Appendix A: Proofs

Proposition 1

Let operator T and spaces B(X, R^2N), ˜B(X, R^2N) be as defined in Section 3.3. If one can show that ˜B(R^N, R^2N) is a complete metric space, that T maps from ˜B(R^N, R^2N) to itself, and that T is a contraction, then the contraction mapping theorem implies that there exists a unique fixed point g^∗ = (θ^∗, ¯θ^∗) of T . The proof then concludes by establishing the properties of this fixed point, most importantly θ^∗ = ¯θ^∗. Let these steps be proved through a series of lemma.

Lemma 1. For any X ⊆ R^l, B(X, R^N) is a Banach space.

Proof. That B(X, R^N) is a normed vector space easily follows from the definition of a normed vector space. It remains to show that B(X, R^N) is complete, so that if {gⁿ}^∞n=1, gn ∈ B(X, R^N) ∀n, is a Cauchy sequence, there exists g ∈ B(X, R^N) such that, for any ǫ > 0, there exists Mǫ such that ||gⁿ− g|| ≤ ǫ, all n ≥ M^ǫ, where || · || is the max-sup norm. This is shown by extending the proof that the space of scalar-valued, bounded continuous functions is complete (see e.g., Stokey and Lucas (1989), Theorem 3.1), and involves three steps: (1) to find a candidate for g, (2) to show {gⁿ} converges to g in the max-sup norm, and (3) to show g ∈ B(X, R^N). First, fix any z ∈ X. Then the sequence of real numbers {gn^j(z)} satisfies

|gn^j(z) − gm^j (z)| ≤ sup

w∈X|g^jn(w) − gm^j (w)| ≤ max_j sup

w∈X|gn^j(w) − gm^j (w)| = ||gⁿ− g^m||, so it satisfies a Cauchy criterion, since {gⁿ} is a Cauchy sequence by assumption, and its convergence to some g^j(z) ∈ R is assured since the space of real numbers is complete. So the candidate for g is g = g¹, g², . . . , g^N
, where g_n^j → g^j pointwise for each j. Next, take any ǫ > 0 and choose Mǫ so that n, m ≥ M^ǫ implies ||gⁿ− g^m|| ≤ ǫ/2. This is possible since {gⁿ} is a Cauchy sequence. Then, for any z ∈ X and m ≥ n ≥ M^ǫ,

|gn^j(z) − g^j(z)| ≤ |g^jn(z) − gm^j (z)| + |g^jm(z) − g^j(z)|

≤ ||gⁿ− g^m|| + |gm^j (z) − g^j(z)|

≤ ǫ/2 + |gm^j (z) − g^j(z)|.

Since g_n^j → g^j pointwise, m can be chosen separately for each z so that |g^jm(z) − g^j(z)| ≤ ǫ/2. Since the choice of z ∈ X was arbitrary, it follows that supz∈X|gn^j(z) − g^j(z)| ≤ ǫ, all n ≥ M^ǫ. Since this holds for any j, it follows that ||gⁿ − g|| ≤ ǫ, all n ≥ M^ǫ. Thus ||gⁿ− g|| → 0 as n → ∞, since the choice of ǫ was arbitrary. Finally, let us show that g ∈ B(X, R^N), which is true if each g^j is bounded, continuous and nondecreasing.

Boundedness is obvious. For continuity, one needs to show that for any ǫ > 0 and any z ∈ X, there exists δ > 0 such that |g^j(z) − g^j(w) | < ǫ if ||z − w||^E < δ, where

|| · ||^E is the Euclidean norm on R^l. For any j, take any ǫ, z and choose k so that sup_w∈X|g^j(w) − g^jk(w)| < ǫ/3. This is possible because, since gⁿ → g in the max-sup norm, each g^j_n → g^j in the sup norm. Then choose δ such that ||z − w||^E < δ implies

|gk^j(z) − gk^j(w) | < ǫ/3. Since gk^j is continuous, such a choice is possible. Then

|g^j(z) − g^j(w) | ≤ |g^j(z) − g^jk(z) | + |gk^j(z) − gk^j(w) | + |gk^j(w) − g^j(w) |

≤ 2 sup

w∈X|g^j(w) − g^jk(w)| + |gk^j(z) − g^jk(w) |

< ǫ,

so g^j is continuous for all j. Therefore g ∈ B(X, R^N), which completes the proof.

