Theory and Measurement - Derivations

A Derivations A.1 Global Productivity Distribution By Theorem 1 in Lind and Ramondo (2022) the joint distribution of productivity across countries at each moment in time is max-stable multivariate Fréchet. Since the proof is constructive, we can calculate the distribution directly as follows. First, we can express the joint distribution of productivity as F (z1 , . . . , zN ; t) ≡ P [Z1t (v) ≤ z1 , . . . , ZN t (v) ≤ zN ] = P [Qi (v)Aio (v) ≤ zo ∀o ∈ Ni (t, v) for all i s.t. t∗i (v) ≤ t]   zo ∗ = P Qi (v) ≤ min for all i s.t. ti (v) ≤ t o∈Ni (t,v) Aio (v)   zo ∗ = P Qi (v) > min for no i s.t. ti (v) ≤ t , o∈Ni (t,v) Aio (v) which yields the result in (9). Next, since applicability is continuously distributed and ni (v) ∈ Ni (t, v), there is a unique origin, o∗i (v, t), that attains the minimum in this expression for each idea known at time t. We then have  zo∗i (v) ∗ for no i s.t. ti (v) ≤ t F (z1 , . . . , zN ; t) = P Qi (v) > Aio∗i (v) (v) " #  ∞ X zo∗i (v) , t∗i (v) ≤ t = exp −E 1 Qi (v) > ∗ (v) (v) A io i i=1 "  # ∞ N XX zo = exp −E 1 Qi (v) > , o∗i (v) = o, t∗i (v) ≤ t . A (v) io i=1 o=1  Next, the expression in (13) follows from  ∞ X N X E 1 Qi (v) > i=1 o=1 ∞ X N X  zo ∗ ∗ , o (v) = o, ti (v) ≤ t Aio (v) i   zo zo zo0 0 ∗ =E 1 Qi (v) > , o ∈ Ni (t, v), ≤ ∀o ∈ Ni (t, v), ti (v) ≤ t Aio (v) Aio (v) Aio0 (v) i=1 o=1   ∞ X N X zo zo0 0 =E 1 i ∈ Io (zi , t, v), ≤ ∀o ∈ Ni (t, v) Aio (v) Aio0 (v) i=1 o=1   N X X zo0 0 Aio (v) ∀o ∈ Ni (t, v) , =E 1 Aio0 (v) ≤ zo o=1 i∈Io (zo ,t,v) for Io (zo , t, v) ≡ {i = 1, 2, . . . | Qi (v)Aio (v) > zo , o ∈ Ni (t, v), t∗i (v) ≤ t} as defined in (11). Note that E ∞ X 1{i ∈ Io (zo , t, v), ni (v) = n, si (v) = s, t∗i (v) ≤ x} i=1 =E ∞ X 1 {Qi (v)Aio (v) > zo , o ∈ Ni (t, v), ni (v) = n, si (v) = s, t∗i (v) ≤ min{t, x}} i=1 ∞ X  zo =E 1 Qi (v) > , o ∈ Ni (t, v), ni (v) = n, si (v) = s, t∗i (v) ≤ t, t∗i (v) ≤ x Aio (v) i=1 Z min{t,x} Z ∞ Z ∞ −σ θq −θ−1 dqde−ao pons (t − t∗ )Λ0ns (t∗ )dt∗ = −∞ Z 0 min{t,x} zo /ao ∞ Z  = −∞ = Γ(1 − 0 θ ) σ zo ao −θ −σ de−ao pons (t − t∗ )Λ0ns (t∗ )dt∗ min{t,x} Z pons (t − t∗ )Λ0ns (t∗ )dt∗ zo−θ , −∞ and hence, E ∞ X 1{i ∈ Io (zo , t, v)} i=1 =E ∞ X S X N X 1{i ∈ Io (zo , t, v), ni (v) = n, si (v) = s, t∗i (v) ≤ t} i=1 s=0 n=1 S X N Z t X = Γ(1 − θ ) σ s=0 for To (t) defined in (12). n=1 −∞ pons (t − t∗ )Λ0ns (t∗ )dt∗ zo−θ = To (t)zo−θ ,  Together, we then have P[ni (v) = n, si (v) = s, t∗i (v) ≤ x | i ∈ Io (zo , t, v)] R min{t,x} pons (t − t∗ )Λ0ns (t∗ )dt∗ zo−θ −∞ = PS PN R t ∗ 0 ∗ ∗ −θ s=0 n=1 −∞ pons (t − t )Λns (t )dt zo R min{t,x} pons (t − t∗ )Λ0ns (t∗ )dt∗ −∞ = PS PN R t , ∗ 0 ∗ ∗ s=0 n=1 −∞ pons (t − t )Λns (t )dt which means that the probability that an idea known to o at time t is from surge s of innovator n is Rt p (t − t∗ )Λ0ns (t∗ )dt∗ −∞ ons , PN R t ∗ )Λ0 (t∗ )dt∗ p (t − t ons ns s=0 n=1 −∞ P[ni (v) = n, si (v) = s | i ∈ Io (zo , t, v)] = PS while the density of discovery times among those ideas is ∂ pons (t − t∗ )Λ0ns (t∗ ) ∗ ∗ . P[t (v) ≤ t | i ∈ I (z , t, v), n (v) = n, s (v) = s] = Rt o o i i ∗ )Λ0 (t∗ )dt∗ ∂t∗ i p (t − t ons ns −∞ Below, we use these two results to take expectations with respect to ni (v), si (v), and t∗i (v) conditional on i ∈ Io (zo , t, v). Conditional on ni (v), si (v), and t∗i (v), 1{o ∈ Ni (t, v)} is a Bernoulli random variable with success probability pons (t−t∗ ) independent across o. That is, for (x1 , . . . , xN ) ∈ {0, 1}N , N Y Pnst∗ [1{o ∈ Ni (t, v)} = xo ∀o = 1, . . . , N ] = (1 − pons (t − t∗ ))1−xo pons (t − t∗ )xo , o=1 where Pnst∗ [·] ≡ P[· | ni (v) = n, si (v) = s, t∗i (v) = t∗ ]. Then,  zo0 P Aio0 (v) ≤ Aio (v) ∀o0 ∈ Ni (t, v) | Aio (v) = ao , i ∈ Io (zi , t, v), ni (v) = n, si (v) = s, t∗i (v) = t∗ zo # " 1{o0 ∈Ni (t,v)} Y  zo0 Aio (v) | Aio (v) = ao , i ∈ Io (zi , t, v) = Enst∗ 1 Aio0 (v) ≤ z o 0 o 6=o "  1{o0 ∈Ni (t,v)} # Y zo0 = Enst∗ 1 Aio0 (v) ≤ ao zo o0 6=o Yh −σ i zo0 = 1 − po0 ns (t − t∗ ) + po0 ns (t − t∗ )e−( zo ao ) , o0 6=o  which uses independence of applicability and conditional independence of 1{o ∈ Ni (t, v)} across countries. We now average this result over applicability levels. The distribution of applicability in o among ideas in Io (zi , t, v) conditional on ni (v), si (v), and t∗i (v) is −σ RxR 0 z /a R∞Ro o 0 zo /ao Pnst∗ [Aio (v) ≤ x | i ∈ Io (zo , t)] = Rx = 0 R∞ 0 −σ dxzo−θ aθo e−ao σa−σ−1 o −σ aθo e−ao σa−σ−1 dao zo−θ o Z = 0 x θq −θ−1 dqde−ao pons (t − t∗ )Λ0ns (t∗ ) −σ θq −θ−1 dqde−ao pons (t − t∗ )Λ0ns (t∗ ) −σ e−ao σaoθ−σ−1 dao . Γ(1 − θ/σ) Therefore,   zo0 0 Hons (z1 , . . . , zN ; t − t ) ≡ Pnst∗ Aio0 (v) ≤ Aio (v) ∀o ∈ Ni (t, v) | i ∈ Io (zi , t, v) zo Z ∞Yh −σ −σ i e−ao σaθ−σ−1 z 0 o ∗ ∗ −( zoo ao ) dao , = 1 − po0 ns (t − t ) + po0 ns (t − t )e Γ(1 − θ/σ) 0 o0 6=o ∗ as in (14). Using the previous results for the distribution of ideas across surges, innovators, and discovery times within Io (zo , t, v) yields   zo0 0 Ho (z1 , . . . , zN ; t) ≡ P Aio0 (v) ≤ Aio (v) ∀o ∈ Ni (t, v) | i ∈ Io (zi , t, v) zo     zo0 0 Aio (v) ∀o ∈ Ni (t, v) | i ∈ Io (zi , t, v) | i ∈ Io (zi , t, v) = E Pnst∗ Aio0 (v) ≤ zo   = E Honi (v)si (v) (z1 , . . . , zN ; t − t∗i (v)) | i ∈ Io (zi , t, v) PS PN R t ∗ ∗ 0 ∗ ∗ s=0 n=1 −∞ Hons (z1 , . . . , zN ; t − t )pons (t − t )Λns (t )dt = dt∗ , R PS PN t ∗ 0 ∗ ∗ s=0 n=1 −∞ , pons (t − t )Λns (t )dt which results in (16). Finally, this result and the previous result on the expected number of ideas in Io (zo , t, v) allow us to compute E N X  zo0 0 Aio (v) ∀o ∈ Ni (t, v) 1 Aio0 (v) ≤ zo  X o=1 i∈Io (zo ,t,v)  X  ∞ N X zo0 0 Aio (v) ∀o ∈ Ni (t, v) | i ∈ Io (zo , t, v) E 1{i ∈ Io (zo , t, v)} = P Aio0 (v) ≤ z o o=1 i=1 = N X Ho (z1 , . . . , zN ; t)To (t)zo−θ . o=1 Therefore, the joint distribution of productivity is " F (z1 , . . . , zN ; t) ≡ P[Z1 (t, v) ≤ z1 , . . . , Zn (t, v) ≤ zN ] = exp − N X # Ho (z1 , . . . , zN ; t)To (t)zo−θ , o=1 as in (15). Note that as zo0 → ∞ for all o0 6= o, we have Hons (z1 , . . . , zN ; t−t∗ ) → 1, Ho (z1 , . . . , zN ; t) → 1, and so P[Zo (t, v) ≤ zo ] = lim zo0 →∞ ∀o0 6=o −θ F (z1 , . . . , zN ; t) = e−To (t)zo . That is, the marginal distribution in o is Fréchet with shape θ and scale To (t). A.2 Trade Shares and Elasticities To get trade shares and elasticities, we apply the results in Proposition 2 of Lind and Ramondo (2018). To do so, first note that the (time-dependent) correlation function implied by (15) is d G (x1 , . . . , xN ; t) = N X o=1
Ho (x1 /T1 (t))−1/θ , . . . , (xN /TN (t))−1/θ ; t xo . Then, Gdo (x1 , . . . , xN ; t) ≡ ∂Gd (x1 , . . . , xN ) ∂xo
= Ho (x1 /T1 (t))−1/θ , . . . , (xN /TN (t))−1/θ ; t
 ∂  Ho (x1 /T1 (t))−1/θ , . . . , (xN /TN (t))−1/θ ; t + xo ∂xo X
 ∂  + xo 0 Ho0 (x1 /T1 (t))−1/θ , . . . , (xN /TN (t))−1/θ ; t ∂xo o0 6=o
= Ho (x1 /T1 (t))−1/θ , . . . , (xN /TN (t))−1/θ ; t N
