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2023阿里巴巴全球数学竞赛预选赛题/决赛部分题个人解 (六)

2023-06-26 09:34 作者:saqatl  | 我要投稿

概率与统计题 2. 对每个正整数 n,找出最大的实数 f(n) 使得以下性质成立:对每个 n%20%5Ctimes%20n 双随机矩阵 M(双随机矩阵是指所有元素均为非负实数且每行每列的元素之和均为 1 的方阵),都存在 %5Bn%5D%20%3D%20%5C%7B1%2C%5Cldots%2Cn%5C%7D 上的置换 %5Cpi 使得 M_%7Bi%2C%5Cpi(i)%7D%20%5Cge%20f(n) 对每个 i%20%5Cin%20%5Bn%5D 都成立。

首先说明答案:f(n)%20%3D%20%5Cmin%5Climits_M%20%5Cmax%5Climits_%7B%5Cpi%7D%20%5Cmin%5Climits_%7Bi%20%5Cin%20%5Bn%5D%7D%20M_%7Bi%2C%5Cpi(i)%7D%20%3D%20%5Cbegin%7Bcases%7D%0A%0A%5Cfrac%7B4%7D%7B(n%2B1)%5E2%7D%2C%20%26%20n%20%5Ctext%7B%20is%20odd%7D%5C%5C%0A%0A%5Cfrac%7B4%7D%7Bn(n%2B2)%7D%2C%20%26%20n%20%5Ctext%20%7B%20is%20even%7D%0A%0A%5Cend%7Bcases%7D

设 t%20%3D%20%5Cmax%5Climits_%7B%5Cpi%7D%20%5Cmin%5Climits_%7Bi%20%5Cin%20%5Bn%5D%7D%20M_%7Bi%2C%5Cpi(i)%7D,并记 A%20%3D%20%5C%7BM_%7Bij%7D%7CM_%7Bij%7D%20%5Cle%20t%5C%7DB%20%3D%20%5C%7BM_%7Bij%7D%7CM_%7Bij%7D%20%3E%20t%5C%7D,则每行、每列的 n 个元素中至少有一个属于集合 A。由 Hall 定理,M 中存在某个 k%20%5Ctimes%20(n%2B1-k) 子矩阵,其中每个元素均属于集合 A(不妨设该子阵位于 M 左上角),则前 k 行后 k-1 列元素之和不超过 k-1。因此前 k 行一定有某一行后 k-1 列元素和不超过 (k-1)%2Fk。从而 t(n%2B1-k)%20%5Cge%201%2Fk,即 t%20%5Cge%201%2Fk(n%2B1-k)%20%5Cge%20f(n)

下面再给出 M 的构造。当 n 为奇数时,取 M_%7Bij%7D%20%3D%20%5Cbegin%7Bcases%7D%0A%0A%5Cfrac%7B4%7D%7B(n%2B1)%5E2%7D%2C%20%26i%2Cj%20%5Cin%20%5C%7B1%2C%5Cldots%2C%5Cfrac%7Bn%2B1%7D%7B2%7D%5C%7D%5C%5C%0A%0A0%2C%20%26i%2Cj%20%5Cin%20%5C%7B%5Cfrac%7Bn%2B3%7D%7B2%7D%2C%20%5Cldots%2C%20n%5C%7D%5C%5C%0A%0A%5Cfrac%7B2%7D%7Bn%2B1%7D%2C%20%26%5Ctext%7Botherwise%7D%0A%0A%5Cend%7Bcases%7D。令 %5Cpi%20%3D%20(n%5C%20%5Cldots%5C%203%5C%202%5C%201) 可得 %5Cmin%5Climits_%7Bi%20%5Cin%20%5Bn%5D%7D%20M_%7Bi%2C%5Cpi(i)%7D%20%3D%20%5Cfrac%7B4%7D%7B(n%2B1)%5E2%7D。若存在 %5Cpi_0 使得 %5Cmin%5Climits_%7Bi%20%5Cin%20%5Bn%5D%7D%20M_%7Bi%2C%5Cpi_0(i)%7D%20%3E%20%5Cfrac%7B4%7D%7B(n%2B1)%5E2%7D,则前 (n%2B1)%2F2 行后 (n%2B1)%2F2 列中至少有 (n%2B1)%2F2 个元素在集合 %5C%7BM_%7Bi%2C%5Cpi_0(i)%7D%5C%7D_%7Bi%20%5Cin%20%5Bn%5D%7D 中,矛盾。因此 %5Cmax%5Climits_%7B%5Cpi%7D%20%5Cmin%5Climits_%7Bi%20%5Cin%20%5Bn%5D%7D%20M_%7Bi%2C%5Cpi(i)%7D%20%3D%20f(n)

