Equilibrium and comparative statics

Chapter IV: Weak and Strong Ties in Social Network

C.2 Equilibrium and comparative statics

function ouside of those neighborhoods of the equilibria of𝑈_∞. This concludes one direction.

Now let us show the other direction. There are three equilibria of𝑈_∞: two trivial and one non trivial. Also, 𝑈_𝑁 has the same two trivial equilibria. Let us prove that there is a non trivial equilibrium of 𝑈_𝑁 in some negihborhood of 𝑝^∗. Let 𝑝1= 𝑝^∗−𝜀and𝑝2= 𝑝^∗+𝜀. For both 𝑝1and𝑝2let us find𝑁

1and𝑁

2such that the maximums of𝑈_∞(𝑝_𝑖, 𝑞_𝑖(𝑝_𝑖), 𝑝1, 𝑞(𝑝1))and𝑈_∞(𝑝_𝑖, 𝑞_𝑖(𝑝_𝑖), 𝑝2, 𝑞(𝑝2))with respect to 𝑝_𝑖 are within the𝛿-neighborhood of the maximums of𝑈_𝑁(𝑝_𝑖, 𝑞_𝑖(𝑝_𝑖), 𝑝1, 𝑞(𝑝1)) and𝑈_𝑁(𝑝_𝑖, 𝑞_𝑖(𝑝_𝑖), 𝑝², 𝑞(𝑝²))correspondingly. Notice, that𝑈_𝑁(𝑝_𝑖, 𝑞_𝑖(𝑝), 𝑝, 𝑞(𝑝)) can not have maximums outside of these neighborhoods as𝑈_∞(𝑝_𝑖, 𝑞_𝑖(𝑝), 𝑝, 𝑞(𝑝))is single-peaked for a fixed𝑝with respect to𝑝_𝑖. Pick𝛿to be equal to min(ℎ(𝑝¹), ℎ(𝑝²)). We can do this because𝑈_𝑁uniformly converges𝑈_∞. Let us pick𝑁

0=max(𝑁

1, 𝑁

2). Then we know that at 𝑝 = 𝑝¹ the maximum of𝑈_𝑁(𝑝_𝑖, 𝑞_𝑖(𝑝_𝑖), 𝑝¹, 𝑞(𝑝¹)) is to the right of𝑝_𝑖 = 𝑝12and at 𝑝 = 𝑝2the maximum of𝑈_𝑁(𝑝_𝑖, 𝑞_𝑖(𝑝_𝑖), 𝑝2, 𝑞(𝑝2))is to the left of𝑝_𝑖 = 𝑝2for all𝑁 > 𝑁

0. Hence, at some point𝑝1 ≤ 𝑝_𝑁 ≤ 𝑝2the maximum of 𝑈_𝑁(𝑝_𝑖, 𝑞_𝑖(𝑝_𝑖), 𝑝, 𝑞(𝑝)) crosses the line of 𝑝_𝑖 = 𝑝. This is a symmetric non trivial equilibrium of𝑈_𝑁 in𝑝^∗’s the neighborhood.

Therefore, there are equilibria of𝑈_𝑁 within the𝜀-neighborhood of (trivial and non trivial) equilibria of 𝑈_∞. Moreover, there are no equilibria of the initial utility function outside of those neighborhoods.

𝜕𝑈_∞

𝜕 𝑝_𝑖 𝑝𝑖=𝑝

=𝑝 𝜋_𝑤²𝐾²𝑁²𝐿²

𝑝²− (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔} − 𝑝²) + +𝑝 𝜋_𝑤(𝐾−1)𝐾 𝑁 𝐿

(𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}− 𝑝²) − 𝑝²

−

−𝜋_𝑤(𝐾 −1)𝐾 𝑁 𝑝³𝐿−𝑝 𝜋_𝑤𝐾 𝑁 𝐿+𝑝(𝐾−1) 1− 𝑝⁴

^𝐾−₂ + + (𝐾−2) (𝐾−1)𝑝³

1− 𝑝² 1− 𝑝^{4 𝐾}−3

=(𝐾−2) (𝐾−1)𝑝³

1− 𝑝² 1− 𝑝^{4 𝐾}−3

+ 𝑝(𝐾 −1) 1− 𝑝⁴

^𝐾−2

+

−𝜋_𝑤𝐾 𝑁 𝐿 𝑝

1− (𝐾 −1) (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}−2𝑝²) +𝜋_𝑤𝐾 𝑁 𝐿(𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔} −𝑝²)

=(𝐾−2) (𝐾−1)𝑝³

1− 𝑝² 1− 𝑝^{4 𝐾−}₃

+ 𝑝(𝐾 −1) 1− 𝑝⁴

^𝐾−₂ +

− 𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

(𝐾−1)𝑝

1− (𝐾−1) (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔} −2𝑝²)+

+ 𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

(𝐾−1) (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}− 𝑝²)

.

Notice, that we can do this if𝑞_𝑖𝑞 > 0, otherwise, all terms that include a weak tie become 0 and we do not differentiate them. When𝑞_𝑖𝑞=0,𝑈_∞it is a trivial equilib- rium of𝑈_∞, as well as of𝑈_𝑁, to choose𝑝_𝑖 =p

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}∀𝑖. Denote(𝜕𝑈_∞/𝜕 𝑝_𝑖) 𝑝_𝑖=𝑝

by𝐹.

𝐹 =(𝐾−2) (𝐾 −1)𝑝³

1− 𝑝² 1− 𝑝⁴ 𝐾−3

+𝑝(𝐾−1)

1−𝑝⁴ 𝐾−2

+

− 𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

(𝐾 −1)𝑝

1− (𝐾 −1) (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}−2𝑝²) + 𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

(𝐾 −1) (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}− 𝑝²)

=𝑇

1(𝑝) +𝑇

2(𝑝). where𝑇

1 is the term on the 1st line and𝑇

2– on the 2nd one. We will prove that 𝐹 has a unique non trivial root.

Under the assumptions of the theorem, 𝐹 has the following form: it is 0 at 𝑝 = 0, has a positive derivative there (when𝑐_{𝑤 𝑒 𝑎 𝑘} > 𝜋_𝑤), it is negative atp

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔} (when 𝑒

−𝐵2

𝐾−1(1− 𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}) < 𝜋_𝑤/𝑐_{𝑤 𝑒 𝑎 𝑘}) and crosses 0 only once on (0,p

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}). Let us show this. It is clear that𝐹(0) =0, so we will skip it.

𝐹⁰(0) =(𝐾−1)

𝜋_𝑤 𝜋_𝑤(𝐾 −1) 3𝑝2− 𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔} 𝑐²

𝑤 𝑒 𝑎 𝑘

+ + −𝑐_{𝑤 𝑒 𝑎 𝑘} 6(𝐾 −1)𝑝2−𝐾 𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔} +𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}

𝑐2 𝑤 𝑒 𝑎 𝑘

+ 1

𝑐2 𝑤 𝑒 𝑎 𝑘

− 1− 𝑝²

𝐾−3 𝑝²+1

𝐾−4

(𝐾 −1) (4𝐾−7)𝑝⁶+ (6𝐾−11)𝑝⁴+ + (5−3𝐾)𝑝²−1

! 𝑝=0

=(𝐾−1)

𝜋_𝑤 (𝐾−1)𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}(𝑐_{𝑤 𝑒 𝑎 𝑘} −𝜋_𝑤) −𝑐_{𝑤 𝑒 𝑎 𝑘} 𝑐2

𝑤 𝑒 𝑎 𝑘

+1

!

