Binary distributions

Chapter IV: Weak and Strong Ties in Social Network

B.1 Binary distributions

A p p e n d i x B

APPENDIX TO CHAPTER 3

Let us think in terms of the likelihood ratio. As we said before, 𝑙_𝑡 is going to be multiplied by𝑞/(1−𝑞)𝑛times and(1−𝑞)/𝑞 𝑘 times. These factors do not depend on𝑙_𝑡, therefore it also does not matter in which order we observe these actions, the final likelihood is going to be equal to

𝑙_𝑡· 𝑞

1−𝑞 ^𝑛

1−𝑞 𝑞

^𝑘

=𝑙_𝑡· 𝑞

1−𝑞 ^𝑛−𝑘

.

Therefore, the public belief is a random walk on a fixed lattice and its position is defined by the difference in the number of times we observed signals𝐻𝑖𝑔 ℎand𝐿 𝑜𝑤 and𝑙

0=1.

Proof of Lemma 11. First we show that once a we stop learning (no prices, 𝑘_𝑡 =1) there is no reason to reenact 𝐿 𝑃 latter. Notice that from the social planner’s perspective this game is stationary. Therefore, if it is optimal to stop at time 𝑡

0 at levels𝑁 and−𝑁then at any time𝑡⁰> 𝑡

0there is no incentives to choose𝑘_𝑡different from 1 as we face the same problem as at time𝑡

0when it is optimal to stop.

Second, suppose we choose to stop upon arrival at either levelℎ > 0 or𝑙 <0. Then the expected utility at 1/2 satisfies the following relation

𝑢(0.5) =𝑎_𝑙+E(𝛿^𝑡)𝑢(𝑙) +𝑎_ℎ+E(𝛿^𝑡)𝑢(ℎ),

where𝑎_ℎand𝑎_𝑙correspond to the utility that we get on the way to the corresponding boundary and does not depend on𝑢. Let us sum up the first two terms and the last two and choose the biggest one. If they are equal - pick one. Let us say it is the second one, that corresponds to level ℎ. Then due to the symmetry of 𝑢 it is profitable for the social planner to choose the lower level to be−ℎinstead of𝑙. This concludes the proof.

Proof of Lemma 12. Suppose at time 𝑡 the public belief 𝑝_𝑡 > 𝑞 and we are still acquiring information/in the learning period. Depending on the private signal we get, our posterior will be equal to

𝜇_𝑡 =













𝑝_𝑡𝑞

𝑝_𝑡𝑞+ (1−𝑞) (1− 𝑝_𝑡) if we get the 𝐻𝑖𝑔 ℎsignal 𝑝_𝑡(1−𝑞)

𝑝_𝑡(1−𝑞) +𝑞(1− 𝑝_𝑡) if we get the 𝐿 𝑜𝑤signal.

Moreover, the former signal happens with probability (𝑝_𝑡𝑞 + (1 − 𝑝_𝑡) (1 − 𝑞)) as with probability 𝑝_𝑡 we are in the 𝐻𝑖𝑔 ℎ state and with (1− 𝑝_𝑡) - in the 𝐿 𝑜𝑤 state. The analogous calculation gives that the latter signal occurs with probability 𝑝_𝑡(1−𝑞) +𝑞(1−𝑝_𝑡). Also, remember when we get the𝐿 𝑜𝑤private signal we take action 0, so our expected utility is 1− 𝜇_𝑡. Therefore, the expected utility today is equal to

𝑝_𝑡𝑞

(𝑝_𝑡𝑞+ (1−𝑞) (1− 𝑝_𝑡))(𝑝_𝑡𝑞+ (1−𝑞) (1− 𝑝_𝑡))+

+

1− 𝑝_𝑡(1−𝑞)

(𝑝_𝑡(1−𝑞) +𝑞(1−𝑝_𝑡))

(𝑝_𝑡(1−𝑞) +𝑞(1− 𝑝_𝑡))=𝑞 . And similarly we get the same expected utility if 𝑝_𝑡 < 1−𝑞.

Now we are ready to prove the main result.

