One of the objectives of this paper is to understand the impact of link addition and deletion on storage availability for those agents who are involved in the link addition/deletion as well as those who are not. The storage availability is determined by the distances between them and their closeness (from Equation (

2)). Therefore, first we study how addition and deletion of a link impacts the shortest distances between pairs of agents and, therefore, their closeness. This analysis provides a base for understanding the effect of link-addition/deletion on agents’ storage availability in

$\mathfrak{g}$.

#### 3.1. Effect of Link Alteration on Closeness

**Lemma** **1.** Suppose $\langle ij\rangle \notin \mathfrak{g}$. Then, ${\mathsf{\Phi}}_{i}(\mathfrak{g}+\langle ij\rangle )>{\mathsf{\Phi}}_{i}(\mathfrak{g})$.

**Proof.** Clearly, ${d}_{ij}(\mathfrak{g}+\langle ij\rangle )<{d}_{ij}(\mathfrak{g})$. As $\langle ij\rangle \notin \mathfrak{g}$, we have, ${d}_{ij}(\mathfrak{g})\ge 2$. Also, ${d}_{ij}(\mathfrak{g}+\langle ij\rangle )=1$. Thus, ${\mathsf{\Phi}}_{i}(\mathfrak{g})$ and ${\mathsf{\Phi}}_{j}(\mathfrak{g})$ increase by at least $\frac{{d}_{ij}(\mathfrak{g})-1}{{d}_{ij}(\mathfrak{g})}$ in $\mathfrak{g}+\langle ij\rangle $. ☐

**Lemma** **2.** Suppose $\langle ij\rangle \in \mathfrak{g}$. Then, ${\mathsf{\Phi}}_{i}(\mathfrak{g}-\langle ij\rangle )<{\mathsf{\Phi}}_{i}(\mathfrak{g})$.

**Proof.** Let us assume there is no path between i and j in $\mathfrak{g}-\langle ij\rangle $, then ${d}_{ij}(\mathfrak{g}-\langle ij\rangle )=\infty $, thus, ${\mathsf{\Phi}}_{i}(\mathfrak{g})$ and ${\mathsf{\Phi}}_{j}(\mathfrak{g})$ decrease by 1 in $\mathfrak{g}-\langle ij\rangle $.

Now, let us assume there exists a path ${\mathcal{P}}_{ij}(\mathfrak{g}-\langle ij\rangle )$ between i and j in $\mathfrak{g}-\langle ij\rangle $, the distance between i and j in $\mathfrak{g}-\langle ij\rangle $ being at least 1 more than that in $\mathfrak{g}$. Thus, ${\mathsf{\Phi}}_{i}(\mathfrak{g})$ and ${\mathsf{\Phi}}_{j}(\mathfrak{g})$ decrease by at least $\frac{{d}_{ij}(\mathfrak{g}-\langle ij\rangle )-1}{{d}_{ij}(\mathfrak{g}-\langle ij\rangle )}$ in $\mathfrak{g}-\langle ij\rangle $. ☐

Lemmas 1 and 2 show that, with respect to closeness, every link benefits agents on either side of the link. An action of link addition or deletion between a pair of agents not only impacts their closeness, but also that of other agents. Now, we study the impact of link addition or deletion between a pair of agents (say, i and j) on the closeness of the other agents $k\in \mathfrak{g}\setminus \{i,j\}$.

**Lemma** **3.** Suppose $\langle ij\rangle \notin \mathfrak{g}$ and $k\in \mathfrak{g}\setminus \{i,j\}$. Then, ${\mathsf{\Phi}}_{k}(\mathfrak{g})={\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$ if and only if ${d}_{kl}(\mathfrak{g})={d}_{kl}(\mathfrak{g}+\langle ij\rangle )$ for all $l\in \mathfrak{g}$.

**Proof.** If

${d}_{kl}(\mathfrak{g})={d}_{kl}(\mathfrak{g}+\langle ij\rangle )$ for all

$l\in \mathfrak{g}$, then by Equation (

1),

${\mathsf{\Phi}}_{k}(\mathfrak{g})={\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$.

