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![Page 1: Duration of courtship effort with memory Robert M Seymour Department of Mathematics & Department of Genetics, Evolution and Environment UCL.](https://reader030.fdocuments.in/reader030/viewer/2022033108/56649e2d5503460f94b1d4e0/html5/thumbnails/1.jpg)
Duration of courtship effort with memory
Robert M Seymour
Department of Mathematics
&
Department of Genetics, Evolution and Environment
UCL
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Peter Sozou
LSE
Acknowledgement to
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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?
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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)
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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
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Arthropods : Hanging fly (Bittacus apicalis)
Thornhill, R.(1976) Am. Nat 110, no. 974, 529-548
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Human courtship can involve a long sequence of outings, gifts….
And …
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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
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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()
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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()
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∗
∗=x
Dm
= equilibrium mating rate
∗(q)Female’s best response curve
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Example 2: increasing impatience
q(t) =q+ηt s(t) =e−qt−12ηt2
F()
η = 0.8 q = 1
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Example 3: fading memory
q(t) =q−η
1+ ts(t) =(1+ t)ηe−qt
F()
η = 0.8 q = 1
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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
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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
⎡⎣⎢
⎤⎦⎥
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q(t) =−′s (t)s(t)
=σ
1−Peσt
‘Perfect’ quitting rate
Tmax0
time t
P = 0.2 = 0.2
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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
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‘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
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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
∞
∫
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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
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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
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> *
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
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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
⎛
⎝⎜
⎞
⎠⎟
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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
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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 .
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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
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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
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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