=(𝐾−1)©

­

­

« 𝜋_𝑤

(𝐾−1)𝑀(𝑐_{𝑤 𝑒 𝑎 𝑘} −𝜋_𝑤) +𝑐_{𝑤 𝑒 𝑎 𝑘} 𝑐𝑤 𝑒 𝑎 𝑘

𝜋_𝑤 −1 𝑐²

𝑤 𝑒 𝑎 𝑘

ª

®

®

¬

=(𝐾−1)©

­

­

« 𝜋_𝑤

(𝑐_{𝑤 𝑒 𝑎 𝑘} −𝜋_𝑤)

(𝐾 −1)𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}+ ^{𝑐𝑤 𝑒 𝑎 𝑘}

𝜋_𝑤

𝑐2 𝑤 𝑒 𝑎 𝑘

ª

®

®

¬

>0, as𝑐_{𝑤 𝑒 𝑎 𝑘} > 𝜋_𝑤.

This implies that within some𝜀-neighborhood of 𝑝 =0, 𝐹 is positive. Now we are going to show that at the other end, at 𝑝=p

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}, it is negative.

𝐹(p

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}) =(𝐾 −1)

𝑀

1 2

𝑠𝑡𝑟 𝑜𝑛𝑔(1− 𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔})^𝐾−2(𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔} +1)^𝐾−3×

× ( (𝐾 −1)𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}+1) − 𝜋_𝑤𝑀

1 2

𝑠𝑡𝑟 𝑜𝑛𝑔

𝑐_{𝑤 𝑒 𝑎 𝑘}

(1+ (𝐾−1)𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔})

=− (𝐾−1) (1+ (𝐾−1)𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔})𝑀

1 2

𝑠𝑡𝑟 𝑜𝑛𝑔

𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

− (1−𝑀²

𝑠𝑡𝑟 𝑜𝑛𝑔)^𝐾−3×

× (1− 𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔})

!

=− (𝐾−1) (1+ (𝐾−1)𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔})𝑀

1 2

𝑠𝑡𝑟 𝑜𝑛𝑔

𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

−

−

1− 𝐵2

(𝐾−1)^{2 𝐾−}3

1− 𝐵 𝐾−1

!

≤ − (𝐾−1) (1+ (𝐾−1)𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔})𝑀

1 2

𝑠𝑡𝑟 𝑜𝑛𝑔

𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

−𝑒

−^{𝐵2(𝐾−3)}

(𝐾−1)2 ×

×

1− 𝐵 𝐾−1

<0, as 𝑐_{𝑤 𝑒 𝑎 𝑘}^𝜋𝑤 > 𝑒

−^{𝐵2(𝐾−3)}

(𝐾−1)2

1− ^𝐵

𝐾−1

. The fact that𝐹(p

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}) < 0,𝐹(𝜀) > 0 for some small𝜀 > 0 and it is a continuous function implies that∃𝑝^∗ ∈ (0,p

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔})such that 𝐹(𝑝^∗) =0. Now we just need to make sure that such non trivial equilibrium 𝑝^∗is unique.

Because our function is positive and increasing near the 0 and is negative near p

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}then, at some point, 𝑝^★, its derivative has to become negative,𝐹⁰(𝑝^★) < 0, before𝐹(𝑝)becomes negative.

𝐹⁰(𝑝) =(𝐾−1) (1−𝑝⁴)^𝐾−4(1− 𝑝²) 1+ (3𝐾−5)𝑝²− (6𝐾−11)𝑝⁴− (𝐾−1)×

× (4𝐾−7)𝑝⁶−

3(𝐾−1)𝑝²

2− ^𝜋𝑤

𝑐𝑤 𝑒 𝑎 𝑘

𝜋_𝑤 𝑐𝑤 𝑒 𝑎 𝑘

(1− 𝑝4)^𝐾−4(1− 𝑝2)

! + + (𝐾−1) 𝜋_𝑤

𝑐_{𝑤 𝑒 𝑎 𝑘}

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}(𝐾−1)

1− 𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

−1

We would like 𝐹⁰(𝑝) to stay negative after the first time it becomes less than 0.