Proof of Theorem 13. For notation convenience we write 𝑢_𝑛 instead of𝑢_𝐻(𝑝(𝑛)). We are going to start with solving (3.6) and then explain why𝑢_𝐿(0.5) from (3.7) is the same. We know that the solution for the recurrence equation

𝑢_𝑛 =(1−𝛿)𝑞+𝛿(𝑞𝑢_𝑛+

1+ (1−𝑞)𝑢_𝑛−

1) is

𝑢_𝑛=𝑐

1

1 2

1 𝛿𝑞

− p

1−4𝑞 𝛿2+4𝑞2𝛿2

𝛿𝑞

! !^𝑛 +𝑐

2

1 2

1 𝛿𝑞

+ p

1−4𝑞 𝛿2+4𝑞2𝛿2

𝛿𝑞

! !^𝑛 +𝑞, for some constants 𝑐

1 and 𝑐

2. It is also straightforward to verify that these 𝑢_𝑛’s indeed satisfy our recurrence equation. Now we need to solve for𝑐

1 and𝑐

2using boundary conditions𝑢_−𝑁 =(1−𝑞)^𝑁/( (1−𝑞)^𝑁+𝑞^𝑁)and𝑢_𝑁 =𝑞^𝑁/( (1−𝑞)^𝑁+𝑞^𝑁). This results in two equations















(1−𝑞)^𝑁

𝑞^𝑁 + (1−𝑞)^𝑁 =𝑐

1𝑎^−𝑁

1 +𝑐

2𝑎^−𝑁

2 +𝑞 𝑞^𝑁

𝑞^𝑁 + (1−𝑞)^𝑁 =𝑐

1𝑎^𝑁

1 +𝑐

2𝑎^𝑁

2 +𝑞 where

𝑎_𝑖 = 1 2

1 𝛿𝑞

± p

1−4𝑞 𝛿2+4𝑞2𝛿2

𝛿𝑞

! ! .

Thus,















 𝑐1=

(1−𝑞)^𝑁

𝑞^𝑁 + (1−𝑞)^𝑁 −𝑞

𝑎^𝑁

1 −𝑐

2𝑎^−𝑁

2 𝑎^𝑁

1

𝑐2

𝑎^𝑁

2 −𝑎^2𝑁

1 𝑎^−𝑁

2

= 𝑞^𝑁

𝑞^𝑁+ (1−𝑞)^𝑁 −𝑞+

𝑞− (1−𝑞)^𝑁 𝑞^𝑁 + (1−𝑞)^𝑁

𝑎^2𝑁

1 . It follows that





















 𝑐2=

𝑞^𝑁

𝑞^𝑁+(1−𝑞)^𝑁 −𝑞

+

𝑞− ^(1−𝑞)

𝑁

𝑞^𝑁+(1−𝑞)^𝑁

𝑎^2𝑁

1

𝑎^𝑁

2 −𝑎^2𝑁

1

𝑎^−𝑁

2

𝑐1=

−

𝑞^𝑁

𝑞^𝑁+(1−𝑞)^𝑁 −𝑞

𝑎^𝑁

1

𝑎^𝑁

2 −

𝑞− ^(1−𝑞)

𝑁

𝑞^𝑁+(1−𝑞)^𝑁

𝑎^−𝑁

2

𝑎^𝑁

1

𝑎^𝑁

2 −𝑎^2𝑁

1 𝑎^−𝑁

2

Now we can plug this into our solution 𝑢0=𝑐

1𝑎⁰

1+𝑐

2𝑎⁰

2+𝑞

=

−

𝑞^𝑁

𝑞^𝑁+(1−𝑞)^𝑁 −𝑞

𝑎^𝑁

1

𝑎^𝑁

2 −

𝑞− ^(1−𝑞)

𝑁

𝑞^𝑁+(1−𝑞)^𝑁

𝑎^𝑁

2

𝑎^𝑁

1

𝑎^𝑁

2 −𝑎^2𝑁

1 𝑎^−𝑁

2

+

+

𝑞^𝑁

𝑞^𝑁+(1−𝑞)^𝑁 −𝑞

+

𝑞− ^(1−𝑞)

𝑁

𝑞^𝑁+(1−𝑞)^𝑁

𝑎^2𝑁

1

𝑎^𝑁

2 −𝑎^2𝑁

1 𝑎^−𝑁

2

= (𝑎^𝑁

2 −𝑎^𝑁

1) ^𝑞

𝑁

𝑞^𝑁+(1−𝑞)^𝑁 −𝑞

−

𝑞− ^(1−𝑞)

𝑁

𝑞^𝑁+(1−𝑞)^𝑁

𝑎^𝑁

1

𝑎^𝑁

2

𝑎2𝑁

2 −𝑎2𝑁 1

=

𝑞^𝑁

𝑞^𝑁+(1−𝑞)^𝑁 −𝑞

−

𝑞− ^(1−𝑞)

