题目
Problem
A continuous random variable X X X
F ( x ) = { 0 for x < 0 4 a x 2 for 0 ⩽ x ⩽ 1 a ( b x 3 − x 4 + 1 ) for 1 < x ⩽ 3 1 for x > 3 \begin{align*}
&\,F(x)= \\[4mm]
&\,\begin{cases}
\begin{aligned}
&\,0 \\[-1mm]
&\,\text{for }x<0
\end{aligned} \\[3mm]
\begin{aligned}
&\,4ax^2 \\[-1mm]
&\,\text{for }0\leqslant x\leqslant1
\end{aligned} \\[3mm]
\begin{aligned}
&\,a\big(bx^3-x^4+1\big) \\[-1mm]
&\,\text{for }1<x\leqslant3
\end{aligned} \\[3mm]
\begin{aligned}
&\,1 \\[-1mm]
&\,\text{for }x>3
\end{aligned}
\end{cases}
\end{align*} F ( x ) = ⎩ ⎨ ⎧ 0 for x < 0 4 a x 2 for 0 ⩽ x ⩽ 1 a ( b x 3 − x 4 + 1 ) for 1 < x ⩽ 3 1 for x > 3 where a a a b b b
(a) Show that b = 4 b = 4 b = 4
(1)
(b) Find the exact value of a a a
(2)
(c) Find P ( X > 2.25 ) P(X > 2.25) P ( X > 2.25 )
(2)
(d) Showing your working clearly,
(i) sketch the probability density function of X X X
(ii) calculate the mode of X X X
(5)
(Total for Question 3 is 10 marks)
题目中文翻译
连续随机变量 X X X
F ( x ) = { 0 for x < 0 4 a x 2 for 0 ⩽ x ⩽ 1 a ( b x 3 − x 4 + 1 ) for 1 < x ⩽ 3 1 for x > 3 \begin{align*}
&\,F(x)= \\[4mm]
&\,\begin{cases}
\begin{aligned}
&\,0 \\[-1mm]
&\,\text{for }x<0
\end{aligned} \\[3mm]
\begin{aligned}
&\,4ax^2 \\[-1mm]
&\,\text{for }0\leqslant x\leqslant1
\end{aligned} \\[3mm]
\begin{aligned}
&\,a\big(bx^3-x^4+1\big) \\[-1mm]
&\,\text{for }1<x\leqslant3
\end{aligned} \\[3mm]
\begin{aligned}
&\,1 \\[-1mm]
&\,\text{for }x>3
\end{aligned}
\end{cases}
\end{align*} F ( x ) = ⎩ ⎨ ⎧ 0 for x < 0 4 a x 2 for 0 ⩽ x ⩽ 1 a ( b x 3 − x 4 + 1 ) for 1 < x ⩽ 3 1 for x > 3 其中 a a a b b b
(a) 证明 b = 4 b=4 b = 4
(1)
(b) 求 a a a
(2)
(c) 求 P ( X > 2.25 ) P(X>2.25) P ( X > 2.25 )
(2)
(d) 清楚写出你的过程，
(i) 画出 X X X
(ii) 求 X X X
(5)
（第 3 题共 10 分）
解答
(a)
解法一
思路
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累积分布函数必须连续。令 x = 1 x=1 x = 1 a a a a a a b = 4 b=4 b = 4
答题过程
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Continuity of F F F x = 1 x=1 x = 1
4 a ( 1 ) 2 = a ( b ( 1 ) 3 − 1 4 + 1 ) . 4a(1)^2=a(b(1)^3-1^4+1). 4 a ( 1 ) 2 = a ( b ( 1 ) 3 − 1 4 + 1 ) . Therefore,
4 a = a b . 4a=ab. 4 a = ab . Since a > 0 a>0 a > 0
b = 4 , \boxed{b=4}, b = 4 , as required.
(b)
解法一
思路
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支持区间的右端点为 3，而 x > 3 x>3 x > 3 F ( x ) = 1 F(x)=1 F ( x ) = 1 x = 3 x=3 x = 3 b = 4 b=4 b = 4 F ( 3 ) = 1 F(3)=1 F ( 3 ) = 1 a a a
答题过程
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Continuity at x = 3 x=3 x = 3
a ( 27 b − 81 + 1 ) = 1. a(27b-81+1)=1. a ( 27 b − 81 + 1 ) = 1. Using b = 4 b=4 b = 4
a ( 108 − 81 + 1 ) = 1 28 a = 1. \begin{align*}
a(108-81+1)=&\,1 \\[4mm]
28a=&\,1.
\end{align*} a ( 108 − 81 + 1 ) = 28 a = 1 1. Hence,
a = 1 28 . \boxed{a=\frac1{28}}. a = 28 1 .
(c)
解法一
思路
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由于 2.25 2.25 2.25 a , b a,b a , b F ( 2.25 ) F(2.25) F ( 2.25 )
答题过程
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Using a = 1 / 28 a=1/28 a = 1/28 b = 4 b=4 b = 4
F ( 2.25 ) = 1 28 [ 4 ( 2.25 ) 3 − ( 2.25 ) 4 + 1 ] = 5359 7168 . \begin{align*}
&\,F(2.25) \\[4mm]
&\,=\frac1{28}\big[4(2.25)^3 \\[4mm]
&\,\hspace{2pt}-(2.25)^4+1\big] \\[4mm]
&\,=\frac{5359}{7168}.
