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IAL 2019 Jan Q2

A Level / Edexcel / S1

IAL 2019 Jan Paper · Question 2

题目

Problem

The discrete random variable XX has the following probability distribution.

xx−2-2−1-1013
P(X=x)P(X=x)0.15aabbaa0.4

(a) Find E(X)E(X).

(2)

Given that E(X2)=4.54E(X^2)=4.54

(b) find the value of aa and the value of bb.

(5)

The random variable Y=3−2XY=3-2X

(c) Find Var⁡(Y)\operatorname{Var}(Y).

(3)

解答

(a)

解法一

思路

展开

直接用期望公式，−a-a 和 +a+a 会抵消。

答题过程

展开 E(X)=(−2)(0.15)+(−1)a+0b+1a+3(0.4).\begin{align*} E(X)=(-2)(0.15)+(-1)a+0b+1a+3(0.4). \end{align*}

So

E(X)=−0.3+1.2=0.9.\begin{align*} E(X)=-0.3+1.2=0.9. \end{align*}

(b)

解法一

思路

展开

先用 E(X2)=4.54E(X^2)=4.54 求 aa，再用概率总和为 1 求 bb。

答题过程

展开 E(X2)=(−2)2(0.15)+(−1)2a+12a+32(0.4).\begin{align*} E(X^2)=(-2)^2(0.15)+(-1)^2a+1^2a+3^2(0.4). \end{align*}

Thus

E(X2)=0.6+2a+3.6=4.2+2a.\begin{align*} E(X^2)=0.6+2a+3.6=4.2+2a. \end{align*}

Given E(X2)=4.54E(X^2)=4.54,

4.2+2a=4.54.\begin{align*} 4.2+2a=4.54. \end{align*}

So

a=0.17.\begin{align*} a=0.17. \end{align*}

The probabilities sum to 1:

0.15+a+b+a+0.4=1.\begin{align*} 0.15+a+b+a+0.4=1. \end{align*}

Therefore

b=1−0.55−2(0.17)=0.11.\begin{align*} b=1-0.55-2(0.17)=0.11. \end{align*}

(c)

解法一

思路

展开

先求 Var⁡(X)=E(X2)−[E(X)]2\operatorname{Var}(X)=E(X^2)-[E(X)]^2，再用 Var⁡(3−2X)=4Var⁡(X)\operatorname{Var}(3-2X)=4\operatorname{Var}(X)。

答题过程

展开 Var⁡(X)=4.54−(0.9)2=3.73.\begin{align*} \operatorname{Var}(X)=4.54-(0.9)^2=3.73. \end{align*}

Since Y=3−2XY=3-2X,

Var⁡(Y)=(−2)2Var⁡(X).\begin{align*} \operatorname{Var}(Y)=(-2)^2\operatorname{Var}(X). \end{align*}

Therefore

Var⁡(Y)=4(3.73)=14.92.\begin{align*} \operatorname{Var}(Y)=4(3.73)=14.92. \end{align*}

解法二

思路

展开

变量分布直接计算法（直接求 YY 的分布）。 对于随机变量的线性变换 Y=3−2XY = 3 - 2X，除了套用方差线性性质公式 Var⁡(aX+b)=a2Var⁡(X)\operatorname{Var}(aX+b) = a^2 \operatorname{Var}(X) 之外，我们也可以直接建立 YY 的概率分布表，并直接计算其方差：

  • 根据 XX 的取值 xi=−2,−1,0,1,3x_i = -2, -1, 0, 1, 3，对应的 YY 取值为 yi=3−2xiy_i = 3 - 2x_i，即 7,5,3,1,−37, 5, 3, 1, -3；
  • 对应的概率保持不变。 由此得到 YY 的概率分布表后，可以直接套用期望和方差公式算得结果。这种方法更为直观，特别适合在想验证线性性质公式的正确性，或者题目要求列出新变量分布的情况下使用。

答题过程

展开

Find the probability distribution of Y=3−2XY = 3 - 2X:

  • For x=−2x = -2: y=3−2(−2)=7y = 3 - 2(-2) = 7, with P(Y=7)=0.15P(Y = 7) = 0.15
  • For x=−1x = -1: y=3−2(−1)=5y = 3 - 2(-1) = 5, with P(Y=5)=a=0.17P(Y = 5) = a = 0.17
  • For x=0x = 0: y=3−2(0)=3y = 3 - 2(0) = 3, with P(Y=3)=b=0.11P(Y = 3) = b = 0.11
  • For x=1x = 1: y=3−2(1)=1y = 3 - 2(1) = 1, with P(Y=1)=a=0.17P(Y = 1) = a = 0.17
  • For x=3x = 3: y=3−2(3)=−3y = 3 - 2(3) = -3, with P(Y=−3)=0.40P(Y = -3) = 0.40

Thus, the probability distribution of YY is:

y7531−3P(Y=y)0.150.170.110.170.40\begin{array}{c|ccccc} y & 7 & 5 & 3 & 1 & -3 \\ \hline P(Y=y) & 0.15 & 0.17 & 0.11 & 0.17 & 0.40 \end{array}

Calculate the expectation E⁡(Y)\operatorname{E}(Y):

E⁡(Y)=  7(0.15)+5(0.17)+3(0.11)+1(0.17)+(−3)(0.40)=  1.05+0.85+0.33+0.17−1.20=  1.20.\begin{align*} \operatorname{E}(Y) =&\,\, 7(0.15) + 5(0.17) + 3(0.11) + 1(0.17) + (-3)(0.40)\\[3mm] =&\,\, 1.05 + 0.85 + 0.33 + 0.17 - 1.20\\[3mm] =&\,\, 1.20. \end{align*}

Calculate the expectation of Y2Y^2, E⁡(Y2)\operatorname{E}(Y^2):

E⁡(Y2)=  72(0.15)+52(0.17)+32(0.11)+12(0.17)+(−3)2(0.40)=  49(0.15)+25(0.17)+9(0.11)+1(0.17)+9(0.40)=  7.35+4.25+0.99+0.17+3.60=  16.36.\begin{align*} \operatorname{E}(Y^2) =&\,\, 7^2(0.15) + 5^2(0.17) + 3^2(0.11) + 1^2(0.17) + (-3)^2(0.40)\\[3mm] =&\,\, 49(0.15) + 25(0.17) + 9(0.11) + 1(0.17) + 9(0.40)\\[3mm] =&\,\, 7.35 + 4.25 + 0.99 + 0.17 + 3.60\\[3mm] =&\,\, 16.36. \end{align*}

Therefore:

Var⁡(Y)=  E⁡(Y2)−[E⁡(Y)]2=  16.36−1.22=  16.36−1.44=  14.92.\begin{align*} \operatorname{Var}(Y) =&\,\, \operatorname{E}(Y^2) - [\operatorname{E}(Y)]^2\\[3mm] =&\,\, 16.36 - 1.2^2\\[3mm] =&\,\, 16.36 - 1.44\\[3mm] =&\,\, 14.92. \end{align*}