Given Lemma 1, B(X, R^N) is a complete metric space with the metric d (f, g) = ||f−g||

for f, g ∈ B(X, R^N), where || · || is the max-sup norm. Thus, being a closed subset of B(X, R^N), ˜B(X, R^N) is also a complete metric space with the max-sup norm. Since this is true for any X ⊆ R^l and N, ˜B(R², R⁴) is a complete metric space.

Lemma 2. T maps ˜B(R², R⁴) to itself.

Proof. Take any g^∗_n such that g_n^∗(θ) = (θ^1∗_n (θ²) , θ^2∗_n (θ¹) , ¯θ_n^1∗(θ²) , ¯θ^2∗_n (θ¹)) ∈ ˜B(R², R⁴).

To prove g_n+1^∗ = T g_n^∗ ∈ ˜B(R², R⁴), one must show that all component functions of g^∗_n+1, namely θ^j∗_n+1 and ¯θ_n+1^j∗ , j = {1, 2}, are bounded, continuous and nondecreasing. Bound-edness is obvious since θ^j∗_n+1, ¯θ_n+1^j∗ ∈ (0, 1) from (21) and (22). For continuity, note from (17) and (18) that if θ^j∗_n and ¯θ^j∗_n are bounded, continuous and nondecreasing for j = {1, 2}, then Γ^j∗n (xi) − ¯Γ^k∗n (xi) and ¯Γ^j∗_n (xi) − Γ^k∗n (xi) are bounded and continuous in both arguments, decreasing in x^j_i, and nondecreasing in x^k_i. Then, (19) and (20) imply that x^j∗_n x^k_i
and ¯x^j∗_n x^k_i
are continuous and nondecreasing, and increasing for the range of x^k_i for which x^j∗_n x^k_i
and ¯x^j∗_n x^k_i
are determined by the indifference condition between attacking j and k under the corresponding beliefs. It then follows from (21) and (22) that θ^j_n+1 and ¯θ^j_n+1 are continuous and nondecreasing (in fact, increasing), hence g_n+1^∗ ∈ B(R˜ ², R⁴).

Lemma 3. T is a contraction.

Proof. The proof proceeds by showing that the Blackwell’s sufficient conditions for a

con-traction are satisfied for each component function. Take any g_n^∗(θ) = (θ^1∗_n (θ²) , θ^2∗_n (θ¹) , ¯θ_n^1∗(θ²) , ¯θ^2∗_n (θ¹)) B(R˜ ², R⁴). For monotonicity, fix j ∈ {1, 2} and define gn^∗∗ ∈ ˜B(R², R⁴) by replacing the

j-th component of g_n^∗, namely θ^j∗_n, by θ^j∗∗_n ≥ θ^j∗n. Further, let Γ^∗∗_n , ¯Γ^∗∗_n , x^∗∗_n, ¯x^∗∗_n , θ^∗∗_n+1,

θ¯^∗∗_n+1 be the functions obtained from g_n^∗∗ by (17)–(22). For any xi = x^j_i, x^−j_i
∈ R², (17) since the choice of a ≥ 0 was arbitrary, discounting also holds. Repeating the argument above for each component function of g_n^∗, it follows that monotonicity and discounting holds for θ^j∗_n and ¯θ_n^j∗, j = {1, 2}. Now, let gn^∗ and g_n+1^∗ be as defined above, take any g_n⁺∈ ˜B(R², R⁴) such that

g_n⁺(θ) = (θ¹⁺_n (θ²) , θ²⁺_n (θ¹) , ¯θ_n¹⁺(θ²) , ¯θ_n²⁺(θ¹)), and let g⁺_n+1 = T g_n⁺. Then, noting θ^j∗_n ≤ θ^j+n + sup

θ^−j∈R^{N −1}|θ^j∗n θ^−j
− θ^j+n θ^−j
| and invoking monotonicity and discounting shown above,

θ^j∗_n+1 ≤ θ^j+n+1+ λ^j sup

θ^−j∈R^{N −1}|θ^j∗n θ^−j
− θ^j+n θ^−j
|.

