Competitive Experimentation and Private Information

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1 Compettve Expermentaton an Prvate Informaton Guseppe Moscarn an Francesco Squntan Omtte Analyss not Submtte for Publcaton Dervatons for te Gamma-Exponental Moel Dervaton of expecte azar rates. By Bayes rule, posteror belefs are π t,t 0( x, y) π ( z x, z j y, t t, t j t 0 ) π () (x ) (y )[ F (t )] [ F j (t 0 )] RΛ π (0 ) (x 0 ) (y 0 )[ F (t 0 )] [ F j (t 0 0 )] 0 e α β e x e y e (ζ +)t e (ζ j +)t0 RΛ e α0 0β 0 e 0 x 0 e 0 y e (ζ +)t e (ζ j +)t0 0 e (α x y+t+t0) β+ Γ (β +)(α x y + t + t 0 ) β, π t,t 0( x, y+) π ( z x, z j y, t t, t j t 0 ) π () (x )[ H(y )] [ F (t )] [ F j (t 0 )] RΛ π (0 ) (x 0 )[ H(y 0 )] [ F (t 0 )] [ F j (t 0 0 )] 0 β e (α+t+t0 x) e (α+t+t0 x y) Γ (β +) (α + t + t 0 x) β (α + t + t 0 x y) β. Takng expectatons of te azar rate wt respect to tese tree posterors yels te followng expressons: E t,t [ x, y] β + 0 α x y + t + t, 0 E t,t [ x, y ] β + 0 α x y + t + t, 0 E t,t 0 [ x, y+] (β +)(α + t + t0 x) β (α + t + t 0 x y) β (α + t + t 0 x) β (α + t + t 0 x y) β. Lemma A.3 In te Gamma-exponental moel, E t,t [ζ + x, y+] satsfes Assumptons an. Proof. We use te followng tecncal result.

2 Teorem A. Let q : Λ < n < +,ntegrablentsfrst argument an fferentable n te remanng n arguments, wt R Λ q(, θ) (0, ). Ten te c..f. efne by: ϕ(l, θ) R L q(, θ) RΛ q 0, θ 0 for every L, s stocastcally strctly ncreasng n every component of θ f q(, θ) s log-supermoular n (, θ ),.e. f log q(, θ)/ θ s strctly ncreasng n. Proof We want to sow tat for every L, usng te clam reas Z Λ R 0 > L q(, θ) θ RΛ q(, θ) Z q(, θ) θ log q(, θ) q(, θ) θ RΛ q 0, > θ 0 R L q(, θ) R L θ RΛ q(, θ) q(, θ) R Λ θ R θ) q(, Λ Z log q(, θ) q(, θ θ) Z L q(, θ) log q(, θ) q(, θ) θ R L q 0, θ 0 Asuffcent conton for te latter nequalty s tat te RHS be strctly ncreasng n L. Snce te RHS s fferentable n L, tsuffces tat 0 < Z L log q(, θ) q(, θ) L θ R L q 0, θ 0 log q(l, θ) q(l, θ) θ R L q 0, q(l, θ) θ 0 or log q(l, Z θ) L > θ Z L log q(, θ) q(, θ) θ R L q 0, θ 0 log q(, θ) q(, θ) θ R L q 0,. θ 0 But, snce R L q(, θ) L q ( 0,, ts follows from te assumpton tat te LHS s strctly θ) 0 ncreasng n. It now suffces to sow tat for every x, y, t, t 0, te c..f. assocate wt te posteror belefs π t,t 0 ( x, y+) are stocastcally strctly ecreasng n t an n t 0 an strctly ncreasng n x an n y. We prove all tese results as corollares of Teorem A. Let θ (x, y, t, t 0 ), q(, θ)π () (x )[ H (y )] [ F (t )] [ F (t 0 )],

