The model : costs and benefits

Male’s cost per unit time of participating in courtship: x Payoff to good male from mating: A_m

Payoff to bad male from mating: D_m ^{Am > Dm > 0}

Male

Female’s cost per unit time of participating in courtship:

Payoff to female from mating with a good male: A_f > 0 Payoff to female from mating with a bad male: - C_f < 0

Female

φ

Mating immediately

The female’s expected payoff from mating immediately is

Π

= A

P − C

(1 − P )

Assume P is sufficiently large so that

Π

> 0

The female gets a positive payoff from mating immediately

The female doesn’t quit first

0 t

Π₀ > 0

female quits

−φt

{

Female gets positive expected payoff from

mating

Either the male will quit first

Or the female will mate while she can still get a positive expected payoff

Either way she doesn’t quit first

Can assume that the female never quits

t_G t_B

Π₀ > 0

t_m

bad male quits

t

t_B^∗

bad male best response

t_m^∗

female best response

Pure strategies

There are no non-trivial equilibria in pure strategies

t_G > t_B >0

The equilibrium mating strategy

At equilibrium a bad male is indifferent between his pure strategies: quitting or not quitting

quit not quit

mate not mate

0 0

−xδt

D_m −xδt

Suppose the female mates with probability p = t

At equilibrium the female’s mating rate is constant

λ = x D_m

Expected payoff from not quitting = ^D_m ^{p −}xδt _{= 0}

A good male never quits

quit not quit

mate not mate

0 0

−xδt A_m −xδt

At equilibrium

Expected payoff from not quitting = ^(Amλ −x)δt

= (A_m − D_m)λδt when λ = x D_m

> 0 since A_m > D_m A good male always gets a positive expected

payoff from not quitting

With and without memory

With memory

Players have an internal clock

They know how much the game has cost them at any time All rounds are distinguished

Without memory

Players cannot track objective time No information is acquired over time All rounds look the same to players

Seymour R.M. & Sozou P.D (2009) Duration of courtship effort as a costly signal. J.

Theor Biol 256, 1 - 13

Bad male quitting strategies

A bad male’s quitting rate q(t) is assumed to be conditioned on time (or equivalently, cost)

Associated probability of survival function is

s(t) = exp − q(τ)τ

0

∫

{ }

s(0)= 0

s(t) > 0 for a t≥0

s(t) is non-increasing in t s(t)→ 0 as t→ ∞

quitting rate q(t) survival probability s(t)

time t time t

The female’s expected payoff

Probability that female mates at time t

payoff Probability

that male is Good

Probability that male is Bad

Probability that female mates at time t, before bad male has quit

payoff

Probability that bad male quits at time t, before female has mated

E^F(λ) = P (A_f −φt)e^−λtλt

0

∫

∫

∫

{ }

payoff

σ = φ C_f

E^F(λ) = P A_f C_f

⎛

⎝⎜ ⎞

⎠⎟ −Pσ

λ ⁻⁽¹⁻P)(λ + σ)sˆ(λ) sˆ(λ) = Laplace transform of s(t) ⁼

∫

E^F(λ) = constant −(1−P)F(λ)

1

C_f E^F → E^F

Scaling transformation:

F(λ) = (λ + σ)sˆ(λ) + Pσ 1−P

⎛

⎝⎜

⎞

⎠⎟

1 λ

The female’s best response

For a given bad male quitting rate function q(t), the female’s best response mating strategy  maximizes her payoff E^F()

dE^F

dλ ^{= 0} Solution ^* of:

which defines a maximum of E^F()

Equivalently

Solution ^* of: F′(λ ) = 0

which defines a minimum of F()

Example 1: no memory

Seymour R.M. & Sozou P.D (2009) Duration of courtship effort as a costly signal. J.

