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.
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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).
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