An iterated logarithm law for the maximum in a stationary gaussian sequence
An iterated logarithm law for the maximum in a stationary gaussian sequence
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平稳高斯序列中最大值的迭代对数定律
DOI：
10.1007/bf00538755
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发表时间：
1969
影响因子：
0.9
通讯作者：
James PickandsIII
中科院分区：
文献类型：
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作者：
James PickandsIII
设{Xn，n=1，2，⋯}是离散坐标平稳高斯随机过程的连续项。在不失一般性的前提下，假设所有n的exn=0且ro=exn2=1.设rn≡ex k Xk+n为协方差函数.如果存在α&>0，使得
$$\mathop{\lim}\Limits_{n\to\Infty}n^\Alpha r_n=0，或\sum\Limits_{n=-\inty}^\inty{r_n^2<；\infty，}$$
然后
$$P\{\mathop{\lim inf}\Limits_{n\to\inty}(2\log n)^{\tfrac{1}{2}}(Z_n-(2\log n)^{\tfrac{1}{2}})/\log n=-\tfrac{1}{2}，\mathop{limsup}\Limits_{n\to\inty}(2\log n)^{\tfrac{1}{2}}(Z_n-(2\log n)^{\tfrac{1}{2}})/\log n=-\tfrac{1}{2}\}=1，$$
哪里
$$Z_n\EQUV\Mathop{Sup}\Limits_{1\leqq k\leqq n}X_k.$$
这是不够的，
$$\Mathop{LIM}\Limits_{n\to\inty}r_n=0.$$
Let {X n , n=1, 2, ⋯} be the successive terms of a discrete coordinate stationary Gaussian stochastic process. Assume, without loss of generality, that EX n =0 and r o =EX n 2 =1 for all n. Let r n ≡EX k X k+n be the covariance function. If either there exists an α>0 such that
$$\mathop {\lim }\limits_{n \to \infty } n^\alpha r_n = 0, or \sum\limits_{n = - \infty }^\infty {r_n^2 < \infty ,}$$
then
$$P\{ \mathop {\lim inf}\limits_{n \to \infty } (2\log n)^{\tfrac{1}{2}} (Z_n - (2\log n)^{\tfrac{1}{2}} )/\log log n = - \tfrac{1}{2}, \mathop {limsup}\limits_{n \to \infty } (2\log n)^{\tfrac{1}{2}} (Z_n - (2\log n)^{\tfrac{1}{2}} )/\log log n = - \tfrac{1}{2}\} = 1,$$
where
$$Z_n \equiv \mathop {Sup}\limits_{1 \leqq k \leqq n} X_k .$$
It is not sufficient that
$$\mathop {lim}\limits_{n \to \infty } r_n = 0.$$