Abstract. Let (X
Testo completo
Anscombe’s theorem (AT) gives conditions for X Nn
For instance, condition (a*) suffices for X Nn
However, under (a*), convergence in distribution of X n is not enough. To get converge in distribution of X Nn
One natural solution is to request stable convergence of X n . This is made precise by a result of Zhang Bo [9] (Theorem 1 in the sequel). According to Theorem 1, X Nn
3.2. Classical Anscombe’s theorem and one of its developments. Let µ be a Borel probability measure on S. According to AT, for X Nn
Soon after its appearance, AT has been investigated and developed in various ways. See e.g. [4], [5], [7], [8], [9] and references therein. To our knowledge, most results preserve the structure of the classical AT, for they lead to convergence of X Nn
Then, X Nn
Or else, X Nn
Theorem 1 is our starting point. Roughly speaking, it can be summarized as follows. Suppose (a*) and (c) hold but (a) fails. If U is discrete, X Nn
3.3. Improving Theorem 1. Suppose conditions (a*)-(b*) hold but U is not necessarily discrete. As implicit in Theorem 1, it may be that (c) holds and yet X Nn
One (crude) possibility is just replacing n with N n in condition (c), that is, (d) inf δ>0 lim sup n P M Nn
M Nn
d(X j , X Nn
In view of (a*), another option is replacing M n (δ) with M [kn
d(X j , X [kn
(e) inf δ>0 lim sup n P M [kn
X Nn
M Nn
d(X j , X Rn
P M Nn
P M Nn
P M Rn
d(X Rn
conditions (a*) and (e) yield d(X Rn
Since U δ is discrete, strictly positive and G-measurable, condition (b*) yields X Rn
for all j ≥ 1. Then, (b*) implies lim n Ef (X Rn
Ef (X [kn
[k n U ] − [k n U δ2
k n (U − U δ2
d(X Rn
≤ P (U ≤ δ) + P M Rn
d(X Rn
d(X Rn
With such a δ, since X Rn
P X Rn
d(X Rn
X Rn
X Rn
As → 0, it follows that lim sup n P X Rn
P M Rn
P M [kn
M Rn
P M Rn
P M [kn
P M Rn
P M Rn
P M Rn
that is, A 1 = [0, log 2), A 2 = [log 2, 1) ∪ [0, (log 3) − 1) and so on. Define also X 0 = 0 and X n = I An
where the r n are non-negative integers such that r n → ∞. Condition (a*) is trivially true. Further, for each n, one obtains {N n = k} ⊂ A k for all k, so that X Nn
Thus, X Nn
P M Nn
P M Nn
P M Nn
I {Zi
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