Costly signalling theory and dishonest signalling

Abstract

We analyze the model of costly signalling theory and show that dishonest signalling is still a possible outcome even for costly indices that cannot be faked. We assume that signallers pay the cost for sending a signal and that the cost correlates negatively with signaller’s quality q and correlates positively with signal’s strength s. We show that for any given function f with continuous derivative, there is a cost function t(s, q) increasing in s and decreasing in q so that when the signaller of quality q optimizes the strength of the signal, it will send the signal of strength f(q). In particular, optimal signals can follow any given function f. Our results can explain the curvilinear relationship between the strength of signals and physical condition of three-spined stickleback (Gasterosteus aculeatus).

Appendices

Appendix A: Proof of the result for a Lipschitz function

Here, we show that for any Lipschitz function f : [0, 1] → (0, 1), there is a cost function t(s, q) satisfying Eqs. 1–3 such that the optimal signal strength for a sender of quality q is given by s^∗(q) = f(q).

Let L be a Lipschitz constant of f, i.e., |f(x) − f(y)|≤ L|x − y| for all x, y ∈ [0, 1]. Since f is continuous, it attains maximum and minimum and thus there is a ∈ (0, 1) such that a ≤ f(q) ≤ 1 − a for each q ∈ [0, 1].

Let h: (0, 1] → (0, 1) be an increasing function satisfying $\max \{q-\frac {a}{2L},\frac {q}{2}\}<h(q)<q$ for each q ∈ (0, 1]. For example, we can take a function

$$ h(q) = \frac12\left( q+\max\left\{q-\frac a{2L},\frac{q}{2}\right\}\right), $$

(10)

(see Fig. 3).

Fig. 3

a Graph of f(q) = q² − q + 0.6. We get $L = \max _{q\in [0,1]} |f^{\prime }(q)| = \max _{q\in [0,1]}| 2q-1| = 1$ and since f(q) ∈ [0.35,0.6], we can set a = 0.35. b Graph of h(q) given by Eq. 10. c Graph of g(q) given by Eq. 11

Further, for q ∈ (0, 1], put

$$ g(q)=\frac{a}{2}\cdot\frac{h(q)(2-h(q))}{2(1-h(q))} $$

(11)

and note that g is increasing, since $x\mapsto \frac {x(2-x)}{1-x}$ is increasing on (−∞, 1), see Fig. 3.

For each s, q ∈ [0, 1] we define two auxiliary functions

$$ \begin{array}{@{}rcl@{}} \alpha(s,q)&=&h(q)+\frac{q-h(q)}{f(q)-\frac{a}{2}}\left( s-\frac{a}{2}\right) \end{array} $$

(12)

$$ \begin{array}{@{}rcl@{}} &=&q+\frac{q-h(q)}{f(q)-\frac{a}{2}}\left( s-f(q)\right) \end{array} $$

(13)

$$ \begin{array}{@{}rcl@{}} \beta(s,q)&=&q+\frac{q-h(q)}{1-\frac{a}{2}-f(q)}\left( f(q)-s\right) \end{array} $$

(14)

$$ \begin{array}{@{}rcl@{}} &=&h(q)+\frac{q-h(q)}{1-\frac{a}{2}-f(q)}\left( 1-\frac{a}{2}-s\right) \end{array} $$

(15)

Note that α(s, q) is increasing in s and β(s, q) is decreasing in s. Both functions are affine in s. We will show further below that both functions are increasing in q.

Also, for each s ∈ [0, 1] and q ∈ (0, 1] we define a function u(s, q) in the following way:

$$ u(s,q)=\left\{\begin{array}{ll} \frac{2}{a}\left( s + g(q) - \sqrt{s^{2} + g(q)^{2}} \right)&\quad\text{for } s\in\left[0,\frac{a}{2}\right],\\ \alpha(s,q) &\quad\text{for } s\in\left[\frac{a}{2},f(q)\right],\\ \beta(s,q) &\quad\text{for } s\in\left[f(q),1 - \frac{a}{2}\right],\\ \frac{2h(q)}a(1 - s)&\quad\text{for } s\in\left[1 - \frac{a}{2},1\right]. \end{array}\right. $$

(16)

It is not too difficult to check that

$$ \begin{array}{@{}rcl@{}} &&\frac{2}{a}\left( \frac{a}{2}+g(q)-\sqrt{\left( \frac{a}{2}\right)^{2}+g(q)^{2}} \right)\\&=&h(q)=\alpha\left( \frac{a}{2},q\right) \end{array} $$

(17)

$$ \begin{array}{@{}rcl@{}} \alpha(f(q),q)&=&q=\beta\left( f(q),q\right) \end{array} $$

(18)

$$ \begin{array}{@{}rcl@{}} \beta\left( 1-\frac{a}{2},q\right) &=& h(q) = \frac{2h(q)}a\left( 1-\left( 1-\frac{a}{2}\right)\right) \end{array} $$

(19)

and so u is a properly defined function that is continuous in s (see Fig. 1) where the re-scaled function $w = \frac {a}{2}u$ is shown.