3 Snce te expressons log q(, θ) t ϕ(l, θ) Z L π t,t 0 ( x, y+). f (t ) F (t ), log q(, θ) t 0 f (t0 ) F (t 0 ) log q(, θ) x 0 (x ) (x ), log q(, θ) (y ) y H (y ) ex e x are strctly monotonc n, all monotoncty results follow from Teorem A. lm t E t,t 0 [ x, y+] 0 follows from: Te lmt lm τ π τ,τ ( >ε x, y+) lm τ R ε R e α β e x e y e τ e 0 α0 0β 0 e 0 x e 0. 0 y e 0 τ 0 Lemma A.4 In te Gamma-exponental moel, Et,t [ζ + x, y+] satsfes Assumpton 3. Proof. It s mmeate to verfy tat t E t,t [ζ 0 + x, y+] x E t,t [ζ 0 + x, y+], as t can be apprecate wt te cange of varable q α +t x, an φ (q, y, β) (β +) q β (q y) β q β (q y) β t E t,t [ζ 0 + x, y+] φ q q t φ q, x E t,t [ζ 0 + x, y+] φ q q x φ q Hence, settng G just a bt larger tan an tus G 0 < / satsfes te nequalty n Assumpton 3. For te frst nequalty, we sow tat t s strct at G 0 / an terefore, by contnuty, for G 0 slgtly larger: y E t,t 0 [ζ + x, y+] < t E t,t 0 [ζ + x, y+] x E t,t 0 [ζ + x, y+]. Algebra yels x E t,t 0 [ζ + x, y+] (β +) (β +) (α +t x) β 3 (α +t x y) β 3 (α +t x) β (α +t x y) β (β +) (α +t x) β +(α +t x y) β (α +t x) β (α +t x y) β 3

4 y E t,t [ζ 0 + x, y+] (β +)(α +t x y) β 3 (α +t x) β (α +t x y) β +(β +)(α +t x y) β (α +t x) β (α +t x y) β (β +) (α +t x) β (α +t x y) β. Hence, x E t,t [ζ 0 + x, y+] + y E t,t [ζ 0 + x, y+] (β +) (α +t x) β 3 (α +t x) β (α +t x y) β (β +) (α +t x) β (α +t x) β (α +t x y) β (β +) (α +t x) β (α +t x y) β (α + t + t 0 x) β (β +) (α + t + t 0 x) β (α + t + t 0 x y) β (α + t + t 0 x) β 3 (β +) (α + t + t 0 x) β (α + t + t 0 x y) β q β β 3 q (β +) (β +) q β β (q y) q β (q y) β We want ts quantty to be strctly negatve. Ts s so f te followng ervatve s strctly postve for all β,y: Ã! q β β β q β (q y) β q β q β q β (q y) β ln q + q β ln q (q y) β ln (q y) + β q β β (q y) q β (q y) β q β (q y) β + β q β (q y) β ln q + q β ln q (q y) β ln (q y) q β (q y) β +(q y) β β (ln q ln (q y)) : χ (β,y) Note tat for y<0 q β χ (β,y) y β ln q ln (q y) β (q y) β+ (y q)(q y) β (q y) ln q) β (ln < 0, β β+ (q y) (q y) β+ 4

5 wle χ (β,0) q β q β + q β β (ln q ln (q)) 0, χ(β, ) q β q β + β q y q β > 0. So nee χ (β,0) > 0 for all β,y. (q y) β +β(ln(q y) ln q) (q y) β Lemma A.5 In te Gamma-exponental moel, tere exsts a unque monotonc fferentable equlbrum. Proof. Proceeng by contracton, suppose tat for some x, y tere exst two optmal stoppng tmes as frst qutter. We can reprase ts as follows: for some τ, tere exst sgnals (x, y) 6 (x 0,y 0 ) suc tat q A E τ,τ [ x, y+],q B E τ,τ [ y, x+],q A E τ,τ [ x 0,y 0 +], q B E τ,τ [ y 0,x 0 +]. Note tat we can take (x, y) < 0, an (x 0,y 0 ) < 0 by contnuty of V,t (τ x) /τ tτ an V,t (τ y) /τ tτ. Smple manpulatons of Eq. (6) ten yel: q A (β +)(α +τ x y)) (β+) (α +τ x y)) (β+) q A (β +)(α +τ x) (β+) (α +τ x) (β+) Conser te functons q B (β +)(α +τ x y)) (β+) (α +τ x y)) (β+) q B (β +)(α +τ y) (β+) (α +τ y) (β+). ϕ (ξ) q (β +)(α +τ ξ) (β+) (α +τ ξ) (β+) for A, B. Note tat ϕ 0 (ξ) q (β +)+ξ α τ, s contnuous n ξ, as at most one zero, ξ 0 α +τ q (β +), an as te same sgn as ξ ξ 0. So, suppose tat x 0 <x.te contons ϕ A (x) ϕ A (x + y) an ϕ A (x 0 )ϕ A (x 0 + y 0 ) requre x 0 + y 0 >x+ y, an ence y 0 >y.but ten, te conton ϕ B (x) ϕ B (x + y) an ϕ B (x 0 )ϕ B (x 0 + y 0 ) requre tat x 0 + y 0 <x+ y, an we ave a contracton. A smlar contracton s obtane wen supposng tat x 0 >x. Te Bnary Pror Moel Suppose tat te prze arrves accorng to an exponental process of ntensty,.e. wt F (t ) e t. Te pror s concentrate on two ponts: p 0 Pr( ) Pr ( 0 ) were > 0 > 0. Contonal on, te prvate sgnal s strbute as Pr (z z ) e z for z 0. Terefore te posteror belef of s π t,t 0 (x, y) p 0 e t e t 0 e x e y ( p 0 ) e 0t e 0t 0 e 0x 0 e 0y + p 0 e t e t 0 e x e y [+q (p 0, 0,,t,t 0,x,y)] 5

