Distributions with polynomial tails

Chapter IV: Weak and Strong Ties in Social Network

A.5 Distributions with polynomial tails

In this appendix we prove Theorem 6, showing that for private log-likelihood dis- tributions with polynomial tails, the expected time to learn is finite.

As in the setting of Theorem 6, assume that the conditional distributions of the private log-likelihood ratio satisfy

𝐺₊(𝑥) =1− 𝑐

𝑥^𝑘 for all𝑥 > 𝑥

0 (A.6)

𝐺₋(𝑥) = 𝑐

(−𝑥)^𝑘 for all𝑥 <−𝑥

0 (A.7)

for some𝑥

0> 0.

We remind the reader that we denote byℓ^∗

𝑡 the log-likelihood ratio of the public belief that results when the first 𝑡 −1 agents take action +1. It follows from Theorem 3 that in this setting, ℓ^∗

𝑡 behaves asymptotically as 𝑡^1/(𝑘+1). Notice also that, by the symmetry of the model, the log-likelihood ratio of the public belief that results when the first𝑡−1 agents take action−1 is−ℓ^∗

𝑡.

We begin with the simple observation that a strong enough bound on the probability of mistake is sufficient to show that the expected time to learn is finite. Formally, we have the following lemma. We remind the reader that P+(·) is shorthand for P(· |𝜃=+1).

Lemma 36. Suppose there exist 𝑘, 𝜀 > 0such that for all 𝑡 ≥ 1, P⁺(𝑎_𝑡 = −1) <

𝑘· ¹

𝑡2+𝜀. ThenE+(𝑇_𝐿)is finite.

Proof. Since𝑇_𝐿 =𝑡 only if𝑎_𝑡−

1=−1,P⁺(𝑇_𝐿 =𝑡) ≤ P⁺(𝑎_𝑡−

1=−1). Thus E⁺(𝑇_𝐿) =

∞

Õ

𝑡=1

𝑡·P⁺(𝑇_𝐿 =𝑡)

≤P+(𝑇_𝐿 =1) +

∞

Õ

𝑡=2

𝑡·P+(𝑎_𝑡−

1 =−1)

≤1+𝑘

∞

Õ

𝑖=2

𝑡 (𝑡−1)^2+𝜀

< ∞.

Accordingly, this section will be primarily devoted to studying the rate of decay of the probability of mistake,P+(𝑎_𝑡 =−1). In order to bound this probability, we will need to make use of the following lemmas, which give some control over how the public belief is updated following an upset.

Lemma 37. For 𝐺₊ and 𝐺₋ as in (A.6) and (A.7), |ℓ_𝑡+

1| ≤ |ℓ_𝑡| whenever |ℓ_𝑡| is sufficiently large and𝑎_𝑡 ≠𝑎_𝑡+

1.

Proof. Assume without loss of generality that𝑎_𝑡 =+1 and𝑎_𝑡+

1 =−1, so that ℓ_𝑡+

1=ℓ_𝑡 +𝐷₋(ℓ_𝑡).

Thus, to prove the claim we compute a bound for 𝐷₋. To do so we first obtain a bound for the left tail of𝐺₊. By assumption, for 𝑥 > 𝑥

0 (with 𝑥

0 as in (A.6) and (A.7)),

𝑔₋(−𝑥) =𝐺⁰₋(−𝑥) = 𝑐 𝑘 𝑥^𝑘+1

and so by (A.1),

𝑔₊(−𝑥) =e^−𝑥𝑔₋(−𝑥) =𝑐 𝑘 e^−𝑥 𝑥^𝑘+1

. Hence,

𝐺₊(−𝑥) =

∫ −𝑥

−∞

𝑔₊(𝜁)d𝜁 =

∫ −𝑥

−∞

𝑐 𝑘 e^𝜁

(−𝜁)^𝑘+1d𝜁 =𝑐 𝑘

∫ ∞ 𝑥

𝜁^−𝑘−1e^−𝜁d𝜁 . For𝜁sufficiently large, 𝜁^−𝑘−1is at least, say, e^−.1𝜁. Thus, for𝑥sufficiently large,

𝐺₊(−𝑥) ≥𝑐 𝑘

∫ ∞ 𝑥

e^−1.1𝜁d𝜁 = 𝑐 𝑘

1.1e^−1.1𝑥.