Conversely, suppose ${\mathsf{\Phi}}_{k}(\mathfrak{g})={\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$.

It is easy to see that, if for some $l\in \mathfrak{g}$, if ${d}_{kl}(\mathfrak{g}+\langle ij\rangle )\ne {d}_{kl}(\mathfrak{g})$, then ${d}_{kl}(\mathfrak{g}+\langle ij\rangle )<{d}_{kl}(\mathfrak{g})$. (Paths in $\mathfrak{g}$ exist in $\mathfrak{g}+\langle ij\rangle $ too).

We have ${d}_{kl}(\mathfrak{g}+\langle ij\rangle )\le {d}_{kl}(\mathfrak{g})$ for all $l\in \mathfrak{g}$ and, if there exists x such that ${d}_{kx}(\mathfrak{g}+\langle ij\rangle )<{d}_{kx}(\mathfrak{g})$, then ${\mathsf{\Phi}}_{k}(\mathfrak{g})<{\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$, a contradiction. ☐

**Lemma** **4.** Suppose $\langle ij\rangle \in \mathfrak{g}$ and $k\in \mathfrak{g}\setminus \{i,j\}$. Then, ${\mathsf{\Phi}}_{k}(\mathfrak{g})={\mathsf{\Phi}}_{k}(\mathfrak{g}-\langle ij\rangle )$ if and only if ${d}_{kl}(\mathfrak{g})={d}_{kl}(\mathfrak{g}-\langle ij\rangle )$ for all $l\in \mathfrak{g}$.

**Proof.** As ${d}_{kl}(\mathfrak{g}-\langle ij\rangle )\ge {d}_{kl}(\mathfrak{g})$ for all l, the proof follows in lines similar to that of Lemma 3. ☐

We now show necessary and sufficient conditions for increase in the closeness of agents who are not involved in link addition or deletion.

**Theorem** **1.** Suppose $\langle ij\rangle \notin \mathfrak{g}$, and let k be an agent distinct from i and j. Then, ${\mathsf{\Phi}}_{k}(\mathfrak{g})<{\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$ if and only if there exists at least one agent $l\in \mathfrak{g}$ such that ${d}_{kl}(\mathfrak{g})\ge 3$ and all shortest paths ${\mathcal{P}}_{kl}(\mathfrak{g}+\langle ij\rangle )$ from k to l in $\mathfrak{g}+\langle ij\rangle $ contain $\langle ij\rangle $.

**Proof.** Let ${\mathsf{\Phi}}_{k}(\mathfrak{g})<{\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$. Then, by Lemma 3, there must be at least one agent, say l, such that ${d}_{kl}(\mathfrak{g})>{d}_{kl}(\mathfrak{g}+\langle ij\rangle )$.

Suppose $i,k,$ and l are all distinct. Note that j may be the same as l.

If possible, let ${d}_{kl}(\mathfrak{g})<{d}_{ki}(\mathfrak{g})+{d}_{ij}(\mathfrak{g})+{d}_{jl}(\mathfrak{g})$ for all $l\in \mathfrak{g}$. Then, ${d}_{kl}(\mathfrak{g})={d}_{kl}(\mathfrak{g}+\langle ij\rangle )$ for all $l\in \mathfrak{g}$. From Lemma 1, ${\mathsf{\Phi}}_{k}(\mathfrak{g})<{\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$, a contradiction. Therefore, there exists an $l\in \mathfrak{g}$ such that ${d}_{kl}(\mathfrak{g})={d}_{ki}(\mathfrak{g})+{d}_{ij}(\mathfrak{g})+{d}_{jl}(\mathfrak{g})$.

As $\langle ij\rangle \notin \mathfrak{g}$, ${d}_{ij}(\mathfrak{g})\ge 2$. As $k\ne i$, ${d}_{ik}(\mathfrak{g})\ge 1$ and $j=l$. Hence, ${d}_{kl}(\mathfrak{g})\ge 3$.

