Spatial competition on 2-dimensional markets and networks when consumers don’t always go to the closest firm

Abstract

We investigate the strategic behavior of firms in a Hotelling spatial setting. The innovation is to combine two important features that are ubiquitous in real markets: (1) the location space is two-dimensional, often with physical restrictions on where firms can locate; (2) consumers with some probability shop at firms other than the nearest. We characterise convergent Nash equilibria (CNE), in which all firms cluster at one point, for several alternative markets. In the benchmark case of a square convex market, we provide a new direct geometric proof of a result by Cox (Am J Political Sci 31:82–108, 1987) that CNE can arise in a sufficiently central part of the market. The convexity of the square space is of restricted realism, however, and we proceed to investigate grids, which more faithfully represent a stylised city’s streets. We characterise CNE, which exhibit several new phenomena. CNE in more central locations tend to be easier to support, echoing the unrestricted square case. However, CNE on the interior of edges differ substantially from CNE at nodes and follow quite surprising patterns. Our results also highlight the role of positive masses of indifferent consumers, which arise naturally in a network setting. In most previous models, in contrast, such masses cannot exist or are assumed away as unrealistic.

Appendix

Proof of Theorem 4

First, let us consider case (i). Consider the point $u=(x-\epsilon ,y)$, where $x>1$ and $y>0$ are integers and $0<\epsilon <1$. Consider a firm deviating to $(x-\epsilon -\eta ,y)$, where $\eta >0$ is small. As in the case where u is the node (x, y), it would be better to deviate to the node $(x-1,y)$, because by moving to the left the firm is gaining from consumers along multiple edges and is losing from only one edge.

Fig. 13

Comparing the move just north of $(x-1,y-1+\epsilon )$ to the move just west of $(x-1-\epsilon ,y-1)$. Thick black lines indicate consumers preferring $u'$ to u; thin black lines: consumers preferring u to $u'$

Upon reaching the node, the firm would then want to continue deviating towards $(x-1,y-1)$ because, again, there are more edges from which the firm is gaining consumers than from which it is losing. However, the firm will continue to gain only until it reaches the point $(x-1,y-1+\epsilon )$, at which point the northwest and southeast corners become indifferent (see Fig. 15a). Therefore, the move that maximises ${K_{u'\succ u}}/{K_{u\succ u'}}$ could be $(x-1,y-1+\epsilon )$, just to the north, or just south of it. However, the latter move is actually dominated since by continuing to move down the edge towards $(x-1,y-1)$ the number of edges from which the firm is gaining consumers exceeds those from which it is losing consumers. Even on reaching $(x-1,y-1)$, since $x>1$, there is a further incentive to deviate west, to the point $(x-1-\epsilon ,y-1)$. Here, there is another discontinuity in the score (see Fig. 15b) so the following points could also potentially be the best deviation: $(x-1-\epsilon ,y-1)$, just to the east, or west of it.

It turns out, however, that of the latter three points, only the point just east of $(x-1-\epsilon ,y-1)$ is not dominated by another deviation. To see this, first consider the deviation to a point inifinitesimally west of $(x-1-\epsilon ,y-1)$. This is dominated by the deviation infinitesimally north of $(x-1,y-1+\epsilon )$. This can clearly be seen by comparing the two panels in Fig. 13.¹⁵

As for the deviation to precisely the point $(x-1-\epsilon , y-1)$, it is dominated by the move precisely to $(x-1,y-1+\epsilon )$. To see this, we show that ${|SW'|}/{|NE'|}>{|SW''|}/{|NE''|}$. That is,

$$\begin{aligned}\frac{|SW|-\epsilon ({M/2}+x-1)}{|NE|-\epsilon ({M/2}-y)}>\frac{|SW|-\epsilon }{|NE|+M-2y+1+\epsilon }. \end{aligned}$$

This can be expressed as

$$\begin{aligned}&|SW|(2+\epsilon ) ({M/2}-y+1)+2\epsilon |NE| >\epsilon ^2({M/2}-y)\nonumber \\&\quad +\epsilon ({M/2}+x-1)(M-2y+1+\epsilon )+\epsilon |NE|({M/2}+x). \end{aligned}$$

(10)

