Advertising to a social network

Abstract

Direct advertising—sending promotional messages to individual customers—is increasingly used by marketers as a result of the recent improvements in consumer reachability. Most current methods to calculate optimal budgets for such advertising campaigns consider customers in isolation and ignore word-of-mouth communication (WOM). When the customer base forms a network (as is the case in telecom or social network databases) ignoring WOM clearly leads to suboptimal advertising budgets. This paper develops a model to help address this challenge. We assume that firms know the network structure formed by customers but do not know (or are not allowed to use) data on individuals’ connections. Under this scenario, we compare the optimal campaign of a monopolist to that of firms competing in simultaneous-move or sequential-move games. The analysis provides two key insights: (i) we show that ignoring the existence of WOM leads to significant profit loss for firms and this is more so under competition. In particular, knowing the “density” of consumer connections is crucial for the design of optimal campaigns. (ii) Competition in direct advertising exhibits strong first-mover advantages and, even in a simultaneous-move game between identical firms, highly asymmetric outcomes are possible. The paper also explores two extensions. First, we study the nature of the trade-off between increasing network size versus increasing the connectivity between existing network members. Second, we investigate how firms’ advertising activity may endogenously grow the membership base.

Appendix

Proof of Proposition 1

Let v be a node in the network and let k be the number of advertising messages sent by the monopolist. As shown in Section 4, the expected surplus from v is

$$ \mathop{\mathrm E}\limits_{|S|=k}\left[r_1(v)\right]=1-\frac{{{n-\textnormal{d}(v)-1} \choose k}}{{n \choose k}}. $$

Summing this up for every node in the network and taking into account the cost of sending messages, we arrive to the following formula for the profit of the monopolist:

$$ \Pi^M(k,c) = \left[\sum\limits_{v \in V}{1-\frac{{{n-\textnormal{d}(v)-1} \choose k}}{{n \choose k}}}\right]-k \cdot c. $$

To see the rest of the claim, we are going to prove that Π ^M(k + 1,c) − Π ^M(k,c) is (weakly) decreasing in k. Whereas it is probably shorter to do this applying basic calculus, we choose a different proof technique as it reveals more general underlying mechanisms that may help gaining a further insight to the problem discussed in this article. We focus on the case when E ≠ ∅, as otherwise the proposition is obvious (for c = 1, see also footnote 4).

First we are going to prove the submodularity¹¹ of the surplus function corresponding to the deterministic selection of the set of targeted consumers. Next, we carry out an averaging argument to see that the submodularity also holds for the surplus from randomized targeting. This property of the surplus function then in turn implies the claims.

For notational convenience, for any set $S \subseteq V$, let r(v,S) be the surplus of the monopolist from node v when exactly the nodes in S are targeted. Let then the total surplus of the monopolist under deterministic selection be $W(S)=\sum\limits_{v \in V}{r(v,S)}$ and furthermore the expected surplus from selecting k nodes at random be

$$R(k)=\mathop{\mathrm E}\limits_{|S|=k}{[W(S)]}=\frac{\sum\limits_{|S|=k}{W(S)}}{{n \choose k}}. $$

(1)

The following lemma states the submodularity of the surplus under deterministic targeting.

Lemma 1

(Kempe et al. 2003)¹² W(·) is submodular, i.e.

$$ W(S \cup \{v\})-W(S) \geq W(T \cup \{v\})-W(T) $$

for all nodes v and all pairs of sets $S,T \subseteq V$ such that $S \subseteq T$.

Proof

For any pair of sets S, T with $S \subseteq T \subseteq V$ and any given nodes x,v ∈ V, we have the following:

• 1. If x is aware of the product after the firm targets all members of S, then it is also aware of the message after the firm targets all members of T. Thus, adding v to either the target set S or to the set T does not generate any additional surplus from x.

• 2. If x is unaware of the product after the firm targets S, then

  • If x becomes aware of the product after the firm targets v, then targeting v generates an additional surplus of 1 for Firm 1 from x when v is added to S. The same 1 unit of surplus is generated when x is unaware of the product after targeting every member of T. However, when targeting T informs x about the product (directly or indirectly), then adding v to T does not generate additional surplus from x.

  • If x remains unaware of the product after the firm targets v, this is because v is not equal to or connected to x. Then adding v to S does not generate any additional surplus for Firm 1 from x, and the same applies when v is added to T.