Reversing the role of θ^j∗_n and θ^j+_n , one obtains θ^j+_n+1 ≤ θ^j∗n+1+ λ^j sup

θ^−j∈R^{N −1}|θ^j∗n θ^−j
− θ^j+n θ^−j
|.

Therefore, sup

θ^−j∈R^{N −1}|θ^j+n+1 θ^−j
− θ^j∗n+1 θ^−j
| ≤ λ^j sup

θ^−j∈R^{N −1}|θ^j∗n θ^−j
− θ^j+n θ^−j
| (34) for all j. A similar argument establishes

sup

which implies that T is a contraction.

Given Lemma 1–3, T has a unique fixed point g^∗ = (θ^∗, ¯θ^∗) from the contraction

Now, let ˜B^′(R², R⁴) denote the space of bounded, continuous, nondecreasing vector-valued functions g : R² → R⁴, g = (g¹, g², g³, g⁴) where g¹= g³and g² = g⁴, equipped with

Proposition 2

The proof is almost identical to the N = 2 case. First, since B(X, R^N) is a Ba-nach space as shown in Lemma 1, ˜B(X, R^N) is a complete metric space, and thus so is B(R˜ ^N, R^2N). That T^′ maps ˜B(R^N, R^2N) to itself is also straightforward, except that this time, x^j∗_n and ¯x^j∗_n may not always be nondecreasing, but x^j∗_{n,N D} and ¯x^j∗_{n,N D} are nondecreas-ing by construction, such that θ^j_n+1 and ¯θ_n+1^j are nondecreasing (indeed, increasing). It remains only to confirm that using x^j∗_{n,N D} and ¯x^j∗_{n,N D}, instead of x^j∗_n and ¯x^j∗_n, to compute θ^j_n+1 and ¯θ_n+1^j does not hamper the satisfaction of the Blackwell’s sufficient conditions for contraction.

For monotonicity, fix j ∈ {1, 2, . . . , N} and define gn^∗∗∈ ˜B(R^N, R^2N) by replacing the j-th component of g_n^∗, namely θ^j∗_n, by θ^j∗∗_n ≥ θ^j∗n. Let Γ^∗∗_n, ¯Γ^∗∗_n, x^∗∗_n, ¯x^∗∗_n, x^∗∗_{n,N D}, ¯x^∗∗_{n,N D}, θ^∗∗_n+1, θ¯^∗∗_n+1be the functions obtained from g^∗∗_n by (17)–(20) and (27)–(32), with c^j = c and β^j = β for all j. Proceeding as in the proof of Lemma 3, it follows that x^j∗∗_n ≥ x^j∗n, and since x^j∗_n ≥ x^j∗n,N D from the definition of x^j∗_{n,N D}, x^j∗_{n,N D} ∈ X^j∗∗n ≡ f ∈ ˜B R^N−1, R
|f ≤ x^j∗∗n . Then, x^j∗∗_{n,N D} ≥ x^j∗n,N D from the definition of x^j∗∗_{n,N D}, and thus θ^j∗∗_n+1 ≥ θ^j∗n+1 from (31), which implies monotonicity.

For discounting, fix j ∈ {1, 2, . . . , N}, take any a ≥ 0 and define gn^# ∈ ˜B(R^N, R^2N) by replacing the j-th component of g^∗_n, namely θ^j∗_n, by θ^j∗_n + a. Let Γ^#_n, ¯Γ^#_n, x^#_n, ¯x^#_n, θ^#_n+1, ¯θ_n+1^# be the functions obtained from g_n^# by (17)–(20) and (27)–(32). Proceeding as in the proof of Lemma 3, x^j#_n ≤ x^j∗n + a, which implies x^j#_{n,N D} − a ≤ x^j#n − a ≤ x^j∗n. Then x^j#_{n,N D}− a ∈ X^j∗n = f ∈ ˜B R^N−1, R
|f ≤ x^j∗n , so from the definition of x^j∗_{n,N D}, x^j#_{n,N D}− a ≤ x^j∗n,N D, or equivalently, x^j#_{n,N D} ≤ x^j∗n,N D+ a. Thus, arguing as in the proof of Lemma 3, θ^j#_n+1(θ^−j) ≤ θ^j∗n+1(θ^−j) + λa, where γ =√

β/√

2π and λ ≡ γ/ (1 + γ) ∈ (0, 1), and since the choice of a ≥ 0 was arbitrary, discounting also holds.

