Comparative statics and heterogeneity

Abstract

This paper elucidates the role played by the heterogeneity of interactions between the endogenous variables of a model in determining the model’s behavior. It is known that comparative statics are well-behaved if these interactions are relatively small, but the formal condition imposed on the Jacobian which typically captures this idea–diagonal dominance–ignores the distribution of the interaction terms. I provide a new condition on the Jacobian—mean positive dominance—which better captures a trade-off between the size and heterogeneity of interaction terms. In accord with Samuelson’s (Foundations of economic analysis, Oxford University Press, London, 1947) correspondence principle, I also show that mean positive dominance yields stability and uniqueness results. I apply the results to provide new, or to generalize known, comparative statics results in the following settings: optimization problems, platform monopoly, normality, differentiable games including Cournot oligopoly, and competitive exchange economies.

Appendix

1.1 Proof of Lemma 1

(1) Since $y=A^{-1}b$ we have for each i,  $y_{i}=\sum _{j=1}^{n}b_{j}\delta _{ij}.$ It follows that

$$\begin{aligned} \sum _{i=1}^{n}y_{i}=\sum _{i=1}^{n}\sum _{j=1}^{n}b_{j}\delta _{ij}=\sum _{j=1}^{n}\left( \sum _{i=1}^{n}\delta _{ij}\right) b_{j}\ge 0, \end{aligned}$$

where the last inequality is valid for all $b\ge 0$ iff $A\in {\mathscr {N}}_{cs}^{-1}.$

(2) This follows from the fact that $y_{i}=\sum _{j=1}^{n}b_{j}\delta _{ij}=b_{i}\delta _{ii}$ when b is positive only in element i.

(3) In this case $b_{i}={\bar{b}}\ge 0$ for all i,  so $y_{i}= \sum _{j=1}^{n}b_{j}\delta _{ij}={\bar{b}}\left( \sum _{j=1}^{n}\delta _{ij}\right) \ge 0,$ where the last inequality is valid for all ${\bar{b}}>0$ iff $A\in {\mathscr {N}}_{rs}^{-1}.$

1.2 Proof of Lemma 3

(1) This follows from the facts that a B-matrix has a strictly positive determinant, the determinant is a continuous function of the matrix entries, and weak inequalities are preserved in the limit.

(2) The proof of this claim is a slight modification of the proof of Proposition 2.5 in Peña (2001). Consider the set of natural numbers $M=\left\{ 1,\ldots ,n\right\} ;$ let $\alpha$ be a subset of M with k elements, and let $\alpha ^{\prime }$ be the complement of $\alpha$ in M,  $\alpha ^{\prime }=M\backslash \alpha$. The elements of $\alpha$ and $\alpha ^{\prime }$ are understood to be arranged in increasing order. Denote by $A[\alpha ]$ any principal submatrix of A with elements $\left( a_{ij}\right)$ with $i,j\in \alpha .$

First we establish that $A[\alpha ]$ has nonnegative row sums. To this end, note that $a_{ii}\ge \sum _{h\in H}\left| a_{ih}\right| ,$ where $H=\left\{ h|1\le h\le n\text { and }a_{ih}<0\right\} .$ This follows from the fact that rearranging (7) implies $a_{ii}-a_{ii}^{+}\ge \sum _{j\ne i}(a_{ii}^{+}-a_{ij})$ and observing that $a_{ii}\ge a_{ii}-a_{ii}^{+},$$a_{ii}^{+}-a_{ij}\ge 0$ for all $i\ne j,$ and $a_{ii}^{+}-a_{ij}\ge \left| a_{ij}\right|$ if $a_{ij}<0.$

To show that $ka_{ij}\le \sum _{s\in \alpha }a_{is}$ for all $i\ne j\in \alpha ,$ assume instead that for some $i\ne j\in \alpha ,$$ka_{ij}>\sum _{s\in \alpha }a_{is}$ to derive a contradiction. Since $ka_{ii}^{+}\ge ka_{ij}$ for all i,  it follows that for some i, 

$$\begin{aligned} na_{ii}^{+}\ge ka_{ii}^{+}+\sum _{r\in \alpha ^{\prime }}a_{ir}>\sum _{s\in \alpha }a_{is}+\sum _{r\in \alpha ^{\prime }}a_{ir}=\sum _{p=1}^{n}a_{ip}, \end{aligned}$$

which contradicts the assumption that A is a $B_{0}$-matrix.

(3)–(4) These are straightforward corollaries of (2).