It follows that for𝑥sufficiently large, 𝐷₋(𝑥) =log

𝐺₊(−𝑥) 𝐺₋(−𝑥) ≥ log

𝑐 𝑘 1.1

−1.1𝑥+𝑘log𝑥 ≥ −1.2𝑥 . Thus, forℓ_𝑡sufficiently large,

ℓ_𝑡+

1 =ℓ_𝑡+𝐷₋(ℓ_𝑡) =ℓ_𝑡+log

𝐺₊(−ℓ_𝑡)

𝐺₋(−ℓ_𝑡) ≥ ℓ_𝑡+1.2(−ℓ_𝑡) =−.2ℓ_𝑡 so in particular,|ℓ_𝑡+

1| < |ℓ_𝑡|.

We will make use of the following lemma, which bounds the range of possible values thatℓ_𝑡 can take.

Lemma 38. For𝐺₊ and𝐺₋ as in(A.6)and(A.7), there exists an𝑀 >0such that for all𝑡 ≥ 0,|ℓ_𝑠| ≤ 𝑀·ℓ^∗

𝑡 for all𝑠 ≤ 𝑡. Proof. For each𝜏 ≥ 0, define

𝑀_𝜏 =max

|ℓ_𝜏| ℓ^∗_𝜏

where the maximum is taken over all outcomes. Note that there are at most 2^𝜏 possible values for this expression, so 𝑀_𝜏 is well-defined and finite. Put

𝑀 =sup

𝜏≥0

𝑀_𝜏.

To establish the claim, we must show that𝑀is finite. To do this, it suffices to show that for𝜏sufficiently large, 𝑀_𝜏+

1 ≤ 𝑀_𝜏.

Now, let𝑢₊(𝑥) =𝑥+𝐷₊(𝑥)and𝑢₋(𝑥) =𝑥+𝐷₋(𝑥). Then as shown in the section about the model, whenever agent 𝜏 takes action +1, ℓ_𝜏+

1 = 𝑢₊(ℓ_𝜏), and whenever agent𝜏takes action−1,ℓ_𝜏+

1=𝑢₋(ℓ_𝜏).

By Lemma 31, 𝑢₊ and 𝑢₋ are eventually monotonic. Thus, there exists 𝑥

0 > 0 such that𝑢₊ is monotone increasing on (𝑥

0,∞) and𝑢₋ is monotone decreasing on (−∞,−𝑥

0).

For 𝜏 sufficiently large, ℓ^∗

𝜏 > 𝑥

0. Further, it follows from Lemma 37 that for 𝜏 sufficiently large, |ℓ_𝜏+

1|< |ℓ_𝜏|whenever𝑎_𝜏 ≠ 𝑎_𝜏+

1and|ℓ_𝜏|> |ℓ^∗

𝜏|. Let (𝑎_𝜏)be any sequence of actions with ^|ℓℓ^𝜏∗+1|

𝜏+1

=𝑀_𝜏+

1. If𝑎_𝜏 ≠𝑎_𝜏+

1

𝑀_𝜏+

1= |ℓ_𝜏+

1| ℓ^∗

𝜏+1

≤ |ℓ_𝜏| ℓ^∗

𝜏

≤ 𝑀_𝜏.

If 𝑎_𝜏 = 𝑎_𝜏+

1, then either 𝑀_𝜏+

1 = 1, in which case 𝑀_𝜏+

1 ≤ 𝑀_𝜏, or 𝑀_𝜏+

1 > 1.

If 𝑀_𝜏+

1 > 1, then since |𝐷₊| and |𝐷₋| are decreasing on (𝑥

0,∞) and (−∞,−𝑥

0) respectively,|ℓ_𝜏+

1−ℓ_𝜏|/|ℓ_𝜏| ≤ |ℓ^∗

𝜏+1−ℓ^∗

𝜏|/|ℓ^∗

𝜏|. So 𝑀_𝜏+

1= |ℓ_𝜏+

1| ℓ^∗

𝜏+1

= |ℓ_𝜏| + |ℓ_𝜏+

1−ℓ_𝜏| ℓ^∗

𝜏+ |ℓ^∗

𝜏+1−ℓ^∗

𝜏| where the second equality follows from the fact thatℓ_𝜏andℓ_𝜏+

1have the same sign.

