Sequential competition and the strategic origins of preferential attachment

Abstract

We analyze whether random network formation processes, such as preferential attachment, can emerge as the outcome of strategic behaviour. We represent network formation as an extensive game in which players sequentially form links as they enter the network. In this setting, we investigate under which conditions subgame perfect equilibria of the game are observationally equivalent with random network formation process. We put forward two structural conditions that are necessary in this respect. First, players must have some form of imperfect information as randomization is purposeful only if its realization is not perfectly observed by the other players. Second, there must be some form of competition between a player and its successors: a player has incentives to reduce the information available to its successors only to the extent that their objectives are in opposition. Accordingly, we put forward a class of games where players compete with their predecessors and their successors for the costs and benefits induced by link formation and show that subgame perfect equilibria of this game are observationally equivalent with random network formation process. In particular, when linkage costs are inversely proportional to the degree of a node, equilibrium play induces a preferential attachment process. This provides a positive answer to the question of the existence of strategic foundations for preferential attachment. However the very specific conditions requiredfor the observational equivalence to hold suggest that preferential attachment can be explained by strategic considerations only in a limited number of situations.

Appendix

Proof of Lemma 3.2

The expected payoff of player T if he uses the action $\alpha \in \Delta (A_{T-1})$ given the action $s \in \Delta (A_{T-1})$ of player $T-1$ and the history of the network up to the end of period $T-2,$ $\tilde{g}_{T-2}$ (in fact the payoff of player T only depends on the history $\tilde{g}_{T-3}$ up to period $T-3$ because he only competes with his predecessor ) is given by:

$$\begin{aligned} -\lambda \left[ \sum _{i=1}^{n_{T-3}} f(\tilde{g}_{T-3},i) s_i \alpha _i +\overline{f}(\tilde{g}_{T-3}) \alpha _{n_{T-2}} \right] . \end{aligned}$$

(6.1)

Given he has no successor, the objective of player T simply is to minimize his linkage cost, i.e. he wants to avoid competition with his predecessor. We can distinguish two cases depending on where the minimum cost is reached: on link to node before $n_{T-3}$ or on $n_{T-2}$.

• if $\min _{i \in A_{T-1}} s_i f(\tilde{g}_{T-3},i) <\overline{f}(\tilde{g}_{T-3}),$ any node, and thus any probabilistic strategy in

  $$\begin{aligned} {\text {argmin}}_{i \in A_{T-1}} \ s_i f(\tilde{g}_{T-3},i) \end{aligned}$$

  is a best response for player T and his expected payoff is $-\lambda \min _{i \in A_{T-1}}s_i f(\tilde{g}_{T-3},i).$

• If $\min _{i \in A_{T-1}} s_i f(\tilde{g}_{T-3},i) \ge \overline{f}(\tilde{g}_{T-3}),$ then one has for all $i=1,\ldots , T-3,$ $p_i \ge \nicefrac {\overline{f}(\tilde{g}_{T-3})}{f(\tilde{g}_{t-3},i) }$ and thus, as $\overline{f}(\cdot )$ is the harmonic mean of the $f(\cdot ,i),$ for every $i\in \{1,\ldots , T-3\},$ $s_i=\nicefrac {\overline{f}(\tilde{g}_{T-3})}{f(\tilde{g}_{T-3},i) }.$ Any node, and thus any probabilistic strategy in $A_{T-1}$ is then a best response for player T and his expected payoff is $- \lambda \overline{f}(\tilde{g}_{T-3})=-\lambda \min _{i \in A_{T-1}} s_i f(\tilde{g}_{T-3},i)$. $\square$