𝑁

𝑞^𝑁+(1−𝑞)^𝑁 1−𝑞

𝑞

^𝑁 𝑎^𝑁

2 +𝑎^𝑁

1

, where in the third equation we multiplied everything by𝑎

2 ≠ 0 and in the last one we used the fact that

𝑎1𝑎

2 = 1 4

1

𝛿2𝑞2 − 1−4𝑞 𝛿2+4𝑞2𝛿2

𝑞2𝛿2

= 1−𝑞 𝑞

·

What is left to explain is why𝑢_𝐿(0.5) would be the same as𝑢_𝐻(0.5)which would save us time solving the second analogous system. Before, in state 𝐻𝑖𝑔 ℎ, we had a random walk with a drift 𝑞 towards the boundary with higher utility, level 𝑁. When we condition on state being𝐿 𝑜𝑤 we have the same drift but in the opposite direction, towards level−𝑁. But notice that utilities of level−𝑁 in state𝐿 𝑜𝑤 and

level 𝑁 in state 𝐻𝑖𝑔 ℎ are equal to each other. The same is true for the other two boundary utilities. Moreover, in both states we have the same underlying lattice for the random walk. Therefore, problem (3.7) is the same problem as (3.6) up to renaming levels. This concludes the proof.

Proof of Lemma 14. This is a classic Gambler’s ruin problem (Feller, 1968)

Proof of Proposition 15 . For notation simplicity we will write 𝑢(𝑛) instead of 𝑢(𝑝(𝑛)).

We know, that the social planner adopts a symmetric strategy of stopping at levels 𝑁or−𝑁. Suppose that if we increase the stopping boundaries from𝑁 to𝑁+1 and from−𝑁to−𝑁−1 correspondingly such that the utility at level𝑁,𝑢(𝑁), increases.

Notice that the utility at level 0, given this strategy, is equal to the utility at the boundary levels multiplied by the expected discount factor plus some utility that we collect on the way to it. The randomness comes from the hitting the boundary levels time. We divide this expression in two parts: for the level𝑁 and−𝑁.

𝑢(0.5) =𝑎_−𝑁 +𝑢(−𝑁)E^−𝑁(𝛿^𝑡) +𝑎_𝑁+𝑢(𝑁)E^𝑁(𝛿^𝑡),

where𝑎_−𝑁,𝑎_𝑁 are constantsE^±𝑁(𝛿^𝑡) is the expected discounted factor until we hit the corresponding boundary.

Recall that 𝑢 is symmetric and so 𝑢(−𝑁) = 𝑢(𝑁), therefore if we increase 𝑢(𝑁) then we increase𝑢(0.5)also and vice versa. This concludes the proof.

Patient planner

Now we would like to see what happens to the optimal𝑁^∗ and the expected utility, when the social planner becomes more patient. In other words, how does the optimal 𝑁^∗ and 𝑢(0.5) behave, when 𝛿 → 1. We first establish a condition for when 𝑁^∗ does not go to∞ when𝛿 → 1 in Proposition 16: precision also has to go to 1 at a certain speed in this case. Secondly, we look at our expected utility as𝛿 increases, but𝑞stays fixed in Proposition 17. There, we stumble upon (1−𝑞)/𝑞factor for the third time.

Proof of Proposition 16. Recall that the expected utility when we start with prior 1/2 (at level𝑁) and stop the random walk upon arrival at levels 2𝑁 or 0 is

𝑢(𝑁)=

−

− ^(1−𝑞)

𝑁

(1−𝑞)^𝑁+𝑞^𝑁 +𝑞 ¹⁻

𝑞 𝑞

^𝑁 +

𝑞^𝑁

𝑞^𝑁+(1−𝑞)^𝑁 −𝑞 𝑎^𝑁

1 +𝑎^𝑁

2

+𝑞, where𝑎

1 < 𝑎

2and

𝑎_𝑖 = 1 2

1 𝛿𝑞

± s

1 (𝛿𝑞)² − 4

𝑞 +4

! .

In order to find bounds on optimal 𝑁 ∈ Rwe are going to focus on the first term of𝑢(𝑁) as the second one is just a constant that does not affect optimal𝑁, and let 𝑞 =1−𝜀, 𝛿=1−𝛾where𝜀, 𝛾 →0.