Then it can not cross 0 more than once. Assume that the constant term above is

positive3. It is a part of the derivative of the second summand of 𝐹(𝑝), 𝑇

2(𝑝). Notice, that𝑇

2(𝑝)has to be negative at the non trivial equilibrium, because𝑇

1(𝑝)is always positive and their sum is 0. Also,𝑇

2(𝑝)is concave,𝑇

2(0) =0 and𝑇⁰

2(0) ≥ 0, therefore, before 𝑇

2 becomes negative, its derivative has to become less than 0.

Denote by𝑝

1a point at which𝑇⁰

2(𝑝

1) =0. Then we can rewrite 𝐹⁰(𝑝)as 𝐹⁰(𝑝) =(𝐾−1) (1− 𝑝⁴)^𝐾−4(1− 𝑝²)𝐴(𝑝),

where

𝐴(𝑝) =𝑎 𝑝²−2𝑝²−𝑏 𝑝⁴−𝑐 𝑝⁶−

𝑎 𝑑(𝑝2− 𝑝2 1) 1−𝑝2

1− 𝑝4𝐾−4 +1,

and𝑎 =3(𝐾−1),𝑏 =(6𝐾−11),𝑐 =(𝐾−1) (4𝐾−7),𝑑 =(2−𝜋_𝑤/𝑐_{𝑤 𝑒 𝑎 𝑘})𝜋_𝑤/𝑐_{𝑤 𝑒 𝑎 𝑘}. It will be sufficient to show that once 𝐴 becomes negative it stays negative as (1−𝑝⁴)^𝐾−4(1−𝑝²)is always positive for𝑝 ∈ [0,p

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}]. Notice that 𝐴(0) > 0, so before it gets negative its derivative has to become negative.

𝐴⁰(𝑝) =2𝑝

−𝑎 𝑑

𝑝²+1 ₃−𝐾

1−𝑝² ₂−𝐾

2(𝐾 −4)𝑝⁴+ 𝑝²

(−2𝐾+7)𝑝

12+1

−

− 𝑝

12+1

+𝑎−2𝑏 𝑝²−3𝑐 𝑝⁴−2

Now, call the term in parentheses𝐵(𝑝), then𝐵⁰is negative for𝑝 > 0.

𝐵⁰(𝑝) =−4𝑝

1− 𝑝⁴ −𝐾

1−𝑝^{4 𝐾}

𝑏+3𝑐 𝑝²

+𝑎 𝑑

𝑝²−1 𝑝²+1 ₂

×

×

− (𝐾 −4) (2𝐾−7)𝑝⁶+ (2𝐾−7)𝑝⁴

(𝐾 −3)𝑝

12−1

+ + 𝑝²

2(𝐾−3)𝑝

12−3𝐾+10

+ (𝐾−3)𝑝

12−1 !

=−4𝑝

1− 𝑝⁴ −𝐾

1−𝑝^{4 𝐾}

𝑏+3𝑐 𝑝²

+𝑎 𝑑

𝑝²−1 𝑝²+1 ₂

×

×

−1− (2𝐾−7)𝑝⁴

(𝐾−4)𝑝²+1− (𝐾−3)𝑝

12

−

− 𝑝²

−2(𝐾 −3)𝑝

12+2𝐾+7

− (𝐾−3) (𝑝²− 𝑝

12) !

<0.

3If it is negative then instead of having𝑝²−𝑝²

1below we are going to have𝑝²+𝑝²

1 which will only help with the analogous proof.

Thus, 𝐵⁰(𝑝)is negative. This implies the following. When 𝐴⁰becomes negative it stays negative. As 𝐴(0) > 0, before 𝐴 becomes negative its derivative, 𝐴⁰, has to become negative. Therefore, once 𝐴starts decreasing it continues decreasing from that point on. So when 𝐴 becomes negative it stays negative and this is what we wanted. Hence, 𝐹(𝑝) crosses 0 only once on (0,p

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}). Let us call this point 𝑝^∗. This is a unique non trivial equilibrium of𝑈_∞.