Proof of Lemma 3.3

Let us consider player $T-1$. We denote by s the action of its predecessor and by $\alpha$ the action played by player $T-1$. One considers the auxiliary linear program for $k=1,\ldots , n_{T-3}$:

$$\begin{aligned} \mathcal {P}_k:= \left\{ \begin{array}{cc}\max &{} -\lambda \left[ \sum _{i=1}^{n_{T-4}} f(\tilde{g}_{T-4},i) s_i \alpha _i + \overline{f}(\tilde{g}_{T-4}) \alpha _{n_{T-3}} \right] + \alpha _k f(\tilde{g}_{T-3},k) \\ s.t &{} \\ &{} \sum _{i=1}^ {n_{T-3}} \ \alpha _i=1 \\ &{} \forall i \in A_{T-1}, \ \alpha _i \ge 0 \\ &{} \forall i \in A_{T-1}, \ \ \alpha _k f(\tilde{g}_{T-3},k) \le \alpha _i f(\tilde{g}_{T-3},i) \end{array}\right. \end{aligned}$$

(6.2)

Letting then on the one hand $\gamma _k:= f(\tilde{g}_{T-3},k) -\lambda f(\tilde{g}_{T-4},k)s_k,$ $\gamma _{n_T-3}:=-\lambda \overline{f}(\tilde{g}_{T-4})$ and for $i \not =k,n_{T-3}$, $\gamma _i:= - \lambda f(\tilde{g}_{T-4},i) s_i$ and on the other hand $\phi _i= f(\tilde{g}_{T-3},i)$ one has:

$$\begin{aligned} \mathcal {P}_k:= \left\{ \begin{array}{cc}\max &{} \sum _{i=1}^{n_{T-3}} \gamma _i \alpha _i \\ s.t &{} \\ &{} \sum _{i=1}^ {n_{T-3}} \ \alpha _i=1 \\ &{} \forall i \in N_{n-2}, \ \alpha _i \ge 0 \\ &{} \forall i \in A_{T-1}, \ \alpha _k \phi _k \le \alpha _i \phi _i \end{array}\right. \end{aligned}$$

(6.3)

Let us denote $I^*= {\text {argmax}}_{i \not = k} \gamma _i$ and $\gamma _{I^*}= {\text {max}}_{i \not = k} \gamma _i.$ A solution $\alpha$ of $\mathcal {P}_k$ must be such that for all $i \not \in I^*,$ $\alpha _i=\alpha _k\nicefrac {\phi _k}{\phi _i}$ and such that $\alpha _k$ is a solution of:

$$\begin{aligned} \left\{ \begin{array}{cc} \max &{} \left( 1- \left( \sum _{i \not \in I^*}\dfrac{\phi (k)}{\phi (i)} \alpha _k \right) \right) \gamma _{I^*} +\sum _{i \not \in I^*}\gamma _i \dfrac{\phi (k)}{\phi (i)} \alpha _k \\ s.t &{} 0 \le \alpha _k\le \dfrac{1}{\sum _{i \in A_{T-1}} \dfrac{\phi _k}{\phi _i}}\end{array}\right. \end{aligned}$$

This is a one-dimensional problem. Its solutions must satisfy either $\alpha _k=0$ or $\alpha _k=\nicefrac {1}{(\sum _{i \in A_{T-1}} \nicefrac {\phi _k}{\phi _i})}.$ Then, one has:

• If $\alpha _k=0,$ any probability distribution with support in $I^*$ is a solution to $\mathcal {P}_k$ and the value is $\gamma _{I^*}:= -\lambda \min \left( \overline{f}(\tilde{g}_{T-4}), \min _{i \not =k} f(\tilde{g}_{T-4},i) s_i\right) .$

• If $\alpha _k>0,$ one must have for all $i \in A_{T-1},$ $\alpha _i=\nicefrac {\phi _k}{\phi _i},$ that is to say $\alpha _i=\nicefrac { f(\tilde{g}_{T-3},k)}{ f(\tilde{g}_{T-3},i)} \alpha _k.$ Thus the solution must be inversely proportional to the $f(\tilde{g}_{T-3},i).$ Hence, one has:

  $$\begin{aligned} \alpha ^*= \mu ^{T-1}_{f}(\tilde{g}_{T-3}):=\left( \dfrac{\overline{f}(\tilde{g}_{T-3})}{f(\tilde{g}_{T-3},1)},\ldots ,\dfrac{\overline{f}(\tilde{g}_{T-3})}{f(\tilde{g}_{T-3},n_{T-3})} \right) \end{aligned}$$

  where

  $$\begin{aligned} \overline{f}(\tilde{g}_{T-3})= \dfrac{1}{\left( { \sum _{i \in A_{T-1}}\dfrac{1}{f(\tilde{g}_{T-3},i)}}\right) }. \end{aligned}$$

