WIP: Devam eden çalışma
Aşağıdaki s. 370 Cramer's 1946 Matematiksel İstatistik Metodlarını tanımlamakΞn=n(1−Φ(Zn)).
Buraya Φ standart normal dağılımın kümülatif dağılım fonksiyonudur, N(0,1). Tanımının bir sonucu olarak,0≤Ξn≤n neredeyse eminim.
Belirli bir gerçekleşmeyi düşünün ω∈Ωbizim örnek alan. O zaman bu anlamdaZn her ikisi de n ve ω, ve Ξn bir işlevi Zn,n, ve ω. Sabit birω, düşünebiliriz Zn belirleyici bir işlevi n, ve Ξn belirleyici bir işlevi Zn ve nböylece sorunu basitleştirir. Neredeyse tamamen herkes için geçerli olan sonuçları göstermeyi hedefliyoruzω∈Ωsonuçlarımızı belirleyici olmayan bir analizden belirleyici olmayan ortama aktarmamıza olanak tanıyor.
Aşağıdaki s. Cramer'in 1946 Matematiksel İstatistik Yöntemleri'nin 374'ü ( şimdilik geri dönmeyi ve daha sonra bir kanıt sunmayı hedefliyorum) varsayalım kiω∈Ω) aşağıdaki asimptotik genişleme muhafazaları (parçalara göre entegrasyon ve Φ):
2π−−√nΞn=1Zne−Z2n2(1+O(1Z2n)) as Zn→∞.(~)
Açıkçası buna sahibiz Zn+1≥Zn herhangi n, ve Zn neredeyse kesinlikle artan bir fonksiyonudur. n gibi n→∞bu nedenle, (hemen hemen hepsi) sabit ω: Zn→∞⟺n→∞.
Dolayısıyla, ∼asimptotik denkliği belirtir ):
2π−−√nΞn∼1Zne−1Z2n as Zn→∞n→∞.
Aşağıdaki adımlarda nasıl ilerlediğimiz esas olarak baskın denge yöntemine bağlıdır ve manipülasyonlarımız aşağıdaki lemma tarafından resmen gerekçelendirilecektir:
Lemma: Varsayınf(n)∼g(n) gibi n→∞, ve f(n)→∞ (Böylece g(n)→∞). Sonra herhangi bir işlev verildihlogaritmaların ve güç yasalarının (esasen herhangi bir " polilog " fonksiyonu) bileşimleri, ilaveleri ve çarpımları ile oluşturulan,n→∞: h(f(n))∼h(g(n)).
Başka bir deyişle, bu "polilog" fonksiyonlar asimtotik eşdeğeri korur .
Bu lemmanın gerçeği Teorem 2.1'in bir sonucudur . burada yazıldığı gibi . Ayrıca, aşağıdakilerin çoğunlukla burada bulunan benzer bir sorunun cevabının genişletilmiş (daha fazla ayrıntı) bir versiyonu olduğunu unutmayın .
Her iki tarafın logaritmasını alarak şunu elde ederiz:
log(2π−−√Ξn)−logn∼−logZn−Z2n2.(1)
Cramer biraz kurnazdır; o sadece "varsayarakΞn "sınırlıdır" diyebiliriz. Ξnhemen hemen hiç önemsiz görünmüyor. Bunun kanıtı, esasen Galambos'un 265-267. Sında tartışılanların bir parçası olabilir, ancak bu kitabın içeriğini hala anlamak için çalıştığımdan emin değilim.
Her neyse, birinin bunu gösterebileceğini varsayarsaklogΞn=o(logn), sonra ( −Z2n/2 terim hakim −logZn terim):
−logn∼−Z2n2⟹Zn∼2logn−−−−−√.
Bu biraz güzel, çünkü göstermek istediğimiz şeylerin çoğu zaten, yine de aslında sadece kutuyu yolda tekmelemek olduğunu belirtmek gerekir, çünkü şimdi neredeyse kesin olarak sınırlı bazı göstermeliyiz Ξn. Diğer yandan,Ξn herhangi bir maksimum iid sürekli rasgele değişkenleri için aynı dağılıma sahiptir, bu nedenle bu izlenebilir olabilir.
Her neyse, eğer Zn∼2logn−−−−−√ olarak, açıkça Zn∼2logn−−−−−√(1+α(n)) herhangi α(n) hangisi o(1) gibi n→∞. Yukarıdaki asimtotik denkliği koruyan polilog fonksiyonları ile ilgili lemmamızı kullanarak bu ifadeyi(1) almak:
log(2π−−√Ξn)−logn∼−log(1+α)−12log2−12loglogn−logn−2αlogn−α2logn.