Now, $\begin{array}{ll}{d}_{kl}(\mathfrak{g}+\langle ij\rangle )& ={d}_{ki}(\mathfrak{g}+\langle ij\rangle )+{d}_{ij}(\mathfrak{g}+\langle ij\rangle )+{d}_{jl}(\mathfrak{g}+\langle ij\rangle )\\ & ={d}_{ki}(\mathfrak{g})+{d}_{ij}(\mathfrak{g}+\langle ij\rangle )+{d}_{jl}(\mathfrak{g})\\ & <{d}_{ki}(\mathfrak{g})+{d}_{ij}(\mathfrak{g})+{d}_{jl}(\mathfrak{g})\\ & ={d}_{kl}(\mathfrak{g})\end{array}$

It follows that every shortest path between k and l in $\mathfrak{g}+\langle ij\rangle $ contains $\langle ij\rangle $. (Note that if there exists a shortest path from k to l in $\mathfrak{g}+\langle ij\rangle $ that does not contain $\langle ij\rangle $, then this shortest path exists in $\mathfrak{g}$ too).

Conversely, let $l\in \mathfrak{g}$ such that ${d}_{kl}(\mathfrak{g})\ge 3$ and all shortest paths ${\mathcal{P}}_{kl}(\mathfrak{g}+\langle ij\rangle )$ from k to l in $\mathfrak{g}+\langle ij\rangle $ contain $\langle ij\rangle $.

Clearly, ${\mathsf{\Phi}}_{k}(\mathfrak{g})\le {\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$.

If possible, let ${\mathsf{\Phi}}_{k}(\mathfrak{g})={\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$. This means for every l in $\mathfrak{g}$ there exists a shortest path from k to l in $\mathfrak{g}+\langle ij\rangle $ that does not contain $\langle ij\rangle $, a contradiction. Therefore, ${\mathsf{\Phi}}_{k}(\mathfrak{g})<{\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$. ☐

**Theorem** **2.** Suppose $\langle ij\rangle \in \mathfrak{g}$, and let k be an agent distinct from i and j. Then, ${\mathsf{\Phi}}_{k}(\mathfrak{g}-\langle ij\rangle )<{\mathsf{\Phi}}_{k}(\mathfrak{g})$ if and only if there exists at least one agent $l\in \mathfrak{g}$ such that ${d}_{kl}(\mathfrak{g})\ge 2$ and all shortest paths ${\mathcal{P}}_{kl}(\mathfrak{g})$ from k to l in $\mathfrak{g}$ contain $\langle ij\rangle $.

We skip the proof as it is similar to the proof of Theorem 1.

In subsequent sections, we present our results due to link addition. We present our results on link deletion in

Appendix B.

#### 3.2. Effect of Closeness on Distances of Agents Not Involved in Link Alteration

In this section, we classify agents whose mutual distances from each other remain the same after link alteration. We use the same to analyze the effect of closeness on distances between agents who are not involved in the link addition or deletion.

Given k such that ${\mathsf{\Phi}}_{k}(\mathfrak{g})<{\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$, we use ${L}_{k}^{+}$ to denote the set of all $l\in \mathfrak{g}$ such that all shortest paths from k to l in $\mathfrak{g}+\langle ij\rangle $ contain $\langle ij\rangle $. We use ${l}_{k}^{+}$ to denote an agent in ${L}_{k}^{+}$.

**Proposition** **1.** Suppose i, j, and k are distinct agents in $\mathfrak{g}$. Suppose l is another agent, distinct from i and k, and suppose ${\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )>{\mathsf{\Phi}}_{k}(\mathfrak{g})$. If ${d}_{ki}(\mathfrak{g}+\langle ij\rangle )<{d}_{kj}(\mathfrak{g}+\langle ij\rangle )\le {d}_{kl}(\mathfrak{g}+\langle ij\rangle )$, then ${d}_{ik}(\mathfrak{g}\phantom{\rule{3.33333pt}{0ex}}+\phantom{\rule{3.33333pt}{0ex}}\langle ij\rangle )\phantom{\rule{3.33333pt}{0ex}}=\phantom{\rule{3.33333pt}{0ex}}{d}_{ik}(\mathfrak{g})$.

