Supplementary Appendix (not for publication) for: The Value of Network Information

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1 Suppementary Appendix not for pubication for: The Vaue of Network Information Itay P. Fainmesser and Andrea Gaeotti September 6, 03 This appendix incudes the proof of Proposition from the paper "The Vaue of Network Information." For competeness, we repeat the statement of the Proposition. Proposition. Suppose that γk max < /. Then, for any of the price discrimination schemes considered, there exists a unique finite price schedue p that soves the monopoy s profit maximization probem. We now prove the Proposition. Notation and definitions Reca that γ > 0 is the network externaity coefficient, that out-degrees are indexed by k and in-degrees by. Reca further that P k and H are the fractions of consumers with out-degree k and in-degree respectivey, and that = Ek ] and = E k] capture the conditiona expected out-degree of a consumer with in-degree and conditiona expected in-degree of a consumer with out-degree k. Let K out = k max K in = k max K in/out = k max + k max where k max is maxima degree among a in- and out-degrees of a individuas. Let ϕ out : {0,..., k max } {0,..., K out } ϕ in : {0,..., k max } {0,..., K in } ϕ in/out : {0,..., k max } {0,..., K in/out } Fainmesser: Department of Economics, Brown University, Providence, RI 09 emai: Itay_Fainmesser@Brown.edu; Gaeotti: Department of Economics, University of Essex, Wivenhoe Park, Cochester CO 3S, United Kingdom emai: agaeo@essex.ac.uk.

2 be the bijections defined by ϕ out i = i ϕ in i = i ϕ in/out x, y = k max + x + y and et qk, HP k. The monopoist s profit Π is given by Π out = k P kpkxk Π in = Π in/out = H k H k P k pxk, P k pk, xk, and the negative of Hessian matrices are H out = Π out ps pt s,t {0,...,k max } H in = Π in ps pt s,t {0,...,k max } H in/out = Π in/out p ϕ in/out s p ϕ in/out t s,t {0,...,K in/out } The corresponding first order condition of the monopoy s profit maximization probem are: where 0 = Π out ps = P s ps + γ p out s P kpkk ] H s γ k k ] 0 = Π in ps = Hs ps + γ p in s s Hp γ k 0 = Π in/out = HyP x y px, y + γ p ] in/out x H pk, kγ p in/out P k px, y γ k γ k px, y k p out = H k P k pk p in = Hp p in/out = H P k pk, k

3 and p out ps = P s H s p in ps = Hss p in/out HyP x yy = px, y Intermediate resuts The proof of the Proposition makes use of the foowing three emmas. Lemma. Let K, Π, ϕ {K i, Π i, ϕ i }, where i {out, in, in/out}. p j, b j, c j j {0,...,K} R 3 + K+ such that for every s, t {0,..., K}, Then there exists Γ R + and Π p ϕ s p ϕ t = p sp t Γ b t c s + b s c t ] + p s {s=t}. Lemma. Let G = p s p t Γ b t c s + b s c t ] + p s {s=t} s,t {0,...,K}. Then the determinant of G is given by detg = K i p i + Γ i b i c i p i Γ p i p j b j c i b i c j ]. i<j Lemma 3. Assume that γk max < /. Then H out, H in, and H in/out are positive definite. Proofs Proof of Proposition. It suffices to show that the first and second order conditions are satisfied. The first order conditions are verified in the Appendix to the paper. To see that the second order conditions hod, note that for each price discrimination scheme, the second order partia derivatives are of the form given in Lemma. A necessary and sufficient condition for maximizing profit is that the Hessian matrix is negative definite, or equivaenty that the negative of the Hessian matrix is positive definite. The determinant of this matrix is given by the formua in Lemma. By Lemma 3, we concude positive definiteness and hence that the second order condition hods. We now compete the proof by proving the three emmas above. 3