1X To (t)1/θ , xo0 Ho0 o (x1 /T1 (t))−1/θ , . . . , (xN /TN (t))−1/θ ; t x−1/θ−1 o θ o0 =1 − where Ho0 o ≡ ∂Ho0 /∂zo . Then, −θ To (t)zo−θ Gdo (T1 (t)z1−θ , . . . , TN (t)zN ; t) N = To (t)zo−θ Ho 1X (z1 , . . . , zN ; t) − To0 (t)zo−θ 0 Ho0 o (z1 , . . . , zN ; t) zo . θ o0 =1 Define M (z1 , . . . , zN ; t) ≡ S X N Z X t Z −∞ s=0 n=1 0 N h ∞Y zo 1 − pons (t − t∗ ) + pons (t − t∗ )e−( x ) −σ i o=1 x−θ−1 dxΛ0ns (t∗ )dt∗ . Then, by a change of variables of ao = zo /x in (14), ∗ Z Hons (z1 , . . . , zN ; t − t ) = 0 ∞ i Yh −σ 1 − po0 ns (t − t∗ ) + po0 ns (t − t∗ )e−(zo0 /x) o0 6=o −σ × e−(zo /x) σzoθ−σ xσ−θ−1 dx, Γ(1 − θ/σ) we have Z ∞ S N Z zo −σ ∂M (z1 , . . . , zN ; t) X X t = pons (t − t∗ )e−( x ) σzo−σ−1 xσ ∂zo s=0 n=1 −∞ 0 Yh zo0 −σ i × 1 − po0 ns (t − t∗ ) + po0 ns (t − t∗ )e−( x ) x−θ−1 dxΛ0ns (t∗ )dt∗ o0 6=o = S X N Z X s=0 n=1 = Γ(1 − t Γ(1 − θ/σ)pons (t − t∗ )zo−θ−1 Hons (z1 , . . . , zN ; t − t∗ )Λ0ns (t∗ )dt∗ −∞ θ/σ)zo−θ−1 Ho (z1 , . . . , zN ; t) S X N Z X s=0 n=1 ∞ pons (t − t∗ )Λ0ns (t∗ )dt∗ 0 = To (t)zo−θ−1 Ho (z1 , . . . , zN ; t). Consequently, ∂ 2 M (z1 , . . . , zN ; t) ∂ = To0 (t)zo−θ−1 Ho0 (z1 , . . . , zN ; t) 0 ∂zo0 ∂zo ∂zo = To0 (t)zo−θ−1 Ho0 o (z1 , . . . , zN ; t). 0 Then, replacing yields −θ To (t)zo−θ Gdo (T1 (t)z1−θ , . . . , TN (t)zN ) N = To (t)zo−θ Ho 1X To0 (t)zo−θ (z1 , . . . , zN ; t) − 0 Ho0 o (z1 , . . . , zN ; t) zo θ o0 =1 N = To (t)zo−θ Ho (z1 , . . . , zN ; t) − 1X To (t)zo−θ Hoo0 (z1 , . . . , zN ; t) zo0 . θ o0 =1 Note that Ho (z1 , . . . , zN ; t) is homogenous of degree zero because Hnos (z1 , . . . , zN ; t− t∗ ) is homogenous degree zero. Then, by Euler’s theorem for homogenous funcP tions, N o0 =1 zo0 Hoo0 (z1 , . . . , zN ; t) = 0, implying that −θ ) = To (t)zo−θ Ho (z1 , . . . , zN ; t) . To (t)zo−θ Gdo (T1 (t)z1−θ , . . . , TN (t)zN Therefore,
Gdo (x1 , . . . , xN ; t) = Ho (x1 /T1 (t))−1/θ , . . . , (xN /TN (t))−1/θ ; t
= To1/θ x−1/θ−1 Mo (x1 /T1 (t))−1/θ , . . . , (xN /TN (t))−1/θ ; t , o since Mo (z1 , . . . , zN ; t) = To (t)zo−θ−1 Ho (z1 , . . . , zN ; t) where Mo ≡ d G (x1 , . . . , xN ; t) = = N X o=1 N X ∂M . ∂zo Then, xo Gdo (x1 , . . . , xN ; t)
(xo /To (t))−1/θ Mo (x1 /T1 (t))−1/θ , . . . , (xN /TN (t))−1/θ ; t , o=1 and 1 Pd (t) = Gd (T1 (t)P1d (t)−θ , . . . , TN (t)PN d (t)−θ )− θ " N #− θ1 X . = Ho (P1d (t), . . . , PN d (t); t) To (t)Pod (t)−θ o=1 1 Note that, here, ζ = Γ(1 − (η − 1)/θ) η−1 , which is the inverse of γ in Proposition −θ 2 of Lind and Ramondo (2018), and To (t)Pod (t)−θ corresponds to Pod in that same proposition. Next, trade shares are Tod (t)Pod (t)−θ Gdo (T1d (t)P1d (t)−θ , . . . , TN d (t)PN d (t)−θ ; t) Gd (T1d (t)P1d (t)−θ , . . . , TN d (t)PN d (t)−θ ; t) Tod (t)Pod (t)−θ Ho (P1d (t), . . . , PN d ; t) , = PN 0 d (t)Po0 d (t)−θ Ho0 (P1d (t), . . . , PN d ; t) T 0 o o =1 πod (t) = yielding (18). We next use this result to derive cross-price elasticities of substitution. Since Ho is homogenous of degree zero, πod (t) = PN To (t)Ho (P1d (t), . . . , PN d (t); t)Pod (t)−θ To0 (t)Ho0 (P1d (t), . . . , PN d (t); t)Po0 d (t)−θ  −θ   Pod (t) P1d (t) PN d (t) = To (t) Ho ,..., ;t . Pd (t) Pd (t) Pd (t) o0 =1 Then, for o0 6= o, ∂πodt ∂ ln Po0 d (t) Pd (t)  = To (t) Pod (t) Pd (t) −θ ∂ ∂ ln Po0 d (t) Pd (t)  Ho  P1d (t) PN d (t) ,..., ;t Pd (t) Pd (t) Note that ∂ Ho (z1 , . . . , zN ; t) ∂ ln zo0  −σ S X N Z ∞Z ∞ X z 0 −σ zo0 −( zoo ) x −θ x = To (t)zo po0 ns (a)e σ zo 0 s=0 n=1 0    Y  zm −σ × 1 − pmns (a) + pmns (a)e−( zo ) x  To (t)zo−θ 0} m∈{o,o / −x ρ−1 pons (a)Λ0 (t − a) x da dx PS PN R ∞ ns 0 Γ(ρ) s=0 n=1 0 pons (a)Λns (t − a)da    −σ S X N Z ∞Z ∞  X Y −σ  z 0 z m o = σzo−θ po0 ns (a) xρ  1 − pmns (a) + pmns (a)e−( zo ) x  z o 0 0 0 s=0 n=1 × e m∈{o,o / } ×e −σ +z −σ zo o0 − −σ zo x dxpons (a)Λ0ns (t − a)da  ! z −σ N Z ∞Z ∞ S X −σ −x̃ ρ X Y − −σm −σ x̃
zo−σ 1−ρ 0 zo zo +z 0 −σ −σ  e x̃ dx̃  o = σρ −σ 1 − p (a) + p (a)e z + z 0 mns mns o o 2 Γ(1 + ρ) (zo + zo−σ 0 ) 0 0} s=0 n=1 0 m∈{o,o / × pons (a)po0 ns (a)Γ(ρ)Λ0ns (t − a)da, with ρ ≡ 1 − θ/σ. Evaluating at zo = Pod (t), we get (20). A.3 Innovation Stocks, Scales, and Sharing Probabilities Since the innovation rate in country n at time t is λn (t) = Λ0n (t) S X 1 φ = γn Λn (t) + λns ν ns s=1  t − µns νns  , we have S X  ∂  −γn t 1 −γn t 0 −γn t e Λn (t) = e Λn (t) − γn e Λn (t) = λns e−γn t φ ∂t νns s=1  t − µns νns  . Integrating from 0 to t yields e −γn t Λn (t) − Λn (0) = S X s=1 Z λns t −γn t∗ e 0 1 φ νns  t∗ − µns νns  dt∗ , and hence, Λn (t) = Λn0 (t) + S X s=1 Λns (t), (A.1) where Λn0 (t) ≡ λn0 e 0 Z γn t λn0 ≡ Λn (0) − λns for e −γn t∗ −∞ and t Z γn (t−t∗ ) Λns (t) ≡ λns e −∞ 1 φ νns  1 φ νns t∗ − µns νns   t∗ − µns νns  dt∗ , dt∗ . We use the following lemma to solve for Λns (t) (s ≥ 1) and for other results in the section. Lemma A.1. Z t a+bx 1 e −∞ s  φ x−m s  dx = e a+bm+b2 s2 /2  Φ t − m − bs2 s  . (A.2) Proof. Z t a+bx 1 e −∞ s  φ x−m s  Z t−m s ea+b(m+sz) φ (z) dx −∞   t − m − bs2 a+bm+b2 