一道大学概率的题 the pair (X,Y) is chosen uniformly from the set of

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一道大学概率的题
the pair (X,Y) is chosen uniformly from the set of pairs of integers (k,l) satisfying k>=1,L>=1,and k+l

先算出有多少种可能：
X=1,Y=1,...,n-1 共n-1种可能
X=2,Y=1,...,n-2 共n-2种可能
...
X=n-2,Y=1,2共2种可能
X=n-1,Y=1共1种可能
故有1+2+...+n-1共n(n-1)/2种可能
a) P(X=k) = (n-k)/ [n(n-1)/2] =2(n-k)/n(n-1)
b) E(XY) = 加总P(X=k) * E(XY|X=k) (加总k=1,2,...,n-1)
=加总 2(n-k)/n(n-1) * k(1+...+n-k)/(n-k) =加总 2/n(n-1) * k(n-k)(n-k+1)/2
= 1/n(n-1) 加总 k[(n-k)^2 + (n-k)]
后面需要展开然后对带k的项求和,利用自然数平方和和自然数立方和公式.
再问：能不能解释一下为什么E(XY|X=k)= k(1+...+n-k)/(n-k)? 非常感谢。。周三考试…………哎
再答： E(XY|X=k)就是给定X=k这一条件，求XY的期望。在给定X=k时，Y只能是1,2,...n-k中一种，且可能性都相等，也就是说每种可能性的概率为1/n-k，所以此时XY的期望就为 (k*1 + k*2 + ... + k*(n-k))/(n-k)

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