4 Lemma Proof. We compute the second partia derivatives. Reca that Cs = for any t, s, x, y, z, w {0,..., k max }, H s = ˆs and et Ct = ˆt. Then, Π out pt ps = γ p P s ps + pt γ k s γ P kpkk ] H s γ k k = P s {s=t} γs p γ k pt γ P kpkk H s γ k pt k = P s {s=t} γs γ k P t H t γ γ k P tt H s γ = P s {s=t} P sp t s H t + t H s γ k γ = P s {s=t} P sp t sct + tcs γ k ] Π in pt ps = Hs ps + γ ps s Hp pt γ k ] = Hs {s=t} + γ Hp γ k pt Π in/out pz, w px, y = = Hs = Hs {s=t} + {s=t} + γ γ k γ γ k γ = Hs {s=t} HsHt γ k HyP x y pz, w = qx, y {z,w=x,y} s p pt s s Htt s ] Htt ] s t Htt + sht t Cs + s Ct γ p px, y + γ k x γ γ k x p pz, w ] ] ] pz, w H k = qx, y {z,w=x,y} γ ] ww zγ xqz, qz, w γ k γ k γ = qx, y {z,w=x,y} qx, yqz, w γ k xw + γ γ k γ = qx, y {z,w=x,y} qx, yqz, w xw + zy γ k,k pk, kγ p P k γ k px, y pk, kγ qk, γ k qx, yy zy p px, y To concude: for the case of discrimination on out-degree, we take p i = P i, b i = i, c i = Ci, and Γ = for discrimination on in-degree, we take p i = P i, b i = i, c i = Ci, and Γ = in- and out-degrees, we take p i = qi = qi, i, b i = i, c i = i, and Γ = γ γ k γ k. γ k ; ; and for discrimination on

5 Lemma Proof. Fix k max = K > 0. For {,..., K}, and for any J {0,..., K} with J =, et L = {0,..., } and M J = G st s L t J. Denote S = J L. We wi write J = {j 0, j,..., j } with j 0 < j < < j. Aso denote Q = J \ L = {m i } 0 i< Q and R = L \ J = {n i } 0 i< R. Note that J = S Q and L = S R. Aso, Q = R since J = = L. Moreover, note that any eement of the set Q exceeds any eement of the set S. We caim that M J = κ, Jψ, J where κ, J = S i J L if R = 0 p i n 0+ S if R = n 0+n if R = and + Γ i b ic i p i + Γ p ip j b i b j c i c j b i c j if Q = 0 Γ b n0 c m0 + b m0 c n0 + ψ, J = Γ r S bm0 c n0 + b n0 c m0 b r c r p r b m0 b n0 p r c r + c m0 c n0 p r b r if Q =. Γ b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0 if Q = 0 if Q 3 Denote Φ = i=0 p i. We wi prove the caim by induction on. First we wi show that the formua is vaid for =. Next we wi show that, if the formua is vaid for < K, then it is aso vaid for +. Base case = : det G st = p 0 p t Γ b t c 0 + b 0 c t ] + p 0 {0=t} p 0 = Γ b 0c 0 ] + p 0 if t = 0, i.e. Q = 0 p 0 p t Γ b t c 0 + b 0 c t ] if t 0, i.e. Q = p 0 + Γ b 0 c 0 p 0 ] if t = 0, i.e. Q = 0 =. p 0 p t Γ b t c 0 + b 0 c t ] if t 0, i.e. Q = Now assume hods for some < K. Let L = L {} = {0,..., }, and et J = J {j} for some j J. Accordingy, et S = J L, Q = J \ L, and R = L \ J. We need to show that hods for +, i.e., that M J + For two sets X and Y, et X Y be the disjoint union of X and Y. = κ +, J ψ +, J. 5

6 By cofactor expansion aong the ast row, we have M J + = i+ M J \{j i } G ji, 0 i so it suffices to show that i+ M J \{j i } G ji = κ +, J ψ +, J. 0 i We consider the foowing cases. Case 0. R = 0 In this case Q = R = and J = S = {0,..., }, so L \ J \ {} = L \ {} \ J \ {} = L \ J = 0 Simiary, for r S \ {}, L \ J \ {r} = J L {r} C C J = L C {r} = L \ J L {r} = {r} = and in particuar R = {r}. Thus r+ p Φ Φ M J \{r} M J \{} We wi show that M J S p Φ S M J p Φ + + G r = Γ b r c + b c r + Γ j,r p p r Γ b c r + b r c ] G = + Γ b i c i p i + Γ p i p j bi b j c i c j b i c j i< = +Γ i {0,...,} b ic i p i + Γ p ip j b i b j c i c j b i c j = r+ r+ Γ b r c + b c r + Γ r< b c r + b r c b j c j p j b b r p j c j + c c r p j b j j,r p Γ b c ] + p.. Since S = +, b c r + b r c b j c j p j b b r p j c j + c c r p j b j 6