s2 /2 =e Φ . s dx = Applying Lemma A.1 to (A.1) yields t  t∗ − µns Λns (t) = λns e dt∗ ν ns −∞   2 t − µns + γn νns 2 /2) γn (t−µns +γn νns = λns e Φ . νns Z γn (t−t∗ ) 1 φ νns  Next, we use the following lemma to calculate scale parameters and the probability that any two countries share an idea. Lemma A.2. Let Λ(t) = λeγ(t−µ+γν Z t (1 − e −∞ −δ(t−t∗ ) 2 /2) Φ  t−µ+γν 2 ν  . Then,    δ t − µ − δν 2 −δ(t−µ−δν 2 /2) )Λ (t )dt = Λ(t) − λe Φ . δ+γ ν (A.3) 0 ∗ ∗ Proof. First, note that t  t∗ − µ + γν 2 e γe Φ dt∗ ν −∞     Z t  ∗  γ t − µ + γν 2 t − µ + γν 2 (δ+γ)t∗ 1 (δ+γ)t ∗ = e e Φ − φ dt . δ+γ ν ν ν −∞ Z δt∗ γt∗  Then, t t   ∗  ∂ t − µ + γν 2 γ(t∗ −µ+γν 2 /2) e Λ (t )dt = e λe Φ dt∗ ∗ ∂t ν −∞ −∞      ∗  Z t γ (δ+γ)t t − µ + γν 2 δ t − µ + γν 2 (δ+γ)t∗ 1 ∗ −δt−γµ+γ 2 ν 2 /2 e Φ + e φ dt . = λe δ+γ ν δ + γ −∞ ν ν Z −δ(t−t∗ ) 0 ∗ Z ∗ −δ(t−t∗ ) From Lemma A.1, Z t (δ+γ)t∗ e −∞ 1 φ ν  t∗ − µ + γν 2 ν  ∗ dt = e (δ+γ)µ+(δ 2 −γ 2 )ν 2 /2  Φ t − µ − δν 2 ν  . Therefore, Z t ∗ (1 − e−δ(t−t ) )Λ0 (t∗ )dt∗ −∞      γ (δ+γ)t t − µ + γν 2 t − µ − δν 2 δ (δ+γ)µ+(δ2 −γ 2 )ν 2 /2 −δt−γµ+γ 2 ν 2 /2 = Λ(t) − λe e Φ e Φ + δ+γ ν δ+γ ν    2 δ t − µ − δν 2 = Λ(t) − λe−δ(t−µ−δν /2) Φ . δ+γ ν We now use Lemma A.2 to calculate the scale in (12) normalized by Γ(ρ): Γ(ρ)−1 To (t) = X 1{o ∈ N }Λ(N , t) N ⊆{1,...,N } = = S X N Z X t s=0 n=1 −∞ S X N Z t X s=0 n=1 = Λo (t) + P [o ∈ Ni (t; v) | si (v) = s, ni (v) = n, t∗i (v) = t∗ ] Λ0ns (t∗ )dt∗ ∗ (1 − e−δons (t−t ) )Λ0ns (t∗ )dt∗ −∞ S X X s=0 n6=o    2 δons t − µns − δons νns 2 /2) −δons (t−µns −δons νns Λns (t) − λns e Φ . δons + γn νns We can also use this result to calculate the probability of two countries sharing an idea. For o 6= o0 , S X N Z X t E [Ioi (t, v)Io0 i (t, v) | ni (v) = n, t∗i (v) = t∗ ] Λ0ns (t∗ )dt∗ s=0 n=1 −∞ S X N Z t X = s=0 n=1 ∗ ∗ (1 − e−δons (t−t ) )(1 − e−δo0 ns (t−t ) )Λ0ns (t∗ )dt∗ −∞ −1 = Γ(ρ) (To (t) + To0 (t)) − S X N Z X s=0 n=1 t ∗ (1 − e−(δons +δo0 ns )(t−t ) )Λ0ns (t∗ )dt∗ −∞ = Γ(ρ)−1 (To (t) + To0 (t)) − Λo (t) − Λo0 (t)    S 2 X X t − µns − δoo0 ns νns δoo0 ns 2 /2) −δoo0 ns (t−µns −δoo0 ns νns Λns (t) − λns e Φ − δoo0 ns + γn νns 0 s=0 n∈{o,o / } S X X    2 δons t − µns − δons νns 2 /2) −δons (t−µns −δons νns = Λns (t) − λns e Φ δ + γn νns s=0 n6=o ons    S X 2 X δo0 ns t − µns − δo0 ns νns 2 /2) −δo0 ns (t−µns −δo0 ns νns + Λns (t) − λns e