题目
Problem
The probability generating function GX(t) of a random variable X is given by
GX(t)=61(t+1)(t2+2).
(a) Use GX(t) to find E(X) and Var(X).
(4)
(b) The random variable Y is the sum of three independent observations of X.
Find P(Y≤2).
(3)
题目中文翻译
随机变量 X 的概率生成函数 GX(t) 为
GX(t)=61(t+1)(t2+2).
(a) 利用 GX(t) 求 E(X) 和 Var(X)。
(4)
(b) 随机变量 Y 是对 X 进行三次相互独立的观测所得观测值的和。
求 P(Y≤2)。
(3)
原卷通用作答要求
Answer all questions. You should use a calculator where appropriate.
You must show all necessary working clearly; no marks will be given for unsupported answers from a calculator.
Give non-exact numerical answers correct to 3 significant figures, or 1 decimal place for angles in degrees, unless a different level of accuracy is specified in the question.
请回答所有题目。适当时应使用计算器。必须清楚展示所有必要的计算过程;仅给出计算器答案而没有过程,不予评分。除非题目另有精度要求,非精确数值答案保留 3 位有效数字,以度为单位的角度保留 1 位小数。
解答
(a)
解法一
思路
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展开生成函数,求两阶导数在 1 处的值,再代入方差公式。
答题过程
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The expanded PGF is GX(t)=(2+2t+t2+t3)/6, so
GX′(t)=62+2t+3t2,GX′′(t)=62+6t.
Hence
GX′(1)=62+2+3=67,GX′′(1)=62+6=34.
Thus E[X]=7/6 and
Var(X)==GX′′(1)+GX′(1)−(GX′(1))234+67−(67)2=3641.
(b)
解法一
思路
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用独立性将生成函数取三次幂。只保留常数、一次项、二次项,不必展开更高次项。
答题过程
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We have GY(t)=GX(t)3=2161(1+t)3(2+t2)3. Up to degree two, the two factors are 1+3t+3t2 and 8+12t2. Thus the coefficients of 1,t,t2 in their product are 8,24,36. Therefore
Pr(Y≤2)=2168+24+36=5417.
解法二
思路
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也可以枚举三个观察值相加不超过 2 的组合,再计算每种排列的概率。
答题过程
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The probabilities for X=0,1,2 are 1/3,1/3,1/6. The allowed triples, up to ordering, are (0,0,0), (1,0,0), (2,0,0) and (1,1,0). Independence gives
=+++=Pr(Y≤2)(31)33(31)3361(31)23(31)35417.