7 p p r Γ b c r + b r c ] Γ i< = + AΓ + BΓ + CΓ 3. b i c i p i + Γ i,j< It suffices to show A = i {0,...,} b ic i p i, B = A = p b c + i< b ic i p i. Next note that B = p p r b c r + b r c ] b r c + b c r r< + p b c ] b i c i p i i< + p i p j bi b j c i c j b i c j i,j< = p p i b c i + b i c ] b i c + b c i + b i c i p i b c p = = = + + i< i,j< p i p j bi b j c i c j b i c j p p i b c i + b i c i b c + b i c ] + bi c i p i b c p i< i,j< i< i< p i p j bi b j c i c j b i c j p p i b c i + b i c i b c b i c ] + p i p bi b c i c b i c + = p i p j bi b j c i c j b i c j. To see that C = 0, note that C = p r< j,r p i p j bi b j c i c j b i c j p Γ b c ] + p ip j b i b j c i c j b i c j, and C = 0. Ceary we have i,j< p p j b b j c c j b c j j< p i p j bi b j c i c j b i c j + i,j< b c r + b r c b j c j p j b b r p j c j + c c r p j b j pr b c r + b r c ] + p i p j bi b j c i c j b i c j b c ] i,j< = i,j< p i p j bi b j c i c j b i c j b c i + b i c b j c j p j b b i p j c j + c c i p j b j pi b c i + b i c ] + p i p j bi b j c i c j b i c j b c ] 7

8 = p i p j b c i + b i c b j c j b b i c j + c c i b j b c i + b i c ] + ] b i b j c i c j b i c j b c ] i,j< Note that for every i, j, the coefficient of p i p j is zero: b c i + b i c b j c j b b i c j + c c i b j b c i + b i c ] ] + b i b j c i c j b i c j b c ] b c j + b j c b i c i b b j c i + c c j b i b c j + b j c ] ] + b i b j c i c j b jc i b c ] = b c i b j c j b i c jc i + c j b i c i b j c ] i c j c b i b j c j c i b jb i + b j b i c i c j b ] i b j = 0. b c + b c b i b j c i c j ] ci b i b j c j b i c j c i b j b i c j + cj b j b i c i c i b j b i c j b j c i ] Case. R = Case a. S In this case, R = L \ J =. Note that M J + = i M J \{j i } G ji j i J = i M J \{j i } G ji j i S Q = i M J \{} G + First, note anaogousy to Case 0 that j i S\{} i M J \{j i } G ji + i M J \{j i } G ji. j i Q L \ J \ {} = L \ J Next consider r S \{}. Note that r J and hence r J C. Since R, we have L \J = R = R \{} = L \ J, so =. L \ J \ {r} = J L {r} C C J = L C {r} = L \ J L {r} = L \ J {r} 8

9 = L \ J {r} = + =. Now consider j Q. Note that j {0,..., }. As before R = R \ {}, so Thus n0+r p m0 p Φ M J \{r} L \ J \ {j} = L J {j} C C J = L C {j} = L {} C J C L {j} = L {} C J C G r = L \ J = L \ {} \ J = L \ J \ {} = L \ J =. = Γ b n0 b m0 c r c b n0 b c r c m0 b r b m0 c n0 c + b r b c n0 c m0 p p r Γ b c r + b r c ] n0+ M J \{} G p m0 Φ = Γ b n0 c m0 + b m0 c n0 + Γ bm0 c n0 + b n0 c m0 b r c r p r b m0 b n0 p r c r + c m0 c n0 p r b r p Γ b c ] + p r S \{} n0+ M J \{m 0 } G m0 p Φ = Γ b n0 c + b c n0 + Γ b c n0 + b n0 c b r c r p r b b n0 p r c r + c c n0 p r b r p p m0 Γ b c m0 + b m0 c ] We wi show that S M J p m0 p Φ + r S \{} = Γ b n0 c m0 + b m0 c n0 + Γ r S bm0 c n0 + b n0 c m0 b r c r p r b m0 b n0 p r c r + c m0 c n0 p r b r. First note that S M J p m0 p Φ + 9

