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S2 2023 Oct Q6

A Level / Edexcel / S2

IAL 2023 Oct Paper · Question 6

题目

Problem

The continuous random variable YY has cumulative distribution function given by

F(y)={0,y<0,y221,0⩽y⩽k,215(6y−y22)−75,k<y⩽6,1,y>6.\begin{aligned} &F(y)=\\[1mm] &\begin{cases} \begin{aligned} &0,\\[-1mm] &\quad y<0, \end{aligned}\\[2mm] \begin{aligned} &\dfrac{y^2}{21},\\[-1mm] &\quad0\leqslant y\leqslant k, \end{aligned}\\[2mm] \begin{aligned} &\dfrac2{15}\Big(6y-\dfrac{y^2}{2}\Big)\\[1mm] &\quad-\dfrac75,\\[1mm] &\quad k<y\leqslant6, \end{aligned}\\[2mm] \begin{aligned} &1,\\[-1mm] &\quad y>6. \end{aligned} \end{cases} \end{aligned}

(a) Find

P(Y<k4∣Y<k)P\Big(Y<\frac{k}{4}\mid Y<k\Big)
(2)

(b) Find the value of kk

(4)

(c) Use algebraic calculus to find E⁡(Y)\operatorname{E}(Y)

(6)

(Total for Question 6 is 12 marks)

题目中文翻译

连续随机变量 YY 的累积分布函数为

F(y)={0,y<0,y221,0⩽y⩽k,215(6y−y22)−75,k<y⩽6,1,y>6.\begin{aligned} &F(y)=\\[1mm] &\begin{cases} \begin{aligned} &0,\\[-1mm] &\quad y<0, \end{aligned}\\[2mm] \begin{aligned} &\dfrac{y^2}{21},\\[-1mm] &\quad0\leqslant y\leqslant k, \end{aligned}\\[2mm] \begin{aligned} &\dfrac2{15}\Big(6y-\dfrac{y^2}{2}\Big)\\[1mm] &\quad-\dfrac75,\\[1mm] &\quad k<y\leqslant6, \end{aligned}\\[2mm] \begin{aligned} &1,\\[-1mm] &\quad y>6. \end{aligned} \end{cases} \end{aligned}

(a) 求

P(Y<k4∣Y<k).P\Big(Y<\frac{k}{4}\mid Y<k\Big).
(2)

(b) 求 kk 的值。

(4)

(c) 使用代数微积分求 E⁡(Y)\operatorname{E}(Y)。

(6)

（第 6 题共 12 分）

解答

(a)

解法一

思路

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由于 k/4<kk/4<k，条件概率的分子就是 P(Y<k/4)P(Y<k/4)。这两个概率都由 CDF 的同一段 F(y)=y2/21F(y)=y^2/21 给出，作比后 kk 会约去。

答题过程

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Since k/4<kk/4<k,

 P(Y<k4∣Y<k) =F(k/4)F(k) =(k/4)2/21k2/21 =116.\begin{align*} &\,P\Big(Y<\frac{k}{4}\mid Y<k\Big)\\[2mm] &\,=\frac{F(k/4)}{F(k)}\\ &\,=\frac{(k/4)^2/21}{k^2/21}\\ &\,=\boxed{\frac1{16}}. \end{align*}

(b)

解法一

思路

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YY 是连续型随机变量，因此其累积分布函数连续，在分段点 y=ky=k 处左右两段取值相等。建立方程后清除分母，所得二次式恰好是完全平方，从而唯一确定 kk。

答题过程

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Since YY is continuous, its cumulative distribution function is continuous at y=ky=k. Hence

k221=215(6k−k22)−75.\frac{k^2}{21} =\frac{2}{15}\Big(6k-\frac{k^2}{2}\Big)-\frac75.

Multiplying by 105 and simplifying,

 5k2=84k−7k2−147, 12k2−84k+147=0, 4k2−28k+49=0, (2k−7)2=0.\begin{align*} &\,5k^2=84k-7k^2-147,\\[2mm] &\,12k^2-84k+147=0,\\[2mm] &\,4k^2-28k+49=0,\\[2mm] &\,(2k-7)^2=0. \end{align*}

Therefore,

k=72.\boxed{k=\frac72}.