Thus, r(x,S ∪ {v}) − r(x,S) ≥ r(x,T ∪ {v}) − r(x,T). The claim follows.□

The next step is to show how the submodularity of W(·) leads to the “diminishing returns” property of R(·). For a given k, let now sum up all the possible submodularity inequalities of the form

$$ W(X_{k-1})-W(X_{k-2}) \geq W(Y_{k})-W(Y_{k-1}), $$

where X _k − 1 = X _k − 2 ∪ {v},Y _k  = Y _k − 1 ∪ {v},|X _k − 2| = k − 2,|Y _k − 1| = k − 1 and X _k − 2 ⊂ Y _k − 1. To describe such an equation, we have to select a (k − 2)-element set X _k − 2 $({n \choose k-2} {\rm possibilities})$, a node v outside this set (n − k + 2 possibilities) and a node u such that {u} = Y _k − 1 ∖ X _k − 2 (n − k + 1 possibilities). Thus, there are

$$ {n \choose k-2}(n-k+2)(n-k+1)=k(k-1){n \choose k} $$

such submodularity inequalities.

Because of the symmetry of these inequalities, for every k-element set S _k , W(S _k ) appears

$$ \frac{k(k-1){n \choose k}}{{n \choose k}}=k(k-1) \text{ times on the right-hand side}. $$

Similarly, for every (k − 1)-element set S _k − 1, W(S _k − 1) appears

$$ \frac{k(k-1){n \choose k}}{{n \choose k-1}}=(k-1)(n-k+1) \text{ times on both sides (with opposite signs).} $$

Finally, for every (k − 2)-element set S _k − 2, W(S _k − 2) appears

$$ \frac{k(k-1){n \choose k}}{{n \choose k-2}}=(n-k+1)(n-k+2) \text{ times on the left-hand side}. $$

Thus, substituting Eq. 1 into the sum of the submodularity inequalities, we get

$$ \begin{array}{rll} &&{\kern-6pt}(k-1)(n-k+1){n \choose k-1} \cdot R(k-1) - (n-k+1)(n-k+2){n \choose k-2} \\ &&{\kern6pt}\cdot\, R(k-2) \geq k(k-1){n \choose k} \cdot R(k) - (k-1)(n-k+1){n \choose k-1} \cdot R(k-1), \end{array} $$

which simplifies to R(k − 1) − R(k − 2) ≥ R(k) − R(k − 1). It follows that the quantities R(k + 1) − R(k) and Π ^M(k + 1,c) − Π ^M(k,c) are (weakly) decreasing in k. Further, if E ≠ ∅ and R(k − 1) − R(k − 2) > 0 then in fact there is at least one sharp submodularity inequality in the above proof and so R(k − 1) − R(k − 2) > R(k) − R(k − 1). Since c > 0, this implies for any k ≥ 0 that if E ≠ ∅ and if Π ^M(k,c) = Π ^M(k + 1,c) then Π ^M(k + 1,c) > Π ^M(k + 2,c), concluding the proof of Proposition 1.□

Proof of Corollary 1

It is straightforward to check that Π ^M(k,c) is decreasing in c and has decreasing differences in (k,c). As c is taking values from the convex interval (0, ∞ ), the convexity of Π ^M(k,c) follows.□

Proof of Proposition 2

First, we analyze how the monopolist may change its strategy when one more edge is added to a network. The following lemma states sufficient conditions under which more WOM leads to less advertising.

Lemma 2

Let G(V,E) be a not complete network with |V| = n with the smallest degree in G being d. Let the cost of advertising be c, and denote the optimal advertising of the monopolist over G by k = k^M*(c). If n + 1 < (k + 1) ·(d + 2), then, when one extra relationship is added to E, the monopolist does not increase its advertising.

Proof

When adding one edge to the network, the degree of the two nodes connected by the new edge will increase by one, while the degree of every other node remains the same. In the proof of Proposition 1 we show that the expected return from any node depends only on the size of the network, advertising intensity and the degree of the node in question. Thus, the return to the same advertising intensity only changes because of the changes to the returns from the two nodes connected by the extra edge. Since we know that the monopolist does not find it beneficial to send the k + 1-st message to the original network, it is therefore sufficient to show that for any given node v ∈ V, increasing the degree of v by one decreases the return on the k + 1-st message sent by the monopolist.

