题目
Problem
The random variable X X X F \mathrm{F} F
F ( x ) = { 0 , x < − 2 , 1 12 ( x 2 + 4 x + 4 ) , − 2 ≤ x < 0 , 1 15 ( x 3 + x + 5 ) , 0 ≤ x ≤ 2 , 1 , x > 2. \mathrm{F}(x)=\begin{cases}
0,&x<-2,\\[2mm]
\frac{1}{12}(x^2+4x+4),&-2\le x<0,\\[2mm]
\frac{1}{15}(x^3+x+5),&0\le x\le2,\\[2mm]
1,&x\gt2.
\end{cases} F ( x ) = ⎩ ⎨ ⎧ 0 , 12 1 ( x 2 + 4 x + 4 ) , 15 1 ( x 3 + x + 5 ) , 1 , x < − 2 , − 2 ≤ x < 0 , 0 ≤ x ≤ 2 , x > 2. (a) Find the exact value of the lower quartile of X X X
(2)
(b) Find Var ( X 2 ) \operatorname{Var}(X^2) Var ( X 2 )
(5)
(c) Find P ( X 2 < 1 ) \mathrm{P}(X^2<1) P ( X 2 < 1 )
(2)
题目中文翻译
随机变量 X X X F \mathrm{F} F
F ( x ) = { 0 , x < − 2 , 1 12 ( x 2 + 4 x + 4 ) , − 2 ≤ x < 0 , 1 15 ( x 3 + x + 5 ) , 0 ≤ x ≤ 2 , 1 , x > 2. \mathrm{F}(x)=\begin{cases}
0,&x<-2,\\[2mm]
\frac{1}{12}(x^2+4x+4),&-2\le x<0,\\[2mm]
\frac{1}{15}(x^3+x+5),&0\le x\le2,\\[2mm]
1,&x\gt2.
\end{cases} F ( x ) = ⎩ ⎨ ⎧ 0 , 12 1 ( x 2 + 4 x + 4 ) , 15 1 ( x 3 + x + 5 ) , 1 , x < − 2 , − 2 ≤ x < 0 , 0 ≤ x ≤ 2 , x > 2. (a) 求 X X X
(2)
(b) 求 Var ( X 2 ) \operatorname{Var}(X^2) Var ( X 2 )
(5)
(c) 求 P ( X 2 < 1 ) \mathrm{P}(X^2<1) P ( X 2 < 1 )
(2)
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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在 0 0 0 1 / 3 1/3 1/3
答题过程
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For the lower quartile q q q ( q + 2 ) 2 / 12 = 1 / 4 (q+2)^2/12=1/4 ( q + 2 ) 2 /12 = 1/4 ( q + 2 ) 2 = 3 (q+2)^2=3 ( q + 2 ) 2 = 3 − 2 ≤ q < 0 -2\le q<0 − 2 ≤ q < 0 q + 2 ≥ 0 q+2\ge0 q + 2 ≥ 0 q = − 2 + 3 q=-2+\sqrt3 q = − 2 + 3
(b)
解法一
思路
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先对两段 CDF 求导，再分别积分 x 2 f X ( x ) x^2f_X(x) x 2 f X ( x ) x 4 f X ( x ) x^4f_X(x) x 4 f X ( x )
答题过程
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The density is ( x + 2 ) / 6 (x+2)/6 ( x + 2 ) /6 − 2 < x < 0 -2<x<0 − 2 < x < 0 ( 3 x 2 + 1 ) / 15 (3x^2+1)/15 ( 3 x 2 + 1 ) /15 0 < x < 2 0<x<2 0 < x < 2
E [ X 2 ] = 1 6 ∫ − 2 0 ( x 3 + 2 x 2 ) d x + 1 15 ∫ 0 2 ( 3 x 4 + x 2 ) d x = [ x 4 24 + x 3 9 ] − 2 0 + [ x 5 25 + x 3 45 ] 0 2 = 2 9 + 328 225 = 42 25 . \begin{align*}
&\,\operatorname E[X^2]\\
=&\,\frac16\int_{-2}^0(x^3+2x^2)\,\mathrm{d}x\\
+&\,\frac1{15}\int_0^2(3x^4+x^2)\,\mathrm{d}x\\
=&\,\Big[\frac{x^4}{24}+\frac{x^3}{9}\Big]_{-2}^0\\
&\,\hspace{2pt}+\Big[\frac{x^5}{25}+\frac{x^3}{45}\Big]_0^2\\
=&\,\frac29+\frac{328}{225}=\frac{42}{25}.
\end{align*} = + = = E [ X 2 ] 6 1 ∫ − 2 0 ( x 3 + 2 x 2 ) d x 15 1 ∫ 0 2 ( 3 x 4 + x 2 ) d x [ 24 x 4 + 9 x 3 ] − 2 0 + [ 25 x 5 + 45 x 3 ] 0 2 9 2 + 225 328 = 25 42 . Similarly,
E [ X 4 ] = 1 6 ∫ − 2 0 ( x 5 + 2 x 4 ) d x + 1 15 ∫ 0 2 ( 3 x 6 + x 4 ) d x = [ x 6 36 + x 5 15 ] − 2 0 + [ x 7 35 + x 5 75 ] 0 2 = 16 45 + 2144 525 = 6992 1575 . \begin{align*}
&\,\operatorname E[X^4]\\
=&\,\frac16\int_{-2}^0(x^5+2x^4)\,\mathrm{d}x\\
+&\,\frac1{15}\int_0^2(3x^6+x^4)\,\mathrm{d}x\\
=&\,\Big[\frac{x^6}{36}+\frac{x^5}{15}\Big]_{-2}^0\\
&\,\hspace{2pt}+\Big[\frac{x^7}{35}+\frac{x^5}{75}\Big]_0^2\\
=&\,\frac{16}{45}+\frac{2144}{525}=\frac{6992}{1575}.
\end{align*} = + = = E [ X 4 ] 6 1 ∫ − 2 0 ( x 5 + 2 x 4 ) d x 15 1 ∫ 0 2 ( 3 x 6 + x 4 ) d x [ 36 x 6 + 15 x 5 ] − 2 0 + [ 35 x 7 + 75 x 5 ] 0 2 45 16 + 525 2144 = 1575 6992 . Hence Var ( X 2 ) = 6992 / 1575 − ( 42 / 25 ) 2 = 63668 / 39375 ≈ 1.62 \operatorname{Var}(X^2)=6992/1575-(42/25)^2=63668/39375\approx1.62 Var ( X 2 ) = 6992/1575 − ( 42/25 ) 2 = 63668/39375 ≈ 1.62
(c)
解法一
思路
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X 2 < 1 X^2<1 X 2 < 1 − 1 < X < 1 -1<X<1 − 1 < X < 1
答题过程
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Since X X X
Pr ( X 2 < 1 ) = F X ( 1 ) − F X ( − 1 ) = 7 15 − 1 12 = 23 60 . \Pr(X^2<1)=F_X(1)-F_X(-1)=\frac7{15}-\frac1{12}=\frac{23}{60}. Pr ( X 2 < 1 ) = F X ( 1 ) − F X ( − 1 ) = 15 7 − 12 1 = 60 23 . 