最大化艾德高斯派的最有力结果是什么？在实践中最常用？

由于$X_1, \ldots, X_n, \ldots \sim \mathscr{N}(0,1)$ IID，考虑随机变量

$$Z_n := \max_{1 \le i \le n} X_i\,.$$

问题：这些随机变量最“重要”的结果是什么？

为了澄清“重要性”，哪个结果具有其他大多数这样的结果是合乎逻辑的结果？在实践中最常使用哪个结果？

更具体地说，似乎是（理论上的）统计学家之间的民俗知识，即至少渐近地“基本上与”。（请参阅此相关问题。）$Z_n$$\sqrt{2 \log n}$

但是，这种类型的结果有很多，而且似乎大多数情况不是等效的，也不是相互暗示的。例如$^*$，

$$\frac{Z_n}{\sqrt{2 \log n}} \overset{a.s.}{\to} 1 \,, \tag{1}$$

如果没有别的，也暗示了概率和分布的相应结果。

但是，它甚至似乎并不暗示也有相关的结果（请参见另一个问题），例如

$$\lim_{n \to \infty} \frac{\mathbb{E}Z_n}{\sqrt{2 \log n}} =1 \,, \tag{2}$$

（这是第49页的练习2.17 ），或另一个民俗结果：$\dagger$

$$\mathbb{E}Z_n = \sqrt{2 \log n} + \Theta(1) \,. \tag{3}$$

非渐近地，还知道对于每个（请参见此处的证明），$n$

$$\sqrt{c \log n} \le \mathbb{E}Z_n \le \sqrt{2 \log n} \tag{4}$$

对于一些小。也可以显示类似的结果 ，因为严重偏右。$c$$|Z_n|$$Z_n$

最后一个结果的证明比其他结果的证明更直接。我的希望是，第一个渐近结果将隐含所有其他渐近结果，以便让我有信心将所有的时间和精力都集中在理解该结果上。但是，再次，这似乎是不正确的，所以现在我不清楚应该关注哪个方面。

$^*$参见Galambos第二版第265-267页，《极端秩序统计的渐近理论》，于1987年出版。第一版中可能也有提及。

$\dagger$ Boucheron，Lugosi，Massart，浓度不平等：非渐近独立理论。旁白：这本书实际上引用了Galambos的结果，但我找不到在Galambos的任何地方提及它-这只是我提到的第一个结果。

在任何概率应用中，最基本的对象是分布，矩和极限性质可以从中得出。因此，最“重要”的结果，在你所描述的意义上说，是完整的分布函数$F_{Z_n}(z) = \Phi^n(z)$（等同于相应的密度函数）。在实践中，这种分布结果可能比您已经列出的一些更基本的渐近性质少了启发。尽管从逻辑上讲暗示了这些渐近结果，但在我看来，当我们改变$n$，这些结果可能更有助于理解极值的变化性质。

从您的问题中可以很清楚地了解，在最大IID标准正态随机变量的情况下，您对极值属性有很好的了解。这些属性在逻辑上都可以从$Z_n$的分布函数导出，因此这是解决此问题的最基本对象。在许多情况下，最基本的对象不一定是最有启发性的对象，因此您可能会发现必须了解所有结果并知道它们可以启发问题的各个方面。

在制品：进行中

继p。克莱姆1946 370 数理统计方法，定义

$$\Xi_n = n(1 - \Phi(Z_n)) \,.$$ 这里$\Phi$是标准正态分布的累积分布函数，$\mathscr{N}(0,1)$。作为其定义的结果，我们保证$0\le \Xi_n \le n$几乎可以肯定。

考虑一个给定的实现$\omega \in \Omega$我们的样本空间。然后，在这个意义上$Z_n$既是功能$n$和$\omega$，和$\Xi_n$的功能$Z_n, n$和$\omega$。对于固定的$\omega$，我们可以考虑$Z_n$的确定性函数$n$，和$\Xi_n$的确定性函数$Z_n$和$n$，从而简化了问题。我们的目标是展示其持有几乎可以肯定所有的结果$\omega \in \Omega$，使我们可以将结果从不确定性分析转移到不确定性设置。