**Proof.** We have ${\mathsf{\Phi}}_{k}(\mathfrak{g})<{\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$. Then, from Theorem 1, there exists $l\in \mathfrak{g}$ such that all shortest paths ${\mathcal{P}}_{kl}(\mathfrak{g}+\langle ij\rangle )$ from k to l in $\mathfrak{g}+\langle ij\rangle $ contain $\langle ij\rangle $.

We consider the two cases $j=l$ and $j\ne l$.

Suppose $j=l$. As ${d}_{ki}(\mathfrak{g}+\langle ij\rangle )<{d}_{kj}(\mathfrak{g}+\langle ij\rangle )$, k observes i before j on all shortest paths ${\mathcal{P}}_{kl}(\mathfrak{g}+\langle ij\rangle )$. This implies ${d}_{ik}(\mathfrak{g}+\langle ij\rangle )={d}_{ik}(\mathfrak{g})$.

Suppose $j\ne l$. As ${d}_{ki}(\mathfrak{g}+\langle ij\rangle )<{d}_{kj}(\mathfrak{g}+\langle ij\rangle )\le {d}_{kl}(\mathfrak{g}+\langle ij\rangle )$, k observes i before j, and j before l, on all shortest paths ${\mathcal{P}}_{kl}(\mathfrak{g}+\langle ij\rangle )$. This implies ${d}_{ik}(\mathfrak{g}+\langle ij\rangle )={d}_{ik}(\mathfrak{g})$. ☐

**Definition** **3.** Suppose $\langle ij\rangle \notin \mathfrak{g}$ and k is an agent such that ${\mathsf{\Phi}}_{k}(\mathfrak{g})<{\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$. A $(k,+ij)$-shortest-path-network, ${\mathfrak{g}}_{ij}^{{k}^{+}}$, is a subnetwork of $\mathfrak{g}+\langle ij\rangle $ that consists of all shortest paths from k to ${l}_{k}^{+}$ in $\mathfrak{g}+\langle ij\rangle $, which contain $\langle ij\rangle $, for all ${l}_{k}^{+}\in {L}_{k}^{+}$.

**Definition** **4.** An (all $k,+ij)$-shortest-path-network, ${\mathfrak{g}}_{ij}^{+}$, is $\bigcup _{{\displaystyle \underset{{\mathsf{\Phi}}_{k}(\mathfrak{g})<{\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )}{k\in \mathfrak{g},}}}}{\mathfrak{g}}_{ij}^{{k}^{+}$, the smallest network consisting of all $(k,+ij)$-shortest-path-networks.

**Definition** **5.** A sub-$(i,+)$-network, ${\mathfrak{g}}_{i}^{+}$ of ${\mathfrak{g}}_{ij}^{+}$, is the induced subnetwork of ${\mathfrak{g}}_{ij}^{+}$ consisting of all agents $k\in {\mathfrak{g}}_{ij}^{+}$ such that ${d}_{ik}(\mathfrak{g})={d}_{ik}(\mathfrak{g}+\langle ij\rangle )$. Similarly, we define the sub-$(j,+)$-network, ${\mathfrak{g}}_{j}^{+}$ of ${\mathfrak{g}}_{ij}^{+}$, as the induced subnetwork of ${\mathfrak{g}}_{ij}^{+}$ consisting of all agents $l\in {\mathfrak{g}}_{ij}^{+}$ such that ${d}_{jl}(\mathfrak{g})={d}_{jl}(\mathfrak{g}+\langle ij\rangle )$.

Refer to

Appendix A for an illustration of the above definitions.

**Proposition** **2.** For all $k,\overline{k}\in {\mathfrak{g}}_{i}^{+}$, ${d}_{k\overline{k}}(\mathfrak{g})={d}_{k\overline{k}}(\mathfrak{g}+\langle ij\rangle )$.