Φ 0 ns + γn δ νns o 0 s=0 n6=o    S 2 X X δoo0 ns t − µns − δoo0 ns νns 2 /2) −δoo0 ns (t−µns −δoo0 ns νns − Λns (t) − λns e Φ , 0 ns + γn δ ν oo ns 0 s=0 n∈{o,o / } where δoo0 ns ≡ δons + δo0 ns and the second to last equality follows from Lemma A.2. P P We get the probability by dividing this result by Ss=0 N n=1 Λns (t). B Additional Results Table B.1: Estimation results: country-level parameters. Country code Country name λn1 µn1 νn1 AUT BLX CAN CHN DEU ESP FRA GBR IND IRL ITA JPN KOR MEX NORD PRT RoW TUR TWN USA Austria 6.6e-5 0.011 0.003 Benelux 0.0 0.101 0.006 Canada 0.0 0.058 0.006 China 0.051 0.0001 0.049 Germany 0.0 0.026 0.005 Spain 0.016 0.001 0.021 France 0.0 0.018 0.002 United Kingdom 0.0 0.036 0.0 India 0.0 0.007 0.0 Ireland 0.054 0.001 0.010 Italy 0.004 0.006 0.016 Japan 0.001 0.054 0.072 Republic of Korea 0.007 0.002 0.009 Mexico 0.0 0.111 0.001 Nordic Countries 3.0e-6 0.038 1.2e-5 Portugal 0.0 0.019 0.0 Rest of World 5.1e-5 0.216 0.048 Turkey 0.0 0.010 0.0 Taiwan 9.6e-5 0.033 0.012 United States 0.0 0.248 7.0e-6 1,987 1,989 1,967 1,999 1,983 1,985 1,980 1,996 2,001 1,998 1,976 1,968 1,997 1,990 1,988 2,006 1,985 2,005 1,984 1,990 29 27 56 6.2 37 16 38 42 49 8.3 23 3.4 6.3 36 46 13 39 18 11 49 γn λn0 Notes: Benelux includes Belgium, Luxembourg, and Netherlands, while Nordic countries include to Denmark, Sweden, and Finland. Estimated parameters of the innovation process in (6) for the one-surge model. µn1 is the (average) year of the innovation surge, s = 1, for each country. Figure B.1: Estimation results: Elasticities. (a) United States, 1962 (b) United States, 2019 (c) China, 1962 (d) China, 2019 Notes: Estimates from the model with S = 1. εoo0 d (t) are calculated using (20). Figure B.2: Estimation results: Probabilities of shared knowledge. (a) Unconditional, 1962 (b) Unconditional, 2019 (c) Conditional, 1962 (d) Conditional, 2019 Notes: Estimates from the model with S = 1. Unconditional sharing probabilities are calculated as Koo0 (t) in (21), relative to To (t) in (12), while conditional probabilities are calculated as Koo0 (t)/Koo (t). Figure B.3: Cross-price elasticities and shared knowledge, by destination country. Notes: Estimates from the model with S = 1. εoo0 d (t) are calculated using (20), by destination country d, while sharing probabilities are calculated using (21).