Market concentration and persuasive advertising: a theoretical approach

Abstract

This paper examines the structural relation between persuasive advertising intensity and market concentration. The interaction of advertising costs with the consumer’s willingness to pay shapes the way markets respond to changes in sunk cost structures. This adjustment may involve firm entry and exit or modifications in advertising levels. It is shown that a non-monotonic association between advertising intensity and concentration may emerge even in the absence of collusion, requiring as a necessary condition that the ratio of operational profits and advertising cost elasticities with respect to a measure of perceived quality be decreasing. This result is robust to changes in both exogenous and endogenous sunk cost parameters. A simple tool is also proposed to empirically assess the behavior of the elasticities ratio. Finally, the model describes how intertemporal general equilibrium mechanisms may skew or even reverse the advertising-concentration relation through scale effects.

Appendix

1.1 Existence and uniqueness of advertising solution

Equation 20 can be rearranged as

$$\begin{aligned} \frac{F_{Y}}{\beta }=-\left\{ \frac{\varphi ^{\prime }(z) [1-\theta (z)]}{\theta ^{\prime }(z)}+\varphi (z)\right\} . \end{aligned}$$

(37)

Denoting the right hand side of this Eq. 37 as $\chi$,

$$\begin{aligned} \frac{d\chi }{dz}=\frac{-\varphi ^{\prime }(z)[1-\theta (z)] \left[ \frac{\varphi ^{\prime \prime }(z) \theta ^{\prime }(z)}{\varphi ^{\prime }(z) }-\theta ^{\prime \prime }(z)\right] }{[\theta ^{\prime }(z)]^{2}}. \end{aligned}$$

(38)

Given that $z\in [0,\infty )$, $\theta (z)\in [0,1]$ and $\theta ^{\prime }(z)<0$, then $\theta (0)=1$ and $\theta (\infty )=0$. Under the sufficient conditions $\theta ^{\prime \prime }(z)>0$ and $\varphi ^{\prime \prime }(z)\ge 0$, added to assumptions $\varphi ^{\prime }(z)>0$ and $\theta ^{\prime }(z)<0$, it follows that $\frac{d\chi }{dz}>0,\forall z>0$.

Next, recall that $\varphi (0)=0$. Hence, $\chi \vert _{z=0}=0$. Given that $\frac{d\chi }{dz}>0,\forall z>0$, it follows that $\chi >0,\forall z>0$. Finally, the fact that $\theta (z)$ converges monotonically and asymptotically to a finite value as $z\rightarrow \infty$ implies that $\lim _{z\rightarrow \infty } \theta ^{\prime }(z)=0^{-}$. Hence, $\chi \vert _{z\rightarrow \infty }=\infty$. In conclusion, $\chi$ ranges monotonically from $0$ to $\infty$, ensuring that for any pair $(F_{Y},\beta )$ of cost parameters there is a unique solution for $z$.

1.2 Proof of Proposition 1

As shown in the previous proof, $\chi$ is increasing in $z$. Hence, as $F_{Y}$ increases in the left hand side of Eq. 37, the same must happen with $z$ on the right hand side. In other words, $\frac{\partial z}{\partial F_{Y}}>0$.

Regarding the relation between $F_{Y}$ and $N$, Eq. 29 yields

$$\begin{aligned} \frac{\partial N}{\partial F_{Y}}=\frac{\beta L(\Lambda _{1}-\Lambda _{2})}{\{\beta \varphi ^{\prime }(z)[ 1-\theta (z)(1-\alpha )\alpha ]\}^{2}}\frac{\partial z}{\partial F_{Y}}, \end{aligned}$$

(39)

where

$$\begin{aligned} \Lambda _{1}=[1-\theta (z)(1-\alpha ) \alpha ][\varphi ^{\prime \prime }(z) \theta ^{\prime }(z)-\theta ^{\prime \prime }( z)\varphi ^{\prime }(z)] \end{aligned}$$

(40)

and

$$\begin{aligned} \Lambda _{2}=[\theta ^{\prime }(z)]^{2}( 1-\alpha ) \alpha \varphi ^{\prime }(z). \end{aligned}$$

(41)

Given that $\frac{\partial z}{\partial F_{Y}}>0$, $\theta (z) \in [0,1]$, $\theta ^{\prime }(z)<0$, $\varphi ^{\prime }(z)>0$, $\theta ^{\prime \prime }(z)>0$ and $\varphi ^{\prime \prime }(z)\ge 0$, it follows that $\Lambda _{1}<0$, $\Lambda _{2}>0$ and $\frac{\partial N}{\partial F_{Y} }<0,\forall F_{Y}>0$.