10 = S \{} i+ n 0+i M J \{j i } G j p m0 p Φ i j i S \{} + + n 0+ M J \{} G + + n 0+ = n 0+ S \{} p m0 p Φ j i S \{} = n 0+ AΓ + BΓ + CΓ 3. M J \{j i } G j i + It suffices to show that A = b n0 c m0 + b m0 c n0, B = and C = 0. First, it is easy to see that and hence A = b n0 c m0 + b m0 c n0. Next, note that M J \{} M J \{m 0 } G m 0 G M J \{m 0 } G m 0 r S bm0 c n0 + b n0 c m0 b r c r p r b m0 b n0 p r c r + c m0 c n0 p r b r, S p m0 p Φ A = S p m0 Φ p b n0 c m0 + b m0 c n0, S p m0 p Φ B = + S p m0 Φ p bm0 c n0 + b n0 c m0 b r c r p r b m0 b n0 p r c r + c m0 c n0 p r b r r S \{} + S p m0 Φ p b c ] b n0 c m0 + b m0 c n0 S p Φ p p m0 b c m0 + b m0 c ] b n0 c + b c n0 B = + bm0 c n0 + b n0 c m0 b r c r p r b m0 b n0 p r c r + c m0 c n0 p r b r r S \{} + p b c ] b n0 c m0 + b m0 c n0 p b c m0 + b m0 c ] b n0 c + b c n0 = + bm0 c n0 + b n0 c m0 b r c r p r b m0 b n0 p r c r + c m0 c n0 p r b r r S \{} + p b n0 c m0 b c + b m0 c n0 b c p b c m0 b n0 c + b c m0 b c n0 + b m0 c b n0 c + b m0 c b c n0 ] = + bm0 c n0 + b n0 c m0 b r c r p r b m0 b n0 p r c r + c m0 c n0 p r b r + = r S \{} bm0 c n0 + b n0 c m0 b c p b m0 b n0 p c + c m 0 c n0 p b bm0 c n0 + b n0 c m0 b r c r p r b m0 b n0 p r c r + c m0 c n0 p r b r r S To see that C = 0, note that C = r S \{} b n 0 b m0 c r c b n0 b c r c m0 b r b m0 c n0 c + b r b c n0 c m0 p p r b c r + b r c ] 0

11 S \{} p b c bm0 c n0 + b n0 c m0 b r c r p r b m0 b n0 p r c r + c m0 c n0 p r b r r S \{} + S \{} p b c m0 + b m0 c ] b c n0 + b n0 c b r c r p r b b n0 p r c r + c c n0 p r b r r S \{} Note that each term in the sum is zero: for each r, b n 0 b m0 c r c b n0 b c r c m0 b r b m0 c n0 c + b r b c n0 c m0 b c r + b r c ] + b c bm0 c n0 + b n0 c m0 b r c r b m0 b n0 c r + c m0 c n0 b r b c m0 + b m0 c ] b c n0 + b n0 c b r c r b b n0 c r + c c n0 b r b m0 c n0 c b c n0 c m0 c b c c m0 c n0 + b c m0 + b m0 c c c n0 = b r + c r b n 0 b m0 c b n0 b c m0 b b c b m0 b n0 + b c m0 + b m0 c b b n0 + b r c r c b n0 b m0 c b n0 b c m0 c b m0 c n0 c b + b c n0 c m0 b + b r c r b m0 c n0 b c + b n0 c m0 b c b r c r b c m0 b c n0 + b c m0 b n0 c + b m0 c b c n0 + b m0 c b n0 c ] = 0. Case b. R First consider j S. Since R = L \ J, we have J. In this case, = R = Q = J \ L. Since j S, we have j L, so J \ {j} \ L = J \ {j} L \ {} = J \ L \ {} = J \ {} \ L \ {} = J \ L Note that R = {}, and hence S = {0,..., }. This impies that for j S, so j+ p m0 Φ M J \{j} = = Γ b j c m0 + b m0 c j + Γ r S \{j} bm0 c j + b j c m0 b r c r p r b m0 b j p r c r + c m0 c j p r b r. Now et j Q. Reca that in this case, L \ J =. Using the facts that j L and J, we have L \ J \ {j} L J = C {j} = L J C L {j}