当 n 为偶数时,取 M_%7Bij%7D%20%3D%20%5Cbegin%7Bcases%7D%0A%0A%5Cfrac%7B4%7D%7Bn(n%2B2)%7D%2C%20%26i%20%5Cin%20%5C%7B1%2C%5Cldots%2C%5Cfrac%7Bn%2B2%7D%7B2%7D%5C%7D%2C%20j%20%5Cin%20%5C%7B1%2C%5Cldots%2C%5Cfrac%7Bn%7D%7B2%7D%5C%7D%5C%5C%0A%0A%5Cfrac%7B2%7D%7Bn%7D%2C%20%26i%20%5Cin%20%5C%7B%5Cfrac%7Bn%2B4%7D%7B2%7D%2C%5Cldots%2Cn%5C%7D%2C%20j%20%5Cin%20%5C%7B1%2C%5Cldots%2C%5Cfrac%7Bn%7D%7B2%7D%5C%7D%5C%5C%0A%0A%5Cfrac%7B2%7D%7Bn%2B2%7D%2C%20%26i%20%5Cin%20%5C%7B1%2C%5Cldots%2C%5Cfrac%7Bn%2B2%7D%7B2%7D%5C%7D%2C%20j%20%5Cin%20%5C%7B%5Cfrac%7Bn%2B2%7D%7B2%7D%2C%5Cldots%2Cn%5C%7D%5C%5C%0A%0A0%2C%20%26%5Ctext%7Botherwise%7D%0A%0A%5Cend%7Bcases%7D。同理可以证明 %5Cmax%5Climits_%7B%5Cpi%7D%20%5Cmin%5Climits_%7Bi%20%5Cin%20%5Bn%5D%7D%20M_%7Bi%2C%5Cpi(i)%7D%20%3D%20f(n)

经过分析不难看出这是一个相异代表系,而对此本人只知道 Hall 定理,所幸这样问题就可以做出来。

概率与统计题 3. n 为正整数。对 (u_1%2C%20%5Cldots%2C%20u_n)%20%5Cin%20%5B0%2C1%5D%5En 定义

S(u_1%2C%20%5Cldots%2C%20u_n)%20%3D%20%5Cmin%5Climits_%7Bi%20%5Cin%20%5Bn%5D%7D%20%5Cfrac%7Bn%7D%7Bi%7Du_%7B(i)%7D%2C%20%5Cquad%20M_r(u_1%2C%20%5Cldots%2C%20u_n)%20%3D%20%5Cleft(%5Cfrac%7B1%7D%7Bn%7D%20%5Csum%5Climits_%7Bi%3D1%7D%5En%20u_i%5Er%5Cright)%5E%7B1%2Fr%7D

其中 u_%7B(i)%7D 是 u_i%2C%5Cldots%2Cu_n 中第 i 个最小的数,实数 r%20%5Cne%200。定义一个标准均匀变量是随机变量 U 满足 %5Cmathbb%7BP%7D(U%20%5Cle%20%5Calpha)%20%3D%20%5Calpha 对所有 %5Calpha%20%5Cin%20(0%2C1) 成立,一个次均匀变量是随机变量 U 满足 %5Cmathbb%7BP%7D(U%20%5Cle%20%5Calpha)%20%5Cge%20%5Calpha 对所有 %5Calpha%20%5Cin%20(0%2C1) 成立。

令 U_1%2C%20%5Cldots%2C%20U_n 为一组独立的标准均匀变量,证明:

(a) S(U_1%2C%20%5Cldots%2C%20U_n) 为标准均匀变量。

(b) M_r(U_1%2C%20%5Cldots%2C%20U_n) 为次均匀变量当且仅当 r%20%5Cle%20-1

本题是对 Bonferroni 校正的优化,(a) 问是经典结论,(b) 问不知道从哪里来的。

(a) 用数学归纳法证明。当 n%20%3D%201 时结论显然成立;当 n%20%3D%202 时,若 S(U_1%2C%20U_2)%20%5Cle%20%5Calpha,则要么二者有其一不大于 %5Calpha%2F2,此时 %5Cmathbb%7BP%7D_1%20%3D%20%5Cfrac%7B%5Calpha%7D%7B2%7D%20%2B%20%5Cleft(1%20-%20%5Cfrac%7B%5Calpha%7D%7B2%7D%5Cright)%20%5Ccdot%20%5Cfrac%7B%5Calpha%7D%7B2%7D%20%3D%20%5Calpha%20-%20%5Cfrac%7B%5Calpha%5E2%7D%7B4%7D;要么二者均大于 %5Calpha%2F2,此时 %5Cmathbb%7BP%7D_2%20%3D%20%5Cleft(%5Cfrac%7B%5Calpha%7D%7B2%7D%5Cright)%5E2%20%3D%20%5Cfrac%7B%5Calpha%5E2%7D%7B4%7D。因此 %5Cmathbb%7BP%7D(S(U_1%2C%20U_2)%20%5Cle%20%5Calpha)%20%3D%20%5Cmathbb%7BP%7D_1%20%2B%20%5Cmathbb%7BP%7D_2%20%3D%20%5Calpha