1.3 Proof of Proposition 2

Given the behavior of $\chi$, as $\beta$ increases in the left hand side of Eq. 37, $z$ must decrease on the right hand side. In other words, $\frac{\partial z}{\partial \beta }<0$.

Regarding the relation between $\beta$ and $N$, Eq. 29 may be combined with 20 to present the equilibrium number of firms as

$$\begin{aligned} N=[F_{Y}+\beta \varphi (z)]^{-1}\frac{[1-\theta (z)]L}{[ 1-\theta (z)(1-\alpha )\alpha ]}. \end{aligned}$$

(42)

From here,

$$\begin{aligned} \frac{\partial N}{\partial \beta }=\Gamma _{1}+\Gamma _{2}, \end{aligned}$$

(43)

where

$$\begin{aligned} \Gamma _{1}=-\frac{\partial [\beta \varphi (z)] }{\partial \beta }\frac{[1-\theta (z)] L}{[F_{Y}+\beta \varphi (z)]^{2}[ 1-\theta (z)(1-\alpha ) \alpha ]} \end{aligned}$$

(44)

and

$$\begin{aligned} \Gamma _{2}=[F_{Y}+\beta \varphi (z)]^{-1} \frac{\theta ^{\prime }(z)[(1-\alpha ) \alpha -1]L}{[1-\theta (z)(1-\alpha )\alpha ]^{2}}\frac{\partial z}{\partial \beta }. \end{aligned}$$

(45)

Given that $\frac{\partial z}{\partial \beta }<0$ and $\theta ^{\prime }(z)<0$, $\Gamma _{2}<0,\forall \beta >0$. Accordingly, $\frac{\partial [\beta \varphi (z)] }{\partial \beta }>0$ is a sufficient condition for $\frac{\partial N}{\partial \beta }<0$. Expanding on $\frac{\partial [\beta \varphi (z)]}{\partial \beta }$ and using again Eq. 20, further manipulation yields

$$\begin{aligned} \frac{\partial N}{\partial \beta }=\Gamma _{3}+\Gamma _{4}, \end{aligned}$$

(46)

where

$$\begin{aligned} \Gamma _{3}=-\frac{\varphi (z)L}{[1-\theta (z)(1-\alpha ) \alpha ][1-\theta (z)]}\left[ \frac{\theta ^{\prime }(z)}{\beta \varphi ^{\prime }(z)}\right] ^{2} \end{aligned}$$

(47)

and

$$\begin{aligned} \Gamma _{4}=-\frac{L(1-\alpha )\alpha [ \theta ^{\prime }(z)]^{2}}{\beta \varphi ^{\prime }(z)[ 1-\theta (z)(1-\alpha )\alpha ]^{2}}\frac{\partial z}{\partial \beta }. \end{aligned}$$

(48)

It is straightforward to show that under partial equilibrium, setting $\alpha =0$, it is always the case that $\frac{\partial N}{\partial \beta }<0$.

Finally, notice that the equilibrium condition 20 may be rearranged as

$$\begin{aligned} \frac{F_{Y}}{\beta \varphi (z)}=-\left\{ \frac{\varphi ^{\prime }(z)[1-\theta (z)]}{\theta ^{\prime }(z) \varphi (z)}+1\right\} . \end{aligned}$$

(49)

Hence, the sufficient condition $\frac{\partial [ \beta \varphi (z)]}{\partial \beta }>0$ implies

$$\begin{aligned} \frac{d\left\{ -\frac{\varphi ^{\prime }(z)[1-\theta (z)] }{\theta ^{\prime }(z)\varphi (z)}\right\} }{d\beta }<0 \Leftrightarrow \frac{d\left\{ -\frac{\theta ^{\prime }(z) \varphi (z)}{\varphi ^{\prime }(z)[ 1-\theta (z)]}\right\} }{d\beta }>0. \end{aligned}$$

(50)

Finally, notice that

$$\begin{aligned} -\frac{\theta ^{\prime }(z)\varphi (z)}{\varphi ^{\prime }(z)[1-\theta (z)]}\equiv \frac{-\frac{d[ \theta (z)]}{dz}\frac{z}{1-\theta (z) }}{\frac{d[\varphi (z)] }{dz}\frac{z}{\varphi (z)}}\equiv \frac{ \varepsilon _{\pi ,z}}{\varepsilon _{\varphi ,z}}, \end{aligned}$$

(51)

where $\varepsilon _{\pi ,z}$ is the elasticity of operational profits with respect to perceived quality and $\varepsilon _{\varphi ,z}$ is the elasticity of advertising costs (conditional on a given $\beta$) with respect to perceived quality.