Let v be a node of degree d(v). Then the (k + 1)-st message brings in

$$ \begin{array}{rll} \frac{{n-d(v)-1 \choose k}}{{n \choose k}} &-& \frac{{n-d(v)-1 \choose k+1}}{{n \choose k+1}}\\ &=&\frac{(n-d(v)-1) \cdot (n-d(v)-2) \cdot \ldots \cdot (n-d(v)-k) \cdot (n-k)}{{n \choose k+1} \cdot (k+1)!}\\ &&-\frac{(n-d(v)-1) \cdot (n-d(v)-2) \cdot \ldots \cdot (n-d(v)-k-1)}{{n \choose k+1} \cdot (k+1)!}\\ &=& \frac{(n-d(v)-1) \cdot (n-d(v)-2) \cdot \ldots \cdot (n-d(v)-k) \cdot (d(v)+1)}{{n \choose k+1} \cdot (k+1)!} \end{array} $$

extra revenue from v on the original network. Below, we show that this quantity is weakly decreasing in d(v), which is equivalent to showing that the numerator is decreasing in d(v). For d(v) + k ≥ n, the product is zero both before and after the extra edge is added. Below, we focus on the case d(v) + k < n.

From the condition in the statement of the lemma, we have that

$$ \begin{array}{rll} n+2 & \leq & (k+1) \cdot (d(v)+2) \\& \Leftrightarrow & n \leq k \cdot (d(v)+2) + d(v) \\& \Leftrightarrow & n-d(v)-k \leq k \cdot (d(v)+1) \\ & \Leftrightarrow &\frac{1}{d(v)+1} \leq \frac{k}{n-d(v)-k} < \sum\limits_{i=1}^{k}\frac{1}{n-d(v)-i}, \end{array} $$

and so

$$ \begin{array}{rll} &&{\kern-6pt}\frac{\partial (n-d(v)-1) \cdot (n-d(v)-2) \cdot \ldots \cdot (n-d(v)-k) \cdot (d(v)+1)}{\partial d(v)} \\ &&{\kern6pt}= (n-d(v)-1) \cdot (n-d(v)-2) \cdot \ldots \cdot (n-d(v)-k) \\ &&{\kern18pt}\cdot \left[ 1 - \sum\limits_{i=1}^{k}{\frac{d(v)+1}{n-d(v)-i}}\right] < 0 \end{array} $$

since the first k terms are all positive. This concludes the proof of Lemma 2.□

Let E _c denote the edges of the complete network: the set of all pairs of nodes from V. For any set of edges E over V, let $\overline{E}=E_c \setminus E$ denote the complement set of E. For any set of edges E ⊂ E _c and an edge $e \in \overline{E}$, let c(E,e) denote the maximum advertising cost for which k ^M*(c) ≥ (n + 1)/(d + 2) − 1, where d denotes the minimum degree in the network G(V,E). Note that for the advertising intensity k, if k + d < n, the minimum degree node is reached with probability less than 1. This guarantees that c(E,e) > 0. Let now $c^{*}=\min\limits_{E \subset E_c, e \in \overline{E}}{c(E,e)}$. (Because E _c is finite, c ^* must exist.) As we can get G ₂(V,E ₂) from G ₁(V,E ₁) in a finite number of steps by adding only one edge to the network in each step, this value clearly fulfills the conditions in Proposition 2.□

Proof of Proposition 3

Without loss of generality, for notational convenience, we prove the statement for $n=3m,\,m \in {\mathbb Z}^{+}$. We show that if the conditions hold, the monopolist sends at least m messages to the network. The fact that the network is connected implies that the minimum degree in the network is at least 1. Together with Lemma 2, this implies the claim.

From Proposition 1 we know that advertising exhibits diminishing returns. Thus, we only need to prove that the expected surplus from the m-th message to the monopolist is at least c, the cost of sending that message. The return on the m-th message from node v with degree d(v) is

$$ \begin{array}{rll} \frac{{3m-d(v)-1 \choose m-1}}{{3m \choose m-1}} - \frac{{3m-d(v)-1 \choose m}}{{3m \choose m}} & = & \frac{\frac{2m+1}{m} \cdot {3m-d(v)-1 \choose m-1} - {3m-d(v)-1 \choose m}}{{3m \choose m}}\\ & = & \frac{\frac{2m+1}{2m-d(v)} \cdot {3m-d(v)-1 \choose m}}{{3m \choose m}}. \end{array} $$