若结论对 1%20%5Cle%20n%20%5Cle%20k 均成立。当 n%20%3D%20k%20%2B%201 时,设 U_1%2C%20U_2%2C%20%5Cldots%2C%20U_n 中有 m 个数不大于 %5Calpha,这一概率为 %5Cmathbb%7BP%7D_%7B1m%7D%20%3D%20%5Cmathrm%7BC%7D_n%5Em%20%5Calpha%5Em%20(1%20-%20%5Calpha)%5E%7Bn-m%7D。将这 m 个数从小到大排序为 U_1'%2C%20%5Cldots%2C%20U_m',则 S(U_1%2C%20%5Cldots%2C%20U_n)%20%5Cle%20%5Calpha 当且仅当存在 1%20%20%5Cle%20i%20%5Cle%20m 使得 U_i'%20%5Cle%20%5Cfrac%7Bi%5Calpha%7D%7Bn%7D

当 m 确定时,令 U_i''%20%3D%20U_i'%20%2F%20%5Calpha,则 U_1''%20%3C%20%5Ccdots%20%3C%20U_m'' 为 %5B0%2C1%5D 上的均匀分布,且存在 1%20%20%5Cle%20i%20%5Cle%20m 使得 U_i''%20%5Cle%20%5Cfrac%7Bi%7D%7Bn%7D。对这 m 个数和 %5Calpha_1%20%3D%20m%2Fn 使用归纳假设可知此时满足 S(U_1%2C%20%5Cldots%2C%20U_n)%20%5Cle%20%5Calpha 的概率为 %5Cmathbb%7BP%7D_%7B2m%7D%20%3D%20%5Cfrac%7Bm%7D%7Bn%7D。故 %5Cmathbb%7BP%7D_m%20%3D%20%5Cmathbb%7BP%7D_%7B1m%7D%20%5Ccdot%20%5Cmathbb%7BP%7D_%7B2m%7D%20%3D%20%5Cmathrm%7BC%7D_n%5Em%20%5Calpha%5Em%20(1%20-%20%5Calpha)%5E%7Bn-m%7D%20%5Cfrac%7Bm%7D%7Bn%7D

因此

%5Cmathbb%7BP%7D(S(U_1%2C%20%5Cldots%2C%20U_n)%20%5Cle%20%5Calpha)%20%3D%20%5Csum%5Climits_%7Bi%3D1%7D%5En%20%5Cmathbb%7BP%7D_i%20%3D%20%5Csum%5Climits_%7Bi%3D1%7D%5En%20%5Cmathrm%20%7BC%7D_n%5Ei%20%5Calpha%5Ei%20(1%20-%20%5Calpha)%5E%7Bn-1%7D%20%5Ccdot%20%5Cfrac%7Bi%7D%7Bn%7D

其中,对于 2%20%5Cle%20j%20%5Cle%20n%5Calpha_j 的系数为

%5Cbegin%7Baligned%7D%0A%0A%26%5Csum%5Climits_%7Bi%20%3D%200%7D%5Ej%20%5Cmathrm%7BC%7D_n%5E%7Bj%20-%20i%7D%20%5Ccdot%20%5Cmathrm%7BC%7D_%7Bn-j%2Bi%7D%5Ei%20%5Ccdot%20%5Cfrac%7Bj-i%7D%7Bn%7D%20%5Ccdot%20(-1)%5Ei%20%5C%5C%0A%0A%3D%20%26%20%5Csum%5Climits_%7Bi%20%3D%200%7D%5E%7Bj-1%7D%20%5Cmathrm%7BC%7D_%7Bn-1%7D%5E%7Bj%20-%20i%20-%201%7D%20%5Ccdot%20%5Cmathrm%7BC%7D_%7Bn-j%2Bi%7D%5Ei%20%5Ccdot%20(-1)%5Ei%5C%5C%0A%0A%3D%20%26%20%5Cmathrm%7BC%7D_%7Bn-1%7D%5E%7Bj-1%7D%20%5Ccdot%20%5Csum%5Climits_%7Bi%20%3D%200%7D%5E%7Bj-1%7D%20(-1)%5Ei%20%5Cmathrm%7BC%7D_%7Bj-1%7D%5Ei%5C%5C%0A%0A%3D%20%260%0A%0A%5Cend%7Baligned%7D

因此 %5Cmathbb%7BP%7D(S(U_1%2C%20%5Cldots%2C%20U_n)%20%5Cle%20%5Calpha)%20%3D%20%5Calpha

(b) 比较奇怪,积分放缩了半天对 -1%20%3C%20r%20%3C%200 出不来。而且这个题的结论对 n%20%3D%201 直接就是错的,不太清楚这题怎么搞。

这专栏写公式太痛苦了!以后不写了,再见


2023阿里巴巴全球数学竞赛预选赛题/决赛部分题个人解 (六)的评论 (共 条)

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