Appendix 1. The sequential probability ratio test
The female obtains a sequence X 1 , X 2 , X 3 ,  of observations on male quality. These
observations are assumed to be independent and identically distributed given the male
type. Let f H denote their common probability density function given that the male is
helpful, and let f N denote their common probability density function given the male is
not helpful. Let
 f (X )
Z i  log  H i 
 fN (Xi )
(A1.1)
be the log-likelihood ratio for the ith observation. For each n  1 set
n
W ( n)   Z i .
(A1.2),
i 1
which is the sum of the log-likelihood ratios for all n observations gathered thus far.
Let a and b be positive constants. Then if the female implements the sequential
probability ratio test with absorbing barriers at –a and b she behaves as follows. She
continues to sample while the process W (n) : n  0 remains between the barriers
(i.e.  a  W (n)  b ). Once the process hits one of the barriers she ceases sampling and
decides on male type: if W (n)  a the female decides that the male is not helpful, if
W ( n)  b the female decides the male is helpful. The error probabilities under this
procedure are
α = Prob(process absorbed at - a | male is helpful)
(A1.3)
and
β = Prob(process absorbed at b | male is non-helpful)
(A1.4).
The continuous time limit. From now on we assume that, if the male is helpful, the X i
have a normal distribution with mean  H and variance  2 . If the male is not helpful they
have a normal distribution with mean  N and variance also  2 . Then the log-likelihood
ratio for the ith observation is
Zi 
1
2 2
(  H   N )[ 2 X i  (  N   H )] .
(A1.5)
Then, given the male is helpful, this random variable has mean
E H Z i  
1
2 2
( H   N ) 2
(A1.6)
and variance
Var H Z i   (  H   N ) 2 /  2 .
(A1.7)
2
The analogous quantities when the male is not helpful are
E N Z i   
1
( H   N ) 2
(A1.8)
Var N Z i   (  H   N ) 2 /  2 .
(A1.9)
2 2
and
We now suppose that there are n observation per unit time. When time t is multiple of
1 / n we can set Y (t )  W (nt ) . This random variable Y summarizes the useful information
up until time t. Define  and  2 by
n
( H   N ) 2
2 2
(A1.10)
and
2 n
( H   N ) 2
2
.
(A1.11)
Then
E H Yt   t
and
Var H Yt    2 t .
(A1.12)
3
Similarly
E N Yt   t and
Var N Yt    2 t .
(A1.13)
We now let n tend to infinity, allowing (  H   N ) 2 /  2 to tend to zero so that the
product of these two terms, v, remains constant. In this limit the process Yt : t  0 tends
to a diffusion process with drift   v / 2 and variance σ2= v. We assume that Yt : t  0
is such a diffusion in what follows.
Error probabilities. Suppose that the process Yt : t  0 has absorbing barriers at –a and
b. If the male is helpful the process is a diffusion process with drift  and diffusion
coefficient equal to . Thus by standard results, the probability of being absorbed at the
barrier at –a is

A  AB
1  AB
(A1.14)
where
A  exp{2a /  2 }  exp{a}
(A1.15)
and
4
B  exp{2b /  2 }  exp{b}
(A1.16)
(e.g. Taylor & Karlin 1998). Similarly

B  AB
.
1  AB
(A1.17)
We can also use these relationships to express A and B in terms of the error probabilities
as
A

1 
(A1.18)
and
B

.
1
(A1.19)
Mean times. The mean time to be absorbed at one of the boundaries can be expressed as
H 
b(1  A)  a(1  B) A
 (1  AB)
(A1.20)
(Taylor & Karlin 1998). Thus by equations (A1.15), (A1.16), (A1.18) and (A1.19)
5
2
v
1  
  
   log 
 .
  
 1   
(A1.21)
2
1  
  
(1   ) log 
   log 
 .

v
  
 1   
(A1.22)
 H  (1   ) log 
Similarly
N 
Taylor, H.M. & Karlin S. 1998 An introduction to stochastic modelling. San Diego, CA
Academic Press.
Appendix 2. Results when females cannot gain information
We consider first the special case when females are unable to gain information on male
type. In this case females do not sample, but just mate with all males encountered. Thus
equations (1) – (8) hold with H = N = 0,  = 0 and  = 1. As we now show, even in this
restricted case both male types can be maintained in the population at evolutionary
stability.
Consider a population in which the proportion of helpful males is p and the proportion of
non-helpful males is 1 – p. In this population the mean time for a male to encounter a
female, 1 /( F ) , is determined by equations (1) – (5). To emphasise the dependence on
6
p we denote this mean time by T(p). Note that this mean time is the same for both types
of male. Before finding the ESS we first investigate dependence of T(p) on p.
Females encounter males at rate s M , where M is the proportion of males that are
searching. Since females mate on encounter, and mating and caring takes unit time, the
proportion of time that females spend searching is
F 
1 / s M
1

.
1  (1 / s M )
1  s M
(A2.1)
A searching male takes mean time
T ( p) 
1
 F
(A2.2)
to encounter a female. Thus by equations (A2.1) and (A2.2)
T ( p)  1  sM .
(A2.3)
Non-helpful males spend all of their time searching. Thus the proportion of males
searching is
 M  (1  p)  p H
(A2.4)
7
where  H is the proportion of time that helpful males spend searching. This proportion is
given by
H 
1 /  F
1  (1 /  F )
.
(A2.5)
Thus by equations (A2.2), (A2.4) and (A2.5)
M  1 
p
.
T ( p)  1
(A2.6)
From equations (A2.3) and (A2.6) T(p) is the positive root of the quadratic equation
T ( p) 2  (1  1  s)T ( p)  (1  s(1  p))  0 ,
(A2.7)
and hence is given by
T ( p) 
(1  1  s) 2  4 ps  (1  1  s)
.
2
(A2.8)
Thus
dT
s

.
dp
(1  1  s ) 2  4 ps
(A2.9)
8
This shows that T decreases as p increases. The intuitive reason for this is as follows. As
the proportion of males that care increases there is a decrease in the proportion of males
that are searching at any time. Thus it takes longer for females to find mates, so that the
number of females that are searching increases. This means that it is easier for any single
male to find a mate, so that search times are decreased.
We now examine which type of male does best in this population, and how this depends
on p. By equations (6) and (7) the reward rates of the two types of male are
 H ( p) 
BH
1  T ( p)
and
 N ( p) 
BN
.
T ( p)
Thus non-helpful males have a higher reward rate if and only if
BN
 R( p) ,
BH
where
9
T ( p)
.
1  T ( p)
R( p) 
Since T(p) is a decreasing function of p , so is R(p). There are thus three possible cases.
Case 1:
BN
 R(0)
BH
In this case we have
BN
 R( p)
BH
for all p
so that non-helpful males always have higher fitness. Thus p* = 0 is the only ESS.
Case 2.
R(1) 
BN
 R(0) .
BH
In this case helpful males do better when rare and can invade. Similarly, non-helpful
males do better when they are rare and can invade. There is a unique p* such that R(p) >
10
BN/BH for p < p* and R(p) < BN/BH for p > p*. Thus the population will always evolve to
the unique equilibrium p*
Case 3.
BN
 R(1)
BH
In this case
BN
 R( p )
BH
for all p.
and the population evolves to p* = 1.
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