Clearly, the above quantity is decreasing in d(v). (It is necessarily positive since from the condition stated in the proposition, we have n = 3m > 0.75 (d + 1)² ≥ 1.5 ·(d + 1), i.e. 2m > d + 1.) Thus, it is enough to show that the return on the m-th advertising message from a node with the maximum degree d is at least c/(3m). Now,

$$ \begin{array}{rll} \frac{\frac{2m+1}{2m-d} \cdot {3m-d-1 \choose m}}{{3m \choose m}} & = & \frac{2m+1}{2m-d} \cdot \frac{3m-d-1}{3m} \cdot \frac{3m-d-2}{3m-1} \cdot \ldots \cdot \frac{2m-d}{2m+1}\\ & > & \left[\frac{2m-d-1}{2m}\right]^m\\ & = & \left[1-\frac{\frac{d+1}{2}}{m}\right]^m, \end{array} $$

which is known to be

$$ \begin{array}{rll} \left[1-\frac{\frac{d+1}{2}}{m}\right]^m & = & e^{-\frac{d+1}{2}} \cdot \left\{1 - \frac{\left(\frac{d+1}{2}\right)^2}{2m} - \frac{\left(\frac{d+1}{2}\right)^3 \cdot (8-3 \cdot \frac{d+1}{2})}{24m^2} + \ldots \right\} \\ & > & e^{-\frac{d+1}{2}} \cdot \left[ 1 - \frac{\left(\frac{d+1}{2}\right)^2}{m} \right] \\ & \geq & \frac{c}{3m}, \end{array} $$

where the last inequality follows from the condition stated in the proposition. This concludes the proof of Proposition 3.□

Proof of Proposition 4

Let v be a node in the network and k ₁ and k ₂ be the actions chosen by the two firms, respectively. By either firm, node v is reached if and only if it is either a target of an advertising message sent by the given firm, or if one of its neighbors are targeted. The corresponding probabilities can be calculated as in the proof of Proposition 1, yielding the following formula for the expected surplus of Firm i from node v:

$$ \mathop{\mathrm E}\limits_{|S_i|=k_i,|S_{-i}|=k_{-i}}\left[r_i(v)\right]=\left[{1-\frac{{{n-\textnormal{d}(v)-1} \choose k_i}}{{n \choose k_i}}}\right] \cdot \left[{\frac{{{n-\textnormal{d}(v)-1} \choose k_{-i}}}{{n \choose k_{-i}}}+p \cdot \left(1-\frac{{{n-\textnormal{d}(v)-1} \choose k_{-i}}}{{n \choose k_{-i}}}\right)}\right], $$

where $S_i,S_{-i} \subseteq V$ are the sets of targeted consumers by Firms i and − i respectively, chosen independently uniformly at random. Summing this up for every node in the network and taking into account the cost of sending messages, we arrive to the following formula for the profit of the firms competing in a duopoly:

$$ \Pi_{i}^D(k_i,k_{-i},p,c) = \sum\limits_{v \in V}\left[{1-\frac{{{n-\textnormal{d}(v)-1} \choose k_i}}{{n \choose k_i}}}\right] \cdot \left[{p+(1-p)\cdot\frac{{{n-\textnormal{d}(v)-1} \choose k_{-i}}}{{n \choose k_{-i}}}}\right]-k_i \cdot c. $$

To see the first half of the proposition, observe that in the above sum, the strategies of the two firms are interacting only through, for every node v, the probabilities that v is reached by the firms, respectively. As these probabilities are (weakly) increasing in the corresponding firm’s actions, we get that $\Pi_{i}^D(k_i,k_{-i},p,c)$ has decreasing differences in (k _i ,k _− i ). That is, the strategies of the competing firms are strategic substitutes.

Let us now formulate a game by reversing the actions of Firm 2, i.e., consider the game where the payoffs to the strategies (k ₁,n − k ₂) are the payoffs from the random targeting game to the strategies (k ₁,k ₂). By the above, the reversed game is supermodular (see also Vives 1990). Thus, by Topkis (1998), its equilibria form a non-empty complete lattice. Since the mapping of the strategy spaces is one-to-one, this proves the existence of a pure-strategy equilibrium in the game of random targeting.