Finally,

𝑀_𝜏+

1= |ℓ_𝜏| ℓ_𝜏^∗

· 1+ |ℓ_𝜏+

1−ℓ_𝜏|/|ℓ_𝜏| 1+ |ℓ^∗

𝜏+1−ℓ_𝜏^∗|/ℓ^∗_𝜏

≤ |ℓ_𝜏| ℓ_𝜏^∗

≤ 𝑀_𝜏.

Thus, for all sufficiently large𝜏,𝑀_𝜏+

1≤ 𝑀_𝜏.

Proposition 39. There exists𝜅 > 0such thatP⁺(𝑎_𝑡 =−1) < 𝜅𝑡⁻2.1for all𝑡 >0.

Proof. Let𝛽=−2.1/log𝛾, where𝛾is as in Proposition 7. To carry out our analysis, we will divide the event that𝑎_𝑡 = −1 into three disjoint events and bound each of them separately:

𝐴= (𝑎_𝑡 =−1) and(Ξ𝑡 > 𝛽log𝑡)

𝐵1= (𝑎_𝑡 =−1) and(Ξ𝑡 ≤ 𝛽log𝑡)and (|{𝑠: 𝑠 < 𝑡 , 𝑎_𝑠 =+1}| ≥ 1 2

𝑡) 𝐵2= (𝑎_𝑡 =−1) and(Ξ𝑡 ≤ 𝛽log𝑡)and (|{𝑠: 𝑠 < 𝑡 , 𝑎_𝑠 =+1}| < 1

2 𝑡). First, by Corollary 34 we have a bound forP⁺(𝐴)

P⁺(𝐴) ≤𝑐· 1 𝑡2.1

. Next, we boundP⁺(𝐵

1). This is the event that the number of upsets so far is small and the majority of agents so far have taken the correct action.

Since there are at most𝛽log𝑡upsets, there are at most¹₂𝛽log𝑡maximal good runs.

Since, furthermore, there are at least ¹₂𝑡 agents who take action+1, there is at least one maximal good run of length at least𝑡/(𝛽log𝑡).

Thus, P+(𝐵

1) is bounded from above by the probability that there are some 𝑠

1 <

𝑠2 < 𝑡 such that there is a good run of length 𝑠

2− 𝑠

1 ≥ 𝑡/(𝛽log𝑡) from 𝑠

1 and 𝑎_𝑠

2 =−1.

For fixed𝑠

1, 𝑠

2, denote by𝐸_𝑠

1,𝑠

2 the event that there is a good run of length𝑠

2−𝑠

1

from𝑠

1. Denote byΓ𝑠

1,𝑠

2 the event (𝐸_𝑠

1,𝑠

2, 𝑎_𝑠

2 =−1). Then P⁺(Γ𝑠

1,𝑠

2) =P⁺(𝑎_𝑠

2 =−1|𝐸_𝑠

1,𝑠

2) ·P⁺(𝐸_𝑠

1,𝑠

2)

≤ P⁺(𝑎_𝑠

2 =−1|𝐸_𝑠

1,𝑠

2). By Lemma 35, there exists a 𝑧 > 0 such that 𝐸_𝑠

1,𝑠

2 implies that ℓ_𝑠

2 ≥ ℓ^∗

𝑠2−𝑠

1−𝑧. Therefore,

P⁺(Γ𝑠

1,𝑠

2) ≤𝐺₊(−ℓ^∗

𝑠2−𝑠₁−𝑧). Since for𝑡sufficiently largeℓ^∗

𝑡 > 𝑡

1

𝑘+2 and since𝐺₊(−𝑥) ≤e^−𝑥by (A.1), P⁺(Γ𝑠

1,𝑠

2) ≤ e^{−𝛼(𝑠2−𝑠1−𝑧)}

𝑘+12 ≤ e^{−𝛼(𝑡/(𝛽log𝑡)−𝑧)}

𝑘+12

.