Theor Biol 256, 1 - 13

q(t) = q a constant s(t) = e^−qt

F(λ) = λ +σ λ + q ⁺

Pσ 1−P

⎛

⎝⎜

⎞

⎠⎟

1 λ

λ^∗(q) = q Pσ

(1 − P)(q−σ ) − Pσ

F()

λ^∗

λ^∗ = x

D_m = equilibrium mating rate

λ^∗(q)

Female’s best response curve

Example 2: increasing impatience

q(t) = q +ηt s(t) = e^{−qt−12ηt2}

F()

 = 0.8 q = 1

Example 3: fading memory

q(t)= q− η

1 +t s(t)= (1 +t)^ηe^−qt

F()

 = 0.8 q = 1

Example 4: ‘perfect’ memory

This is equivalent the female being indifferent between all her constant mating strategies 

Suppose the female is indifferent between all her pure strategies (mating times t_m) in response to a bad male quitting rate q(t)

dE^F

dλ ^{= 0}  _{F′(λ ) = 0} ^{for all  > 0}

(λ +σ)sˆ(λ) + Pσ 1−P

⎛

⎝⎜

⎞

⎠⎟

1

λ ⁼ K ^{s(t) = − P}

1−P⁺ K + P 1−P

⎧⎨

⎩

⎫⎬

⎭e^−σt



Solution with initial condition s(0) = 1 has K = 1

s(t) = e^−σt −P 1−P

A bad male will definitely have quit when s(t) = 0

This gives a maximum endurance time for a bad male

Maximum endurance time for bad male

T_max = 1 σ ^n

1 P

⎡

⎣⎢

⎤

⎦⎥

q(t) = −s′(t)

s(t) = σ 1−Pe^σt

‘Perfect’ quitting rate

T_max

0 time t

P = 0.2 s = 0.2

Maximum length of memory

Length of memory = T

For equilibrium to be possible the memory cannot be too long

There are no viable equilibria with ^T_max ^< T Viable equilibria require ^T_max ^≥T

T

0 t

bad male has definitely quit

female can safely mate

t_B^∗

T_max

‘Completing’ a perfect memory

q(t) = σ

1−e^σtP ^for t≤T

q(t) = f(t−T) for t>T with f(t) a positive function defined for t  0

T_max T

q(t) s(t)

F() F() is monotonically decreasing and is minimized at  = 

Female’s best response is to mate immediately

s(ˆ λ) = 1 1−P

1

λ +σ ⁽¹⁻e^−(λ+σ)T)−P

λ ⁽¹⁻e^−λT)

⎧⎨

⎩

⎫⎬

⎭+e^−λTs(T)sˆ₁(λ)

where is the Laplace transform of s^ˆ₁(λ)

s₁(t) = exp − f(τ)τ

0

∫

{ }

0 < λs^ˆ(λ) = λe^−λts(t)t

0

∫

∫

sˆ₀(λ) < sˆ(λ) < sˆ₀(λ) + e^−λTs(T) 1 λ

sˆ₀(λ) = 1 1−P

1

λ +σ ⁽¹⁻e^−(λ+σ)T)−P

λ ⁽¹⁻e^−λT)

⎧⎨

⎩

⎫⎬

⎭ sˆ(λ) = s(t)e^−λtt

0

∫

∫

Bounds for F(  )

F₀(λ) = 1

1−P (1−e^−(λ+σ)T)−P(1−e^−λT) + Pσ λ e^−λT

⎧⎨

⎩

⎫⎬

⎭ F₀(λ) < F(λ) < F₀(λ) + e^−λTs(T) 1+ σ

λ

⎛

⎝⎜

⎞

⎠⎟

lower bound upper bound

mating rate  0

Minimum of F

(  )

This occurs at

λ^∗(T) = 2

T −1 + 1 + 4 e^{−σ T} −P Pσ T

⎧⎨

⎪

⎩⎪

⎫⎬

⎪

⎭⎪

0 T₀^∗ T₁^∗ T_max

mating rate

memory length T

equilibrium

mating frequency λ^∗= x D_m

 > ^*

Bad male wants to decrease his quitting rate

 < ^*

Bad male wants to increase his quitting rate

λ^∗

T^∗

T λ

best response curve ^*(T) probability that bad male quits during the perfect memory phase