Now, assume that𝜋_𝑤 ≥ 𝑐_{𝑤 𝑒 𝑎 𝑘} then𝜋_𝑤/𝑐_{𝑤 𝑒 𝑎 𝑘} ≥ 1> (1−𝑀2

𝑠𝑡𝑟 𝑜𝑛𝑔)^𝐾−3(1−𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}). Furthermore, it means that 𝐹(0) = 0 and 𝐹⁰ is negative on (0,p

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}). Thus, there is only one equilibrium of𝑈_∞: ∀𝑖 𝑝_𝑖 =0 and𝑞_𝑖 =𝑀_{𝑤 𝑒 𝑎 𝑘}. When𝜋_𝑤 > 𝑐_{𝑤 𝑒 𝑎 𝑘} weak links are more beneficial at both distance 1 and 2 and so no one wants to invest in friends.

On the other hand, if𝜋_𝑤/𝑐_{𝑤 𝑒 𝑎 𝑘} < (1− 𝑀²

𝑠𝑡𝑟 𝑜𝑛𝑔)^𝐾−3(1−𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔})then𝜋_𝑤 < 𝑐_{𝑤 𝑒 𝑎 𝑘}. It means that 𝐹(p

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}) > 0. Hence, 𝐹(𝑝) =0 does not have a solution and is positive on (0,p

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}]. So without the two conditions of the theorem we have only trivial symmetric equilibria.

To finish the proof we will show that we found the maximum and not the minimum.

𝜕²𝑈_∞

𝜕 𝑝²

𝑖

𝑝_𝑖=𝑝

=− (𝐾−2) (𝐾−1)𝑝⁴

1− 𝑝^{2 𝐾}−3

𝑝²+1 ^𝐾−4

(𝐾−1)𝑝²+2

<0.

Thus, the extremum we found is indeed the maximum. This concludes the proof.

Proof of Proposition 24. To prove this proposition, let us calculate derivatives of 𝐹(𝑝) with respect to 𝑐_{𝑤 𝑒 𝑎 𝑘}, 𝜋_𝑤 and 𝐾. We know from Theorem 23 the derivative of 𝐹 at 𝑝^∗ is negative. Hence, if we know the derivatives of 𝐹 and their signs with respect to those parameters we will be able to calculate the corresponding comparative statics. Notice, that only the second summand of𝐹 depends on these 3 parameters, which simplifies calculations.

1) 𝜕 𝐹(𝑝^∗)

𝜕 𝑐_{𝑤 𝑒 𝑎 𝑘}

=𝜕 − 𝜋_𝑤(𝐾−1) 𝑐_{𝑤 𝑒 𝑎 𝑘}

𝑝^∗

1− (𝐾 −1) (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}−2𝑝^∗

2) + + 𝜋_𝑤

𝑐_{𝑤 𝑒 𝑎 𝑘}

(𝐾 −1) (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}− 𝑝^∗

2)

!

/𝜕 𝑐_{𝑤 𝑒 𝑎 𝑘}

=

𝜋_𝑤𝑝^∗𝑐_{𝑤 𝑒 𝑎 𝑘}

1− (𝐾−1) (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔} −2𝑝^∗2) 𝑐³

𝑤 𝑒 𝑎 𝑘

+

+𝜋_𝑤𝑝^∗

2𝜋_𝑤(𝐾 −1)

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}− 𝑝^∗2 𝑐3

𝑤 𝑒 𝑎 𝑘

·

Remember, that the second summand of𝐹is negative at the equilibrium,𝑇

2(𝑝^∗) < 0 from Theorem 23, therefore

𝜕 𝐹(𝑝^∗)

𝜕 𝑐_{𝑤 𝑒 𝑎 𝑘}

=

−𝑇

2(𝑝^∗) + ^{𝜋2𝑤𝑝}

∗(𝐾−1) (𝑀_{𝑠𝑡 𝑟 𝑜𝑛𝑔}−𝑝^∗2) 𝑐²

𝑤 𝑒 𝑎 𝑘

𝑐_{𝑤 𝑒 𝑎 𝑘}

>0.