Note that s↦u(s, q) is affine on each of the intervals $[\frac {a}{2},f(q)]$, $[f(q),1-\frac {a}{2}]$, and $[1-\frac {a}{2},1]$, with $u\left (\frac {a}{2}, q\right ) = h(q)$, u(f(q),q) = q, $u(1-\frac {a}{2},q)=h(q)$, and u(1,q) = 0.

As a last step for each q ∈ (0, 1] we set

$$ \begin{array}{@{}rcl@{}} t(s,q)&=&\frac{a}{2}\frac{u(s,q)}s \text{ for } s\in(0,1] \text{ and} \end{array} $$

(20)

$$ \begin{array}{@{}rcl@{}} t(0,q)& =&\lim\limits_{s\to0+}t(s,q)=1. \end{array} $$

(21)

(see Fig. 1.)

Now, we will show that for each q ∈ (0, 1], the function s↦u(s, q) attains it maximum at s = f(q). Fix q ∈ (0, 1]. It is easily seen that s↦u(s, q) is increasing on $\left [0,\frac {a}{2}\right ]$. As h(q) < q, it readily follows that s↦u(s, q) is increasing on [0,f(q)] and decreasing on [f(q), 1], with attaining the maximum value of q at f(q).

Next, we show that for a fixed s ∈ (0, 1), the function q↦u(s, q) is increasing. For $s\in (0,\frac {a}{2}]$ this follows from the fact that both $x\mapsto s+x-\sqrt {s^{2}+x^{2}}$ and g are increasing. For $s\in \left [1-\frac {a}{2},1\right )$ this is easily seen from the fact that h is increasing. It remains to deal with $s\in \left (\frac {a}{2},1-\frac {a}{2}\right )$. First, notice that if we consider two affine functions γ₁, γ₂ on an interval [c, d], then γ₁ < γ₂ on [c, d] if and only if γ₁(c) < γ₂(c) and γ₁(d) < γ₂(d). Now, fix any q₁,q₂ ∈ (0, 1], q₁ < q₂. In the case that f(q₂) ≥ f(q₁), we have

$$ \alpha\left( \frac{a}{2},q_{1}\right)=h(q_{1})<h(q_{2})=\alpha\left( \frac{a}{2},q_{2}\right) $$

(22)

and

$$ \begin{array}{@{}rcl@{}} \alpha\left( f(q_{2}),q_{1}\right)&=&q_{1}+\frac{q_{1}-h(q_{1})}{f(q_{1})-\frac{a}{2}}\left( f(q_{2})-f(q_{1})\right) \end{array} $$

(23)

$$ \begin{array}{@{}rcl@{}} &\le& q_{1}+\frac{q_{1}-h(q_{1})}{\frac{a}{2}}\left( f(q_{2})-f(q_{1})\right) \end{array} $$

(24)

$$ \begin{array}{@{}rcl@{}} &\le& q_{1}+\frac{q_{1}-h(q_{1})}{\frac{a}{2}}L(q_{2}-q_{1}) \end{array} $$

(25)

$$ \begin{array}{@{}rcl@{}} &<&q_{2}=\alpha\left( f(q_{2}),q_{2}\right), \end{array} $$

(26)

where the last inequality follows from the properties of h. Hence, α(s, q₁) < α(s, q₂) whenever $s\in \left [\frac {a}{2},f(q_{2})\right ]$. Also, β(s, q₁) ≤ q₁ ≤ α(s, q₁) whenever s ≥ f(q₁) and so u(s, q₁) < u(s, q₂) whenever $s\in [\frac {a}{2},f(q_{2})]$. On the other hand,

$$ \begin{array}{@{}rcl@{}} \beta(f(q_{2}),q_{1}) &\le& q_{1}<q_{2}=\beta(f(q_{2}),q_{2}), \text{ and} \end{array} $$

(27)

$$ \begin{array}{@{}rcl@{}} \beta(1-\frac{a}{2},q_{1}) &=&h(q_{1})<h(q_{2})=\beta\left( 1-\frac{a}{2},q_{2}\right) \end{array} $$

(28)

and so u(s, q₁) < u(s, q₂) whenever $s\in \left [f(q_{2}),1-\frac {a}{2}\right ]$. In the case f(q₂) < f(q₁), the reasoning is analogous.

It remains to check that the function t(s, q) satisfies the properties (1)–(3). Since u(1,q) = 0, we get that t(1,q) = 0. By Eq. 21, t(0,q) = 1 and thus Eq. 1 holds. Also, since the function q↦u(s, q) is increasing, t(s, q) satisfies Eq. 3. Finally, it is easily checked separately on each of the intervals $\left [0,\frac {a}{2}\right ]$, $\left [\frac {a}{2},f(q)\right ]$, $\left [f(q),1-\frac {a}{2}\right ]$, and $[1-\frac {a}{2},1]$ that s↦t(s, q) is decreasing. Indeed, note that $s\mapsto \frac {ks+r}s$ is decreasing whenever r > 0. In our situation, this is always the case, since

$$ h(q)-\frac{a}{2}\frac{q-h(q)}{f(q)-\frac{a}{2}}\ge h(q)-(q-h(q))=2h(q)-q>0. $$

(29)

Consequently, t(s, q) satisfies Eq. 2.