  Accordingly, the value in that case is:

  $$\begin{aligned} \sum _{i \in A_{T-1}} \alpha ^*_i \gamma _i=\overline{f}(\tilde{g}_{T-3})\left[ 1-\lambda \overline{f}(\tilde{g}_{T-4}) -\lambda \sum _{i \in A_{T-2}} \dfrac{f(\tilde{g}_{T-4},i)}{f(\tilde{g}_{T-3},i)}s_i \right] \end{aligned}$$

Now, the best-response of player $T-1$ corresponds to the solutions of the problems $\mathcal {P}_k$ that have the largest value. In this respect, it is clear that under Assumption 3.1, one has $\overline{f}(\tilde{g}_{T-4})\le 1$ and $\sum _{i \in A_{T-2}} \dfrac{f(\tilde{g}_{T-4},i)}{f(\tilde{g}_{T-3},i)}s_i\le 2$ so that

$$\begin{aligned} \overline{f}(\tilde{g}_{T-3})\left[ 1-\lambda \overline{f}(\tilde{g}_{T-4}) -\lambda \sum _{i \in A_{T-2}} \dfrac{f(\tilde{g}_{T-4},i)}{f(\tilde{g}_{T-3},i)}s_i \right] \ge \overline{f}(\tilde{g}_{T-3}) \left( 1- 3 \lambda \right) \end{aligned}$$

(6.4)

Hence for $\lambda < \nicefrac {1}{3},$ one has for all $(p,\tilde{g}_{T-3},\tilde{g}_{T-4}) \in \Delta (A_{T-2}) \times \mathcal {H} \times \mathcal {H}:$

$$\begin{aligned}&\overline{f}(\tilde{g}_{T-3})\left[ 1-\lambda \overline{f}(\tilde{g}_{T-4}) -\lambda \sum _{i \in A_{T-2}} \dfrac{f(\tilde{g}_{T-4},i)}{f(\tilde{g}_{T-3},i)}s_i \right] \ge \overline{f}(\tilde{g}_{T-3}) \left( 1- 3 \lambda \right) \nonumber \\&\quad > 0 \ge -\lambda \min \left( \overline{f}(\tilde{g}_{T-4}), \min _{i \not =k} f(\tilde{g}_{T-4},i) s_i\right) . \end{aligned}$$

(6.5)

Hence, $\mu ^{T-1}_{f}(\tilde{g}_{T-3})$ is the solution of all the $\mathcal {P}_k$ and the best-response of player $T-1$ is to play proportionally to f independently of the action of its predecessor.$\square$

Proof of Lemma 3.4

The property has been proven for player T in Lemma 3.2 and $T-1$ in Lemma 3.3. Then, an immediate backward induction shows that:

• The expected payoff of a player $t \in U_T:={\mathbb {N}}^* \cap \{T-2k,\ k \in {\mathbb {N}}\}$ using a strategy $\alpha \in \Delta (A_{t})$ given the predecessor action $s \in \Delta (A_{t-1})$ is of the form:

  $$\begin{aligned} -\lambda \left[ \sum _{i=1}^{n_{t-3}} f(\tilde{g}_{t-3},i) s_i \alpha _i +\overline{f}(\tilde{g}_{t-3}) \alpha _{n_{t-2}} \right] + {\kappa _t(\tilde{g}_{t-2},s)} \end{aligned}$$

  (6.6)

  where $\kappa _t(\tilde{g}_{t-2},s)$ is independent of $\alpha$. Indeed, player $t+1 \in V_T$ is following a one-lag subgame perfect equilibrium. Hence, by the induction assumption, he plays as a function of the graph at the end of stage $t-1$ hence independently of the action $\alpha$. The best-reply of player t is then determined similarly to the best-reply of player T : 