Therefore, as we increase𝑐_{𝑤 𝑒 𝑎 𝑘}, 𝑝^∗also increases.

2) 𝜕 𝐹(𝑝^∗)

𝜕 𝐻

=− 𝑝^∗

1− (𝐾−1) (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔} −2𝑝^∗2) +2𝑐𝑤 𝑒 𝑎 𝑘^𝜋𝑤

(𝐾−1)

𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}− 𝑝^∗2 𝑐_{𝑤 𝑒 𝑎 𝑘}

<0.

We get the last inequality in the same way we argued in 1) that𝑇

2(𝑝^∗)/𝜋_𝑤 minus something negative is negative at 𝑝^∗.

3) 𝜕 𝐹⁰(𝑝)

𝜕 𝐾

=𝑝

1− 𝑝² 1− 𝑝^{4 𝐾−}₃

(𝐾−1)𝑝²+1

log

1− 𝑝⁴

+𝑝²

−

− 𝑝²

2− 𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

!

Notice, that log(1−𝑝4) < 0. Furthermore, at the approximate equilibrium 𝑝^∗ we have

𝐹(𝑝^∗) =0

0=(𝐾 −1)𝑝^∗

1− 𝑝^∗

2^𝐾−2 𝑝^∗

2 +1 ^𝐾−3

(𝐾−1)𝑝^∗

2 +1

−

− 𝜋_𝑤(𝐾−1) 𝑐_{𝑤 𝑒 𝑎 𝑘}

𝑝^∗

1− (𝐾−1) (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔} −2𝑝^∗

2) + 𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

(𝐾−1) (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔} − 𝑝^∗

2)

. Hence,

1− 𝑝^∗

2

1− 𝑝^∗

4^𝐾−₃

= 𝜋_𝑤

1− (𝐾 −1) (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}−2𝑝^∗2) 𝑐_{𝑤 𝑒 𝑎 𝑘} (𝐾−1)𝑝^∗2+1 + +

𝜋2 𝑤

𝑐𝑤 𝑒 𝑎 𝑘

(𝐾−1) (𝑀− 𝑝^∗2) 𝑐_{𝑤 𝑒 𝑎 𝑘} (𝐾−1)𝑝^∗2+1

Let us substitute this into𝜕 𝐹⁰(𝑝)/𝜕 𝐾, but skip the term with the logarithm and the 𝑝term outside the parentheses.

𝜋_𝑤

1− (𝐾−1) (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}−2𝑝^∗2) + ^𝜋𝑤

𝑐_{𝑤 𝑒 𝑎 𝑘}(𝐾 −1) (𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}− 𝑝^∗2)

𝑐_{𝑤 𝑒 𝑎 𝑘} (𝐾 −1)𝑝^∗2 +1 𝑝^∗

2−

− 𝑝^∗

2

2− 𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

= 𝜋_𝑤𝑝^∗2

𝑐_{𝑤 𝑒 𝑎 𝑘}( (𝐾 −1)𝑝^∗2 +1) 1− (𝐾−1)𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}

1− 𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

+ 𝑝^∗

2(𝐾−1)

2− 𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

−

−

2− 𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

( (𝐾−1)𝑝^∗2+1)

!

= 𝜋_𝑤𝑝2

𝑐_{𝑤 𝑒 𝑎 𝑘}( (𝐾 −1)𝑝2+1) −1+ 𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

− (𝐾−1)𝑀_{𝑠𝑡𝑟 𝑜𝑛𝑔}

1− 𝜋_𝑤 𝑐_{𝑤 𝑒 𝑎 𝑘}

!