**Proof.** If $k,\overline{k}\in {\mathfrak{g}}_{i}^{+}$ then, from Definition 5, ${d}_{ik}(\mathfrak{g})={d}_{ik}(\mathfrak{g}+\langle ij\rangle )$ and ${d}_{i\overline{k}}(\mathfrak{g})={d}_{i\overline{k}}(\mathfrak{g}+\langle ij\rangle )$. As $k,\overline{k}\in {\mathfrak{g}}_{ij}^{+}$ as well, there exists l and $\overline{l}$ such that ${d}_{kl}(\mathfrak{g})>{d}_{kl}(\mathfrak{g}+\langle ij\rangle )$ and ${d}_{\overline{k}\overline{l}}(\mathfrak{g})>{d}_{\overline{k}\overline{l}}(\mathfrak{g}+\langle ij\rangle )$.

It is sufficient to show that, given $\overline{k}$, $\overline{l}$ can never be k.

If possible, let $\overline{l}=k$. Then, from Definition 5, ${d}_{\overline{k}i}(\mathfrak{g})={d}_{\overline{k}i}(\mathfrak{g}+\langle ij\rangle )$ implies $\overline{k}$ observes i first, and subsequently j to reach k, on all shortest paths ${\mathcal{P}}_{\overline{k}k}(\mathfrak{g}+\langle ij\rangle )$ from $\overline{k}$ to k in $\mathfrak{g}+\langle ij\rangle $. Then, ${d}_{ik}(\mathfrak{g})\ne {d}_{ik}(\mathfrak{g}+\langle ij\rangle )$. This is because, if $k=j$, ${d}_{ik}(\mathfrak{g})<{d}_{ik}(\mathfrak{g}+\langle ij\rangle =1$. Therefore, $k\notin {\mathfrak{g}}_{i}^{+}$, which is a contradiction. Now, if $k\ne j$, then k must first visit j, and later i, to reach $\overline{k}$ on all shortest paths ${\mathcal{P}}_{k\overline{k}}(\mathfrak{g}+\langle ij\rangle )$ from $\overline{k}$ to k. This implies ${d}_{ik}(\mathfrak{g})\ne {d}_{ik}(\mathfrak{g}+\langle ij\rangle )$ and hence, $k\notin {\mathfrak{g}}_{i}^{+}$, again, a contradiction. Thus, $k\ne \overline{l}$. ☐

We discuss our results on shortest distances due to link deletion in

Appendix B.1.

#### 3.3. Effect of Link Alteration on Storage Availability

Our aim here is to analyze under what conditions agents’ chance of obtaining storage space in the network increases or decreases by adding a new link. We present our results in the case of link deletion in

Appendix B.2.

**Lemma** **5.** Suppose agent i and j add a direct link in $\mathfrak{g}$ and let $k\notin {\mathfrak{g}}_{ij}^{+}$. Then, ${\alpha}_{ik}(\mathfrak{g})={\alpha}_{ik}(\mathfrak{g}+\langle ij\rangle )$ and ${\alpha}_{jk}(\mathfrak{g})={\alpha}_{jk}(\mathfrak{g}+\langle ij\rangle )$.

**Proof.** If agent

$k\notin {\mathfrak{g}}_{ij}^{+}$ then

${\mathsf{\Phi}}_{k}(\mathfrak{g})={\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$. Thus,

${d}_{ki}(\mathfrak{g})={d}_{ki}(\mathfrak{g}+\langle ij\rangle )$. Therefore, from Equation (

2),

${\alpha}_{ik}(\mathfrak{g})={\alpha}_{ik}(\mathfrak{g}+\langle ij\rangle )$. A similar proof holds for

j too. ☐

**Lemma** **6.** Suppose agents i, j, k, and l are such that $i\ne j$, $j\ne k$, $i\ne l$, and $k\ne l$. (Agents i and k may be the same, and agents j and l may be the same). Suppose $\langle ij\rangle \notin \mathfrak{g}$, $k\in {\mathfrak{g}}_{i}^{+}$, and $l\in {\mathfrak{g}}_{j}^{+}$. Then,

${\alpha}_{kl}(\mathfrak{g})<{\alpha}_{kl}(\mathfrak{g}+\langle ij\rangle )$, and

$i\ne k$ implies that ${\alpha}_{ik}(\mathfrak{g})>{\alpha}_{ik}(\mathfrak{g}+\langle ij\rangle )$. Similarly, if $j\ne l$, then ${\alpha}_{jl}(\mathfrak{g})>{\alpha}_{jl}(\mathfrak{g}+\langle ij\rangle )$.