  • if $\min _{i \in A_{t-1}} s_i f(\tilde{g}_{t-3},i) <\overline{f}(\tilde{g}_{t-3}),$ any node and thus any probabilistic strategy in

    $$\begin{aligned} {\text {argmin}}_{i \in A_{T-1}} \ s_i f(\tilde{g}_{t-3},i) \end{aligned}$$

    is a best response for player T and his expected payoff is $-\lambda \min _{i \in A_{T-1}}s_i f (\tilde{g}_{T-3},i).$

  • If $\min _{i \in A_{T-1}} s_i f(\tilde{g}_{t-3},i) \ge \overline{f}(\tilde{g}_{t-3}),$ one then has for all $i=1,\ldots , t-3,$ $s_i \ge \nicefrac {\overline{f}(\tilde{g}_{t-3})}{f(\tilde{g}_{t-3},i) }$ and thus, as $\overline{f}(\cdot )$ is the harmonic mean of the $f(\cdot ,i),$ for every $i\in \{1,\ldots , t-3\},$ $s_i=\nicefrac {\overline{f}(\tilde{g}_{t-3})}{f(\tilde{g}_{t-3},i) }.$ Any node, and thus any probabilistic strategy in $A_{t}$ is then a best response for player t and his expected payoff is $- \lambda \overline{f}(\tilde{g}_{t-3})$.

• The expected payoff of a player $t \in V_T:={\mathbb {N}}^* \cap \{T-(2k+1), \ k \in {\mathbb {N}}\},$ using an action $\alpha \in \Delta (A_{t})$ given the predecessor action $s \in \Delta (A_{t-1})$ is of the form:

  $$\begin{aligned} -\lambda \left[ \sum _{i=1}^{n_{t-3}} f(\tilde{g}_{t-3},i) s_i \alpha _i+ \overline{f}(\tilde{g}_{t-3})\alpha _{n_{t-2}}\right] +\min _{i=1,\ldots , n_t-2} \alpha _i f(\tilde{g}_{t-2},i)\end{aligned}$$

  (6.7)

  Indeed, player $t+1 \in U_T$ is following a one-lag subgame perfect equilibrium hence play independently of the future to minimize the payoff of player t. Using Lemma 3.3, it is clear that the best-response of player t is to play proportionally to f independently of the action of its predecessor: hence to play $\mu ^{t-1}_{f}(\tilde{g}_{t-2})$.$\square$

Proof of Theorem 3.6

The theorem is an immediate consequence of Lemma 3.4. We define the strategy of player t as a function of $\tilde{g}_{t-2}$ the graph at stage $t-2$ and $s^{t-1}$ the action of player $t-1$. The first one can be computed from $g_0$ and the sequence of choices of nature. The second one is directly observed by player t. Then

$$\begin{aligned} \phi ^*(\tilde{g}_{t-2})= \mu ^{T-1}_{f}(\tilde{g}_{t-2}), \end{aligned}$$

form a one-lag subgame perfect equilibrium. The second statement is a reformulation of the second part of Lemma 3.4.$\square$

Proof of Theorem 3.7

The second statement is an immediate corollary of the second statement in Theorem 3.6. Let $\varepsilon >0$ and $n_*\in {\mathbb {N}}$ such that $\frac{1}{n_*} \le \varepsilon /2$.