To see the final statement of Proposition 4, w.l.o.g., let us show that if the condition is met then $k_1^{*} \neq 0$. Let $\text{BR}_i(k_j,p,c)$ denote the best response of Firm i when the other firm sends k _j messages at parameter values p (reduced surplus after jointly reached consumers) and c (cost per message). Consider $k^\prime=\text{BR}_2(0,p,c).$ It is obvious that k ^′ = k ^M*(c), and so we have $\text{BR}_1(\text{BR}_2(0,p,c),p,c) = \text{BR}_1(k^{M*}(c),p,c).$ Now, if

$$ \sum\limits_{v \in V}{\left[\frac{\textnormal{d}(v)+1}{n}\right]\cdot\left[p+(1-p)\cdot\frac{{{n-\textnormal{d}(v)-1} \choose {k^{M*}(c)}}}{{n \choose {k^{M*}(c)}}}\right]}>c, $$

then $\text{BR}_1(k^{M*}(c),p,c)>0$ (since the left-hand-side of the inequality is exactly the surplus of Firm 1 from sending one message at random) and so any equilibrium $k_1^{*}$ (and, symmetrically, $k_2^{*}$) is different from 0. This concludes the proof of Proposition 4.□

Proof of Corollary 2

Let the expected surplus for selecting k _i nodes when the competitor selects k _− i nodes (the two firms selecting their targets independently, uniformly at random) be

$$ R_i(k_1,k_2,p)=\frac{\sum\limits_{|S_1|=k_1}\sum\limits_{|S_2|=k_2}W_i(S_1,S_2,p)}{{n \choose k_1}\cdot{n \choose k_2}}. $$

Further, for any set $S_2 \subseteq V$, let

$$ R_1^{\prime}(k_1,S_2,p)=\frac{\sum\limits_{|S_1|=k_1}W_1(S_1,S_2,p)}{{n \choose k_1}}. $$

From Proposition 1 it follows that there is a k ^* such that for the surplus of a monopolist, R(k ^*) − R(k ^* − 1) > 0 but for any z ∈ ℕ, R(k ^* + z + 1) − R(k ^* + z) = 0. Moreover, we also have that for any 0 < k < k ^*, R(k) − R(k − 1) ≥ R(k ^*) − R(k ^* − 1).

Clearly, for any firm, independently from the competitor’s behavior, the surplus from a yet unreached node when reached is at least p (p or 1). Thus, with the above claims we get that for any $0< k_1 \leq k^{*}$ and any set $S_2 \subseteq V$,

$$ R_1^\prime(k_1,S_2,p)-R_1^\prime(k_1-1,S_2,p) \geq p \cdot (R(k^{*})-R(k^{*}-1)), $$

which, averaging over |S ₂| = k ₂, yields that for any k ₂,

$$ R_1(k_1,k_2,p)-R_1(k_1-1,k_2,p) \geq p \cdot (R(k^{*})-R(k^{*}-1)). $$

Similarly, for any $k_1 > k^{*}$ and any set k ₂, we have

$$ R_1(k_1,k_2,p)-R_1(k_1-1,k_2,p)=0. $$

Fix now any p ^* > 0 and let c ^* = p ^* ·(R(k ^*) − R(k ^* − 1)). Clearly, for any p, p ^* < p ≤ 1 and any c, 0 ≤ c < c ^*, the inequality c < p · (R(k ^*) − R(k ^* − 1)) holds and so the state (k ^*,k ^*) is the unique equilibrium of the random targeting game.□

Proof of Proposition 5

Our proof is essentially an application of Theorem 3 in Villas-Boas (1997). For the reason that a few conditions slightly differ in our case, we here re-prove the theorem, applying it to our context. Throughout this proof, we implicitly rely on the transitivity and reflexivity of the standard component-wise order over ℕ × ℕ.

Let T ₁,T ₂: ℕ × ℕ → ℕ × ℕ be the mappings defined as T ₁(k ₁,k ₂) = $\left(\textnormal{BR}_2(k_1,\!p^\prime\!,\!c^\prime),\textnormal{BR}_1(k_2,p^\prime\!,c^\prime)\right)$ and $T_2(k_1\!,k_2)\!=\!\left(\textnormal{BR}_2(k_1,\!p,\!c),\textnormal{BR}_1(k_2,\!p,c)\right)$. Applying considerations detailed in the proofs of Results 1 and 2 we can easily see the following.

