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 quitaccept

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: Am

Payoff to bad male from mating: Dm

Am > Dm > 0

Male

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

Payoff to female from mating with a good male: Af > 0

Payoff to female from mating with a bad male: - Cf < 0

Female

φ

Mating immediately

The female’s expected payoff from mating immediately is

Π0 = A f P − C f (1 − P)

Assume P is sufficiently large so that

Π0 > 0

The female gets a positive payoff from mating immediately

The female doesn’t quit first

t0

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

tGtB

Π0 > 0

tm

bad male quits

ttB∗

bad male best

response

tm∗

female best response

Pure strategies

There are no non-trivial equilibria in pure strategies

tG > tB >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

−xδtDm −xt

Suppose the female mates with probability p = t

At equilibrium the female’s mating rate is constant

=x

Dm

Expected payoff from not quitting = Dm p −xt = 0

A good male never quits

quit

not quit

mate not mate

0 0

−xδtAm −xt

At equilibrium

Expected payoff from not quitting = (Am −x)t

=(Am − Dm )λδt when =x

Dm

> 0 since Am > Dm

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 )dτ0

t

∫{ }

s(0) =0s(t) > 0 for all t≥0s(t) is non-increasing in ts(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

EF () =P (Af −φt)e−tdt0

∞

∫ −(1−P) (C f +φt)e−ts(t)dt0

∞

∫ + φte−ts(t)q(t)dt0

∞

∫{ }

payoff

σ =φ

C f

EF () =PAf

C f

⎛

⎝⎜

⎞

⎠⎟−

Pσ

−(1−P)( +σ )s()

s() = Laplace transform of s(t) = e−λ t s(t)dt0

∞

∫

EF () =constant −(1−P)F()

1

C f

EF → EFScaling transformation:

F() =( +σ )s() +Pσ1−P

⎛⎝⎜

⎞⎠⎟1

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 EF()

dEF

d=0Solution * of:

which defines a maximum of EF()

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() = +σ +q

+Pσ1−P

⎛⎝⎜

⎞⎠⎟1

∗(q) =q Pσ

(1 − P)(q − σ ) − Pσ

F()

∗

∗=x

Dm

= equilibrium mating rate

∗(q)Female’s best response curve

Example 2: increasing impatience

q(t) =q+ηt s(t) =e−qt−12ηt2

F()

η = 0.8 q = 1

Example 3: fading memory

q(t) =q−η

1+ ts(t) =(1+ t)ηe−qt

F()

η = 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 tm) in response to a bad male quitting rate q(t)

dEF

d=0 for all > 0 ′F (λ ) = 0

( +σ )s() +Pσ1−P

⎛⎝⎜

⎞⎠⎟1=K s(t) =−

P1−P

+ K +P

1−P⎧⎨⎩

⎫⎬⎭e−σt

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

s(t) =e−σt −P1−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

Tmax =1σ

ln1P

⎡⎣⎢

⎤⎦⎥

q(t) =−′s (t)s(t)

=σ

1−Peσt

‘Perfect’ quitting rate

Tmax0

time t

P = 0.2 = 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 Tmax <T

Viable equilibria require Tmax ≥T

Tt

0

bad male has definitely quit

female can safely mate

tB∗Tmax

‘Completing’ a perfect memory

q(t) =σ

1−eσtPfor t≤T

q(t) = f(t−T ) for t >Twith f(τ) a positive function defined for 0

TmaxT

q(t) s(t)

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

Female’s best response is to mate immediately

s() =1

1−P1

+σ(1−e−(+σ )T )−

P(1−e−T )⎧

⎨⎩

⎫⎬⎭+ e−Ts(T )s1()

where is the Laplace transform of s1()

s1(t) =exp − f(τ )dτ0

t

∫{ }

0 < s() = e−ts(t)dt0

∞

∫ < e−tdt0

∞

∫ =1

s0 () < s() < s0 () + e−Ts(T )1

s0 () =1

1−P1

+σ(1−e−(+σ )T )−

P(1−e−T )⎧

⎨⎩

⎫⎬⎭

s() = s(t)e−tdt0

T

∫ + s(t)e−tdtT

∞

∫

Bounds for F()

F0 () =1

1−P(1−e−(+σ )T )−P(1−e−T ) +

Pσ

e−T⎧⎨⎩

⎫⎬⎭

F0 () < F() < F0 () + e−Ts(T ) 1+σ

⎛⎝⎜

⎞⎠⎟

lower bound

upper bound

mating rate 0

Minimum of F0()

This occurs at

∗(T ) =2

T −1 + 1 + 4e−σ T − P

Pσ T

⎧⎨⎪

⎩⎪

⎫⎬⎪

⎭⎪

0 TmaxT0∗ T1

∗

mat

ing

rate

memory length T

equilibrium mating frequency ∗=

x

Dm

> *

Bad male wants to decrease his quitting rate

< *

Bad male wants to increase his quitting rate

∗

T ∗

T

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

∗

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

′Q (T ) > 0

& = ∗(T ) − λ( ) BF (λ ,T )

&T = (λ ∗ − λ )BM (λ ,T )

J(∗,T∗) = −BF∗ ∗′(T∗)BF

∗

−BM∗ 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

EF (t) = Expected payoff to female from the pure strategy: mate at time t

EF () =Expected payoff (at time t = 0) to female from the mixed strategy

EF () = EF (t)e−tdt0

∞

∫

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

EF (t) =Π a constant (independent of t)

Hence

EF () =Π e−tdt0

∞

∫ =Π

is constant, independent of .

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

Conversely

EF () = EF (t)e−tdt0

∞

∫ =EF ()

where is the Laplace transform of EF () EF (t)

Hence, if EF() = , a constant (independent of ), then

EF () =Π

Therefore, taking inverse Laplace transforms

EF (t) =Π

is constant, independent of t.

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

Q(T ,) = e−ts(t)q(t)dt0

T

∫ =− e−t ′s (t)dt0

T

∫

=σ

1− Pe−λ te−σ tdt

0

T

∫ =1

1− P

σ

λ + σ⎛⎝⎜

⎞⎠⎟

1− e−(λ +σ )T( )

∂Q

∂T=

σ

1 − Pe−(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

′h (q) = −T(1+Tq) +T{ }e−qT =−T 2qe−qT < 0h(0) =0 ⇒ h(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 tG or tB

Female pure strategy: mating time tm

tG tB

Π0 > 0

tm

good male has quit

tm

any male has quit

tm

no male has quit

t

In all cases the female does better to mate immediately

0 < tG tB