To simplify, we further bound this last expression to arrive at, for some𝑐 > 0, P+(Γ𝑠

1,𝑠

2) ≤ 𝑐e^−𝑡

𝑘+13

for all𝑡. Since𝐵

1is covered by fewer than𝑡²events of the formΓ𝑠

1,𝑠

2 (as𝑠

1and𝑠

2

are less than𝑡), it follows that P⁺(𝐵

1) < 𝑐𝑡²e^−𝑡

1 𝑘+3

< 1 𝑡2.1

for all𝑡 large enough.

Finally we boundP⁺(𝐵

2). This is the event that the number of upsets so far is small and the majority of agents so far have taken the wrong action. As in 𝐵

1, there is a maximal bad run of length at least𝑡/(𝛽log(𝑡)).

Denote by𝑅the event that there is at least one bad run of length𝑡/(𝛽log(𝑡))before time𝑡 and by 𝑅_𝑠 the event that agents 𝑠through𝑠+𝑡/(𝛽log𝑡) −1 take action−1.

Since 𝐵

2is contained in 𝑅, and since 𝑅 is contained in the union∪^𝑡

𝑠=1𝑅_𝑠, we have that

P⁺(𝐵

2) ≤P⁺(𝑅) ≤

𝑡

Õ

𝑠=1

P⁺(𝑅_𝑠).

Taking the maximum of all the addends in the right hand side, we can further bound the probability of𝐵

2:

P⁺(𝐵

2) ≤ 𝑡· max

1≤𝑠≤𝑡P⁺(𝑅_𝑠).

Conditioned onℓ_𝑠, the probability of 𝑅_𝑠 is P⁺(𝑅_𝑠|ℓ_𝑠)=

𝑠+𝑡/(𝛽log𝑡)−1

Ö

𝑟=𝑠

𝐺₊(−ℓ_𝑟). By Lemma 38, there exists 𝑀 > 0 such that |ℓ_𝑟| ≤ 𝑀 ℓ^∗

𝑡, for all𝑟 ≤ 𝑡. Therefore, since𝐺₊ is monotone,

P+(𝑅_𝑠) ≤𝐺₊(𝑀 ℓ^∗

𝑡)^{𝑡/(𝛽log𝑡)}. It follows that

P⁺(𝐵

2) ≤𝑡·𝐺₊(𝑀 ℓ^∗

𝑡)^{𝑡/(𝛽log𝑡)}. Since𝐺₊(𝑥) =1−𝑐·𝑥^−𝑘 for𝑥large enough, and sinceℓ^∗

𝑡 is asymptotically at most 𝑡^1/(𝑘+0.5), we have that

log𝐺₊(𝑀 ℓ^∗

𝑡) ≤ −𝑐 𝑀^−𝑘·𝑡^{−𝑘/(𝑘+0.5)}. Thus

P⁺(𝐵

2) ≤ 𝑡·exp

−𝑐 𝑀^−𝑘 ·𝑡^1/(2𝑘+1)/(𝛽log𝑡)

≤ 𝑡^−2.1,

for all𝑡 large enough. This concludes the proof, because P⁺(𝑎_𝑡 = −1) = P⁺(𝐴) + P⁺(𝐵

1) +P⁺(𝐵

2) ≤ 𝜅 ¹

𝑡2.1 for some constant𝜅.

Given this bound on the probability of mistakes, the proof of the main theorem of this section follows easily from Lemma 36.

Proof of Theorem 6. By Proposition 39, there exists𝜅 > 0 such thatP(𝑎_𝑡 =−1|𝜃= +1) < 𝜅 ¹

𝑡2.1 for all𝑡 ≥ 1. Hence, by Lemma 36E(𝑇_𝐿 |𝜃 =+1) < ∞. By a symmetric argument the same holds conditioned on𝜃 = −1. Thus, the expected time to learn is finite.

A p p e n d i x B

APPENDIX TO CHAPTER 3