Concerning the first statement, we consider the following profile of strategies denoted by $(\phi ^*_t)_{1\le t\le T}$:

• if player $t \in U_T={\mathbb {N}}^* \cap \{T-2k,\ k \in {\mathbb {N}}\}$ and he faces the predecessor action s at state $\tilde{g}_{n_t-3}$. Let $\theta =\min _{i \in A_{t-1}} s_i f(\tilde{g}_{t-3},i)$ and $\Theta =argmin_{i \in A_{t-1}} s_i f(\tilde{g}_{t-3},i)$. Player t plays the restriction of $\mu _f^t(\tilde{g}_{t-2})$ to $\Theta$, i.e. the action i with probability

  $$\begin{aligned} \alpha _i={\left\{ \begin{array}{ll} \frac{\frac{1}{f(\tilde{g}_{t-2},i)}}{\sum _{j\in \Theta } \frac{1}{f(\tilde{g}_{t-2},j)}} {\text { if }} i\in \Theta ,\\ 0 {\text { otherwise}}. \end{array}\right. } \end{aligned}$$

• if player $t \in V_T={\mathbb {N}}^* \cap \{T-(2k+1), \ k \in {\mathbb {N}}\},$ then independently of the predecessor action $s\in \Delta (A_{t-1}),$ he plays following $\mu _f^{t}(\tilde{g}_{t-2}) \in \Delta (A_{t}).$

Let us check that the previous profile of strategies is a $\varepsilon$-one-lag subgame perfect equilibrium. First, since the payoffs of players in $V_T$ are the same in the original game and the auxiliary game, it is clear that the inequalities are satisfied.

We want now to prove that players in $U_T={\mathbb {N}}^* \cap \{T-2k,\ k \in {\mathbb {N}}\}$ can not gain more $\varepsilon$ by deviating. Let $t\in U_T$ and let $s\in A_{t-1}$ be its predecessor action. For all $i\in \{1,\ldots ,n_{t-3}\}$, the payoff of player t if he plays action i is

$$\begin{aligned} -\lambda f(\tilde{g}_{t-3},i) s_i +\sum _{j=1}^{n_{t-3}} f\big (\tilde{g}_{t-2},i\big )s_j \mu _f^{t+1}(\tilde{g}_{t-2}+(n_{t-1},j),i), \end{aligned}$$

(6.8)

where $\tilde{g}_{t-2}+(n_{t-1},j)$ is the graph $\tilde{g}_{t-2}$ augmented by an additional link from node $n_{t-1}$ to node j. By construction, we know that Player $t+1$ is in $V_T$ and follows

$$\begin{aligned} \mu _f^{t+1}(\tilde{g}_{t-2}+(n_{t-1},j),i)=\frac{\frac{1}{f(\tilde{g}_{t-2}+(n_{t-1},j),i)}}{\sum _{l\in A_{t+1}}(\frac{1}{f(\tilde{g}_{t-2}+(n_{t-1},j),l)})}. \end{aligned}$$

Hence, the payoff become

$$\begin{aligned} -\lambda f(\tilde{g}_{t-3},i) s_i +\sum _{j=1}^{n_{t-3}} s_j \left( \frac{\frac{f(\tilde{g}_{t-2},i)}{f(\tilde{g}_{t-2}+(n_{t-1},j),i)}}{\sum _{l\in A_{t+1}}\left( \frac{1}{f(\tilde{g}_{t-2}+(n_{t-1},j),l)}\right) }\right) . \end{aligned}$$

(6.9)

Let us compare the payoff in the original game and in the auxiliary game. The difference of payoff yields:

$$\begin{aligned}&|\pi '_{f,\lambda }(i,\phi ^*_{-t})-\pi _{f,\lambda }(i,\phi ^*_{-t})| \\&\quad =\left| \left( \sum _{j=1}^{n_{t-3}}\frac{s_j}{\sum _{l\in A_{t+1}}\left( \frac{1}{f(\tilde{g}_{t-2}+(n_{t-1},j),l)}\right) } \left( \frac{f(\tilde{g}_{t-2},i)}{f(\tilde{g}_{t-2}+(n_{t-1},j),i)}-1\right) \right) \right| , \\&\quad = \sum _{j=1}^{n_{t-3}} \left| \frac{s_j}{\sum _{l\in A_{t+1}}\left( \frac{1}{f(\tilde{g}_{t-2}+(n_{t-1},j),l)}\right) } \left( \frac{1}{f(\tilde{g}_{t-2}+(n_{t-1},j),i)}-\frac{1}{f(\tilde{g}_{t-2},i)}\right) f(\tilde{g}_{t-2},i)\right| , \\&\quad \le \sum _{j=1}^{n_{t-3}} \frac{s_j}{\sum _{l\in A_{t+1}}\left( \frac{1}{f(\tilde{g}_{t-2}+(n_{t-1},j),l)}\right) }. \end{aligned}$$

