题目
Problem
The continuous random variable X X X F \mathrm{F} F
F ( x ) = { 0 , x < 0 , − 1 108 x 3 + 1 12 x 2 , 0 ≤ x ≤ 3 , − 1 32 x 2 + 7 16 x + a , 3 < x ≤ b , 1 , x > b . \mathrm{F}(x)=\begin{cases}
0,&x<0,\\[2mm]
-\frac{1}{108}x^3+\frac{1}{12}x^2,&0\le x\le3,\\[2mm]
-\frac{1}{32}x^2+\frac{7}{16}x+a,&3<x\le b,\\[2mm]
1,&x\gt b.
\end{cases} F ( x ) = ⎩ ⎨ ⎧ 0 , − 108 1 x 3 + 12 1 x 2 , − 32 1 x 2 + 16 7 x + a , 1 , x < 0 , 0 ≤ x ≤ 3 , 3 < x ≤ b , x > b . (a) Verify that the median of X X X 3 3 3
(1)
(b) Show that a = − 17 32 a=-\frac{17}{32} a = − 32 17 b b b
(3)
(c) Find the probability density function of X X X
(2)
(d) Find E ( 1 X ) \mathrm{E}\big(\frac{1}{X}\big) E ( X 1 )
(3)
题目中文翻译
连续随机变量 X X X F \mathrm{F} F
F ( x ) = { 0 , x < 0 , − 1 108 x 3 + 1 12 x 2 , 0 ≤ x ≤ 3 , − 1 32 x 2 + 7 16 x + a , 3 < x ≤ b , 1 , x > b . \mathrm{F}(x)=\begin{cases}
0,&x<0,\\[2mm]
-\frac{1}{108}x^3+\frac{1}{12}x^2,&0\le x\le3,\\[2mm]
-\frac{1}{32}x^2+\frac{7}{16}x+a,&3<x\le b,\\[2mm]
1,&x\gt b.
\end{cases} F ( x ) = ⎩ ⎨ ⎧ 0 , − 108 1 x 3 + 12 1 x 2 , − 32 1 x 2 + 16 7 x + a , 1 , x < 0 , 0 ≤ x ≤ 3 , 3 < x ≤ b , x > b . (a) 验证 X X X 3 3 3
(1)
(b) 证明 a = − 17 32 a=-\frac{17}{32} a = − 32 17 b b b
(3)
(c) 求 X X X
(2)
(d) 求 E ( 1 X ) \mathrm{E}\big(\frac{1}{X}\big) E ( X 1 )
(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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中位数的累计概率应为 1 / 2 1/2 1/2 x = 3 x=3 x = 3
答题过程
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At x = 3 x=3 x = 3
F X ( 3 ) = − 27 108 + 9 12 = 1 2 . F_X(3)=-\frac{27}{108}+\frac9{12}=\frac12. F X ( 3 ) = − 108 27 + 12 9 = 2 1 . Thus 3 3 3
(b)
解法一
思路
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连续型变量的 CDF 在 x = 3 x=3 x = 3 a a a 1 1 1
答题过程
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Continuity at 3 3 3
− 9 32 + 21 16 + a = 1 2 , -\frac9{32}+\frac{21}{16}+a=\frac12, − 32 9 + 16 21 + a = 2 1 , so a = − 17 / 32 a=-17/32 a = − 17/32 b b b − b 2 / 32 + 7 b / 16 − 17 / 32 = 1 -b^2/32+7b/16-17/32=1 − b 2 /32 + 7 b /16 − 17/32 = 1 b 2 − 14 b + 49 = ( b − 7 ) 2 = 0 b^2-14b+49=(b-7)^2=0 b 2 − 14 b + 49 = ( b − 7 ) 2 = 0 b = 7 b=7 b = 7
(c)
解法一
思路
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对两个非恒定分段求导，并将支撑区间外的密度设为零。
答题过程
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A density is
f X ( x ) = { x / 6 − x 2 / 36 , 0 ≤ x ≤ 3 , ( 7 − x ) / 16 , 3 < x ≤ 7 , 0 , otherwise . f_X(x)=
\begin{cases}
x/6-x^2/36, & 0\le x\le3,\\
(7-x)/16, & 3<x\le7,\\
0, & \text{otherwise}.
\end{cases} f X ( x ) = ⎩ ⎨ ⎧ x /6 − x 2 /36 , ( 7 − x ) /16 , 0 , 0 ≤ x ≤ 3 , 3 < x ≤ 7 , otherwise . The two formulas give the same value 1 / 4 1/4 1/4 3 3 3
(d)
解法一
思路
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将每段密度除以 x x x x x x 1 / x 1/x 1/ x
答题过程
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The expectation is
E [ 1 X ] = ∫ 0 3 ( 1 6 − x 36 ) d x + 1 16 ∫ 3 7 ( 7 x − 1 ) d x = [ x 6 − x 2 72 ] 0 3 + 1 16 [ 7 ln x − x ] 3 7 = 3 8 + 7 ln ( 7 / 3 ) − 4 16 = 2 + 7 ln ( 7 / 3 ) 16 ≈ 0.496. \begin{align*}
&\,\operatorname E\Big[\frac1X\Big]\\
=&\,\int_0^3\Big(\frac16-\frac{x}{36}\Big)\,\mathrm{d}x\\
+&\,\frac1{16}\int_3^7\Big(\frac7x-1\Big)\,\mathrm{d}x\\
=&\,\Big[\frac{x}{6}-\frac{x^2}{72}\Big]_0^3\\
&\,\hspace{2pt}+\frac1{16}\big[7\ln x-x\big]_3^7\\
=&\,\frac38+\frac{7\ln(7/3)-4}{16}\\
=&\,\frac{2+7\ln(7/3)}{16}\approx0.496.
\end{align*} = + = = = E [ X 1 ] ∫ 0 3 ( 6 1 − 36 x ) d x 16 1 ∫ 3 7 ( x 7 − 1 ) d x [ 6 x − 72 x 2 ] 0 3 + 16 1 [ 7 ln x − x ] 3 7 8 3 + 16 7 ln ( 7/3 ) − 4 16 2 + 7 ln ( 7/3 ) ≈ 0.496. The first integrand has a finite limit at 0 0 0
