概率生成函数
随机变量
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X 的概率生成函数为：
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F(z)=\sum_{i=0}^\infty P(X=i)z^i
F(z)=i=0∑∞P(X=i)zi
均值与方差
E ( X ) = F ′ ( 1 ) E(X)=F'(1) E(X)=F′(1)
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V(X)=E(X^2)-E(X)^2 =F''(1)+F'(1)-F'(1)^2
V(X)=E(X2)−E(X)2=F′′(1)+F′(1)−F′(1)2
Proof:
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E(X^2)=\sum_{i=0}^\infty P(X=i)i^2=\sum_{i=0}^\infty P(X=i)i(i-1)z^{i-2}+iz^{i-1}=F''(1)+F'(1)
E(X2)=i=0∑∞P(X=i)i2=i=0∑∞P(X=i)i(i−1)zi−2+izi−1=F′′(1)+F′(1)
然后就可以做题了…
1
十连测：http://zhengruioi.com/contest/269/problem/637
期望是一个无穷级数的形式，所以考虑概率生成函数。设第 i 号桶中年份的概率生成函数为 F i ( z ) = ∑ i = 0 ∞ P ( X = i ) z i F_i(z)=\sum_{i=0}^\infty P(X=i)z^i Fi(z)=∑i=0∞P(X=i)zi。我们想要求方差，由于 V ( X ) = E ( X 2 ) − E ( X ) 2 V(X)=E(X^2)-E(X)^2 V(X)=E(X2)−E(X)2，并且根据题意年份为 i 的酒价值为 i·z^i，因此 E ( X ) E(X) E(X) 就等于 F ′ ( z ) ⋅ z F'(z)·z F′(z)⋅z （z 是常数），并且 E ( X 2 ) = ∑ i = 0 ∞ P ( X = i ) i 2 z 2 i = ∑ i = 0 ∞ P ( X = i ) i ( i − 1 ) z 2 i + i z 2 i = F ′ ′ ( z 2 ) z 4 ⋅ F ′ ( z 2 ) ⋅ z 2 E(X^2)=\sum_{i=0}^\infty P(X=i)i^2z^{2i}=\sum_{i=0}^\infty P(X=i)i(i-1)z^{2i}+iz^{2i}=F''(z^2)z^4·F'(z^2)·z^2 E(X2)=∑i=0∞P(X=i)i2z2i=∑i=0∞P(X=i)i(i−1)z2i+iz2i=F′′(z2)z4⋅F′(z2)⋅z2。
那么我们就要求出 F F F。记每一年流出第 i 个桶的酒量为 g，那么有 F i = F i ( 1 − g ) z + ∑ k < i F k ⋅ c k , i ⋅ z F_i=F_i(1-g)z+\sum_{k<i}F_k·c_{k,i}·z Fi=Fi(1−g)z+∑k<iFk⋅ck,i⋅z。注意到 1 号桶的生成函数是可以直接求出封闭形式的，那么就可以推到第二个桶，以此类推。而且，我们只需要知道 F n ′ ( z ) , F n ′ ( z 2 ) , F n ′ ′ ( z 2 ) F_n'(z),F_n'(z^2),F_n''(z^2) Fn′(z),Fn′(z2),Fn′′(z2)，因此可以一开始就把 z 带入递推，时刻维护这些值即可。
附一些求导法则：
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(u(x)v(x))'=u'(x)v(x)+u(x)v'(x)\\ ~\\ \left(\frac {u(x)}{v(x)}\right)'=\frac{u'(x)v(x)-u(x)v'(x)}{v^2(x)}\\ ~\\ u(v(x))'=v(x)'u'(v(x))
(u(x)v(x))′=u′(x)v(x)+u(x)v′(x) (v(x)u(x))′=v2(x)u′(x)v(x)−u(x)v′(x) u(v(x))′=v(x)′u′(v(x))