Notice, that𝑢(𝑁)is single-peaked in 𝑁, therefore, if𝑁 maximizes it overRand𝑁^∗ maximizes overNthen 𝑁^∗ is either d𝑁e or b𝑁c. Thus, bounds on 𝑁 are going to give very tight bounds on 𝑁^∗. First, consider𝑎_𝑖:

𝑎_𝑖 = 1 2

1±p

1−4𝛿2𝑞+4𝑞2𝛿2−4𝑞 𝛿+4𝑞 𝛿 𝛿𝑞

= 1 2

1±p

(2𝑞 𝛿−1)²+4𝑞 𝛿(1−𝛿) 𝛿𝑞

= 1 2

1± (2𝑞 𝛿−1+2𝑞 𝛿𝛾 𝑐

1) +𝑜(𝛾) 𝛿𝑞

, where the thirds equality comes from Taylor series expansion of

√

𝑎2+𝑥around 0.

Therefore,

𝑎1= 1 𝛿𝑞

−1−𝑂(𝛾) 𝑎2=1+𝑂(𝛾) Now we can take derivative of𝑔.

𝑔⁰(𝑁) =

(1−𝑞)^𝑁ln(1−𝑞) ( (1−𝑞)^𝑁+𝑞^𝑁)−(1−𝑞)^𝑁( (1−𝑞)^𝑁ln(1−𝑞)+𝑞^𝑁ln𝑞) ( (1−𝑞)^𝑁+𝑞^𝑁)²

(1−𝑞)^𝑁

𝑞^𝑁 +𝑚

1

1−𝑞 𝑞

^𝑁 ln_1−𝑞^𝑞

(𝑎^𝑁

1 +𝑎^𝑁

2)² +

+

𝑞^𝑁ln𝑞( (1−𝑞)^𝑁+𝑞^𝑁)−𝑞^𝑁( (1−𝑞)^𝑁ln(1−𝑞)+𝑞^𝑁ln𝑞) ( (1−𝑞)^𝑁+𝑞^𝑁)² (𝑎^𝑁

1 +𝑎^𝑁

2) − (𝑎^𝑁

1 ln𝑎

1+𝑎^𝑁

2 ln𝑎

2)𝑚

2

(𝑎^𝑁

1 +𝑎^𝑁

2)²

,

where 𝑞 ≥ 𝑚

1=

− (1−𝑞)^𝑁 (1−𝑞)^𝑁 +𝑞^𝑁

+𝑞

≥ 2𝑞−1 1−𝑞 ≥ 𝑚

2=−

− (1−𝑞)^𝑁 (1−𝑞)^𝑁+𝑞^𝑁

+𝑞 1−𝑞 𝑞

^𝑁 +

𝑞^𝑁

𝑞^𝑁 + (1−𝑞)^𝑁 −𝑞

≥ 01

In order to satisfy F.O.C. nominator should be equal to 0

−

(1−𝑞)^𝑁𝑞^𝑁ln_1−𝑞^𝑞 ( (1−𝑞)^𝑁+𝑞^𝑁)²

(1−𝑞)^𝑁 𝑞^𝑁

+𝑚

1

1−𝑞 𝑞

^𝑁 ln

𝑞 1−𝑞

+

𝑞^𝑁(1−𝑞)^𝑁ln_1−𝑞^𝑞 ( (1−𝑞)^𝑁 +𝑞^𝑁)²

! (𝑎^𝑁

1 +𝑎^𝑁

2)−

− (𝑎^𝑁

1 ln𝑎

1+𝑎^𝑁

2 ln𝑎

2)𝑚

2=0

Notice that the left-hand side is smaller than

≤ 𝑚

1

1−𝑞 𝑞

^𝑁 ln

𝑞 1−𝑞

! (𝑐

1𝑎^𝑁

2) − (𝑎^𝑁

2 ln(1+𝑂(𝛾)))𝑚

2

≤ 𝑚

1𝑐

1

1−𝑞 𝑞

^𝑁 ln

𝑞 1−𝑞

! 𝑎^𝑁

2 −𝑎^𝑁

2𝛾 𝑚

2, where𝑐_𝑖 > 1.

In order for this to be non negative we need𝑁 to satisfy the following constraint

𝑁 ≥ ln

𝑚2𝛾 𝑐1𝑚

1ln(𝑞/(1−𝑞)) ln1−𝑞

𝑞

≥ 𝑟

2ln𝛾 ln𝜀

,

where𝑟

2goes to 1 as 𝛾and𝛿 go to 0. At the same time, notice that LHS is bigger than

≥ 𝑚

1

1−𝑞 𝑞

^𝑁 ln

𝑞 1−𝑞

𝑎^𝑁

2 −𝑎^𝑁

2𝛾 𝑐

3𝑚

2, where 𝑐

3 > 1. In order for this to be non positive 𝑁 should satisfy the following constraint

𝑁 ≤ ln

𝛾 𝑐3𝑚

2

𝑚1ln(𝑞/(1−𝑞)) ln1−𝑞

𝑞

·

1As𝑁increases𝑚

1gets very close to𝑞and𝑚

2to 1−𝑞.

