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2023阿里巴巴全球數(shù)學(xué)競賽預(yù)選賽題/決賽部分題個人解 (六)

2023-06-26 09:34 作者:saqatl  | 我要投稿

概率與統(tǒng)計題 2.?對每個正整數(shù)?n,找出最大的實數(shù)?f(n)?使得以下性質(zhì)成立:對每個?n%20%5Ctimes%20n?雙隨機(jī)矩陣?M(雙隨機(jī)矩陣是指所有元素均為非負(fù)實數(shù)且每行每列的元素之和均為?1?的方陣),都存在?%5Bn%5D%20%3D%20%5C%7B1%2C%5Cldots%2Cn%5C%7D?上的置換?%5Cpi?使得?M_%7Bi%2C%5Cpi(i)%7D%20%5Cge%20f(n)?對每個?i%20%5Cin%20%5Bn%5D?都成立。

首先說明答案: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。

設(shè)?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,并記?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,則每行、每列的?n?個元素中至少有一個屬于集合?A。由 Hall?定理,M?中存在某個?k%20%5Ctimes%20(n%2B1-k)?子矩陣,其中每個元素均屬于集合?A(不妨設(shè)該子陣位于?M?左上角),則前?k?行后?k-1?列元素之和不超過?k-1。因此前?k?行一定有某一行后?k-1?列元素和不超過?(k-1)%2Fk。從而?t(n%2B1-k)%20%5Cge%201%2Fk,即?t%20%5Cge%201%2Fk(n%2B1-k)%20%5Cge%20f(n)。

下面再給出?M?的構(gòu)造。當(dāng)?n?為奇數(shù)時,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,則前?(n%2B1)%2F2?行后?(n%2B1)%2F2?列中至少有?(n%2B1)%2F2?個元素在集合?%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)。

當(dāng)?n?為偶數(shù)時,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。同理可以證明?%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)。

經(jīng)過分析不難看出這是一個相異代表系,而對此本人只知道 Hall?定理,所幸這樣問題就可以做出來。

概率與統(tǒng)計題 3.?n?為正整數(shù)。對?(u_1%2C%20%5Cldots%2C%20u_n)%20%5Cin%20%5B0%2C1%5D%5En?定義

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?個最小的數(shù),實數(shù)?r%20%5Cne%200。定義一個標(biāo)準(zhǔn)均勻變量是隨機(jī)變量?U?滿足?%5Cmathbb%7BP%7D(U%20%5Cle%20%5Calpha)%20%3D%20%5Calpha?對所有?%5Calpha%20%5Cin%20(0%2C1)?成立,一個次均勻變量是隨機(jī)變量?U?滿足?%5Cmathbb%7BP%7D(U%20%5Cle%20%5Calpha)%20%5Cge%20%5Calpha?對所有?%5Calpha%20%5Cin%20(0%2C1)?成立。

令?U_1%2C%20%5Cldots%2C%20U_n?為一組獨立的標(biāo)準(zhǔn)均勻變量,證明:

(a)?S(U_1%2C%20%5Cldots%2C%20U_n)?為標(biāo)準(zhǔn)均勻變量。

(b)?M_r(U_1%2C%20%5Cldots%2C%20U_n)?為次均勻變量當(dāng)且僅當(dāng)?r%20%5Cle%20-1。

本題是對 Bonferroni?校正的優(yōu)化,(a) 問是經(jīng)典結(jié)論,(b) 問不知道從哪里來的。

(a) 用數(shù)學(xué)歸納法證明。當(dāng)?n%20%3D%201?時結(jié)論顯然成立;當(dāng)?n%20%3D%202?時,若?S(U_1%2C%20U_2)%20%5Cle%20%5Calpha,則要么二者有其一不大于?%5Calpha%2F2,此時?%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,此時?%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。

若結(jié)論對?1%20%5Cle%20n%20%5Cle%20k?均成立。當(dāng)?n%20%3D%20k%20%2B%201?時,設(shè)?U_1%2C%20U_2%2C%20%5Cldots%2C%20U_n?中有?m?個數(shù)不大于?%5Calpha,這一概率為?%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。將這?m?個數(shù)從小到大排序為?U_1'%2C%20%5Cldots%2C%20U_m',則?S(U_1%2C%20%5Cldots%2C%20U_n)%20%5Cle%20%5Calpha?當(dāng)且僅當(dāng)存在?1%20%20%5Cle%20i%20%5Cle%20m?使得?U_i'%20%5Cle%20%5Cfrac%7Bi%5Calpha%7D%7Bn%7D。

當(dāng)?m?確定時,令?U_i''%20%3D%20U_i'%20%2F%20%5Calpha,則?U_1''%20%3C%20%5Ccdots%20%3C%20U_m''?為?%5B0%2C1%5D?上的均勻分布,且存在?1%20%20%5Cle%20i%20%5Cle%20m?使得?U_i''%20%5Cle%20%5Cfrac%7Bi%7D%7Bn%7D。對這?m?個數(shù)和?%5Calpha_1%20%3D%20m%2Fn?使用歸納假設(shè)可知此時滿足?S(U_1%2C%20%5Cldots%2C%20U_n)%20%5Cle%20%5Calpha?的概率為?%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,對于?2%20%5Cle%20j%20%5Cle%20n,%5Calpha_j?的系數(shù)為

%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) 比較奇怪,積分放縮了半天對?-1%20%3C%20r%20%3C%200?出不來。而且這個題的結(jié)論對?n%20%3D%201?直接就是錯的,不太清楚這題怎么搞。

這專欄寫公式太痛苦了!以后不寫了,再見


2023阿里巴巴全球數(shù)學(xué)競賽預(yù)選賽題/決賽部分題個人解 (六)的評論 (共 條)

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