Long-term behavior of public belief

Chapter II: The speed of sequential asymptotic learning

2.6 Long-term behavior of public belief

the distribution of𝐿_𝑡is identical to that of𝑠_𝑡, and so𝐺₊ =𝐹₊ and𝐺₋ =𝐹₋. By our definition of𝐹₋, we have that for𝑥 >0

𝐺₋(−𝑥) =𝐶⁻¹·𝑄( ⌈𝑥⌉ −1). (2.5) Now, by Lemma 8, we know that

(1−𝜖) ·𝐺₋(−𝑥) < 𝐷₊(𝑥) < (1+𝜖) ·𝐺₋(−𝑥), for any𝜖 >0 and all𝑥large enough. It follows that

lim inf

𝑥→∞

𝐷₊(𝑥)

𝑄(𝑥) =lim inf

𝑥→∞

𝐺₋(−𝑥) 𝑄(𝑥) , which, by (2.5) equals

lim inf

𝑥→∞

𝐶⁻¹𝑄( ⌈𝑥⌉ −1)

𝑄(𝑥) ≥ 𝐶⁻¹.

□

Lemma 10. Assume that 𝐴: R^>0 → R^>0 is a continuous function with a convex differentiable tail, and that 𝐴(𝑥)goes to0as𝑥goes to∞. Let (𝑎_𝑡)be any sequence satisfying the recurrence equation𝑎_𝑡+1=𝑎_𝑡+𝐴(𝑎_𝑡), and suppose there is a function

𝑓 :R^>0→ R^>0with 𝑓^′(𝑡) = 𝐴(𝑓(𝑡)) for all sufficiently large𝑡. Then

𝑡→∞lim 𝑓(𝑡)

𝑎_𝑡

=1.

Given these lemmas, we are ready to prove our theorem.

Proof of Theorem 6. Let (𝑎_𝑡) be any sequence inR^>0satisfying:

𝑎_𝑡+1=𝑎_𝑡 +𝐺₋(−𝑎_𝑡).

Then by Lemma 10, the sequence(𝑎_𝑡)is well approximated by 𝑓(𝑡), the solution to the corresponding differential equation:

lim

𝑡→∞

𝑎_𝑡 𝑓(𝑡) =1.

Now, conditional on 𝜃 = +1, all agents take action +1 from some point on with probability 1. Thus, with probability 1,

ℓ_𝑡+1=ℓ_𝑡+𝐷₊(ℓ_𝑡) for all sufficiently large𝑡. Further, by Lemma 8,

𝑥→∞lim

𝐷₊(𝑥) 𝐺₋(−𝑥) =1. So by Lemma 9,

𝑡→∞lim ℓ_𝑡 𝑎_𝑡

=1 with probability 1. Thus, we have

𝑡→∞lim ℓ_𝑡

𝑓(𝑡) = lim

𝑡→∞

ℓ_𝑡 𝑎_𝑡

· 𝑎_𝑡 𝑓(𝑡) =1

with probability 1. □

Proofs of Lemmas 9 and 10

Proof of Lemma 9. We prove the claim in two steps. First, we show that for every 𝜀 > 0 there are infinitely many times𝑡such that

(1−𝜀)𝑎_𝑡 ≤ 𝑏_𝑡 ≤ (1+𝜀)𝑎_𝑡. (2.6) Second, we show that if (2.6) holds for some 𝑡 large enough, then it holds for all 𝑡^′> 𝑡, proving the claim.

We start with step 1. Assume without loss of generality that𝑎_𝑡 ≤ 𝑏_𝑡 for infinitely many values of𝑡. Fix 𝜀 > 0. To show that (1−𝜀)𝑎_𝑡 ≤ 𝑏_𝑡 ≤ (1+𝜀)𝑎_𝑡 holds for infinitely many values of𝑡, let𝑥₀> 1 be such that for all𝑥 > 𝑥₀it holds that 𝐴and 𝐵are monotone decreasing,

𝐴(𝑥), 𝐵(𝑥) < 𝜀 < 1 and

(1−𝜀/2)𝐴(𝑥) < 𝐵(𝑥) < (1+𝜀/2)𝐴(𝑥). (2.7) Assume that𝑎_𝑡, 𝑏_𝑡 > 𝑥₀; this will indeed be the case for𝑡large enough, since 𝐴and 𝐵are positive and continuous, and so both𝑎_𝑡 and 𝑏_𝑡 are monotone increasing and tend to infinity. So