Repeating the argument above for each component functions of g^∗_n, with a minor modification for the 2j-th component functions, monotonicity and discounting holds for θ^j∗_n and ¯θ^j∗_n, j = {1, 2, . . . , N}. Thus, proceeding as in the Proof of Lemma 3, T^′ is shown to be a contraction with the factor of contraction λ, so by the contraction mapping theorem, T^′ has a unique fixed point, denoted as θ^∗, ¯θ^∗
.

Proposition 3

Given Proposition 2, it remains to establish the properties of the fixed point θ^∗, ¯θ^∗
, such as θ^∗ = ¯θ^∗ ≡ θ^∗, and this is where one must resort to symmetry. Take g_n^∗ = (θ^∗_n, ¯θ_n^∗) ∈ B(R˜ ^N, R^2N) such that θ^∗_n = ¯θ^∗_n ≡ θn^∗ = θ^1∗_n , θ^2∗_n , . . . , θ_n^N∗
, where θ_n^j∗ is symmetric across j. From (17) and (18), Γ^j∗_n = ¯Γ^j∗_n ≡ Γ^j∗n. Moreover, Γ^j∗_n is symmetric across j, and Γ^j∗_n and Γ^j∗_n − Γ^k∗n are continuous and decreasing in x^j_i, so (19) and (20) imply x^j∗_n = ¯x^j∗_n ≡ x^j∗n,

where x^j∗_n is symmetric across j.

Furthermore, x^j∗_n is nondecreasing, as shown below. Clearly x^j∗_n is nondecreasing when it equals −∞, so suppose x^j∗n is finite. Pick any j, k ∈ {1, 2, . . . , N} , j 6= k. Fix x^−j_i ∈ R^N−1, and let x^j,k_n x^−j_i
be the value of x^j_i such that

Γ^j∗_n (xi) = Γ^k∗_n (xi) . (36) Since Γ^j∗_n is symmetric across j, (36) is satisfied for x^j_i = x^k_i, and since Γ^j∗_n (xi) − Γ^k∗n (xi) is decreasing in x^j_i, no other x^j_i satisfies (36). Therefore, x^j,k_n x^−j_i
= x^k_i, which is continuous and increasing in x^k_i, and is independent of other elements of x^−j_i . Now, with a slight abuse of notation, let x^j,0_n x^−j_i
be the value of x^j_i such that

Γ^j∗_n (xi) = 0. (37)

Since Γ^j∗_n (xi) is continuous, decreasing in x^j_i and nondecreasing in x^−j_i , x^j,0_n x^−j_i
is uniquely determined, and is continuous and nondecreasing in x^−j_i . Then, (19) (or equiv-alently, (20)) implies

Finally, let ˜B^′(R^N, R^2N) denote the space of bounded, continuous, nondecreasing vector-valued functions g : R^N → R^2N, g = g¹, g², . . . , g^2N
, where g^j is symmet-ric across j and g^j = g^2j for j = {1, 2, . . . , N}, equipped with the max-sup norm

||g|| ≡ max^Nj=1(sup_z∈R|g^j(z) |). Further, let ˜B^′′(R^N, R^2N) ⊆ ˜B^′(R^N, R^2N) be the space in which ‘nondecreasing’ in the definition of ˜B^′(R^N, R^2N) is replaced by ‘increasing’, and g^j ∈ (0, 1) for all j. Then, ˜B^′(R^N, R^2N) is a closed subset of ˜B(R^N, R^2N), and the ar-gument above implies T ˜B^′(R^N, R^2N)
⊆ ˜B^′′(R^N, R^2N) ⊆ ˜B(R^N, R^2N), so it follows that g^∗ ∈ ˜B^′′(R^N, R^2N) (Stokey and Lucas (1989), Theorem 3.2, Corollary 1). Therefore, θ^∗ = ¯θ^∗ ≡ θ^∗ = θ^1∗, θ^2∗, . . . , θ^N∗
, where θ^j∗ is symmetric across j, θ^j∗ ∈ (0, 1) and θ^j∗ is continuous and increasing. Then, at this fixed point, the expected payoff functions and the threshold signal functions must be such that Γ^j∗ = ¯Γ^j∗ ≡ Γ^j∗ and x^j∗ = ¯x^j∗ ≡ x^j∗, where x^j∗ is continuous and nondecreasing. That x^j∗ x^−j_i