**Lemma** **7.** Let k and $\overline{k}$ be agents in ${\mathfrak{g}}_{i}^{+}$. Then, ${\alpha}_{k\overline{k}}(\mathfrak{g})={\alpha}_{k\overline{k}}(\mathfrak{g}+\langle ij\rangle )$ and ${\alpha}_{\overline{k}k}(\mathfrak{g})={\alpha}_{\overline{k}k}(\mathfrak{g}+\langle ij\rangle )$.

**Proof.** The proof follows from Proposition 2. ☐

**Theorem** **3.** Suppose agents i and j are such $i\ne j$, and $\langle ij\rangle \notin \mathfrak{g}$. Then, ${\gamma}_{i}(\mathfrak{g})<{\gamma}_{i}(\mathfrak{g}+\langle ij\rangle )$ if and only if $\frac{{\displaystyle \prod _{k\in {\mathfrak{g}}_{i}^{+}}}(1-{\alpha}_{ik}(\mathfrak{g}+\langle ij\rangle ))}{{\displaystyle \prod _{l\in {\mathfrak{g}}_{j}^{+}}}(1-{\alpha}_{il}(\mathfrak{g}))}<\frac{{\displaystyle \prod _{k\in {\mathfrak{g}}_{i}^{+}}}(1-{\alpha}_{ik}(\mathfrak{g}))}{{\displaystyle \prod _{l\in {\mathfrak{g}}_{j}^{+}}}(1-{\alpha}_{il}(\mathfrak{g}+\langle ij\rangle ))}$.

Additionally, ${\gamma}_{i}(\mathfrak{g})<{\gamma}_{i}(\mathfrak{g}+\langle ij\rangle )$ if and only if $\frac{{\displaystyle \prod _{k\in {\mathfrak{g}}_{i}^{+}}}({\alpha}_{ik}(\mathfrak{g}+\langle ij\rangle ))}{{\displaystyle \prod _{l\in {\mathfrak{g}}_{j}^{+}}}({\alpha}_{il}(\mathfrak{g}))}>\frac{{\displaystyle \prod _{k\in {\mathfrak{g}}_{i}^{+}}}({\alpha}_{ik}(\mathfrak{g}))}{{\displaystyle \prod _{l\in {\mathfrak{g}}_{j}^{+}}}({\alpha}_{il}(\mathfrak{g}+\langle ij\rangle ))}$.

**Proof.** The proof follows from Lemmas 5, 6, and 7. ☐

#### 3.4. Externalities

In this section, we study externalities, that is, how a link that is added between a pair of agents affects the utility of others. (Refer to Definition 6). The particular form of externalities (positive, negative, or none) is crucial in determining which network is likely to evolve and the conditions under which it will lead to a stable and efficient network.

**Definition** **6.** [32] Consider a network, $\mathfrak{g}$, with agents $i,j\in \mathfrak{g}$ such that $i\ne j$ and $\langle ij\rangle \notin \mathfrak{g}$. Suppose agents i and j form a direct link $\langle ij\rangle $. Then, agent $k\in \mathfrak{g}\setminus \{i,j\}$ experiences Positive externalities if ${u}_{k}(\mathfrak{g}+\langle ij\rangle )>{u}_{k}(\mathfrak{g})$;

Negative externalities if ${u}_{k}(\mathfrak{g}+\langle ij\rangle )<{u}_{k}(\mathfrak{g})$;

No externalities if ${u}_{k}(\mathfrak{g}+\langle ij\rangle )={u}_{k}(\mathfrak{g})$.

We now show that the type of externalities an agent $k\in \mathfrak{g}$ experiences, can be determined using conditions on the storage availability, independent of the data loss rate and the value that agents associate with their data.