𝐵(𝑏_𝑡) < (1+𝜀/2)𝐴(𝑏_𝑡) ≤ (1+𝜀/2)𝐴(𝑎_𝑡),

where the first inequality follows from (2.7), and the second follows from the fact that 𝐴 is monotone decreasing and 𝑎_𝑡 < 𝑏_𝑡. Since 𝐵(𝑏(𝑡)) = 𝑏_𝑡+1 − 𝑏(𝑡) and 𝐴(𝑎_𝑡) =𝑎_𝑡+1−𝑎(𝑡) we have shown that

𝑏_𝑡+1−𝑏_𝑡 < (1+𝜀/2) (𝑎_𝑡+1−𝑎_𝑡),

and so eventually 𝑏_𝑡 ≤ (1+𝜀)𝑎_𝑡. Also, notice that the first time this obtains, we also have that the left inequality in (2.6) holds at the same moment:

𝑏_𝑡 > 𝑏_𝑡−1> 𝑎_𝑡−1=𝑎_𝑡 − (𝑎_𝑡 −𝑎_𝑡−1) > 𝑎_𝑡−𝜀 > 𝑎_𝑡−𝜀 𝑎_𝑡 =(1−𝜀)𝑎_𝑡.

This completes the first step. Now we go to step 2. Here we show that if (2.6) holds for large enough𝑡 then it holds for all𝑡^′> 𝑡.

Fix𝜀 > 0, and let𝑥₀be defined as above. Suppose that(1−𝜀)𝑎_𝑡 < 𝑏_𝑡 < (1+𝜀)𝑎_𝑡, with 𝑎_𝑡, 𝑏_𝑡 > 𝑥₀. Assume without loss of generality that 𝑏_𝑡 ≥ 𝑎_𝑡. Then our assumptions and (2.7) imply

𝑏_𝑡+1=𝑏_𝑡+𝐵(𝑏_𝑡)

< (1+𝜀)𝑎_𝑡+ (1+𝜀)𝐴(𝑏_𝑡). Because𝑎_𝑡 ≤ 𝑏_𝑡and 𝐴is decreasing we have

𝑏_𝑡+1< (1+𝜀)𝑎_𝑡+ (1+𝜀)𝐴(𝑎_𝑡)

= (1+𝜀)𝑎_𝑡+1. For the other direction, note first that

𝑏_𝑡+1> 𝑏_𝑡 ≥ 𝑎_𝑡,

by assumption. We can write 𝑎_𝑡 = (1 − 𝜀)𝑎_𝑡 + 𝜀 𝑎_𝑡, and since 𝑎_𝑡 > 𝑥₀ > 1, 𝜀 𝑎_𝑡 > (1−𝜀)𝜀, and so

𝑏_𝑡+1 > (1−𝜀)𝑎_𝑡+ (1−𝜀)𝜀 . Now,𝜀 > 𝐴(𝑎_𝑡) since𝑎_𝑡 > 𝑥₀, and so

𝑏_𝑡+1 > (1−𝜀)𝑎_𝑡+ (1−𝜀)𝐴(𝑎_𝑡)

=(1−𝜀)𝑎_𝑡+1. Thus

(1−𝜀)𝑎_𝑡+1 < 𝑏_𝑡+1< (1+𝜀)𝑎_𝑡+1, (2.8) as required.

□ Proof of Lemma 10. We restrict the domain of 𝑓 to the interval(𝑡₀,∞)such that for 𝑡 > 𝑡₀ it already holds that 𝑓^′(𝑡) = 𝐴(𝑓(𝑡)). Since 𝐴is continuous, lim𝑡→∞ 𝑓(𝑡) =

∞, and so we can also assume that in the interval(𝑓(𝑡₀),∞)it holds that𝐴is convex and differentiable.

Since 𝑓 is strictly increasing in (𝑡₀,∞), it has an inverse 𝑓⁻¹. For𝑥 large enough define𝐵(𝑥) = 𝑓(𝑓⁻¹(𝑥) +1) −𝑥.

Now, let(𝑏_𝑡)be any sequence satisfying the recurrence relation 𝑏_𝑡+1=𝑏_𝑡+𝐵(𝑏_𝑡).

In order to apply Lemma 9, we will first show that

𝑥→∞lim 𝐵(𝑥) 𝐴(𝑥) =1.