≤ min^k6=jx^k_i also follows

Proposition 4

Let us first consider the N = 2 case. Among the equations (17)–(22) that define the operator T , (17) and (18) are modified as follows. This time, conditional on receiving x^j_i, θ^j follows N α^jy^j + β^jx^j_i
/ (α^j+ β^j) , 1/ (α^j+ β^j)
such that holds as in the model with only private information. For discounting, take any g_n^∗ = (θ^1∗_n (θ²) , θ^2∗_n (θ¹) , ¯θ^1∗_n (θ²) , ¯θ^2∗_n (θ¹)) ∈ ˜B(R², R⁴). Then fix j ∈ {1, 2}, take any a ≥ 0 and

(21) and (43),

θ^j##_n+1 θ^−j
− θ^j∗n+1 θ^−j

≤ Z

ǫ^−j_i ∈R^{N −1}

φ˜^−j ǫ^−j_i
γ^jh

a α^j+ β^j
/β^j−

θ^j##_n+1 θ^−j
− θ^j∗n+1 θ^−j
i dǫ^−j_i

≤ γ^jh

a α^j + β^j
/β^j−

θ^j##_n+1 θ^−j
− θ^j∗n+1 θ^−j
i

≤ aλ^j α^j + β^j
/β^j,

where λ^j ≡ γ^j/ (1 + γ^j) ∈ (0, 1), so θ^j#n+1(θ^−j) ≤ θ^j##n+1 (θ^−j) ≤ θ^j∗n+1(θ^−j)+aλ^j(α^j + β^j) /β^j. But since α^j <pβ^j√

2π by assumption,

λ^j α^j+ β^j
/β^j = pβ^j/√ 2π 1 +pβ^j/√

2π

α^j + β^j

β^j < pβ^j

√2π +pβ^j

√2π +pβ^j

pβ^j = 1, (44)

and since the choice of a ≥ 0 was arbitrary, discounting also holds. The rest of the proof follows that of Proposition 1.

For the symmetric N > 2 case, among the equations (17)–(20) and (27)–(32) that define the operator T^′, (17) and (18) are modified as (41) and (42), and c^j = c, β^j = β in all equations. It is again straightforward to check that T^′ : ˜B(R^N, R^2N) → ˜B(R^N, R^2N) and that monotonicity holds, and arguing as in the N = 2 case above, discounting follows given α/√

β <√

2π. The rest of the proof follows that of Proposition 3.

Proposition 5

The argument requires only minor modifications to that for Propositions 1–3. When signals ǫi = ǫ¹_i, ǫ²_i, . . . , ǫ^N_i
= ǫ^j_i, ǫ^−j_i
are drawn from a joint pdf ϕ, one can write, for example,

Pr θ^j ≤ θ^j∗n θ^−j
|x^ji, x^−j_i
= Pr x^j_i − ǫ^ji ≤ θ^j∗n x^−j_i − ǫ^−ji

= Z

ǫ^−j_i ∈R^{N −1}

Z ∞

x^j_i−θ^j∗_n(^x−ji −ǫ^−j_i )ϕ (ǫi) dǫ^j_idǫ^−j_i . (45)

Thus, the equations (17)–(22) that define the operator T are rewritten as Suppose N = 2. Lemma 1 clearly does not hinge on these equations. Lemma 2 follows from the same argument as when noises are independent. Regarding Lemma 3, monotonicity can be shown as before. For discounting, proceed as before to obtain

θ^j##_n+1 θ^−j
= the proof follows that of Proposition 1.

For the symmetric N > 2 case, rewrite the equations that define the operator T^′, (17)–(20) and (27)–(32), using ϕ. Then discounting can be shown as for the N = 2 case above, and the rest of the proof follows that of Propositions 2 and 3.