12 Then for j Q, we have Thus Φ M J \{j} S M J p m0 p Φ = S p m0 p Φ = Γ b ji c m0 + b m0 c ji + Γ j i S p ji Γ b c ji + b ji c ] + + Γ i = L J C = L \ J = L \ {} \ J = L \ J \ {} = R = = 0. = + Γ b i c i p i + Γ p i p j bi b j c i c j b i c j. i L i,j L i+ i+ M J \{j i } G j i + + M J \{m 0 } j i S b i c i p i + Γ = + AΓ + BΓ + CΓ 3 r S \{j i } p i p j bi b j c i c j b i c j It suffices to show that A = b c m0 + b m0 c, B = and C = 0. First, it is easy to see that A = b c m0 + b m0 c. Next, note that bm0 c ji + b ji c m0 b r c r p r b m0 b ji p r c r + c m0 c ji p r b r Γ b c m0 + b m0 c ] r S bm0 c + b c m0 b r c r p r b m0 b p r c r + c m0 c p r b r, B = b ji c m0 + b m0 c ji p ji b c ji + b ji c ] + b i c i p i b c m0 + b m0 c ] j i S i = b ic m0 + b m0 c i p i b c i + b i c ] + b i c i p i b c m0 + b m0 c ] i< = p i b ic m0 b c i + b i c m0 b i c + b m0 c i b c i + b m0 c i b i c + b i c i b c m0 + b i c i b m0 c i< = p i b i c m0 b c i + b m0 c i b i c b i c m0 b i c b m0 c i b c i i< = p i cm0 b b i c i + b m0 c b i c i c m0 c b i b m0 b c i i<

13 = i< bm0 c + b c m0 b i c i p i b m0 b p i c i + c m0 c p i b i, from which the caim foows since S = {0,..., }. To see that C = 0, note that C = j i S r S \{j i } + bm0 c ji + b ji c m0 b r c r p r b m0 b ji p r c r + c m0 c ji p r b r pji b c ji + b ji c ] p i p j bi b j c i c j b i c j b c m0 + b m0 c ] = bm0 c i + b i c m0 b j c j p j b m0 b i p j c j + c m0 c i p j b j pi b c i + b i c ] + p i p j bi b j c i c j b i c j b c m0 + b m0 c ] ] = p i p j bm0 c i + b i c m0 b j c j b m0 b i p j c j + c m0 c i p j b j b c i + b i c ] + b i b j c i c j b i c j b c m0 + b m0 c ] ] We wi show that, for each i, j with i j, the coefficient of p i p j is zero: b m0 c i + b i c m0 b j c j b m0 b i c j + c m0 c i b j b c i + b i c ] + b i b j c i c j b i c j b c m0 + b m0 c ] b m0 c j + b j c m0 b i c i b m0 b j c i + c m0 c j b i b c j + b j c ] + b i b j c i c j b jc i b c m0 + b m0 c ] = b ci bm0 c i + b i c m0 b j c j b m0 b i c j + c m0 c i b j + cm0 bi b j c i c j b i c ] j + c bi bm0 c i + b i c m0 b j c j b m0 b i c j + c m0 c i b j + bm0 bi b j c i c j b i c ] j + b cj bm0 c j + b j c m0 b i c i b m0 b j c i + c m0 c j b i + cm0 bi b j c i c j b jc ] i + c bj bm0 c j + b j c m0 b i c i b m0 b j c i + c m0 c j b i + bm0 bi b j c i c j b jc ] i = 0. Hence C = 0. Case. R = Case a. S Reca that M J + = i M J \{j i } G ji j i J = i M J \{j i } G ji j i S Q = i M J \{} G + j i S \{} i M J \{j i } G ji + i M J \{j i } G ji. j i Q 3