Let𝑡 = 𝑓⁻¹(𝑥). Such a𝑡 exists and is unique for all sufficiently large𝑥, because 𝑓 is monotone. Notice that by the definitions of𝐵(𝑥)and 𝑓^′(𝑥)

𝐵(𝑥)= 𝑓(𝑓⁻¹(𝑥) +1) −𝑥

= 𝑓(𝑓⁻¹(𝑥) +1) −𝑥− 𝑓^′(𝑓⁻¹(𝑥)) + 𝑓^′(𝑓⁻¹(𝑥))

= 𝑓(𝑡+1) − 𝑓(𝑡) − 𝑓^′(𝑡) +𝐴(𝑓(𝑡)),

where in the last equality we substitute 𝑡 = 𝑓⁻¹(𝑥). Because 𝑓^′ is positive and decreasing (𝑓 is concave) then 𝑓(𝑡 +1) − 𝑓(𝑡) ≥ 𝑓^′(𝑡+1), and so

𝐵(𝑥) ≥ 𝑓^′(𝑡+1) − 𝑓^′(𝑡) +𝐴(𝑓(𝑡)). By the definition of 𝑓, 𝑓^′(𝑡) = 𝐴(𝑓(𝑡)), and so

𝐵(𝑥) ≥ 𝐴(𝑓(𝑡+1)) −𝐴(𝑓(𝑡)) +𝐴(𝑓(𝑡))= 𝐴(𝑓(𝑡+1)).

Again, due to concavity of 𝑓 we have 𝑓(𝑡+1) ≤ 𝑓(𝑡) + 𝑓^′(𝑡)and as𝐴is decreasing and convex we get

𝐵(𝑥) ≥ 𝐴(𝑓(𝑡) + 𝑓^′(𝑡))

≥ 𝐴^′(𝑓(𝑡))𝑓^′(𝑡) +𝐴(𝑓(𝑡))

= 𝐴^′(𝑓(𝑡))𝐴(𝑓(𝑡)) +𝐴(𝑓(𝑡)). We now substitute back𝑥 = 𝑓(𝑡):

𝐵(𝑥) ≥ 𝐴^′(𝑥)𝐴(𝑥) +𝐴(𝑥)

= 𝐴(𝑥) (𝐴^′(𝑥) +1) so in particular, since 𝐴^′(𝑥) →0 as𝑥 → ∞,

lim inf

𝑥→∞

𝐵(𝑥) 𝐴(𝑥) ≥ 1.

Now we are going to show that lim sup_𝑥→∞ ^𝐵(𝑥)_𝐴(𝑥) ≤ 1 which will conclude the proof.

By the definitions of 𝑓⁻¹(𝑥)and𝐵(𝑥)

𝐵(𝑥) =𝐵(𝑓(𝑡))= 𝑓(𝑡+1) − 𝑓(𝑡) =

∫ ^𝑡+1 𝑡

𝑓^′(𝜁)d𝜁 . As 𝑓^′is decreasing it follows that

𝐵(𝑥) ≤

∫ 𝑡+1 𝑡

𝑓^′(𝑡)d𝜁 = 𝑓^′(𝑡) = 𝐴(𝑓(𝑡)) = 𝐴(𝑥). Therefore,

lim sup

𝑥→∞

𝐵(𝑥) 𝐴(𝑥) ≤ 1. Hence, from these two inequalities we get that

lim

𝑥→∞

𝐵(𝑥) 𝐴(𝑥) =1.

Now notice that, by construction, 𝑓(𝑡+1) = 𝑓(𝑡) +𝐵(𝑓(𝑡)). Thus, by Lemma 9, lim

𝑛→∞

𝑓(𝑡) 𝑎_𝑡

=1.

□

Monotonicity of solutions to a differential equation

We now prove a general lemma regarding differential equations of the form𝑎^′(𝑡) = 𝐴(𝑎(𝑡)). It shows that the solutions to this equation are monotone in 𝐴. This is useful for calculating approximate analytic solutions whenever it is impossible to find analytic exact solutions, as is the case of Gaussian signals, in which we use this lemma.

Lemma 11. Let 𝐴, 𝐵: R^>0→R^>0be continuous, and let𝑎, 𝑏: R^>0→R^>0satisfy 𝑎^′(𝑡) = 𝐴(𝑎(𝑡))and𝑏^′(𝑡) =𝐵(𝑏(𝑡))for all sufficiently large𝑡.