Observation 1

T₁ is weakly higher than T₂, i.e., for any k₁,k₂ ∈ ℕ, T₁(k₁,k₂) ≥ T₂(k₁,k₂) where the order is the standard component-wise order.

Furthermore, by the statement of Proposition 5, we know that $(k_1^{*},k_2^{*})$ is a fixed point of T ₂ and $(k_1^{\prime},k_2^{\prime})$ is a fixed point of T ₁. Finally, T ₁ is a weakly decreasing mapping, i.e., for any (k ₁,k ₂) ≥ (l ₁,l ₂), we have T ₁(l ₁,l ₂) ≥ T ₁(k ₁,k ₂).

Suppose now that $(k_1^{*},k_2^{*}) > (k_1^{\prime},k_2^{\prime})$. Because T ₁ is weakly higher than T ₂, we have

$$ T_1(k_1^{*},k_2^{*}) \geq T_2(k_1^{*},k_2^{*}) = (k_1^{*},k_2^{*}) > (k_1^{\prime},k_2^{\prime}) = T_1(k_1^{\prime},k_2^{\prime}). $$

But, because T ₁ is weakly decreasing, we have $T_1(k_1^{\prime},k_2^{\prime}) \geq T_1(k_1^{*},k_2^{*}),$ a contradiction.□

Proof of Proposition 6

Since the action sets of both players are finite, the existence of the Stackelberg equilibrium is trivial. Now assume that $k_1^{S*}<k_1^{*}$. Then $k_2^{S*}=\textnormal{BR}_2(k_1^{S*}(p,c)) \geq \textnormal{BR}_2(k_1^{*},p,c)=k_2^{*}$, and so, using the formula presented in the proof of Proposition 4 describing the profits of the two firms, in particular that Firm 1’s profits decrease in the action of Firm 2, we get that

$$ \begin{array}{rll} \Pi_1^D(k_1^{S*},k_2^{S*},p,c) &\leq \Pi_1^D(k_1^{S*},k_2^{*},p,c) \\ &\leq \Pi_1^D(\textnormal{BR}_1(k_2^{*},p,c),k_2^{*},p,c) = \Pi_1^D(k_1^{*},k_2^{*},p,c), \end{array} $$

a contradiction. Thus, $k_1^{S*} \!\geq \!k_1^{*}$, and so $k_2^{S*}\!=\!\textnormal{BR}_2(k_1^{S*}(p,\!c)) \!\leq\! \textnormal{BR}_2(k_1^{*}(p,\!c))\! =$ $k_2^{*}$ easily follows.

To see the second claim, let us examine what happens when Firm 1 sends advertising to all consumers in the network. In this case, Firm 2 only receives p surplus after each consumer reached. Normalizing Firm 2’s problem to have the reduced form presented in Section 3, we get that $\textnormal{BR}_2(n,p,c)=k^{M*}(c/p)$. Thus, the profits of Firm 1 in this case would be

$$ \sum\limits_{v \in V}{\left[p+(1-p)\cdot \frac{{{n-d(v)-1} \choose {k^{M*}(c/p)}}}{{{n} \choose {k^{M*}(c/p)}}}\right]}-n \cdot c. $$

Comparing this level to the profits obtained when sending k ^M*(c) messages, we find that Firm 1 prefers saturating the entire market over sending only k ^M*(c) messages. Since the cost of the latter action is less than that of sending n messages, we also get that

$$ R(k^{M*}(c),\textnormal{BR}_2(k^{M*}(c),p,c),p) \leq R(n,k^{M*}(c/p),p). $$

Now, using that R(k ₁ + 1,k ₂,p) − R(k ₁,k ₂,p) is decreasing in k ₁ (which follows exactly as the decreasing property of R(k + 1) − R(k) in the proof of Proposition 1), we may conclude that for any 0 ≤ k ≤ k ^M*, we have $R(k,\textnormal{BR}_2(k,\,p,\,c),\,p)\ \leq \ R\,(k,\,\textnormal{BR}_2(k^{M*}(c),\,p,\,c),\,p) \,\leq \,R(k^{M*}(c),$ $\textnormal{BR}_2(k^{M*}(c),p,c),p)$. By the condition stated in the proposition, it then follows that Firm 1 prefers the action n over every action no larger than k ^M*(c). Hence, $k_1^{S*}(p,c)>k^{M*}(c)$.□