(5) From Eq. (7) it follows that if $a_{ii}^{+}=0,$ then $a_{ii}\ge 0.$ If $a_{ii}^{+}>0,$ then since $\sum _{j=1}^{n}a_{ij}\le a_{ii}+(n-1)a_{ii}^{+},$ Eq. (7) implies $a_{ii}+(n-1)a_{ii}^{+}\ge na_{ii}^{+},$ or $a_{ii}\ge a_{ii}^{+}.$

(6) $A+A^{T}$ is a symmetric $B(B_{0})$-matrix by Lemma 2, so it is positive (semi)definite from Peña (2001) and part (4). A is positive (semi)definite if and only if $A+A^{T}$ is positive (semi)definite, and similarly for $A^{T}.$

(7) Again, A is positive (semi)definite if and only if $A+A^{T}$ is positive (semi)definite, and similarly for $A^{T}.$

(8) This follows from Proposition 2.3 and Theorem 4.3 in Peña (2001). In an effort to keep this paper reasonably self-contained, I will state the key parts of these results here. Let $A=(a_{ik})_{1\le i,k\le n}$ be a real matrix. Define, for each $i=1,\ldots ,n,$

$$\begin{aligned} r_{i}^{+}=\max \left\{ 0,a_{ij}|j\ne i\right\} \text { and }c_{i}^{+}=\max \left\{ 0,a_{ij}|i\ne j\right\} . \end{aligned}$$

Then define, for each $i=1,\ldots ,n,$

$$\begin{aligned} \theta _{i}=\min \left\{ a_{ii}-r_{i}^{+}-\sum _{k\ne i}\left( r_{i}^{+}-a_{ik}\right) , a_{ii}-c_{i}^{+}-\sum _{k\ne i}\left( c_{i}^{+}-a_{ki}\right) \right\} \!. \end{aligned}$$

Letting $\lambda$ be an eigenvalue of A, the first part of Peña’s (2001) Theorem 4.3 implies

$$\begin{aligned} {\hbox {Re}}(\lambda )\ge \min \left\{ \theta _{1},\ldots ,\theta _{n}\right\} . \end{aligned}$$

Proposition 2.3 in Peña (2001) states that the real matrix A is a B-matrix if and only if for all $i=1,\ldots ,n,$

$$\begin{aligned} a_{ii}-r_{i}^{+}-\sum _{k\ne i}\left( r_{i}^{+}-a_{ik}\right) >0. \end{aligned}$$

It follows that $A^{T}$ is a B-matrix if and only if for all $i=1,\ldots ,n,$

$$\begin{aligned} a_{ii}-c_{i}^{+}-\sum _{k\ne i}\left( c_{i}^{+}-a_{ki}\right) >0. \end{aligned}$$

Consequently, $\min \left\{ \theta _{1},\ldots ,\theta _{n}\right\} >0$ if A and $A^{T}$ are B-matrices.

1.3 Proof of Proposition 3

The necessary first-order conditions for a maximum are

$$\begin{aligned} \frac{\partial \pi }{\partial p_{i}}=\left( \sum p-c\right) \frac{\partial Q }{\partial p_{i}}+Q=0\text { for all }i. \end{aligned}$$

To apply Proposition 2, we need to show that the Hessian of $-\pi$ is mean positive dominant. The first-order conditions imply that $\frac{\partial Q}{\partial p}\equiv \frac{\partial Q}{\partial p_{i}}$ for all i at a maximum, so the terms along the main diagonal of the Hessian of $-\pi$ are $-\frac{\partial ^{2}\pi }{\partial p_{i}^{2}} =-\left( \sum p-c\right) \frac{\partial ^{2}Q}{\partial p_{i}^{2}}-2\frac{ \partial Q}{\partial p}.$ The off-diagonal terms are $-\frac{\partial ^{2}\pi }{\partial p_{i}\partial p_{j}}=-\left( \sum p-c\right) \frac{\partial ^{2}Q}{\partial p_{i}\partial p_{j}}-2\frac{\partial Q}{\partial p}.$ The remainder of the proof exploits the fact that $\frac{\partial ^{2}\pi }{\partial p_{i}\partial p_{j}}=\frac{\partial ^{2}\pi }{\partial p_{j}\partial p_{i}}.$

Since $-\frac{\partial ^{2}\pi }{\partial p_{i}\partial p_{j}}\ge 0$ for some $j\ne i,$the Hessian (of $-\pi )$ is mean positive dominant if, for $i=1,\ldots ,n,$

$$\begin{aligned}&\left( \sum p-c\right) \left( \frac{\partial ^{2}Q}{\partial p_{i}^{2}} +\sum _{j\ne i}\frac{\partial ^{2}Q}{\partial p_{j}\partial p_{i}}\right) +2n \frac{\partial Q}{\partial p}\le n\left( \sum p-c\right) \frac{\partial ^{2}Q}{\partial p_{j}\partial p_{i}}\\&\quad +\,n2\frac{\partial Q}{\partial p} \forall j\ne i, \end{aligned}$$

which simplifies to $\frac{\partial ^{2}Q}{\partial p_{i}^{2}}+\sum _{j\ne i} \frac{\partial ^{2}Q}{\partial p_{j}\partial p_{i}}\le n\max \left\{ \frac{ \partial ^{2}Q}{\partial p_{j}\partial p_{i}}|j\ne i\right\} ,$ as desired. We have now shown that the profit function is B-concave under the conditions in Proposition 3, so Proposition 2 applies.