by Assumption 3.1. By assumption on the size of the original networks, we know that every set of actions has more than $n_0$ actions, hence, we have

$$\begin{aligned} \frac{1}{\sum _{l\in A_{t+1}}\left( \frac{1}{f(\tilde{g}_{t-2}+(n_{t-1},j),l)}\right) }\le \frac{1}{n_0} \le \frac{\varepsilon }{2}. \end{aligned}$$

and

$$\begin{aligned} |\pi '_{f,\lambda }(i,\phi ^*_{-t})-\pi _{f,\lambda }(i,\phi ^*_{-t})|&\le \frac{\varepsilon }{2}. \end{aligned}$$

Since it is true for every pure action of Player t, it is also true for every probabilistic action by linearity of the payoff function. Since the strategy $\phi ^*_{t}$ is optimal in the auxiliary game, we obtain that for every $1\le t\le T$ and for every history $h_t\in H_t$, we have

$$\begin{aligned} \forall \alpha '\in \Delta (A_t),\ \pi _{f,\lambda }(\alpha ',\phi ^*_{-t})&\le \pi '_{f,\lambda }(\alpha ',\phi ^*_{-t})+\frac{\varepsilon }{2},\\&\le \pi '_{f,\lambda }(\phi ^*_t,\phi ^*_{-t})+\frac{\varepsilon }{2},\\&\le \pi _{f,\lambda }(\phi ^*_t,\phi ^*_{-t})+\varepsilon . \end{aligned}$$

Hence, the profile of strategies is a lagged subgame perfect $\varepsilon$-equilibrium. $\square$

Proof of Proposition 3.10

As $\sum _{i \in A_{T-1}} \xi _i$ is constant with respect to the degree distribution $(d_j)_{j \in A_{T-1}}$ and each of the $(\xi _i)_{i \in A_{T-1}}$ ought to be equal, irrespectively of the degree distribution, it must be that $\xi _i$ is constant with respect to the degree distribution. Then, differentiating Equation 3.9 with respect to $d_j$ for $j \not =i$ yields

$$\begin{aligned} \frac{\partial f_T}{ \partial d_j} (d_i,d_j) d_j + f_T(d_i,d_j) =0 \end{aligned}$$

This partial differential equation implies there exists $\phi : {\mathbb {R}}_+ \rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} f_T(d_i,d_j)=\dfrac{\phi (d_i)}{d_j} \end{aligned}$$

(6.10)

Then $\xi _i$ can be rewritten as

$$\begin{aligned} \xi _i=\sum _{j \in A_{T-1}/\{i\} } \dfrac{\phi (d_i)}{d_j} \dfrac{d_j}{\overline{d}}+g_T(d_i) \dfrac{d_i}{\overline{d}} \end{aligned}$$

(6.11)

and equivalently

$$\begin{aligned} \xi _i=\sum _{j \in A_{T-1}/\{i\} } \dfrac{\phi (d_i)}{\overline{d}}+g_T(d_i) \dfrac{d_i}{\overline{d}} \end{aligned}$$

(6.12)

Using the fact that $\xi _i$ is constant with respect to the degree distribution and differentiating with respect to $d_i$ yields

$$\begin{aligned} \left[ \sum _{j \in A_{T-1}/\{i\} } \phi '(d_i) \right] + g'_T(d_i)d_i +g_T(d_i)=0 \end{aligned}$$

(6.13)

This differential equation implies there exists $k \in {\mathbb {R}}$ such that

$$\begin{aligned} g_T(d_i):= \dfrac{k- \sum _{j \in A_{T-1}/\{i\} } \phi (d_i)}{d_i} \end{aligned}$$

(6.14)

This ends the proof $\square$