14 Consider r S \ {}. Note as in Case a that This impies M J \{r} = 0 for r S \ {}, so M J + L \ J \ {r} = L \ J {r} = + = 3. = i M J \{} G + i M J \{j i } G ji j i Q Consider j Q {}. Note aso as in Case a that L \ J \ {j} = L \ J =. Denote Q = {m 0, m } and R \ {} = R = {n 0, n }. By, M J \{m } n 0+n S Φ r J \S p r Γ G m = p p m0 b n0 b c n c m0 b n0 b m0 c n c b n b c n0 c m0 + b n b m0 c n0 c p p m Γ b c m + b m c ] M J \{m 0 } n 0+n S Φ r J \S p r Γ G m0 = p p m b n0 b c n c m b n0 b m c n c b n b c n0 c m + b n b m c n0 c p p m0 Γ b c m0 + b m0 c ] M J \{} n 0+n S Φ r J \S p r Γ G = p m0 p m b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0 p Γ b c ] + p The sum of these three terms reduces to n 0+n S Φ r J \S p r Γ j i Q {} i+ M J \{j i } G ji = + p p m0 b n0 b c n c m0 b n0 b m0 c n c b n b c n0 c m0 + b n b m0 c n0 c p p m Γ b c m + b m c ] + + p p m b n0 b c n c m b n0 b m c n c b n b c n0 c m + b n b m c n0 c p p m0 Γ b c m0 + b m0 c ] + + p m0 p m b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0 p Γ b c ] + p = p p m0 b n0 b c n c m0 b n0 b m0 c n c b n b c n0 c m0 + b n b m0 c n0 c p p m Γ b c m + b m c ] p p m b n0 b c n c m b n0 b m c n c b n b c n0 c m + b n b m c n0 c p p m0 Γ b c m0 + b m0 c ] + p m0 p m b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0 p Γ b c ] + p = b c m b n0 b c n c m0 b n0 b m0 c n c b n b c n0 c m0 + b n b m0 c n0 c p p m 0 p m Γ

15 b c m0 b n0 b c n c m b n0 b m c n c b n b c n0 c m + b n b m c n0 c p p m 0 p m Γ + b c b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0 p p m 0 p m Γ + b m c b n0 b c n c m0 b n0 b m0 c n c b n b c n0 c m0 + b n b m0 c n0 c p p m 0 p m Γ b m0 c b n0 b c n c m b n0 b m c n c b n b c n0 c m + b n b m c n0 c p p m 0 p m Γ + b c b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0 p p m 0 p m Γ + b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0 p = p b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0. Hence M J = j i Q {} = S Φ p r Γ r J \S = S p Φ i M J \{j i } G ji j i Q {} i M J \{j i } G ji r J \S p r Γ b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0. Case b. R First consider j S. Note as in Case b that Now et j Q. As in case b, we have Thus r J \S p r J \ {j} \ L = J \ L =. L \ J \ {j} = L \ J \ {} = R = =. M J = i M J \{j i } G ji j i J = S Φ i+ n0+i Γ b n0 b m0 c ji c m j i J \Q b n0 b m c ji c m0 b ji b m0 c n0 c m + b ji b m c n0 c m0 G j i + S Φ + n 0+ Γ b n0 c m + b m c n0 r J \S \m 0 p r 5

