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Convergence of normals Shiu-Tang Li Finished: March 23, 2013 Last updated: November 2, 2013 Theorem Let Xn ∼ N (µn , σn2 ). If Xn → X weakly, then X ∼ N (µ, σ 2 ), and µn → µ, σn → σ. 1 2 2 Proof. Since Xn → X weakly, we have eiµn t− 2 σn t converge to φ(t) for all t, where φ is the ch.f. of X. Now we choose arbitrary s > 0, and we have 1 2 2 1 2 2 eiµn 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. 1 2 2 Since e− 2 σn s > 0 for all s ∈ R, n ∈ N, eiµn t converges for all t ∈ R. If there exists two subsequences {µ1,j }j , {µ2,j }j of {µn }n so that {µ1,j }j → a and {µ2,j }j → b, a 6= b, then eiµn t converges to eiat and eibt for every t ∈ R, which is absurd. If µn → ∞ or µn → −∞, 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 µn converges to some finite µ, and 1 2 2 1 2 2 we have eiµn t− 2 σn t → eiµt− 2 σ t for all t ∈ R. Remark If we take µ1 := 2π, µn = l.c.m(2, 3, · · · , n) · (2π) + 1, then we would find that eiµn 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. 1