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Proof that asymptotic normality implies consistency Without loss of generality, we’ll assume that σ(θ) = 1. We have P (|Tn − τ (θ)| < ²) = P (−² < Tn − τ (θ) < ²) = P (Tn − τ (θ) < ²) − P (Tn − τ (θ) ≤ −²) = √ P ( n(Tn − τ (θ)) < √ n²)− √ √ P ( n(Tn − τ (θ)) ≤ − n²). Need to show that for 1 > δ > 0, there exists n0 such that for all n > n0 P (|Tn − τ (θ)| < ²) > 1 − δ. Let z > 0 be such that Φ(z)−Φ(−z) = 2Φ(z)− 1 = 1 − δ/2. 1 For √ n² > z, or n > (z/²)2, √ √ P ( n(Tn − τ (θ)) < n²)− √ √ P ( n(Tn − τ (θ)) ≤ − n²) ≥ √ P ( n(Tn − τ (θ)) ≤ z)− √ P ( n(Tn − τ (θ)) ≤ −z). Because of the asymptotic normality of {Tn}, the last expression tends to 1 − δ/2. So, there exists n1 such that for n > n1 √ P ( n(Tn − τ (θ)) ≤ z)− √ P ( n(Tn − τ (θ)) ≤ −z) > 1 − δ. The proof is finished upon defining n0 = max(n1, (z/²)2). 2