16 + Γ r S bm c n0 + b n0 c m b r c r p r b m b n0 p r c r + c m c n0 p r b r ] G m 0 + S Φ r J \S \m p r + n 0+ Γ b n0 c m0 + b m0 c n0 + Γ r S bm0 c n0 + b n0 c m0 b r c r p r b m0 b n0 p r c r + c m0 c n0 p r b r ] G m = +n 0 S Φ r J \S p r b n0 b m0 c ji c m b n0 b m c ji c m0 j i J \Q Γ b ji b m0 c n0 c m + b ji b m c n0 c m0 p p ji Γ b c ji + b ji c ] + +n 0 S Φ Γ b n0 c m + b m c n0 + Γ b m c n0 + b n0 c m b r c r p r r S r J \S \m 0 p r b m b n0 p r c r + c m c n0 p r b r ] p p m0 Γ b c m0 + b m0 c ] +n 0 S Φ r J \S \m p r Γ b n0 c m0 + b m0 c n0 + Γ r S b m0 c n0 + b n0 c m0 b r c r p r b m0 b n0 p r c r + c m0 c n0 p r b r ] p p m Γ b c m + b m c ] = AΓ 3 + +n 0 S p Φ p r BΓ r J \S First we wi show that B = b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0 : B = b n0 c m + b m c n0 b c m0 + b m0 c ] b n0 c m0 + b m0 c n0 b c m + b m c ] = b n0 c m b c m0 + b n0 c m b m0 c + b m c n0 b c m0 + b m c n0 b m0 c b n0 c m0 b c m + b n0 c m0 b m c + b m0 c n0 b c m + b m0 c n0 b m c = b n0 b m0 c c m b n0 b m c c m0 b b m0 c n0 c m + b b m c n0 c m0 Now we wi show that A = 0: S p Φ r Q p A = b n0 b m0 c r c m b n0 b m c r c m0 b r b m0 c n0 c m + b r b m c n0 c m0 b c r + b r c ] r r S bm c n0 + b n0 c m b r c r b m b n0 c r + c m c n0 b r b c m0 + b m0 c ] r S + r S bm0 c n0 + b n0 c m0 b r c r b m0 b n0 c r + c m0 c n0 b r b c m + b m c ] It suffices to show that each term is zero. Indeed for every r we have b n0 b m0 c r c m b n0 b m c r c m0 b r b m0 c n0 c m + b r b m c n0 c m0 b c r + b r c ] 6

17 b m c n0 + b n0 c m b r c r b m b n0 c r + c m c n0 b r b c m0 + b m0 c ] + b m0 c n0 + b n0 c m0 b r c r b m0 b n0 c r + c m0 c n0 b r b c m + b m c ] = b n0 b m0 c r c m b n0 b m c r c m0 b r b m0 c n0 c m + b r b m c n0 c m0 b c r ] + b n0 b m0 c r c m b n0 b m c r c m0 b r b m0 c n0 c m + b r b m c n0 c m0 b r c ] b m c n0 + b n0 c m b r c r b c m0 + b m0 c ] + b m b n0 c r + c m c n0 b r b c m0 + b m0 c ] + b m0 c n0 + b n0 c m0 b r c r b c m + b m c ] b m0 b n0 c r + c m0 c n0 b r b c m + b m c ] ] ] = c r b n0 b m0 c m b b n0 b m c m0 b + b m b n0 b c m0 + b m0 c b m0 b n0 b c m + b m c ] ] + b r b m0 c n0 c m c + b m c n0 c m0 c + c m c n0 b c m0 + b m0 c c m0 c n0 b c m + b m c + b r c r b m0 c n0 c m b + b m c n0 c m0 b + b n0 b m0 c m c b n0 b m c m0 c = 0. ] ] b m c n0 + b n0 c m b c m0 + b m0 c + b m0 c n0 + b n0 c m0 b c m + b m c Case 3. R = 3 Case 3a. S For any j J, we have L \ J \ {j} = L \ J {j} C = L J {j} C C J C {j} = L Since S J, we have L \ J = L \ J and hence By, this impies M J \{j} = L J C L {j} L \ J \ {j} L J C = L \ J = L \ J = 3. = 0 for every j J and hence M J + = 0. 7