That inequality (10) is true follows from the following three facts: (a) $2|SW|({M/2}-y+1)> \epsilon |NE| ({M/2}+x)$; (b) $\epsilon |SW| ({M/2}-y+1)> \epsilon ({M/2}+x-1)(M-2y+1+\epsilon )$; and, (c) $2\epsilon |NE|>\epsilon ^2({M/2}-y).$ Inequality (a) can be written as

$$\begin{aligned}4({M/2}\!+\!x)({M/2}\!+\!y)({M/2}\!-\!y\!+\!1)>2\epsilon ({M/2}-x\!+\!1)({M/2}\!-\!y+1)({M/2}+x)\end{aligned}$$

which, cancelling terms, is clearly true. Inequality (b) can be rewritten as

$$\begin{aligned}2({M/2}+x)({M/2}+y)({M/2}-y+1)> 2({M/2}+x-1)({M/2}-y+{(1+\epsilon )/2}).\end{aligned}$$

Again, this is clearly true. Finally, (c) states that $2\epsilon ({M/2}-x+1)({M/2}-y+1)>\epsilon ^2({M/2}-y)$, also true.

Hence, there are three possibilities for the best deviation from u: the move to $(x-1,y-1+\epsilon )$, just north of it, or just east of $(x-1-\epsilon ,y-1)$. Thus, K(u) will be given by Eq. (6) and we have proven case (i).

Fig. 14

Possible deviation from a point u. Thick black lines indicate consumers preferring $u'$ to u; thin black lines: consumers preferring u to $u'$

For case (ii), where $x=1$ and $y>0$, the argument is similar to above. Two possibilities for the best deviation are the point $(x-1,y-1+\epsilon )$ and the point just north of it. The third possibility is the point ${\overline{u}}=(x-1,y-1)$—the firm would not continue west this time because it would gain and lose consumers from the same number of edges (see Fig. 14e). In case (ii), therefore, we have equations (7) and (8).

For case (iii), $u=(x-\epsilon ,0)$, where $x>0$ and $0<\epsilon <1$. We show the move that maximises $\nicefrac {K_{u'\succ u}}{K_{u\succ u'}}$ is to the point $(x-1,0)$. Consider a deviation by i to a point further along the same edge $(x-\epsilon -\eta ,0)$ for very small $\eta >0$. This is not the best deviation because i could continue further towards the node $(x-1,0)$ and gain customers from all the parallel horizontal edges, while only losing customers from the edge on which it is located. It follows that the best deviation would be to $u'=(x-1,0)$. In this case,

$$\begin{aligned} K(u) =\frac{|K_{u'\succ u}|}{|K_{u\succ u'}|}&=\frac{({M/2}+x-1)(M+1)+({M/2}+x)M+M-(M-1)(1-\epsilon )/2}{({M/2}-x+1)M+({M/2}-x)(M+1)+1+(M-1)(1-\epsilon )/2}\\ {}&=\frac{M^2+(2M+1)x-{1/2}+{\epsilon (M-1)/2}}{M^2+2M-(2M+1)x+{1/2}-{\epsilon (M-1)/2}}. \end{aligned}$$

$\square$

Fig. 15

Possible deviations from a point u in the interior of an edge. Thick black lines indicate consumers preferring $u'$ to u; thin black lines: consumers preferring u to $u'$; dotted lines: consumers indifferent between $u'$ and u

Proof of Corollary 1

Case (i) follows by simply differentiating (5) with respect to M. For (ii), we first show that each of the arguments in the maximum function in equation (6) is decreasing in M. From this, the case where u is a node follow by setting $\epsilon =0$ because (6) reduces to (3) (in fact, in the following proof (6) would be decreasing in M even allowing $x=1$, which corresponds to nodes where $x=1$).