Q(T)= Q′(T) > 0

best response curve ^*(T)  > ^*

Bad male wants to decrease his quitting rate

 < ^*

Bad male wants to increase his quitting rate

T^∗

T λ

λ^∗

probability that bad male quits during the perfect memory phase Q(T)=

Q′(T) > 0

&

λ = (λ^∗(T) −λ )B_F(λ,T)

&

T = (λ^∗−λ)B_M(λ,T)

J(λ^∗,T^∗) = −B_F^∗ λ^∗′(T^∗)B_F^∗

−B_M^∗ 0

⎛

⎝⎜ ⎞

⎠⎟

Conclusions

• There are extended courtship equilibria in which participants can condition their behaviour on time

• There are no equilibria in pure strategies

• In any such equilibrium neither the female nor a good male quits, and the game ends in mating

• The female’s equilibrium strategy is a constant mating rate

• There is a ‘perfect’ memory equilibrium in which the female is indifferent between her (pure) mating strategies (constant mating rates)

• In this equilibrium a bad male will quit for sure in a finite time

• There is a stable equilibrium in which a bad male follows the perfect memory quitting strategy for a finite time, and then adopts some other (possibly memoryless) strategy

• There is a high probability that a bad male will quit before the female mates during the perfect memory phase

Female indifference between pure strategies

E^F(t) = Expected payoff to female from the pure strategy: mate at time t E^F(λ) = Expected payoff (at time t = 0) to female from the mixed strategy 

E^F(λ) = E^F(t)e^−λtλt

0

∫

If the female is indifferent between all her pure strategies (mating times) then

E^F(t)= Π a constant (independent of t)

Hence

E^F(λ) = Π e^−λtλt

0

∫

is constant, independent of .

That is, the female is indifferent between all her mixed strategies .

Conversely

E^F(λ) = E^F(t)e^−λtλt

0

∫∞ ^{= λEˆF(λ)}

where is the Laplace transform of Eˆ^F(λ) E^F(t)

Hence, if E^F() = P, a constant (independent of ), then

Eˆ^F(λ) = Π λ

Therefore, taking inverse Laplace transforms E^F(t)= Π

is constant, independent of t.

That is, the female is indifferent between all her pure strategies

Q(T,λ) = e^−λts(t)q(t)t

0

∫T ^{= −} 0 ^{e−λts′(t)t}

∫T

= σ

1 − P e^{−λ t}e^{−σ t}dt

0

∫T ⁼ _{1 −}¹_P^⎛_⎝⎜λ ^σ_+σ ^⎞_⎠⎟^{(1 −e−(λ+σ)T )}

∂Q

∂T = σ

1 − P e^{−(λ+σ )T} > 0  Q is an increasing function of T

∂Q

∂λ = σ 1 − P

1 λ +σ

⎛

⎝⎜

⎞

⎠⎟

2

−1 + 1 + (λ( + σ )T )e^−(λ+σ)T

{ } = σ

1 − P

1 λ + σ

⎛

⎝⎜

⎞

⎠⎟

2

h(λ + σ )

 Q is a decreasing function of 

h(q)= −1 + (1 +Tq)e^−qT

′(q) ={−T(1 +Tq) +T}e^−qT = −T²qe^−qT < 0

(0) = 0 ⇒ (q) < 0 for q< 0

⎫

⎬⎪

⎭⎪

Probability that bad male quits during the perfect memory phase

Pure strategies in the memory game

Male pure strategy: quitting time t_G or t_B Female pure strategy: mating time t_m

t_G t_B

Π₀ > 0

t_m

good male has quit

t_m

any male has quit

t_m

no male has quit

t

In all cases the female does better to mate immediately 0 < t_G  t_B