⟹−log(Ξn2π−−√)∼log(1+α)+12log2+12loglogn+2αlogn+α2logn.
Burada daha da ileri gitmek zorunda ve farzlogΞn=o(loglogn) as n→∞neredeyse eminim . Yine, tüm Cramer "Ξn Ama herkes herkes hakkında a priori diyebilir Ξn bu mu 0≤Xin≤n çünkü, kişinin Ξn=O(1) neredeyse şüphesiz ki bu Cramer'in iddiasının özünü oluşturuyor.
Ama yine de, birinin buna inandığını varsayarsak, o zaman α dır-dir 12loglogn. Dan beriα=o(1), bunu takip eder α2=o(α)ve açıkça log(1+α)=o(α)=o(o(αlogn)), so the dominant term containing α is 2αlogn. Therefore, we can rearrange and (dividing everything by 12loglogn or 2αlogn) find that
−12loglogn∼2αlogn⟹α∼−loglogn4logn.
Therefore, substituting this back into the above, we get that:
Zn∼2logn−−−−−√−loglogn22logn−−−−−√,
again, assuming we believe certain things about Ξn.
We rehash the same technique again; since Zn∼2logn−−−−−√−loglogn22logn√, then it also follows that
Zn∼2logn−−−−−√−loglogn22logn−−−−−√(1+β(n))=2logn−−−−−√(1−loglogn8logn(1+β(n))),
when β(n)=o(1). Let's simplify a little before substituting directly back into (1); we get that:
logZn∼log(2logn−−−−−√)+log(1−loglogn8logn(1+β(n)))log(O(1))=o(logn)∼log(2logn−−−−−√).
Z2n2∼logn−12loglogn(1+β)+(loglogn)28logn(1β)2o((1+β)loglogn)∼logn−12(1+β)loglogn.
Substituting this back into (1), we find that:
log(2π−−√Ξn)−logn∼−log(2logn−−−−−√)−logn+12(1+β)loglogn⟹β∼log(4πΞ2n)loglogn.
Therefore, we conclude that almost surely
Zn∼2logn−−−−−√−loglogn22logn−−−−−√(1+log(4π)+2log(Ξn)loglogn)=2logn−−−−−√−loglogn+log(4π)22logn−−−−−√−log(Ξn)2logn−−−−−√.
This corresponds to the final result on p.374 of Cramer's 1946 Mathematical Methods of Statistics except that here the exact order of the error term isn't given. Apparently applying this one more term gives the exact order of the error term, but anyway it doesn't seem necessary to prove the results about the maxima of i.i.d. standard normals in which we are interested.
Given the result of the above, namely that almost surely:
Zn∼2logn−−−−−√−loglogn+log(4π)22logn−−−−−√−log(Ξn)2logn−−−−−√⟹Zn=2logn−−−−−√−loglogn+log(4π)22logn−−−−−√−log(Ξn)2logn−−−−−√+o(1).(†)
2. Then by linearity of expectation it follows that:
EZn=2logn−−−−−√−loglogn+log(4π)22logn−−−−−√−E[log(Ξn)]2logn−−−−−√+o(1)⟹EZn2logn−−−−−√=1−E[logΞn]2logn+o(1).
Therefore, we have shown that
limn→∞EZn2logn−−−−−√=1,
as long as we can also show that
E[logΞn]=o(logn).
This might not be too difficult to show since again Ξn has the same distribution for every continuous random variable. Thus we have the second result from above.
1. Similarly, we also have from the above that almost surely:
Zn2logn−−−−−√=1−log(Ξn)2logn+o(1),.
Therefore, if we can show that:
log(Ξn)=o(logn) almost surely,(*)
then we will have shown the first result from above. Result (*) would also clearly imply a fortiori that E[log(Ξn)]=o(logn), thereby also giving us the first result from above.
Also note that in the proof above of (†) we needed to assume anyway that Ξn=o(logn) almost surely (or at least something similar), so that if we are able to show (†) then we will most likely also have in the process needed to show Ξn=o(logn) almost surely, and therefore if we can prove (†) we will most likely be able to immediately reach all of the following conclusions.
3. However, if we have this result, then I don't understand how one would also have that EZn=2logn−−−−−√+Θ(1), since o(1)≠Θ(1). But at the very least it would seem to be true that EZn=2logn−−−−−√+O(1).