Consider $|SW'|/|NE'|$. Differentiating with respect to M, the derivative is negative if and only if

$$\begin{aligned} (M-x&-y-{\epsilon /2}+2)(2({M/2}+x)({M/2}+y) -\epsilon ({M/2}+x-1))\\ {}&-(M+x+y-{\epsilon /2})(2({M/2}-x+1)({M/2}-y+1)-\epsilon ({M/2}-y))>0. \end{aligned}$$

This can be rewritten as

$$\begin{aligned} 2(x+y-1)(M(M+2)-\epsilon (M-x+y+3) +2(x+y-2xy)+{\epsilon ^{2}/2})>0, \end{aligned}$$

which is equivalent to

$$\begin{aligned}(x+y-1)(M^2+2M-\epsilon M+\epsilon x-\epsilon y-3\epsilon +2x+2y-4xy+{\epsilon ^{2}/2})>0.\end{aligned}$$

By assumption $x+y-1> 0$ and the term in the second set of brackets can be written as

$$\begin{aligned} (M^2-4xy) +M(2-\epsilon )+y(1-\epsilon )+(2x+y-3\epsilon )+\epsilon x+{\epsilon ^{2}/2}. \end{aligned}$$

That this is positive follows from the facts that $1\le x, y\le {M/2}$ and $0\le \epsilon <1$.

Next, consider ${(|NW'|+|SW'|)/(|NE'|+|SE'|)}$. This can be simplified to

$$\begin{aligned}\frac{({M/2}+x)(2M+1)-{M/2}+y-1+\epsilon ({M/2}-y)}{({M/2}-x+1)(2M+1)-{M/2}-y-\epsilon ({M/2}-y)}.\end{aligned}$$

Differentiating with respect to M, the derivative is negative if and only if

$$\begin{aligned}M^2(4x+\epsilon -2)+(4M+2)(x+y(1-\epsilon )-1)>0,\end{aligned}$$

which is certainly true.

Similarly, consider $(|SW''|+|SE''|)/(|NW''|+|NE''|)$. This can be simplified to

$$\begin{aligned}\frac{M^2+2My+x+y-\epsilon -1}{M^2-2My+2M-x-y+1+\epsilon }.\end{aligned}$$

Differentiating with respect to M, the derivative is negative if and only if

$$\begin{aligned} (2y-1)M^2+2M(x-\epsilon )+2M(y-1)+x+y-1-\epsilon >0. \end{aligned}$$

Given $y>0$, $x>1$ and $0\le \epsilon <1$, this is certainly true.

Next consider the case where $u=(x-\epsilon , y)$ where $y>0$ and $x=1$. It is straightforward to show that, in a similar fashion to the above, (8) is decreasing in M. In particular, its derivative is negative if and only if

$$\begin{aligned}(4y-2)M^2+4My+2y(\epsilon +1)-\epsilon >0,\end{aligned}$$

which is evidently true in this case.

Last, we consider the case where, without loss of generality, $y=0$. The derivative of (9) will be negative if and only if

$$\begin{aligned}M^2(4x+\epsilon -2)+(2M+1)(2x-1-\epsilon )>0,\end{aligned}$$

which is always true.

To prove (iii), if $x>0$, it is easy to see that $|SW'|$, $|NW'|+|SW'|$ and $|SW''|+|SE''|$ are all increasing in x, while $|NW'|$, $|NE'|+|SE'|$ and $|NW''|+|NE''|$ are decreasing in x. Thus, the ratios in (6) are all increasing. This is true even when $\epsilon =0$ and when $x=1$, so the case of nodes follows. The same holds true when y increases. We also need to check the case where $x=1$, i.e., $u=(1-\epsilon , y)$. Two of the arguments in the maximum in (7) are the same as in the previous case, hence increasing in x, and it can also be shown that (8) is less than ${(|SW''(v)|+|SE''(v)|)/(|NW''(v)|+|NE''(v)|)}$ for $v=(2-\epsilon ,y)$. Thus, (8) will be less than K(v).

When $y=0$, it is clear that (9) is increasing in x. Finally, we need to show that K(u) is increasing when $x=0$ and $y>1$. By symmetry, consider instead $u=(x-\epsilon , 0)$ and $v=(x-\epsilon ,1)$. We show that $K(u)< K(v)$. Consider a deviation from v to just north of $v'=(x-1, y-1+\epsilon )$. For this move,

$$\begin{aligned}\frac{K_{v'\succ v}}{K_{v\succ v'}}=\frac{|NW'(v)|+|SW'(v)|}{|NE'(v)|+|SE'(v)|}\le K(v).\end{aligned}$$

Using Eq. (9), after a bit of algebra it can be shown that $K(u)<K(v)$. $\square$