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= 1+ 1 2∂ 2 f(e 1 , e 0 )∂e i = −(1/N) 2 , ∀i, j1 ∂ej 0 e1,e 0=0 ∂ 2 f(e 1 , e 0 )∂e i = ∂2 f(e 1 , e 0 )1 ∂ej 1 e1,e 0=0 ∂e i = (1/N) 2 , ∀i ≠ j0 ∂ej 0 e1,e 0=0 ∂ 2 f(e 1 , e 0 )∂(e i =1e1,e ∂2 f(e 1 , e 0 ))2 0=0∂(e i =0e1,e 1 1 )2 0=0N N − 1 , ∀iSo our approximation evaluates to" 1Nso we have thatX 2 ei 1 X 1 − (ei 1 X N 1 ) 2 − 2 ei 1 X N 1 ei 1 X 2 N 0 − ei 1 X #N 0 + (eiN 0 ) 2= 1 + 1 2P J (p 0 , p 1 ) ≈ P D (p 0 , p 1 ) 1 X (ei 1 XN 0 ) 2 − (eiN 1 ) 21 + 1 V ar ei20 − V ar ei1Proof. Proof of proposition 4.This proof consists of two stages. First we show that the solution to the maximum entropyproblem (8) subject to the constraints that preferences satisfy GARP is w i t = 1/N for all i, t.Then we show that constant and equal budget shares are consistentwith GARP, and so the additional constraint in this problem is not binding.The solution to max w0,w 1− P t w′ t ln w t subject to P t w t = 1 ismaxw 0,w 1− X twhere λ is the Lagrange multiplier.w ′ t ln w t − λ( X tw t − 1)=⇒ w i t1w i t+ ln w i 1 − λ = 0=⇒ ln w i t = λ − 1, ∀i, t=⇒ w i t = exp(λ − 1), ∀i, twhich implies that budget shares are constant across i and t. Combining this with constrainttells us that the entropy maximising budget shares will be 1/N. This completes the first stageof our proof.To prove the second stage our strategy will be to show that a violation of GARP is impossiblewith equal budget shares. A violation of GARP implies that there exist two periods t and s,when the consumer chooses quantities q t and q s such that:andp ′ tq s ≤ x t23