# Proof that asymptotic normality implies consistency Without loss of ```Proof that asymptotic normality
implies consistency
Without loss of generality, we’ll assume that
σ(θ) = 1.
We have
P (|Tn − τ (θ)| &lt; &sup2;) =
P (−&sup2; &lt; Tn − τ (θ) &lt; &sup2;) =
P (Tn − τ (θ) &lt; &sup2;) − P (Tn − τ (θ) ≤ −&sup2;) =
√
P ( n(Tn − τ (θ)) &lt;
√
n&sup2;)−
√
√
P ( n(Tn − τ (θ)) ≤ − n&sup2;).
Need to show that for 1 &gt; δ &gt; 0, there exists
n0 such that for all n &gt; n0 P (|Tn − τ (θ)| &lt; &sup2;) &gt;
1 − δ.
Let z &gt; 0 be such that Φ(z)−Φ(−z) = 2Φ(z)−
1 = 1 − δ/2.
1
For
√
n&sup2; &gt; z, or n &gt; (z/&sup2;)2,
√
√
P ( n(Tn − τ (θ)) &lt; n&sup2;)−
√
√
P ( n(Tn − τ (θ)) ≤ − n&sup2;) ≥
√
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 &gt; n1
√
P ( n(Tn − τ (θ)) ≤ z)−
√
P ( n(Tn − τ (θ)) ≤ −z) &gt; 1 − δ.
The proof is finished upon defining
n0 = max(n1, (z/&sup2;)2).
2
```