Suppose that

lim inf

𝑥→∞

𝐴(𝑥) 𝐵(𝑥) > 1. Then𝑎(𝑡) > 𝑏(𝑡) for all sufficiently large𝑡.

Proof. Notice that 𝑎(𝑡) and 𝑏(𝑡) are eventually monotone increasing and tend to infinity as𝑡 tends to infinity. Thus for all𝑥 greater than some𝑥₀ > 0 large enough, 𝑎and𝑏have inverses that satisfy the following differential equations:

d d𝑥

𝑎⁻¹(𝑥) = 1 𝐴(𝑥) d

d𝑥

𝑏⁻¹(𝑥) = 1 𝐵(𝑥)·

Since lim inf𝑥 𝐴(𝑥)/𝐵(𝑥) > 1, we can furthermore choose𝑥₀so that for all𝑥 ≥ 𝑥₀, 𝐴(𝑥) > (1+𝜀)𝐵(𝑥) for some𝜀 >0. Thus, for𝑥 > 𝑥₀

𝑎⁻¹(𝑥)=𝑎⁻¹(𝑥₀) +

∫ ^𝑥

𝑥₀

1 𝐴(𝑥) d𝑥 𝑏⁻¹(𝑥)=𝑏⁻¹(𝑥₀) +

∫ ^𝑥

𝑥₀

1 𝐵(𝑥)d𝑥 and so

𝑎⁻¹(𝑥) < 𝑎⁻¹(𝑥₀) + 1 1+𝜀

∫ ^𝑥

𝑥₀

1 𝐵(𝑥) d𝑥

=𝑎⁻¹(𝑥₀) + 1 1+𝜀

(𝑏⁻¹(𝑥) −𝑏⁻¹(𝑥₀)) and thus

𝑎⁻¹(𝑥) −𝑏⁻¹(𝑥) <− 𝜀 1+𝜀

𝑏⁻¹(𝑥) +

𝑎⁻¹(𝑥₀) − 1 1+𝜀

𝑏⁻¹(𝑥₀)

.

Since𝑏⁻¹(𝑥)tends to infinity as𝑥tends to infinity, it follows that for all sufficiently large𝑥, 𝑎⁻¹(𝑥) < 𝑏⁻¹(𝑥). Thus, for all sufficiently large𝑡

𝑡 =𝑎⁻¹(𝑎(𝑡)) < 𝑏⁻¹(𝑎(𝑡)), and so, since𝑏(𝑡) is monotone increasing,

𝑏(𝑡) < 𝑎(𝑡).

□

Eventual monotonicity of public belief update

We end this section with a lemma that shows that under some technical conditions on the left tail of 𝐺₋, the function 𝑢₊(𝑥) = 𝑥 + 𝐷₊(𝑥) (i.e., the function that determines how the public log-likelihood ratio is updated when the action +1 is taken) is eventually monotone increasing.

Lemma 12. Suppose 𝐺₋ has a convex and differentiable left tail. Then the map 𝑢₊(𝑥) =𝑥+𝐷₊(𝑥)is monotone increasing for all sufficiently large𝑥.

Proof. Recall that

𝐷₊(𝑥)=log1−𝐺₊(−𝑥) 1−𝐺₋(−𝑥)·

Since 𝐺₋ has a differentiable left tail, it has a derivative 𝑔₋(−𝑥) for all 𝑥 large enough. It then follows from (2.4) that𝐺₊ also has a derivative in this domain, and

𝑢^′₊(𝑥) =1+ 𝑔₊(−𝑥)

1−𝐺₊(−𝑥) − 𝑔₋(−𝑥) 1−𝐺₋(−𝑥)

=1+ e^−𝑥𝑔₋(−𝑥)

1−𝐺₊(−𝑥) − 𝑔₋(−𝑥) 1−𝐺₋(−𝑥)· Since 1−𝐺₋(−𝑥)and 1−𝐺₊(−𝑥)tend to 1 as𝑥tends to infinity,

lim

𝑥→∞

𝑢^′₊(𝑥) = lim

𝑥→∞1+e^−𝑥𝑔₋(−𝑥) −𝑔₋(−𝑥). Since𝐺₋ is eventually convex,𝑔₋(−𝑥) tends to zero, and therefore

𝑥→∞lim

𝑢^′₊(𝑥) =1.

In particular, 𝑢^′₊(𝑥) is positive for 𝑥 large enough, and hence 𝑢₊(𝑥) is eventually

monotone increasing. □

2.7 Gaussian private signals