Proof of Proposition 7

Let w ∈ V ^′ and k ₁ and k ₂ be the actions chosen by the two firms, respectively. By either firm, node w is reached if and only if one of its neighbors (in V by our assumption) are targeted. The corresponding probabilities can be calculated as in the proof of Proposition 1, yielding the following formula for the expected surplus of Firm i from node w:

$$ \mathop{\mathrm E}\limits_{|S_i|=k_i,|S_{-i}|=k_{-i}}\left[r_i(w)\right]=\left[{1-\frac{{{n-\textnormal{d}(w)} \choose k_i}}{{n \choose k_i}}}\right] \cdot \left[{\frac{{{n-\textnormal{d}(w)} \choose k_{-i}}}{{n \choose k_{-i}}}+p \cdot \left(1-\frac{{{n-\textnormal{d}(w)} \choose k_{-i}}}{{n \choose k_{-i}}}\right)}\right], $$

where $S_i,S_{-i} \subseteq V$ are the sets of targeted consumers by Firms i and − i respectively, chosen independently uniformly at random. Summing this up for every node in V ^′ and the expected surplus for members of V (exactly as in the proof of Proposition 4) and taking into account the cost of sending messages, we arrive to the following formula for the profit of the firms competing in a duopoly with endogenously determined market size:

$$ \begin{array}{rll} \Pi_{i}^D(k_i,k_{-i},p,c) & = & \sum\limits_{v \in V}\left[{1-\frac{{{n-\textnormal{d}(v)-1} \choose k_i}}{{n \choose k_i}}}\right] \cdot \left[{p+(1-p)\cdot\frac{{{n-\textnormal{d}(v)-1} \choose k_{-i}}}{{n \choose k_{-i}}}}\right] \\ & & +\sum\limits_{w \in V^\prime}\left[{1-\frac{{{n-\textnormal{d}(w)} \choose k_i}}{{n \choose k_i}}}\right] \cdot \left[{p+(1-p)\cdot\frac{{{n-\textnormal{d}(w)} \choose k_{-i}}}{{n \choose k_{-i}}}}\right] - k_i \cdot c. \end{array} $$

Just as in the proof of Proposition 4, the strategies of the two competing firms are strategic substitutes. The existence of the pure-strategy equilibrium follows exactly as in the proof of Proposition 4.

Now, to see the second claim, we use the same technique as in the proof of Proposition 5. Let T ₁,T ₂: ℕ ×ℕ →ℕ ×ℕ be the mappings defined as $T_1(k_1,k_2)=\left(\textnormal{BR}_2(k_1,p^\prime,c^\prime),\textnormal{BR}_1(k_2,p^\prime,c^\prime)\right)$ on G ^′ and $T_2(k_1,k_2)=\left(\textnormal{BR}_2(k_1,p,c),\textnormal{BR}_1(k_2,p,c)\right)$ on G. Since the expected return to the same advertising campaign cannot be lower on G ^′ than on G, it is easy to see the following.

Table 3 Payoff matrices of the four networks displayed on Fig. 3 in a simultaneous-move duopoly competition where p = 0.1 and c = 0.3. Best responses shown in bold

Observation 2

T₁ is weakly higher than T₂, i.e., for any k₁,k₂ ∈ ℕ, T₁(k₁,k₂) ≥ T₂(k₁,k₂) where the order is the standard component-wise order.

Furthermore, by the statement of the proposition, we know that $(k_1^{*},k_2^{*})$ is a fixed point of T ₂ and $(k_1^{\prime},k_2^{\prime})$ is a fixed point of T ₁. Finally, T ₁ is a weakly decreasing mapping, i.e., for any (k ₁,k ₂) ≥ (l ₁,l ₂), we have T ₁(l ₁,l ₂) ≥ T ₁(k ₁,k ₂).

Suppose now that $(k_1^{*},k_2^{*}) > (k_1^{\prime},k_2^{\prime})$. Because T ₁ is weakly higher than T ₂, we have

$$ T_1(k_1^{*},k_2^{*}) \geq T_2(k_1^{*},k_2^{*}) = (k_1^{*},k_2^{*}) > (k_1^{\prime},k_2^{\prime}) = T_1(k_1^{\prime},k_2^{\prime}). $$

But, because T ₁ is weakly decreasing, we have $T_1(k_1^{\prime},k_2^{\prime}) \geq T_1(k_1^{*},k_2^{*}),$ a contradiction.□