6 were q (p 0, 0,,t,t 0,x,y) p 0 e ( 0 )(t+t 0 x y) p 0 0 So E t,t [ρ 0 (τ ) x, y] 0q (p 0, 0,,t,t 0,x,y)+ q (p 0, 0,,t,t 0,x,y)+. Ts s fferentable n x, y, t, t 0. Snce q s ecreasng n x, y an ncreasng n t, t 0 an te expecte azar rate s ecreasng n q, ten te expecte azar rate s ncreasng n x, y an ecreasng n t, t 0. Next E t,t [ρ 0 (τ ) x, y ] 0q (p 0, 0,,t,t 0,x,y)+ q (p 0, 0,,t,t 0,x,y)+ were q (p 0, 0,,t,t 0,x,y) p 0 e ( 0 )(t+t 0 x y) 0 p 0 Ts as te same propertes as E t,t [ρ 0 (τ ) x, y] w.r. to x, y, t, t 0. Fnally E t,t [ρ 0 (τ ) x, y+] 0q + (p 0, 0,,t,t 0,x,y)+ q + (p 0, 0,,t,t 0,x,y)+. were q + (p 0, 0,,t,t 0,x,y) p 0 e ( 0 )(t+t 0 x) e0y p 0 e y 0 Ts as te same propertes as E t,t 0 [ρ (τ ) x, y] w.r. to x, t, t 0.Because q 0 q + 0 q + (q +) < 0, e ( 0 )(t+t 0 x y) ecreases n x an y an ncreases n t, because e 0 y e y e y 0 0 e y 0 ( e y ) y e y ( e y ) an y e y e ey ( e y +y) ( e y ) e y e y 0e y0 e y 0, > 0, tfollowstatassumptonsanaresatsfe. Because x E t,t [ρ (t ) x, y+] t E t,t [ρ (t ) x, y+], settng G 0 /, te nequaltes n Assumpton 3 are satsfe agan f: 0 < y E t,t [ρ (t ) x, y+] < x E t,t [ρ (t ) x, y+] 6

7 Te frst nequalty s alreay verfe. Te secon nequalty follows because: e ( 0 )(t x) e 0y e ( 0 )(t x) e 0y x e y y e y 0e y 0 e y 0 e y 0 e y + e y 0 e y ( e y ) e ( 0 )(t x) ( 0 ) e 0y e y e( 0 )(t x) e y 0 0 ( e y ) ( e y ) < 0 Fnally, te followng 0 < c <p 0 +( p 0 ) 0 b s suffcent but not even necessary to satsfy Assumpton 4. Specalzng Propostons an 3, player, observng sgnal x anuponobservngtat te opponentasleftattmeτ, wll leave te race at tme: an were log σ, (x, τ) max τ,x+ g j (τ) τ + K + 0 K σ,(x, 0) max 0,x+ x j + K ª 0 log Ã! c b p 0 c b 0 p 0 0 s a measure of te pror bas n favor of. From Proposton 5, we obtan te stoppng tme of a player wt sgnal x, contonal on te opponent j stll beng n te game s: ( "!#) Ã e σ 0,j (σ, (x)) σ,(x) max 0, x + K +( 0 ) log e 0σ,j (σ, (x) ) In te symmetrc payoff moel gj σ, (x) σ (σ (x)) x, so we obtan a close-form soluton for te frst quttng tme: ½ σ (x) max 0, x + K +( 0 ) log e x e 0x ¾. Fnally, from Proposton 7, we obtan tat n any equlbrum of te publc nformaton game, for every par of sgnals x, y, tefrst frm quts te race at tme T (x, y) (x + y + Q) an were Q T (x, y) x + y + Q T, Ã ( 0 ) log c /b p 0 c /b 0 p 0 0!. 7