We note that if we want our function to satisfy Eq. 1’ instead of Eq. 1, we need to replace $\frac {2h(q)}a(1-s)$ in Eq. 16 by a linear equation L that satisfies L(1 − a/2) = h(q) and L(1) = t_min. Moreover, we can consider the function T(s, q) = t(s, q) ∗ t_max; the optima of w(s, q) = s ∗ t(s, q) and W(s, q) = s ∗ T(s, q) are same and T(0,q) = t_max.

Appendix B: Proof of the result for decreasing function

Let $d(q):[0,1]\to \left [\frac {1}{2}, \frac {3}{4}\right ]$ be a decreasing function. Define

$$ t_{d} (s,q)=\min \left\{1-\left( 2-\frac{1}{d(q)}\right)s,2(1-s)\right\} $$

(30)

We will show that the optimal signal strength s^∗(q) for this cost function is given by d(q).

Since $d(q)\geq \frac {1}{2}$, it follows from Eq. 30 that

$$ t_{d} (s,q)= \left\{\begin{array}{lll} 1-\left( 2-\frac{1}{d(q)}\right)s, &\text{if } s\leq d(q),\\ 2(1-s), &\text{if } s\geq d(q). \end{array}\right. $$

(31)

When we set w_d(s, q) = st_d(s, q) and differentiate with respect to s, we get

$$ \frac{\partial w_{d}}{\partial s}= \left\{\begin{array}{lll} 1-\frac{4d(q)-2}{d(q)} s, & \text{ if } s< d(q),\\ (2-4s), & \text{ if }s> d(q). \end{array}\right. $$

(32)

Consequently,

$$ \frac{\partial w_{d}}{\partial s} \left\{\begin{array}{lll} >0, & \text{ for } s<\min\left\{d(q),\frac{d(q)}{4d(q)-2}\right\},\\ <0, & \text{ for } \frac{d(q)}{4d(q)-2} <s<d(q),\\ <0, & \text{ for } s>d(q)\geq \frac{1}{2}. \end{array}\right. $$

(33)

Since $d(q)\leq \frac {d(q)}{4d(q)-2}$ if and only if $d(q)\leq \frac {3}{4}$, we get that the optimal signal size, $s_{d}^{*} (q)$, is given by

$$ s_{d}^{*} (q)= \left\{\begin{array}{lll} d(q), & \text{ if } d(q)\leq 3/4,\\ \frac{d(q)}{4d(q)-2}, & \text{ if } d(q)\geq \frac{3}{4}. \end{array}\right. $$

(34)

Thus, $s_{d}^{*} (q) = d(q)$.

Appendix C: General condition on $\frac {d}{dq}s^{*}(q)>0$

For simplicity, we will assume that that the function t has continuous second derivatives. Because s^∗(q) is such that w(s^∗(q),q) is maximal, we have

$$ \frac{\partial}{\partial s} w(s^{*}(q),q) = 0 $$

(35)

Differentiating (35) with respect to q we get

$$ \begin{array}{@{}rcl@{}} 0 = \frac{d}{dq} \frac{\partial}{\partial s} w(s^{*}(q),q) &=&\frac{\partial^{2}}{\partial s^{2}} w(s^{*}(q),q) \cdot \frac{d}{dq} s^{*}(q)\\ &&+ \frac{\partial^{2}}{\partial s\partial q} w(s^{*}(q),q) \end{array} $$

(36)

If the function t is generic (Broom and Rychtář 2013, p. 21), we may assume that $\frac {\partial ^{2}}{\partial s^{2}} w(s,q) \neq 0$ when $\frac {\partial }{\partial s} w(s,q) = 0$, and thus

$$ \frac{d}{dq} s^{*}(q) = - \frac{\frac{\partial^{2} w}{\partial s\partial q}} { \frac{\partial^{2} w}{\partial s^{2}}} (s^{*}(q),q) $$

Since the maximum is attained at s^∗(q), we have $\frac {\partial ^{2} w}{\partial s^{2}}(s^{*}(q),q)<0$. Consequently,

$$ \begin{array}{@{}rcl@{}} \frac{d}{dq} s^{*}(q) >0 && \text{ if and only if } \frac{\partial^{2} w}{\partial s\partial q} (s^{*}(q),q)>0 \end{array} $$

(37)

$$ \begin{array}{@{}rcl@{}} && \text{ if and only if } \frac{\partial t}{\partial q} (s^{*}(q),q)\\ &&+ s^{*}(q) \frac{\partial^{2} t}{\partial s\partial q} (s^{*}(q),q)>0. \end{array} $$

(38)

We note that condition $\frac {\partial ^{2} w}{\partial s\partial q} (s^{*}(q),q)>0$ depends on the function s^∗(q). Nevertheless, since $\frac {\partial }{\partial s} w(s^{*}(q),q) = 0$, the condition is satisfied if

$$ \frac{\partial^{2} w}{\partial s\partial q} (s,q) > - c \left| \frac{\partial}{\partial s} w(s,q) \right| $$

(39)

for some constant c and all s, q ∈ (0, 1).