< 0.

Thus, the whole𝜕 𝐹⁰(𝑝)/𝜕 𝐾 is also negative at𝑝 = 𝑝^∗. So the probability of having a strong friend is decreasing as we increase𝐾. This concludes our proof.

Proof of Lemma 25. First we are going to show that 𝑝^∗2 > 𝑞^∗2 in equilibrium.

Remember that from the proof of Theorem 23 𝑝^∗

2

> 𝑀

𝑐_{𝑤 𝑒 𝑎 𝑘} −𝜋_𝑤 2𝑐_{𝑤 𝑒 𝑎 𝑘} −𝜋_𝑤

− 𝑐_{𝑤 𝑒 𝑎 𝑘}

(𝐾 −1) (2𝑐_{𝑤 𝑒 𝑎 𝑘} −𝜋_𝑤) = (𝐵−1)𝑐_{𝑤 𝑒 𝑎 𝑘} −𝐵𝜋_𝑤 (𝐾 −1) (2𝑐_{𝑤 𝑒 𝑎 𝑘} −𝜋_𝑤)

as the𝑇

2(𝑝) term has to be negative. Hence, 𝑞^∗

2 = 𝐿(𝑀− 𝑝^∗

2) <

(𝐾−1) 𝐾 𝑁 𝑐_{𝑤 𝑒 𝑎 𝑘}

𝐵(2𝑐_{𝑤 𝑒 𝑎 𝑘} −𝜋_𝑤) − (𝐵−1)𝑐_{𝑤 𝑒 𝑎 𝑘} +𝐵𝜋_𝑤 (𝐾 −1) (2𝑐_{𝑤 𝑒 𝑎 𝑘} −𝜋_𝑤) −

= (𝐵+1) 𝐾 𝑁(2𝑐_{𝑤 𝑒 𝑎 𝑘} −𝜋_𝑤)· Let us look at their difference

𝑝^∗

2 −𝑞^∗

2

>

(𝐵−1)𝑐_{𝑤 𝑒 𝑎 𝑘} −𝐵𝜋_𝑤

(𝐾−1) (2𝑐_{𝑤 𝑒 𝑎 𝑘} −𝜋_𝑤) − (𝐵+1) 𝐾 𝑁(2𝑐_{𝑤 𝑒 𝑎 𝑘} −𝜋_𝑤)

= (𝐵−1)𝑐_{𝑤 𝑒 𝑎 𝑘}𝐾 𝑁−𝐵𝜋_𝑤𝐾 𝑁− (𝐵+1) (𝐾−1) (𝐾−1)𝐾 𝑁(2𝑐_{𝑤 𝑒 𝑎 𝑘} −𝜋_𝑤) · For this difference to be greater than 0 we need to have

𝑐_{𝑤 𝑒 𝑎 𝑘} >

𝐵

(𝐵−1)𝜋_𝑤+ (𝐵+1) (𝐾−1) (𝐵−1)𝐾 𝑁

·

Now let us remember a well-known fact about Erdős - Rényi random graphs: the golbal clustering coefficient for friends’ network is equal to 𝑝^∗2+𝑂 (𝐾 𝑁)^−0.5

and the global𝐶 𝐶 for acquanintances’ network is 𝑞^∗2 +𝑂 (𝐾 𝑁)^−0.5

. Using this fact and the inequality that we got above we can conclude that when 𝑐_{𝑤 𝑒 𝑎 𝑘} is not too close to 𝜋_𝑤 or when 𝑁 is big enough (so 𝑝^∗ > 0 and𝑞^∗ < 𝑝^∗ for big enough 𝑁) the𝐶 𝐶 of the friends’ network is bigger than the𝐶 𝐶 of the acquaintances’. This concludes the proof.