继p。克莱默的1946年的374 数理统计的方法，假定此时（我的目标是回来后提供一个证明），我们能够证明（对于任何给定的$\omega \in \Omega$）以下的渐近展开抱（使用分部积分法和$\Phi$的定义：

$$\frac{\sqrt{2\pi}}{n}\Xi_n = \frac{1}{Z_n}e^{-\frac{Z_n^2}{2}}\left( 1 + O \left( \frac{1}{Z_n^2} \right) \right) \quad ~~ as ~~ Z_n \to \infty \,. \tag{~}$$

很明显，我们有$Z_{n+1} \ge Z_n$任何$n$，和$Z_n$几乎是肯定的增函数$n$作为$n\to \infty$，因此我们在整个后续内容为（几乎肯定全部）固定权利要求$\omega$：

$$Z_n \to \infty \quad \iff \quad n \to \infty \,.$$

因此，我们得出（其中$\sim$表示渐近等价）：

$$\frac{\sqrt{2\pi}}{n} \Xi_n \sim \frac{1}{Z_n} e^{-\frac{1}{Z_n^2}} \quad ~~ as ~~ Z_n \to \infty \quad n \to \infty \,.$$

我们如何进行后续工作基本上等于主导平衡的方法，而我们的操作将通过以下引理形式正式证明：

引理：假定$f(n) \sim g(n)$作为$n \to \infty$，和$f(n) \to \infty$（因此$g(n) \to \infty$）。然后给定通过对数和幂定律的组成，加法和乘积形成的任何函数$h$（本质上是任何“ polylog ”函数），我们还必须具有$n \to \infty$：

$$h(f(n)) \sim h(g(n)) \,.$$ 换句话说，这样的“ polylog”函数保留了渐近等价。

这个引理的真实性是定理2.1的结果。如这里所写。还请注意，以下内容基本上是此处找到的类似问题的答案的扩展版本（更多详细信息）。

取双方的对数，我们得出：

$$\log ( \sqrt{2\pi} \Xi_n ) - \log n \sim -\log Z_n - \frac{Z_n^2}{2} \,. \tag{1}$$

这是Cramer有点笼统的地方。他只是说：“假设$\Xi_n$为界”，我们可以得出结论等等等等等等。但表示$\Xi_n$适当界几乎肯定似乎是个有点不平凡的。看来，这基本上可能是Galambos 265-267页所讨论内容的一部分，但是由于我仍在努力理解该书的内容，因此我不确定。

无论如何，假设一个可显示$\log \Xi_n = o(\log n)$，那么它遵循（因为$-Z_n^2/2$项支配$-\log Z_n$术语）的是：

$$- \log n \sim - \frac{Z_n^2}{2} \quad \implies \quad Z_n \sim \sqrt{2 \log n} \,.$$

这是不是很好，因为它已经是我们要显示什么最，虽然这又是值得的注意的是它基本上只能拖延时间在路上，因为现在我们必须表现出一定的几乎处处有界的$\Xi_n$。在另一方面，$\Xi_n$具有用于任何最大IID连续随机变量的相同的分布，所以这可能是易处理的。

无论如何，如果$Z_n \sim \sqrt{2 \log n}$如，那么显然可以还得出这样的结论$Z_n \sim \sqrt{2 \log n}(1 + \alpha(n))$对于任何$\alpha(n)$其是$o(1)$作为$n \to \infty$。使用上面关于保留对数等价的polylog函数的引理，我们可以将此表达式代入$(1)$以得到：

$$\log(\sqrt{2 \pi} \Xi _n)- \log n \sim -\log (1 + \alpha) - \frac{1}{2}\log 2 - \frac{1}{2}\log \log n - \log n - 2 \alpha \log n - \alpha^2 \log n \,.$$

$$\implies -\log(\Xi_n \sqrt{2 \pi}) \sim \log(1 + \alpha) + \frac{1}{2} \log 2 + \frac{1}{2} \log \log n + 2\alpha \log n + \alpha^2 \log n \,.$$

