题目
Problem
George throws two coins, A and B, at the same time. Coin A is biased so that the probability of obtaining a head is a. Coin B is biased so that the probability of obtaining a head is b, where b<a. The probability generating function of X, the number of heads obtained by George, is GX(t). The coefficients of t and t2 in GX(t) are 125 and 121 respectively.
(a) Find the value of a.
[2]
The random variable Y is the sum of two independent observations of X.
(b) Find the probability generating function of Y, giving your answer as a polynomial in t.
[3]
(c) Find Var(Y).
[3]
题目中文翻译
George 同时抛掷两枚硬币 A 和 B。硬币 A 是不均匀的,得到正面的概率为 a。硬币 B 是不均匀的,得到正面的概率为 b,其中 b<a。随机变量 X 表示 George 得到的正面数,其概率母函数为 GX(t)。在 GX(t) 中,t 和 t2 的系数分别为 125 和 121。
(a) 求 a 的值。
(2)
随机变量 Y 是 X 的两个独立观测值之和。
(b) 求 Y 的概率母函数,并将答案表示为 t 的多项式。
(3)
(c) 求 Var(Y)。
(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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将两个正面指示变量的生成函数相乘。比较一次项与二次项系数,得到 a+b 和 ab。
答题过程
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The generating function is ((1−a)+at)((1−b)+bt). Thus ab=1/12 and a+b−2ab=5/12, giving a+b=7/12. Therefore a and b are roots of
=u2−127u+121(u−31)(u−41)=0.
Since b<a, we have a=1/3 and b=1/4.
(b)
解法一
思路
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由概率和为 1 得出常数项,再把一次观察的生成函数平方并展开。
答题过程
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The constant term of GX is 1−5/12−1/12=1/2. Using independence,
=GY(t)=GX(t)2=144(6+5t+t2)214436+60t+37t2+10t3+t4.
(c)
解法一
思路
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用 (b) 的生成函数求均值和二阶阶乘矩,再代入方差公式。
答题过程
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Differentiating the generating function gives
GY′(t)=GY′′(t)=14460+74t+30t2+4t3,14474+60t+12t2.
The derivatives at 1 give
GY′(1)=14460+74+30+4=67,GY′′(1)=14474+60+12=7273.
Using Var(Y)=GY′′(1)+GY′(1)−(GY′(1))2, we obtain Var(Y)=73/72+7/6−(7/6)2=59/72.
解法二
思路
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也可以把 Y 看成四次独立抛掷的正面总数。两枚硬币各投两次,因此分别求指标方差再相加。
答题过程
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For each of the two throws of coin A the variance is (1/3)(2/3), and for each throw of coin B it is (1/4)(3/4). Independence gives
Var(Y)=2(92+163)=7259.