18 Case 3b. R First consider j S. Note that R impies J. Since j S L, we have This impies M J \{j} = 0 for j S. Hence Now et j Q. As in case b, we have J \ {j} \ L = J \ {j} L M J + = J \ L = J \ {} \ L = J \ {} L = J \ L = 3. = M j i Q i J \{j i } G ji. L \ J \ {j} = L \ J \ {} = R = 3 =. Denote R \ {} = {n 0, n } and Q = {m 0, m, m } with n 0 < n and m 0 < m < m. By, we have M J \{m 0 } S Φ r S p r Γ G m0 = p m p m b n0 b m c n c m b n0 b m c n c m b n b m c n0 b m + b n b m c n0 c m p p m0 Γ b c m0 + b m0 c ] M J \{m } S Φ r S p r Γ G m = p m0 p m b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 b m + b n b m c n0 c m0 p p m Γ b c m + b m c ] M J \{m } S Φ r S p r Γ G m = p m0 p m b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 b m + b n b m c n0 c m0 p p m Γ b c m + b m c ] Thus p m0 p m p m S Φ r S p r Γ i+ M J \{j i } G ji j i Q = b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0 b c m + b m c b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0 b c m + b m c + b n0 b m c n c m b n0 b m c n c m b n b m c n0 c m + b n b m c n0 c m b c m0 + b m0 c = b c m b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0 b c m b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0 8

19 so M J + + b c m0 b n0 b m c n c m b n0 b m c n c m b n b m c n0 c m + b n b m c n0 c m = 0, = 0. + b m c b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0 b m c b n0 b m0 c n c m b n0 b m c n c m0 b n b m0 c n0 c m + b n b m c n0 c m0 + b m0 c b n0 b m c n c m b n0 b m c n c m b n b m c n0 c m + b n b m c n0 c m Case. R > 3 In this case, R = L \ J = L J C. Since we have By, this impies M J \{j} L \ J \ {j} = L \ J {j} C = L J {j} C C = L J C {j} = L J C L {j}, L \ J \ {j} L J C L J C = 3. = 0 for every j J, and hence M J + = 0. Lemma 3 Proof. First we wi consider the case of discrimination on in-degree. The case of discrimination on out-degree is identica. Using the determinant formua given in Lemma, it suffices to show that Γ i<j p i p j b j c i b i c j ] < + Γ i b ip i c i. Reca from the proof of Lemma that p i = P i, b i = i, c i = Ci, and Γ = γ k. Since γkmax < by assumption, we have γ k <. This impies γ k > and hence γ k <, so Γ = γ γ k < k max = k max. Reca that for any random variabe X, we have EX ] = x px x max xpx = x max EX. Thus i<j p i p j b j c i b i c j ] = i<j pi p j i c j ijc i c j + j c i pi p j i c j ijc i c j = i i j i pi p j i c j ijc i c j j 9

20 pi p j i c j + p i p j ijc i c j i j i j i p i p j c j + p i ic i i j i = E k ] E + k max p i c i i = E k ] ] E + k max Ec] = E ] k ] E + k max = E ] k ] E ] + k E max k max. Since 0 < Γ < k max, we have Γ pi p j b j c i b i c j ] < pi k max p j b j c i b i c j ] i<j < < + Γ i i<j b i p i c i as caimed. Now we wi consider the case of discrimination on in- and out-degrees. Again by Lemma it suffices to show that Γ i j p i p j b j c i b i c j ] < + Γ i b ip i c i. As before, we have <. This impies γ k Γ = γ γ k < k max = k max. Reca that = i,i i qi, i, ˆ = i,i i qi, i, and ˆ = due to consistency. Note that i j i j i j + i j < i j k max + i j k max, Using the parameters p i = qi = qi, i, b i = i, c i = i, and Γ = have i j qiqj i j i j ] < qiqj k max i j + j i i j < k max i,j γ obtained in the proof of Lemma, we γ k i j qiqj 0

21 = k max i qi j qj = k max ˆ = k max. i,i j,j Since 0 < Γ <, we have kmax Γ i j qiqj i j i j ] < k max qiqj i j i j ] i j < < + Γ i i i qi. This shows that det H > 0 where the size of the matrix K is arbitrary. Note that the K K upper-eft sub-matrix of H is defined exacty by the same formua as H. Then, since the resut hods for the sub-matrix of any size K, we concude that H is positive definite.