1.4 Proof of Proposition 3

(1) Note that for $i,j=\left\{ S,B\right\}$ and $j\ne i,$

$$\begin{aligned} \frac{\partial ^{2}Q}{\partial p_{i}^{2}}=\frac{\partial ^{2}D^{i}}{\partial p_{i}^{2}}D^{j}\text { and }\frac{\partial ^{2}Q}{\partial p_{i}\partial p_{j}}=\frac{\partial D^{i}}{\partial p_{i}}\frac{\partial D^{j}}{\partial p_{j}}. \end{aligned}$$

The first-order conditions are $\left( \sum p-c\right) \frac{\partial Q}{ \partial p_{i}}+Q=0$ for all i which implies $D^{i}D^{j}=-\left( \sum p-c\right) \frac{\partial D^{i}}{\partial p_{i}}D^{j}$ for $i,j\in \left\{ B,S\right\}$ and $i\ne j.$ Thus, $\frac{D^{S}}{\frac{\partial D^{S}}{ \partial p_{S}}}=\frac{D^{B}}{\frac{\partial D^{B}}{\partial p_{B}}}$ in equilibrium. It follows that $\frac{\partial ^{2}Q}{\partial p_{i}^{2}}\le \frac{\partial ^{2}Q}{\partial p^{i}\partial p^{j}}$ for $i\ne j\in \left\{ B,S\right\}$ if and only if, for $i\in \left\{ B,S\right\} ,$

$$\begin{aligned} \frac{\partial ^{2}D^{i}}{\partial p_{i}^{2}}D^{j}\le \frac{\partial D^{i}}{ \partial p_{i}}\frac{\partial D^{j}}{\partial p_{j}},\text { or }\frac{ \partial ^{2}D^{i}}{\partial p_{i}^{2}}D^{i}\le \left( \frac{\partial D^{i} }{\partial p_{i}}\right) ^{2}. \end{aligned}$$

This last condition is precisely log-concavity of the quasi-demand functions.

(2) $\frac{\partial \pi }{\partial p_{B}\partial p_{S}}=\frac{\partial D^{B} }{\partial p_{B}}D^{S}+D^{B}\frac{\partial D^{S}}{\partial p_{S}}+\left( \sum p-c\right) \frac{\partial D^{B}}{\partial p_{B}}\frac{\partial D^{S}}{ \partial p_{S}}=D^{B}\frac{\partial D^{S}}{\partial p_{S}}\le 0,$ where the second equality follows from the first-order conditions.

(3) With captive buyers, we have $\frac{\partial ^{2}\pi }{\partial p_{B}\partial t}=D^{S}>0$ and $\frac{\partial ^{2}\pi }{\partial p_{S}\partial t}=\left( \sum p-c\right) \frac{\partial D^{S}}{\partial p_{s}} +D^{S}=0$ by the first-order condition, so we conclude that $p^{B}+p^{S}$ and $p^{B}$ increase with more captive buyers.

With marquee buyers, $\frac{\partial ^{2}\pi }{\partial p_{B}\partial t}=- \frac{\partial D^{S}(p_{S}-t)}{\partial p_{S}}\left( D^{B}(p_{B})+\left( \sum p-c\right) \frac{\partial D^{B}}{\partial p_{B}}\right) =0$ using the first-order conditions. Once again using the first-order conditions and the assumption that $D^{S}$ is log concave, we have

$$\begin{aligned} \frac{\partial ^{2}\pi }{\partial p_{S}\partial t}= & {} -D^{B}(p_{B})\left( \frac{\partial D^{S}(p-t)}{\partial p_{S}}+\left( \sum _{i}^{n}p_{i}-c\right) \frac{\partial ^{2}D^{S}(p-t)}{\partial p_{S}^{2}}\right) \\= & {} -D^{B}(p_{B})\left( \frac{\partial D^{S}(p-t)}{\partial p_{S}}-\frac{D^{S} }{\frac{\partial D^{S}}{\partial p_{S}}}\frac{\partial ^{2}D^{S}(p-t)}{ \partial p_{S}^{2}}\right) \\\ge & {} 0. \end{aligned}$$

Thus, $p^{B}+p^{S}$ and $p^{S}$ increase with the addition of marquee buyers.³³