1.4 Proof of Lemma 1

Given $\varphi (z)=z^{\gamma }$ and $\theta (z)=(a+z)^{-1}a$, Eq. 49 becomes

$$\begin{aligned} \frac{F_{Y}}{\beta \varphi (z)}=\frac{\gamma (a+z)}{a}-1. \end{aligned}$$

(52)

This is increasing in $z$. Since $\frac{\partial z}{\partial \beta }<0$, it follows that $\frac{\partial [ \beta \varphi (z)]}{\partial \beta }>0$.

Given $\varphi (z)=z^{\gamma }$ and $\theta (z)=e^{-bz}$, Eq. 49 becomes

$$\begin{aligned} \frac{F_{Y}}{\beta \varphi (z)}=\frac{\gamma (e^{bz}-1)}{bz}-1. \end{aligned}$$

(53)

Denoting the second hand side of 53 by $\Upsilon _{1}$,

$$\begin{aligned} \frac{d\Upsilon _{1}}{dz}=\frac{\gamma b}{(bz)^{2}}[ e^{bz}(bz-1)+1]. \end{aligned}$$

(54)

Notice that $[e^{bz}(bz-1)+1]$ is monotonically increasing in $z$ and this function takes the value $0$ when $z=0$. Hence, $\frac{d\Upsilon _{1}}{dz}>0,\forall z>0$. As in the previous case, this once again implies $\frac{\partial [\beta \varphi (z)]}{\partial \beta }>0$.

Finally, given $\varphi (z)=z^{\gamma }$ and $\theta (z)=[ \ln (1+z)+1]^{-c}$, Eq. 49 becomes

$$\begin{aligned} \frac{F_{Y}}{\beta \varphi (z)}=\frac{[\ln (1+z)+1] \{[\ln (1+z) +1]^{c}-1\}(1+z) \gamma }{cz}-1. \end{aligned}$$

(55)

Denoting the second hand side of 55 by $\Upsilon _{2}$,

$$\begin{aligned} \frac{d\Upsilon _{2}}{dz}=\frac{\gamma }{cz^{2}}( \Gamma _{5}+\Gamma _{6}), \end{aligned}$$

(56)

where

$$\begin{aligned} \Gamma _{5}=z\{[\ln (1+z) +1]^{c}-1\}-[\ln (1+z)]\{ [\ln (1+z)+1]^{c}-1\} \end{aligned}$$

(57)

and

$$\begin{aligned} \Gamma _{6}=(cz-1)[\ln (1+z)+1]^{c}+1. \end{aligned}$$

(58)

Notice that $z>\ln (1+z),\forall z>0$. This ensures that $\Gamma _{5}>0$. Next,

$$\begin{aligned} \frac{\partial \Gamma _{6}}{\partial z}=c[\ln (1+z)+1]^{c}\left\{ 1+\frac{cz-1}{(1+z)[\ln (1+z)+1]}\right\} . \end{aligned}$$

(59)

Since $(1+z)[\ln (1+z)+1]>1>cz-1,\forall z\in (0,c^{-1}]$ and $cz\ge 0,\forall z\ge c^{-1}$, it follows that $\frac{\partial \Gamma _{6}}{\partial z}>0,\forall z>0$. Finally, this condition may be combined with the fact that $\Gamma _{6}\vert _{z=0}=0$ to obtain $\Gamma _{6}>0,\forall z>0$. In conclusion, $\frac{d\Upsilon _{2}}{dz}>0,\forall z>0,$ implying once again $\frac{\partial [\beta \varphi (z)] }{\partial \beta }> 0$.

