# Convergence of normals Shiu-Tang Li Finished: March 23, 2013

```Convergence of normals
Shiu-Tang Li
Finished: March 23, 2013
Last updated: November 2, 2013
Theorem Let Xn ∼ N (&micro;n , σn2 ). If Xn → X weakly, then X ∼ N (&micro;, σ 2 ),
and &micro;n → &micro;, σn → σ.
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2 2
Proof. Since Xn → X weakly, we have ei&micro;n t− 2 σn t converge to φ(t) for all t,
where φ is the ch.f. of X. Now we choose arbitrary s &gt; 0, and we have
1 2 2
1 2 2
ei&micro;n s− 2 σn s → φ(s), which implies e− 2 σn s → |φ(s)|. Therefore, |φ(s)| =
1 2 2
e− 2 σ s , where σn → σ.
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Since e− 2 σn s &gt; 0 for all s ∈ R, n ∈ N, ei&micro;n t converges for all t ∈ R. If
there exists two subsequences {&micro;1,j }j , {&micro;2,j }j of {&micro;n }n so that {&micro;1,j }j → a
and {&micro;2,j }j → b, a 6= b, then ei&micro;n t converges to eiat and eibt for every t ∈ R,
which is absurd. If &micro;n → ∞ or &micro;n → −∞, then P (Xn ∈ (c, d)) → 0 for
every finite interval (c, d) (here we also use the fact we’ve proved: σn → σ),
and this contradicts the fact that Xn converges weakly to some X.
Therefore, the only possibility is that &micro;n converges to some finite &micro;, and
1 2 2
1 2 2
we have ei&micro;n t− 2 σn t → ei&micro;t− 2 σ t for all t ∈ R.
Remark If we take &micro;1 := 2π, &micro;n = l.c.m(2, 3, &middot; &middot; &middot; , n) &middot; (2π) + 1, then we
would find that ei&micro;n t converges to eit for all t ∈ Q. However, the convergence
does not hold for every t ∈ R, for otherwise this would be a contradiction to
the above theorem.
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