复习笔记Day108

2023-02-26 20:53 作者:间宫_卓司 0人读过 | 我要投稿

这几天在看应坚钢的概率论，一开始选这本书的原因倒不是因为我知道这本书的观点比较高（相较于我看过的其他概率论课本来说），不过看看倒也无所谓···因为概率论放在复试的面试部分，所以我不怎么打算做题，主要是基础知识要梳理清楚。过几天如果我能读的下去的话，我会在读完这本书后把这本书的大体框架写成笔记发上来，不过应该会跳着读了，一些完全不可能问到的地方就跳过了。

虽然说不怎么做题，但是因为在习题里面看到了之前做过的数分题，所以还是打算试一下

108.1 利用 $%5Ctext%7BBernoulli%7D$ 的大数定律证明 $%5Ctext%7BWeierstrass%7D$ 定理：设 $f$ 是 $%5B0%2C1%5D$ 上的连续函数，定义 $%5Ctext%7BBernstein%7D$ 多项式

$f_n%5Cleft(%20x%20%5Cright)%20%3D%5Csum_%7Bk%3D0%7D%5En%7B%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09n%5C%5C%0A%09k%5C%5C%0A%5Cend%7Barray%7D%20%5Cright)%20f%5Cleft(%20%5Cfrac%7Bk%7D%7Bn%7D%20%5Cright)%7Dx%5Ek%5Cleft(%201-x%20%5Cright)%20%5E%7Bn-k%7D%2Cx%5Cin%20%5Cleft%5B%200%2C1%20%5Cright%5D%20$

则 $f_n$ 在 $%5B0%2C1%5D$ 上一致收敛于 $f$

首先回顾一下 $%5Ctext%7BBernoulli%7D$ 的大数定律

此外还可以知道， $%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20_n%3Dk%20%5Cright)%20%3D%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09n%5C%5C%0A%09k%5C%5C%0A%5Cend%7Barray%7D%20%5Cright)%20p%5Ek%5Cleft(%201-p%20%5Cright)%20%5E%7Bn-k%7D$ ，为了凑出 $%5Ctext%7BBernstein%7D$ 多项式的形式，先试试看把 $p$ 换成 $x$ 。这个时候， $%5Ctext%7BBernstein%7D$ 多项式就变成了

$f_n%5Cleft(%20x%20%5Cright)%20%3D%5Csum_%7Bk%3D0%7D%5En%7B%5Cleft(%20%5Cbegin%7Barray%7D%7Bc%7D%0A%09n%5C%5C%0A%09k%5C%5C%0A%5Cend%7Barray%7D%20%5Cright)%20%5Cfrac%7Bk%7D%7Bn%7D%7Dx%5Ek%5Cleft(%201-x%20%5Cright)%20%5E%7Bn-k%7D%2Cx%5Cin%20%5Cleft%5B%200%2C1%20%5Cright%5D%20$

这就是 $%5Cfrac%7B%5Cmathbb%7BE%7D%20%5Cxi%20_n%7D%7Bn%7D$ ，那么现在来估计一下 $%5Cleft%7C%20%5Cfrac%7B%5Cmathbb%7BE%7D%20%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C$

$%5Cbegin%7Baligned%7D%0A%09%26%5Cleft%7C%20%5Cfrac%7B%5Cmathbb%7BE%7D%20%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C%5Cle%20%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft%7C%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C%20%5Cright)%5C%5C%0A%09%26%3D%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft%7C%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C%3B%5Cleft%5C%7B%20%5Cleft%7C%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C%5Cle%20%5Cvarepsilon%20%5Cright%5C%7D%20%5Cright)%20%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cleft%7C%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C%5Cle%20%5Cvarepsilon%20%5Cright)%20%0A%5C%5C%26%2B%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft%7C%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C%3B%5Cleft%5C%7B%20%5Cleft%7C%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C%3E%5Cvarepsilon%20%5Cright%5C%7D%20%5Cright)%20%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cleft%7C%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C%3E%5Cvarepsilon%20%5Cright)%5C%5C%0A%09%26%5Cle%20%5Cvarepsilon%20%2B%5Cfrac%7BM%7D%7B4%5Cvarepsilon%20%5E2n%7D%5C%5C%0A%5Cend%7Baligned%7D$

这里的 $%5Cmathbb%7BE%7D(X%3BA)$ 代表把样本空间限制在 $A$ 上对随机变量 $X$ 去求期望

其中的 $M%3D%5Cmax%20%5Cleft(%20%5Cleft%7C%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C%20%5Cright)%20$ ，分母的估计来源于 $%5Ctext%7BBernoulli%7D$ 大数定律的证明，从上面的估计可以看出，对于与 $x$ 无关的充分大的 $n$ ，会成立 $%5Cleft%7C%20%5Cfrac%7B%5Cmathbb%7BE%7D%20%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C%5Cle%202%5Cvarepsilon%20$ ，也就是 $%5Cleft%7C%20f_n%5Cleft(%20x%20%5Cright)%20-x%20%5Cright%7C%3C2%5Cvarepsilon%20$

对于一般的情况来说，从上面的证明可以看出来，实际上就是要估计

$%5Cleft%7C%20%5Cmathbb%7BE%7D%20%5Cleft(%20f%5Cleft(%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D%20%5Cright)%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20%5Cright%7C$

为了估计这个数，试着把它往上面的式子的形式上凑。因为 $f(x)$ 在 $%5B0%2C1%5D$ 上连续，所以它在 $%5B0%2C1%5D$ 上一致连续，也就是说 $%5Cforall%20%5Cvarepsilon%20%3E0$ ， $%5Cexists%20%5Cdelta%20%3E0$ ， $%5Cforall%20%5Cleft%20%7Cx-y%20%5Cright%7C%3C%5Cdelta$ 且 $x%2Cy%5Cin%5B0%2C1%5D$ ，都成立 $%5Cleft%7C%20f%5Cleft(%20x%20%5Cright)%20-f%5Cleft(%20y%20%5Cright)%20%5Cright%7C%3C%5Cvarepsilon%20$ ，那么