**Proposition** **3.** In an SSSC $\mathfrak{g}$, an agent $k\in \mathfrak{g}$ experiences

Positive externalities if ${\gamma}_{k}(\mathfrak{g}+\langle ij\rangle )>{\gamma}_{k}(\mathfrak{g})$;

Negative externalities if ${\gamma}_{k}(\mathfrak{g}+\langle ij\rangle )<{\gamma}_{k}(\mathfrak{g})$;

No externalities if ${\gamma}_{k}(\mathfrak{g}+\langle ij\rangle )={\gamma}_{k}(\mathfrak{g})$.

**Proof.** By Definition 6, ${u}_{k}(\mathfrak{g}+\langle ij\rangle )>{u}_{k}(\mathfrak{g})$

$\Rightarrow \beta (1-\lambda )+\beta \lambda {\gamma}_{k}(\mathfrak{g}+\langle ij\rangle )-\varsigma {\eta}_{k}(\mathfrak{g}+\langle ij\rangle )>\beta (1-\lambda )+\beta \lambda {\gamma}_{k}(\mathfrak{g})-\varsigma {\eta}_{k}(\mathfrak{g})$.

As agent k does not pay the cost for link $\langle ij\rangle $, we have $\varsigma {\eta}_{k}(\mathfrak{g}+\langle ij\rangle )=\varsigma {\eta}_{k}(\mathfrak{g})$.

Thus, $\beta \lambda {\gamma}_{k}(\mathfrak{g}+\langle ij\rangle )>\beta \lambda {\gamma}_{k}(\mathfrak{g})\Rightarrow {\gamma}_{k}(\mathfrak{g}+\langle ij\rangle )>{\gamma}_{k}(\mathfrak{g})$.

For Cases 2 and 3, the proof is similar to that of Case 1. ☐

The following results provide a necessary and sufficient condition under which an agent $k\in \mathfrak{g}$ experiences positive or negative externalities.

**Lemma** **8.** Let i, j, and k be distinct agents in $\mathfrak{g}$. Suppose $k\notin {\mathfrak{g}}_{ij}^{+}$. Then, k experiences only negative externalities.

**Proof.** If agents

i and

j add a direct link in

$\mathfrak{g}$, then, from Lemma 1,

${\mathsf{\Phi}}_{i}(\mathfrak{g})<{\mathsf{\Phi}}_{i}(\mathfrak{g}+\langle ij\rangle )$. If

$k\notin {\mathfrak{g}}_{ij}^{+}$, then, from Theorem 1,

${\mathsf{\Phi}}_{k}(\mathfrak{g})={\mathsf{\Phi}}_{k}(\mathfrak{g}+\langle ij\rangle )$, thus,

${d}_{kl}(\mathfrak{g})={d}_{kl}(\mathfrak{g}+\langle ij\rangle )$ for all

$l\in \mathfrak{g}$. Therefore,

${\alpha}_{ki}(\mathfrak{g}+\langle ij\rangle )<{\alpha}_{ki}(\mathfrak{g})$, by Equation (

2). Now, for all

$l\in \mathfrak{g}$,

${\mathsf{\Phi}}_{l}(\mathfrak{g})\le {\mathsf{\Phi}}_{l}(\mathfrak{g}+\langle ij\rangle )$. If

${\mathsf{\Phi}}_{l}(\mathfrak{g})={\mathsf{\Phi}}_{l}(\mathfrak{g}+\langle ij\rangle )$, then

${\alpha}_{kl}(\mathfrak{g}+\langle ij\rangle )={\alpha}_{kl}(\mathfrak{g})$ and, if

${\mathsf{\Phi}}_{l}(\mathfrak{g})<{\mathsf{\Phi}}_{l}(\mathfrak{g}+\langle ij\rangle )$, then

${\alpha}_{kl}(\mathfrak{g}+\langle ij\rangle )<{\alpha}_{kl}(\mathfrak{g})$. Thus,