Notice, that if(ln𝑐

3𝛾) is smaller or proportional to ln(−ln( (1−𝑞)/𝑞))then 𝑁 is always finite and the lower bound is satisfied when𝑞, 𝛿→ 1, as

ln

−ln¹⁻

𝑞 𝑞

ln^1−𝑞𝑞

→ 0

Otherwise, we get that

𝑁 ≤ ln𝛾 𝑟1ln𝜀

, for some constant𝑟

1. Moreover, if𝑁satisfies these constraints then 𝑁^∗ satisfies ln𝛾

𝑟1ln𝜀 𝑟

≥ 𝑁^∗ ≥ 𝑟

2ln𝛾 ln𝜀

. This concludes the proof.

Proof of Proposition 17. Recall that

𝑢(𝑁) =

−

− ^(1−𝑞)

𝑁

(1−𝑞)^𝑁+𝑞^𝑁 +𝑞 ^1−𝑞

𝑞

^𝑁 +

𝑞^𝑁

𝑞^𝑁+(1−𝑞)^𝑁 −𝑞 𝑎^𝑁

1 +𝑎^𝑁

2

+𝑞 . 𝑎1= 1

(1−𝛾)𝑞

−1−𝑂(𝛾) 𝑎2=1+𝑂(𝛾)

After plugging expressions for𝑎

1and𝑎

2in𝑢(𝑁)we get

𝑢(𝑁) −𝑞 =

−

− ^(1−𝑞)

𝑁

(1−𝑞)^𝑁+𝑞^𝑁 +𝑞 ^1−𝑞

𝑞

^𝑁 +

𝑞^𝑁

𝑞^𝑁+(1−𝑞)^𝑁 −𝑞 1−𝑞

𝑞 −𝑂(𝛾)^𝑁

+ (1+𝑂(𝛾))^𝑁

=

−

− ^(1−𝑞)

𝑁

(1−𝑞)^𝑁+𝑞^𝑁 +𝑞 ¹⁻

𝑞 𝑞

^𝑁 +

𝑞^𝑁

𝑞^𝑁+(1−𝑞)^𝑁 −𝑞 1−𝑞

𝑞

^𝑁

−𝑜(𝛾) +1+𝑂(𝛾 𝑁)

=

−

− ^(1−𝑞)

𝑁

(1−𝑞)^𝑁+𝑞^𝑁 +𝑞 ¹⁻

𝑞 𝑞

^𝑁 +

𝑞^𝑁

𝑞^𝑁+(1−𝑞)^𝑁 −𝑞 1−𝑞

𝑞

^𝑁 +1

−𝑂(𝛾 𝑁).

From the proof of Proposition 16 we know that when𝑞is fixed𝑁behaves as𝑂(ln𝛾). Now, let us transform the first term into a more clear form

𝑢(𝑁) −𝑞 =

𝑞^𝑁

𝑞^𝑁 + (1−𝑞)^𝑁 −𝑞

−

𝑞^𝑁−(1−𝑞)^𝑁 𝑞^𝑁+(1−𝑞)^𝑁

1−𝑞 𝑞

^𝑁 1−𝑞

𝑞

^𝑁 +1

−𝑂(𝛾 𝑁)

As𝑁 → ∞and𝑞 > ¹

2, ( (1−𝑞)/𝑞)^𝑁 →0. Suppose( (1−𝑞)/𝑞)^𝑁 =𝜈then 𝑞^𝑁

𝑞^𝑁 + (1−𝑞)^𝑁 ≥ 1−𝜈

and

𝑞^𝑁−(1−𝑞)^𝑁 𝑞^𝑁+(1−𝑞)^𝑁

1−𝑞 𝑞

^𝑁 1−𝑞

𝑞

^𝑁 +1

≤ 𝜈

𝜈+1

≤ 𝜈 . From the proof of Theorem 16𝜈 =𝑂(𝛾). Hence,

𝑢(𝑁) −𝑞 ≥ 1−𝑞−2𝜈−𝑂(𝛾 𝑁) =1−𝑞−𝑂(𝛾ln𝛾).

It means that as𝑁 → ∞,𝑢(𝑁)goes to its maximal value as𝐶( (1−𝑞)/𝑞)^𝑁·𝑁 → 0, which is quicker than ( (1−𝑞)/𝑞)^{𝑁 𝑘}, for any𝑘 < 1. Also, as optimal𝑁^∗increases then absorbing beliefs are further away from 1/2.