Intertemporal Substitution and Recursive Smooth Ambiguity ...

Lemma 5 We have the homeomorphic relation:H ≃ A S .Proof. Define ξ : H → A S byξ(h)(s) = (h 0 (s), h 1 (s), h 2 (s), · · · ).It follows from Lemma 2 that ξ(h) ∈ A.ξ is one to one: Suppose ξ(h) = ξ(h ′ ). By definition of ξ, we have (h 0 (s), h 1 (s), h 2 (s), · · · ) =(h ′ 0(s), h ′ 1(s), h ′ 2(s), · · · ) for all s ∈ S, which implies h = h ′ .ξ is onto: Take any ˜h ∈ A S . By definition,˜h(s) = (˜h0 (s), ˜h 1 (s), ˜h 2 (s), · · · ) ∈∞∏∆(C × ∆(H t−1 )),t=0for each s ∈ S. Then ξ −1 (˜h) = (h 0 , h 1 , h 2 , · · · ) ∈ H ∗ satisfies h t (s) = ˜h t (s) for each t and s.By Lemmas 2-3, the sequence (h 0 , h 1 , h 2 , · · · ) is coherent and hence ξ −1 (˜h) ∈ H.ξ and ξ −1 are continuous: Immediate from the nature of the product topology.LetP ∗ ={(p t ) ∈(∞∏ t∏)∆ H ττ=0t=0: mrg ∏ tτ=0 H τ p t+1 = p t}.Lemma 6 For any (p t ) ∈ P ∗ , there exists a unique p ∈ ∆(H ∗ ) such thatmrg ∏ tτ=0 H τ p = p t.Moreover, there exists a homeomorphism χ : P ∗ → ∆(H ∗ ).Proof. It follows from Lemma 1 in Brandenberger and Dekel (1993).LetH t =for each t ≥ 0, and{(h 0 , · · · , h t ) ∈}t∏H τ : h τ = ρ τ (h τ+1 ), τ = 0, · · · , t − 1 ,τ=0P = { (p t ) ∈ P ∗ : p t (H t ) = 1, t ≥ 0 } .Lemma 7 χ(P ) = ∆(H). As a result, P ≃ ∆(H) holds through χ.41