在这里，我们必须走得更远，并假设$\log \Xi_n = o( \log \log n) ~~ as ~~ n \to \infty$几乎可以肯定。同样，所有Cramer说是“假定$\Xi_n$为界”。但是，因为所有的人可以说先天约$\Xi_n$是$0 \le Xi_n \le n$因为，它几乎似乎很清楚，应该有$\Xi_n = O(1)$几乎可以肯定，这似乎是克莱默的要求的物质。

但是无论如何，假设有人认为，那么可以得出结论，不包含$\alpha$的主导项是$\frac{1}{2} \log \log n$。由于$\alpha = o(1)$，它遵循$\alpha^2 = o(\alpha)$，并清楚地$\log ( 1 + \alpha) = o (\alpha) = o(o(\alpha \log n))$，因此含有主项$\alpha$是$2 \alpha \log n$。因此，我们可以重新排列和（将所有内容除以$\frac{1}{2}\log\log n$或$2 \alpha \log n$）发现

$$- \frac{1}{2} \log \log n \sim 2 \alpha \log n \quad \implies \quad \alpha \sim - \frac{\log \log n}{4 \log n} \,.$$

因此，将其代入上面，我们得到：

$$Z_n \sim \sqrt{2 \log n}- \frac{\log\log n}{2 \sqrt{2 \log n}} \,,$$

再次，假设我们认为对某些事情$\Xi_n$。

我们再次重新提出同样的技术。因为$Z_n \sim \sqrt{2 \log n} - \frac{\log \log n}{2 \sqrt{2 \log n}}$，那么它也遵循

$$Z_n \sim \sqrt{2 \log n} - \frac{\log \log n}{2 \sqrt{2 \log n}} (1 + \beta(n)) = \sqrt{2 \log n} \left( 1 - \frac{\log \log n}{8 \log n}(1 + \beta(n)) \right) \,,$$

当$\beta(n)=o(1)$。在直接替换为（1）之前，让我们简化一下；我们得到：

$$\log Z_n \sim \log(\sqrt{2 \log n}) + \underbrace{\log \left(1 - \frac{\log \log n}{8 \log n}(1 + \beta(n)) \right) }_{\log(O(1)) = o(\log n)} \sim \log (\sqrt{2 \log n}) \,.$$

$$\frac{Z_n^2}{2} \sim \log n - \frac{1}{2} \log \log n (1 + \beta) + \underbrace{\frac{(\log \log n)^2}{8 \log n} ( 1 \beta)^2}_{o((1+ \beta) \log \log n)} \sim \log n - \frac{1}{2} (1 + \beta) \log \log n \,.$$

将其代入（1），我们发现：

$$\log ( \sqrt{2 \pi} \Xi_n) - \log n \sim - \log(\sqrt{2 \log n}) - \log n + \frac{1}{2}(1 + \beta) \log \log n \quad \implies \quad \beta \sim \frac{\log (4 \pi \Xi_n^2)}{\log \log n} \,.$$

Therefore, we conclude that almost surely

$$Z_n \sim \sqrt{2 \log n} - \frac{\log \log n}{2 \sqrt{2 \log n}} \left(1 + \frac{\log(4 \pi) + 2 \log( \Xi_n)}{\log \log n} \right)\\ = \sqrt{2 \log n} - \frac{\log \log n + \log (4 \pi)}{ 2 \sqrt{2 \log n} } - \frac{\log (\Xi_n)}{\sqrt{2 \log n}} \,.$$

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:

$$Z_n \sim \sqrt{2 \log n} - \frac{\log \log n + \log (4 \pi)}{2 \sqrt{2 \log n}} - \frac{\log (\Xi_n)}{\sqrt{2 \log n}} \quad \implies \\ Z_n = \sqrt{2 \log n} - \frac{\log \log n + \log (4 \pi)}{2 \sqrt{2 \log n}} - \frac{\log (\Xi_n)}{\sqrt{2 \log n}} + o(1)\,. \tag{$\dagger$}$$

