投掷一枚均匀硬币 n 次,如果第 i 次投掷和第 j 次投掷出现同一面,则令 Xij=1, 否则令 Xij=0。证明:Xij(i<j)两两独立但不相互独立。
时间: 2023-06-09 16:08:27 浏览: 103
好的,我明白了,您的第一个问题是有关硬币投掷的概率论问题。现在,我将针对这个问题作出解答。
首先,显然,Xij只能取0或1。设X =(Xij)为n×n矩阵,它的对角线上元素都是0。那么X的非对角线上的元素为随机变量,我们需要证明这些随机变量是两两独立的。
我们假设i < p < q < j,并分别考虑Xi p, Xpq和Xqj的相关性。显然,如果Xi p = 0,则Xpq和Xqj必须都是0;如果Xi p = 1,则Xpq和Xqj必须不相等(即一个为0,一个为1)。因此,我们可以得到以下三个条件概率:
P(Xpq = 0 | Xi p = 0) = 1
P(Xqj = 0 | Xpq = 0, Xi p = 0) = 1
P(Xqj ≠ Xpq | Xi p = 1) = 1
由此可以得到:
P(Xpq = 0, Xqj = 0 | Xi p = 0) = P(Xpq = 0 | Xi p = 0) × P(Xqj = 0 | Xpq = 0, Xi p = 0) = 1
P(Xpq = 1, Xqj = 0 | Xi p = 0) = P(Xpq = 1 | Xi p = 0) × P(Xqj = 0 | Xpq = 1, Xi p = 0) = 0
P(Xpq = 0, Xqj = 1 | Xi p = 0) = P(Xpq = 0 | Xi p = 0) × P(Xqj = 1 | Xpq=0, Xi p = 0) = 0
P(Xpq = 1, Xqj = 1 | Xi p = 0) = P(Xpq = 1 | Xi p = 0) × P(Xqj ≠ Xpq | Xi p = 0) = 0
P(Xpq = 0, Xqj = 0 | Xi p = 1) = P(Xpq = 0 | Xi p = 1) × P(Xqj = 0 | Xpq = 0, Xi p = 1) = 1/4
P(Xpq = 1, Xqj = 0 | Xi p = 1) = P(Xpq = 1 | Xi p = 1) × P(Xqj = 0 | Xpq = 1, Xi p = 1) = 1/4
P(Xpq = 0, Xqj = 1 | Xi p = 1) = P(Xpq = 0 | Xi p = 1) × P(Xqj = 1 | Xpq = 0, Xi p = 1) = 1/4
P(Xpq = 1, Xqj = 1 | Xi p = 1) = P(Xpq = 1 | Xi p = 1) × P(Xqj ≠ Xpq | Xi p = 1) = 1/4
从上面的概率可以看出,Xpq和Xqj在已知Xi p的条件下是不相互独立的。但是,当我们去掉Xi p的条件时,Xpq和Xqj是独立的。因此,Xij(i<j)两两独立但不相互独立。
这就是该问题的证明过程。希望我的解答能够满足您的需要。如果您有任何其他问题,欢迎随时提出,我将尽我所能给出解答。