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Cobweb Theorems with production lags and price forecasting

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In [7] there is a proof that if ρ(E(A1 ⊗ A1)) < 1, then ρ(E(A1)) < 1 as well, from which<br />

we could derive (30). That proof is not very intuitive, however, <strong>and</strong> we propose a more direct<br />

argument. Let x ∈ C N <strong>and</strong> {An} an i.i.d. sequence of N × N matrices. For any complex Y it<br />

is elementary that E|Y | 2 ≥ |EY | 2 . If M = An · · · A1 <strong>and</strong> Y = x T Mx then<br />

E|Y | 2 = E(x T Mxx T M T x)<br />

= E[(x T ⊗ x T )vec(Mxx T M T )]<br />

= (x T ⊗ x T )E(M ⊗ M)vec(xx T ).<br />

Now, since (N1N2) ⊗ (N1N2) = (N1 ⊗ N1)(N2 ⊗ N2),<br />

E(M ⊗ M) = E[(An · · · A1) ⊗ (An · · · A1)] = E<br />

<strong>and</strong> thus<br />

n�<br />

(An−j+1 ⊗ An−j+1) = [E(A1 ⊗ A1)] n ,<br />

E|Y | 2 = (x T ⊗ x T )[E(A1 ⊗ A1)] n vec(xx T ) ≥ |EY | 2 = |Ex T Mx| 2 = |x T (EA1) n x| 2 .<br />

If x is a non-zero eigenvector of E(A1) <strong>and</strong> λ the corresponding eigenvalue, then<br />

j=1<br />

(x T ⊗ x T )[E(A1 ⊗ A1)] n vec(xx T ) ≥ |λ| 2n |x| 4 e.<br />

Divide both sides by σ n > ρ(E(A1⊗A1)) n <strong>and</strong> then let n tend to infinity, to find 0 ≤ |λ| 2 /σ < 1,<br />

or |λ| 2 < σ. This is true for all eigenvalues of EA1 <strong>and</strong> all σ > ρ(E(A1 ⊗ A1)), which gives<br />

ρ(E(A1)) 2 < σ, <strong>and</strong> finishes the proof. �<br />

Note that it is not always true that γ({An}) ≤ log ρ(EA1) for matrices {An} that have both<br />

positive <strong>and</strong> negative entries; this happens in the second numerical example below.<br />

Theorem 11. (a) If<br />

ρ(E|A1|) < 1, E|B|e < ∞<br />

then (27) is stable. Moreover, EXt is finite <strong>and</strong> satisfies<br />

EXt = E(A1)EXt−1 + EBt. (31)<br />

lim<br />

t→∞ EXt = (Iℓ+m,ℓ+m − E(A1)) −1 EB1 = 0. (32)<br />

(b) (Conlisk [7]) Suppose {(An, Bn)} are i.i.d. <strong>and</strong> have finite second moments. Then a sufficient<br />

condition for the system (27) to be stable is<br />

ρ(E(A1 ⊗ A1)) < 1.<br />

When this is the case, the first <strong>and</strong> second moments of Xt are finite, the first moments satisfy<br />

(32) <strong>and</strong> second moments satisfy<br />

vec E(XtX T t ) = E(An ⊗ At) vec E(Xt−1X T t−1) + [E(Bt ⊗ At) + E(At ⊗ Bt)]vec EXt−1 + vec E(BtB T t )<br />

lim<br />

t→∞ vec E(XtX T t ) = (I (ℓ+m) 2 ,(ℓ+m) 2 − E(A1 ⊗ A1)) −1 vec E(B1B T 1 ).<br />

Part (b) of the theorem shows that the second-order conditions<br />

log ρ(E(A1 ⊗ A1)) < 1, E|ɛ 2 1| < ∞,<br />

ensure that Theorem 9 applies. Part (a) shows that the corresponding first-order conditions have<br />

the same consequence.<br />

24

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