So then it seems that we can focus on answering the question of how to show that Ξn=o(logn) almost surely.
We will also need to do the grunt work of providing a proof for (~), but to the best of my knowledge that is just calculus and involves no probability theory, although I have yet to sit down and try it yet.
First let's go through a chain of trivialities in order to rephrase the problem in a way which makes it easier to solve (note that by definition Ξn≥0):
Ξn=o(logn)⟺limn→∞Ξnlogn=0⟺∀ε>0,Ξnlogn>ε only finitely many times⟺∀ε>0,Ξn>εlogn only finitely many times.
One also has that:
Ξn>εlogn⟺n(1−F(Zn))>εlogn⟺1−F(Zn)>εlognn⟺F(Zn)<1−εlognn⟺Zn≤inf{y:F(y)≥1−εlognn}.
Correspondingly, define for all n:
u(ε)n=inf{y:F(y)≥1−εlognn}.
Therefore the above steps show us that:
Ξn=o(logn) a.s.⟺P(Ξn=o(logn))=1⟺P(∀ε>0,Ξn>εlogn only finitely many times)=1⟺P(∀ε>0,Zn≤u(ε)n only finitely many times)=1⟺P(∀ε>0,Zn≤u(ε)n infinitely often)=0.
Notice that we can write:
{∀ε>0,Zn≤u(ε)n infinitely often}=⋂ε>0{Zn≤u(ε)n infinitely often}.
The sequences u(ε)n become uniformly larger as ε decreases, so we can conclude that the events {Zn≤u(ε)n infinitely often}
are decreasing (or at least somehow monotonic) as ε goes to 0. Therefore the probability axiom regarding monotonic sequences of events allows us to conclude that:
P(∀ε>0,Zn≤u(ε)n infinitely often)=P(⋂ε>0{Zn≤u(ε)n infinitely often})=P(limε↓0{Zn≤u(ε)n infinitely often})=limε↓0P(Zn≤u(ε)n infinitely often).
Therefore it suffices to show that for all ε>0,
P(Zn≤u(ε)n infinitely often)=0
because of course the limit of any constant sequence is the constant.
Here is somewhat of a sledgehammer result:
Theorem 4.3.1., p. 252 of Galambos, The Asymptotic Theory of Extreme Order Statistics, 2nd edition. Let X1,X2,… be i.i.d. variables with common nondegenerate and continuous distribution function F(x), and let un be a nondecreasing sequence such that n(1−F(un)) is also nondecreasing. Then, for un<sup{x:F(x)<1}, P(Zn≤un infinitely often)=0 or 1
according as
∑j=1+∞[1−F(uj)]exp(−j[1−F(uj)])<+∞ or =+∞.
The proof is technical and takes around five pages, but ultimately it turns out to be a corollary of one of the Borel-Cantelli lemmas. I may get around to trying to condense the proof to only use the part required for this analysis as well as only the assumptions which hold in the Gaussian case, which may be shorter (but maybe it isn't) and type it up here, but holding your breath is not recommended. Note that in this case ω(F)=+∞, so that condition is vacuous, and n(1−F(n)) is εlogn thus clearly non-decreasing.
Anyway the point being that, appealing to this theorem, if we can show that:
∑j=1+∞[1−F(u(ε)j)]exp(−j[1−F(u(ε)j)])=∑j=1+∞[εlogjj]exp(−εlogj)=ε∑j=1+∞logjj1+ε<+∞.
Note that since logarithmic growth is slower than any power law growth for any positive power law exponent (logarithms and exponentials are monotonicity preserving, so loglogn≤αlogn⟺logn≤nα and the former inequality can always be seen to hold for all n large enough due to the fact that logn≤n and a change of variables), we have that:
∑j=1+∞logjj1+ε≤∑j=1+∞jε/2j1+ε=∑j=1+∞1j1+ε/2<+∞,
since the p-series is known to converge for all p>1, and ε>0 of course implies 1+ε/2>1.
Thus using the above theorem we have shown that for all ε>0, P(Zn≤u(ε)n i.o.)=0, which to recapitulate should mean that Ξn=o(logn) almost surely.
We need to show still that logΞn=o(loglogn). This doesn't follow from the above, since, e.g.,
1nlogn=o(logn),−logn+loglogn≠o(logn).
However, given a sequence xn, if one can show that xn=o((logn)δ) for arbitrary δ>0, then it does follow that log(xn)=o(loglogn). Ideally I would like to be able to show this for Ξn using the above lemma (assuming it's even true), but am not able to (as of yet).