2. Then by linearity of expectation it follows that:

$$\mathbb{E}Z_n = \sqrt{2 \log n} - \frac{\log \log n + \log (4 \pi)}{2 \sqrt{2 \log n}} - \frac{\mathbb{E}[\log (\Xi_n)]}{\sqrt{2 \log n}} + o(1) \quad \implies \\ \frac{\mathbb{E}Z_n}{\sqrt{2 \log n}} = 1 - \frac{\mathbb{E}[\log \Xi_n]}{2 \log n} + o(1) \,.$$

Therefore, we have shown that

$$\lim_{n \to \infty } \frac{\mathbb{E} Z_n}{\sqrt{2 \log n}} = 1 \,,$$

as long as we can also show that

$$\mathbb{E}[\log \Xi_n] = o(\log n) \,.$$

This might not be too difficult to show since again $\Xi_n$ 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:

$$\frac{Z_n}{\sqrt{2 \log n}} = 1 - \frac{\log(\Xi_n)}{2 \log n} +o(1),.$$

Therefore, if we can show that:

$$\log(\Xi_n) = o(\log n) \text{ almost surely}, \tag{*}$$

$\mathbb{E}[\log (\Xi_n)] = o(\log n)$, thereby also giving us the first result from above.

Also note that in the proof above of ($\dagger$) we needed to assume anyway that $\Xi_n = o(\log n)$ almost surely (or at least something similar), so that if we are able to show ($\dagger$) then we will most likely also have in the process needed to show $\Xi_n = o(\log n)$ almost surely, and therefore if we can prove $(\dagger)$ 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 $\mathbb{E}Z_n = \sqrt{2 \log n} + \Theta(1)$, since $o(1) \not= \Theta(1)$. But at the very least it would seem to be true that

$$\mathbb{E}Z_n = \sqrt{2 \log n} + O(1) \,.$$

---

So then it seems that we can focus on answering the question of how to show that

$$\Xi_n = o(\log n) \text{ almost surely.}$$

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 $\Xi_n \ge 0$):

$$\Xi_n = o(\log n) \quad \iff \quad \lim_{n \to \infty} \frac{\Xi_n}{\log n} = 0 \quad \iff \quad \\ \forall \varepsilon > 0, \frac{\Xi_n}{\log n} > \varepsilon \text{ only finitely many times} \quad \iff \\ \forall \varepsilon >0, \quad \Xi_n > \varepsilon \log n \text{ only finitely many times} \,.$$

One also has that:

$$\Xi_n > \varepsilon \log n \quad \iff \quad n(1 - F(Z_n)) > \varepsilon \log n \quad \iff \quad 1 - F(Z_n) > \frac{\varepsilon \log n}{n} \\ \iff \quad F(Z_n) < 1 - \frac{\varepsilon \log n}{n} \quad \iff \quad Z_n \le \inf \left\{ y: F(y) \ge 1 - \frac{\varepsilon \log n}{n} \right\} \,.$$

Correspondingly, define for all $n$:

$$u_n^{(\varepsilon)} = \inf \left\{ y: F(y) \ge 1 - \frac{\varepsilon \log n}{n} \right\} \,.$$

Therefore the above steps show us that:

$$\Xi_n = o(\log n) \text{ a.s.} \quad \iff \quad \mathbb{P}(\Xi_n = o(\log n)) = 1 \quad \iff \quad \\ \mathbb{P}(\forall \varepsilon > 0 , \Xi_n > \varepsilon \log n \text{ only finitely many times}) = 1 \\ \iff \mathbb{P}(\forall \varepsilon > 0, Z_n \le u_n^{(\varepsilon)} \text{ only finitely many times}) = 1 \\ \iff \mathbb{P}(\forall \varepsilon >0, Z_n \le u_n^{(\varepsilon)} \text{ infinitely often}) =0 \,.$$

Notice that we can write:

$$\{ \forall \varepsilon >0, Z_n \le u_n^{(\varepsilon)} \text{ infinitely often} \} = \bigcap_{\varepsilon > 0} \{ Z_n \le u_n^{(\varepsilon)} \text{ infinitely often} \} \,.$$