解法二

思路

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本题还可以先令两段累积函数的斜率相等，得到一个候选衔接点，再代回原来的两段函数确认其取值相等。这样只需求解一次方程；这里并不是把“所有概率密度函数都连续”当作一般性质。

答题过程

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Differentiating the two non-constant expressions gives

ddy(y221)=2y21\frac{\mathrm{d}}{\mathrm{d}y}\Big(\frac{y^2}{21}\Big) =\frac{2y}{21}

and

 ddy[215(6y−y22)−75] =215(6−y).\begin{align*} &\,\frac{\mathrm{d}}{\mathrm{d}y} \Big[\frac2{15}\Big(6y-\frac{y^2}{2}\Big)-\frac75\Big]\\[2mm] &\,=\frac2{15}(6-y). \end{align*}

Equating the two slopes gives a candidate joining point:

2k21= 215(6−k),15k= 21(6−k),15k= 126−21k,36k= 126,k= 72.\begin{align*} \frac{2k}{21}=&\,\frac2{15}(6-k),\\[2mm] 15k=&\,21(6-k),\\ 15k=&\,126-21k,\\ 36k=&\,126,\\[2mm] k=&\,\frac72. \end{align*}

At this value, the first expression gives

(7/2)221=712,\frac{(7/2)^2}{21}=\frac7{12},

and the second gives

 215(6⋅72−(7/2)22)−75 =215⋅1198−75 =11960−8460 =712.\begin{align*} &\,\frac2{15}\Big(6\cdot\frac72-\frac{(7/2)^2}{2}\Big)-\frac75\\[2mm] &\,=\frac2{15}\cdot\frac{119}{8}-\frac75\\[2mm] &\,=\frac{119}{60}-\frac{84}{60}\\[2mm] &\,=\frac7{12}. \end{align*}

The two expressions agree, confirming the continuous join. Therefore,

k=72.\boxed{k=\frac72}.

(c)

解法一

思路

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先对 CDF 的两个非恒定分段求导，得到概率密度函数。再把 E⁡(Y)=∫yf(y) dy\operatorname{E}(Y)=\int yf(y)\,\mathrm{d}y 按分段点 k=7/2k=7/2 拆成两个积分，保留代数计算过程并化成精确分数。

答题过程

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Differentiating the cumulative distribution function gives

f(y)={2y21,0<y⩽72,215(6−y),72<y⩽6,0,otherwise.f(y)= \begin{cases} \begin{aligned} &\dfrac{2y}{21},\\[-1mm] &\quad0<y\leqslant\dfrac72, \end{aligned}\\[2mm] \begin{aligned} &\dfrac{2}{15}(6-y),\\[-1mm] &\quad\dfrac72<y\leqslant6, \end{aligned}\\[2mm] \begin{aligned} &0,\\[-1mm] &\quad\text{otherwise}. \end{aligned} \end{cases}

Therefore,

 E⁡(Y) =221∫07/2y2 dy +215∫7/26(6y−y2) dy =221[y33]07/2 +215[3y2−y33]7/26.\begin{align*} &\,\operatorname{E}(Y)\\[2mm] &\,=\frac2{21}\int_0^{7/2}y^2\,\mathrm{d}y\\ &\,\hspace{2pt}+\frac2{15}\int_{7/2}^{6}(6y-y^2)\,\mathrm{d}y\\[2mm] &\,=\frac2{21}\Big[\frac{y^3}{3}\Big]_0^{7/2}\\ &\,\hspace{2pt}+\frac2{15}\Big[3y^2-\frac{y^3}{3}\Big]_{7/2}^{6}. \end{align*}

Substituting the limits in the two terms separately,

 221[y33]07/2 =221(34324−0) =4936.\begin{align*} &\,\frac2{21}\Big[\frac{y^3}{3}\Big]_0^{7/2}\\[2mm] &\,=\frac2{21}\Big(\frac{343}{24}-0\Big)\\[2mm] &\,=\frac{49}{36}. \end{align*}  215[3y2−y33]7/26 =215[(108−72) −(1474−34324)] =215(36−53924) =215(32524) =6536.\begin{align*} &\,\frac2{15}\Big[3y^2-\frac{y^3}{3}\Big]_{7/2}^{6}\\[2mm] &\,=\frac2{15}\Big[(108-72)\\ &\,\hspace{2pt}-\Big(\frac{147}{4}-\frac{343}{24}\Big)\Big]\\[2mm] &\,=\frac2{15}\Big(36-\frac{539}{24}\Big)\\[2mm] &\,=\frac2{15}\Big(\frac{325}{24}\Big)\\[2mm] &\,=\frac{65}{36}. \end{align*}

Therefore,

E⁡(Y)= 4936+6536 =11436 =196.\begin{align*} \operatorname{E}(Y)=&\,\frac{49}{36}+\frac{65}{36}\\[2mm] &\,=\frac{114}{36}\\[2mm] &\,=\boxed{\frac{19}{6}}. \end{align*}

Thus E⁡(Y)≈3.17\operatorname{E}(Y)\approx3.17 to 3 significant figures.