$%5Cbegin%7Baligned%7D%0A%09%26%5Cleft%7C%20%5Cmathbb%7BE%7D%20%5Cleft(%20f%5Cleft(%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D%20%5Cright)%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20%5Cright%7C%3D%5Cleft%7C%20%5Cmathbb%7BE%7D%20%5Cleft(%20f%5Cleft(%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20%5Cright)%20%5Cright%7C%5C%5C%0A%09%26%5Cle%20%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft%7C%20f%5Cleft(%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20%5Cright%7C%20%5Cright)%5C%5C%0A%09%26%3D%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft%7C%20f%5Cleft(%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20%5Cright%7C%3B%5Cleft%5C%7B%20%5Cleft%7C%20f%5Cleft(%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20%5Cright%7C%5Cle%20%5Cvarepsilon%20%5Cright%5C%7D%20%5Cright)%20%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cleft%7C%20f%5Cleft(%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20%5Cright%7C%5Cle%20%5Cvarepsilon%20%5Cright)%5C%5C%0A%09%26%2B%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft%7C%20f%5Cleft(%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20%5Cright%7C%3B%5Cleft%5C%7B%20%5Cleft%7C%20f%5Cleft(%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20%5Cright%7C%3E%5Cvarepsilon%20%5Cright%5C%7D%20%5Cright)%20%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cleft%7C%20f%5Cleft(%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20%5Cright%7C%3E%5Cvarepsilon%20%5Cright)%5C%5C%0A%09%26%5Cle%20%5Cvarepsilon%20%2B%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft%7C%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C%3B%5Cleft%5C%7B%20%5Cleft%7C%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C%3E%5Cdelta%20%5Cright%5C%7D%20%5Cright)%20%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cleft%7C%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C%3E%5Cdelta%20%5Cright)%5C%5C%0A%09%26%5Cle%20%5Cvarepsilon%20%2B%5Cfrac%7BM%7D%7B4%5Cdelta%20%5E2n%7D%5C%5C%0A%5Cend%7Baligned%7D$

然后和上面一样可以导出结论，其中第二个不等号是因为

$%5Cleft%5C%7B%20%5Cleft%7C%20f%5Cleft(%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D%20%5Cright)%20-f%5Cleft(%20x%20%5Cright)%20%5Cright%7C%3E%5Cvarepsilon%20%5Cright%5C%7D%20%5Csubset%20%5Cleft(%20%5Cleft%7C%20%5Cfrac%7B%5Cxi%20_n%7D%7Bn%7D-x%20%5Cright%7C%3E%5Cdelta%20%5Cright)%20$

这个问题的数分证法见陈纪修，比这个复杂很多

108.2 设随机变量 $%5Cxi$ 是标准化的，证明：

对 $x%3E0$ ， $%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20%5Cge%20x%20%5Cright)%20%5Cle%20%5Cfrac%7B1%7D%7B1%2Bx%5E2%7D$ 。

此不等式不能再改进，即存在标准化的随机变量 $%5Cxi$ ，使得 $%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20%5Cge%20x%20%5Cright)%20%3D%5Cfrac%7B1%7D%7B1%2Bx%5E2%7D$

（提示：先利用 $%5Ctext%7BChebyshev%7D$ 的方法找到适当的函数证明 $%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20%3Ex%20%5Cright)%20%3D%5Cfrac%7B1%2Ba%5E2%7D%7B%5Cleft(%20x%2Ba%20%5Cright)%20%5E2%7D$ ，然后取适当的 $a$ ）

按照提示探索了一段时间后，可知

$%5Cmathbb%7BP%7D%20%5Cleft(%20%5Cxi%20%3Ex%20%5Cright)%20%3D%5Cmathbb%7BE%7D%20%5Cleft(%201%3B%5Cleft%5C%7B%20%5Cxi%20%3Ex%20%5Cright%5C%7D%20%5Cright)%20%5Cle%20%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft(%20%5Cfrac%7B%5Cxi%20%2Ba%7D%7Bx%2Ba%7D%20%5Cright)%20%5E2%3B%5Cleft%5C%7B%20%5Cxi%20%3Ex%20%5Cright%5C%7D%20%5Cright)%20%5Cle%20%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft(%20%5Cfrac%7B%5Cxi%20%2Ba%7D%7Bx%2Ba%7D%20%5Cright)%20%5E2%20%5Cright)%20$

计算可知 $%5Cmathbb%7BE%7D%20%5Cleft(%20%5Cleft(%20%5Cfrac%7B%5Cxi%20%2Ba%7D%7Bx%2Ba%7D%20%5Cright)%20%5E2%20%5Cright)%20%3D%5Cfrac%7B1%2Ba%5E2%7D%7B%5Cleft(%20x%2Ba%20%5Cright)%20%5E2%7D$ （这里之前一直把标准化的随机变量的期望当成1了，结果半天出不来这个结果）

接下来，因为 $%5Cfrac%7B%5Cpartial%7D%7B%5Cpartial%20a%7D%5Cln%20%5Cleft(%20%5Cfrac%7B1%2Ba%5E2%7D%7B%5Cleft(%20x%2Ba%20%5Cright)%20%5E2%7D%20%5Cright)%20%3D%5Cfrac%7B2a%7D%7B1%2Ba%5E2%7D-%5Cfrac%7B2%7D%7Bx%2Ba%7D$ ，易知 $xa%3D1$ 时这个关于 $a$ 的函数有最小值，带入可得