The sequences $u_n^{(\varepsilon)}$ become uniformly larger as $\varepsilon$ decreases, so we can conclude that the events

$$\{ Z_n \le u_n^{(\varepsilon)} \text{ infinitely often} \}$$ are decreasing (or at least somehow monotonic) as $\varepsilon$ goes to $0$. Therefore the probability axiom regarding monotonic sequences of events allows us to conclude that:

$$\mathbb{P}(\forall \varepsilon >0, Z_n \le u_n^{(\varepsilon)} \text{ infinitely often}) = \\ \mathbb{P} \left( \bigcap_{\varepsilon > 0} \{ Z_n \le u_n^{(\varepsilon)} \text{ infinitely often} \} \right) = \\ \mathbb{P} \left( \lim_{\varepsilon \downarrow 0} \{ Z_n \le u_n^{(\varepsilon)} \text{ infinitely often} \} \right) = \\ \lim_{\varepsilon \downarrow 0} \mathbb{P}(Z_n \le u_n^{(\varepsilon)} \text{ infinitely often}) \,.$$

Therefore it suffices to show that for all $\varepsilon >0$,

$$\mathbb{P}(Z_n \le u_n^{(\varepsilon)} \text{ infinitely often}) = 0$$

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 $X_1, X_2, \dots$ be i.i.d. variables with common nondegenerate and continuous distribution function $F(x)$, and let $u_n$ be a nondecreasing sequence such that $n(1 - F(u_n))$ is also nondecreasing. Then, for $u_n < \sup \{ x: F(x) <1 \}$,

$$\mathbb{P}(Z_n \le u_n \text{ infinitely often}) =0 \text{ or }1$$ according as $$\sum_{j=1}^{+\infty}[1 - F(u_j)]\exp(-j[1-F(u_j)]) < +\infty \text{ or }=+\infty \,.$$

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 $\omega(F)=+\infty$, so that condition is vacuous, and $n(1-F(n))$ is $\varepsilon \log n$ thus clearly non-decreasing.

Anyway the point being that, appealing to this theorem, if we can show that:

$$\sum_{j=1}^{+\infty}[1 - F(u_j^{(\varepsilon)})]\exp(-j[1-F(u_j^{(\varepsilon)})]) = \sum_{j=1}^{+\infty}\left[ \frac{\varepsilon \log j}{j} \right]\exp(-\varepsilon \log j) = \varepsilon \sum_{j=1}^{+\infty} \frac{ \log j}{j^{1 + \varepsilon}} < + \infty \,.$$

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 $\log \log n \le \alpha \log n \iff \log n \le n^{\alpha}$ and the former inequality can always be seen to hold for all $n$ large enough due to the fact that $\log n \le n$ and a change of variables), we have that:

$$\sum_{j=1}^{+\infty} \frac{\log j}{j^{1 + \varepsilon}} \le \sum_{j=1}^{+\infty} \frac{j^{\varepsilon/2}}{j^{1 + \varepsilon}} = \sum_{j=1}^{+\infty} \frac{1}{j^{1 + \varepsilon/2}} < +\infty \,,$$

since the p-series is known to converge for all $p>1$, and $\varepsilon >0$ of course implies $1 + \varepsilon/2 > 1$.

Thus using the above theorem we have shown that for all $\varepsilon >0$, $\mathbb{P}(Z_n \le u_n^{(\varepsilon)} \text{ i.o.}) = 0$, which to recapitulate should mean that $\Xi_n = o(\log n)$ almost surely.

We need to show still that $\log \Xi_n = o(\log \log n)$. This doesn't follow from the above, since, e.g.,

$$\frac{1}{n} \log n = o(\log n) \,, - \log n + \log \log n \not= o(\log n) \,.$$

However, given a sequence $x_n$, if one can show that $x_n = o( (\log n)^{\delta})$ for arbitrary $\delta >0$, then it does follow that $\log(x_n) = o(\log \log n)$. Ideally I would like to be able to show this for $\Xi_n$ using the above lemma (assuming it's even true), but am not able to (as of yet).