1.5 Proof of Lemma 2

Given $\varphi (z)=z^{\gamma }$ and $\theta (z)=(a+z)^{-1}a$, Eq. 33 becomes

$$\begin{aligned} \psi =\frac{az}{\gamma (a+z)^{2}}. \end{aligned}$$

(60)

The impact of changes in fixed operational costs is

$$\begin{aligned} \frac{\partial \psi }{\partial F_{Y}}=\frac{a[a-z(F_{Y},\beta )]}{\gamma [a+z(F_{Y},\beta )]^{3}}\frac{\partial z(F_{Y},\beta )}{\partial F_{Y}}. \end{aligned}$$

(61)

Since $\frac{\partial z(\cdot )}{\partial F_{Y}}>0$, Eq. 61 is positive for $z(F_{Y},\beta )<a$. This occurs for low levels of $F_{Y}$ and concentration. With higher levels of $F_{Y}$ and concentration, perceived quality monotonically increases up to where $z(F_{Y},\beta )>a$, making Eq. 61 negative. An inverted U-shape relation arises in this way.

Changes in marginal advertising costs yield similar conclusions, using

$$\begin{aligned} \frac{\partial \psi }{\partial \beta }=\frac{a[a-z(F_{Y},\beta )] }{\gamma [a+z(F_{Y},\beta )]^{3}}\frac{\partial z(F_{Y},\beta )}{\partial \beta }. \end{aligned}$$

(62)

In this case, $\frac{\partial z(\cdot ) }{\partial \beta }<0$. Equation 62 is then positive for $z(F_{Y},\beta )>a$, which occurs for low levels of $\beta$ and concentration. When $z(F_{Y},\beta )<a$ (that is, with a higher $\beta$ and concentration) the derivative becomes negative. Once again, an inverted U-shape relation arises.

Given $\varphi (z)=z^{\gamma }$ and $\theta (z)=e^{-bz}$, Eq. 33 becomes

$$\begin{aligned} \psi =\frac{b}{\gamma }e^{-bz}z. \end{aligned}$$

(63)

The impact of changes in fixed operational costs is now

$$\begin{aligned} \frac{\partial \psi }{\partial F_{Y}}=\frac{b}{\gamma }e^{-bz( F_{Y},\beta )}[1-bz(F_{Y},\beta )] \frac{\partial z(F_{Y},\beta )}{\partial F_{Y}}. \end{aligned}$$

(64)

The proof is identical to the previous case. The same applies when $\beta$ changes.

Finally, given $\varphi (z)=z^{\gamma }$ and $\theta (z)=[\ln (1+z)+1]^{-c}$, Eq. 33 becomes

$$\begin{aligned} \psi =\frac{c[\ln (1+z)+1]^{-c-1}z}{\gamma (1+z)}. \end{aligned}$$

(65)

The impact of changes in fixed operational costs is

$$\begin{aligned} \frac{\partial \psi }{\partial F_{Y}}=\frac{c\{\ln [ 1+z(F_{Y},\beta )]+1-(c+1)z(F_{Y},\beta )\}}{\gamma [1+z( F_{Y},\beta )]^{2}\{\ln [1+z(F_{Y},\beta )]+1\}^{c+2}}\frac{\partial z(F_{Y},\beta )}{\partial F_{Y}}. \end{aligned}$$

(66)

The sign of this derivative depends on the sign of

$$\begin{aligned} \Gamma _{7}=\ln [1+z(F_{Y},\beta )] +1-(c+1)z(F_{Y},\beta ). \end{aligned}$$

(67)

$\Gamma _{7}$ is decreasing on $z$. In addition, $\Gamma _{7}\vert _{z=0}>0$. Next, the first term in $\Gamma _{7}$ increases at a decreasing rate, whereas the last one increases linearly in absolute value. Hence, $\exists z^{*}$ $s.t.$ $\Gamma _{7}<0,\forall z>z^{*}$. The proof for the behavior of $\frac{\partial \psi }{\partial F_{Y}}$ and $\frac{\partial \psi }{\partial \beta }$ follows along the same lines of the previous cases.

1.6 Proof of Proposition 4

Combining Eqs. 29 and 33 yields

$$\begin{aligned} \frac{1}{N}=\frac{\beta \varphi (z)[ 1-\theta (z)(1-\alpha )\alpha ] }{L\psi (z)}. \end{aligned}$$

(68)

Advertising intensity $(\psi )$ or perceived quality $(z)$ do not depend on general equilibrium effects or the sensitivity of rates of return, that is, $(1-\alpha )\alpha$. However, as shown in Eq. 68, the higher this sensitivity is, the lower the concentration level associated with each value of $\psi$.