${\gamma}_{k}(\mathfrak{g}+\langle ij\rangle )<{\gamma}_{k}(\mathfrak{g})$. ☐

**Theorem** **4.** Suppose agents i, j, k, $\overline{k}$, and l are such that $i\ne j$, $i\ne k$, $i\ne l$, $j\ne k$, $k\ne \overline{k}$, and $k\ne l$. (Agents i and $\overline{k}$ may be the same, and agents j and l may be the same). Suppose $\langle ij\rangle \notin \mathfrak{g}$, $\overline{k}\in {\mathfrak{g}}_{i}^{+}$ and $l\in {\mathfrak{g}}_{j}^{+}$. Then, agent k experiences positive externalities if and only if $k\in {\mathfrak{g}}_{ij}^{+}$ and $\frac{{\displaystyle \prod _{\overline{k}\in {\mathfrak{g}}_{i}^{+}}}(1-{\alpha}_{k\overline{k}}(\mathfrak{g}+\langle ij\rangle ))}{{\displaystyle \prod _{l\in {\mathfrak{g}}_{l}^{+}}}(1-{\alpha}_{kl}(\mathfrak{g}))}<\frac{{\displaystyle \prod _{\overline{k}\in {\mathfrak{g}}_{i}^{+}}}(1-{\alpha}_{k\overline{k}}(\mathfrak{g}))}{{\displaystyle \prod _{l\in {\mathfrak{g}}_{j}^{+}}}(1-{\alpha}_{kl}(\mathfrak{g}+\langle ij\rangle ))}$, otherwise k experiences negative externalities.

**Proof.** From Lemma 8, it is required to increment in agent k’s closeness. It is straightforward to observe that $\frac{{\displaystyle \prod _{\overline{k}\in {\mathfrak{g}}_{i}^{+}}}(1-{\alpha}_{k\overline{k}}(\mathfrak{g}+\langle ij\rangle ))}{{\displaystyle \prod _{l\in {\mathfrak{g}}_{l}^{+}}}(1-{\alpha}_{kl}(\mathfrak{g}))}<\frac{{\displaystyle \prod _{\overline{k}\in {\mathfrak{g}}_{i}^{+}}}(1-{\alpha}_{k\overline{k}}(\mathfrak{g}))}{{\displaystyle \prod _{l\in {\mathfrak{g}}_{j}^{+}}}(1-{\alpha}_{kl}(\mathfrak{g}+\langle ij\rangle ))}$, then ${\gamma}_{k}(\mathfrak{g}+\langle ij\rangle )>{\gamma}_{k}(\mathfrak{g})$. Thus, k experiences positive externalities.

Conversely, let ${\gamma}_{k}(\mathfrak{g}+\langle ij\rangle )<{\gamma}_{k}(\mathfrak{g})$, then either from Proposition 8, ${d}_{ki}(\mathfrak{g})={d}_{ki}(\mathfrak{g}+\langle ij\rangle )$, for all $i\in \mathfrak{g}$ or $\frac{{\displaystyle \prod _{\overline{k}\in {\mathfrak{g}}_{i}^{+}}}(1-{\alpha}_{k\overline{k}}(\mathfrak{g}+\langle ij\rangle ))}{{\displaystyle \prod _{l\in {\mathfrak{g}}_{l}^{+}}}(1-{\alpha}_{kl}(\mathfrak{g}))}<\frac{{\displaystyle \prod _{\overline{k}\in {\mathfrak{g}}_{i}^{+}}}(1-{\alpha}_{k\overline{k}}(\mathfrak{g}))}{{\displaystyle \prod _{l\in {\mathfrak{g}}_{j}^{+}}}(1-{\alpha}_{kl}(\mathfrak{g}+\langle ij\rangle ))}$. ☐

Lemma 8 and Theorem 4 show that an increase in the closeness of an agent (who is not involved in the link formation) is necessary in order for that agent to experience positive externalities. Although we have provided a necessary and sufficient condition for positive and negative externalities by performing a microscopic analysis of externalities, it is hard to obtain a general characterization of networks where agents experience only positive externalities. This leads us to the following question. At least for specific network structures, can we show positive (or negative) externalities? For instance, we can argue that in a two diameter network, agents never experience positive externalities.