\end{align*} F ( 2.25 ) = 28 1 [ 4 ( 2.25 ) 3 − ( 2.25 ) 4 + 1 ] = 7168 5359 . Since F ( 3 ) = 1 F(3)=1 F ( 3 ) = 1 F ( 3 ) − F ( 2.25 ) F(3)-F(2.25) F ( 3 ) − F ( 2.25 )
P ( X > 2.25 ) = 1 − F ( 2.25 ) = 1 − 5359 7168 = 1809 7168 = 0.252371 … . \begin{align*}
&\,P(X>2.25) \\[4mm]
&\,=1-F(2.25) \\[4mm]
&\,=1-\frac{5359}{7168} \\[4mm]
&\,=\frac{1809}{7168} \\[4mm]
&\,=0.252371\ldots.
\end{align*} P ( X > 2.25 ) = 1 − F ( 2.25 ) = 1 − 7168 5359 = 7168 1809 = 0.252371 … . Therefore, P ( X > 2.25 ) = 0.252 P(X>2.25)=\boxed{0.252} P ( X > 2.25 ) = 0.252
(d)(i)
解法一
思路
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概率密度函数是累积分布函数的导数。分别对两个非恒定分段求导：第一段得到从原点上升的直线；第二段得到一条在 x = 1 x=1 x = 1 x = 3 x=3 x = 3
答题过程
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Using a = 1 / 28 a=1/28 a = 1/28 b = 4 b=4 b = 4 F F F
f ( x ) = d d x ( 4 a x 2 ) f ( x ) = 8 a x = 2 7 x . \begin{align*}
&\,f(x)=\frac{\mathrm{d}}{\mathrm{d}x}(4ax^2) \\[4mm]
&\,\hphantom{f(x)}=8ax=\frac27x.
\end{align*} f ( x ) = d x d ( 4 a x 2 ) f ( x ) = 8 a x = 7 2 x . For the second nonconstant part, applying the power rule gives
f ( x ) = a ( 12 x 2 − 4 x 3 ) = 3 7 x 2 − 1 7 x 3 . \begin{align*}
f(x)=&\,a(12x^2-4x^3) \\[4mm]
=&\,\frac37x^2-\frac17x^3.
\end{align*} f ( x ) = = a ( 12 x 2 − 4 x 3 ) 7 3 x 2 − 7 1 x 3 . The constant parts have derivative zero. Thus, over the whole range,
f ( x ) = { 2 7 x for 0 ⩽ x ⩽ 1 3 7 x 2 − 1 7 x 3 for 1 < x ⩽ 3 0 otherwise . \begin{align*}
&\,f(x)= \\[4mm]
&\,
\begin{cases}
\begin{aligned}
&\,\dfrac27x \\[-1mm]
&\,\text{for }0\leqslant x\leqslant1
\end{aligned} \\[3mm]
\begin{aligned}
&\,\dfrac37x^2-\dfrac17x^3 \\[-1mm]
&\,\text{for }1<x\leqslant3
\end{aligned} \\[3mm]
\begin{aligned}
&\,0 \\[-1mm]
&\,\text{otherwise}.
\end{aligned}
\end{cases}
\end{align*} f ( x ) = ⎩ ⎨ ⎧ 7 2 x for 0 ⩽ x ⩽ 1 7 3 x 2 − 7 1 x 3 for 1 < x ⩽ 3 0 otherwise . The straight-line section joins ( 0 , 0 ) (0,0) ( 0 , 0 ) ( 1 , 2 / 7 ) (1,2/7) ( 1 , 2/7 ) ( 1 , 2 / 7 ) (1,2/7) ( 1 , 2/7 ) x = 2 x=2 x = 2 x = 3 x=3 x = 3
(d)(ii)
解法一
思路
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第一段密度随 x x x x = 1 x=1 x = 1 2 / 7 2/7 2/7 1 < x < 3 1<x<3 1 < x < 3 x = 2 x=2 x = 2
答题过程
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For 1 < x < 3 1<x<3 1 < x < 3
f ( x ) = 3 7 x 2 − 1 7 x 3 . f(x)=\frac37x^2-\frac17x^3. f ( x ) = 7 3 x 2 − 7 1 x 3 . Differentiating,
f ′ ( x ) = 6 7 x − 3 7 x 2 = 3 7 x ( 2 − x ) . \begin{align*}
f'(x)=&\,\frac67x-\frac37x^2 \\[4mm]
=&\,\frac37x(2-x).
\end{align*} f ′ ( x ) = = 7 6 x − 7 3 x 2 7 3 x ( 2 − x ) . Thus f ′ ( x ) = 0 f'(x)=0 f ′ ( x ) = 0 x = 0 x=0 x = 0 x = 2 x=2 x = 2 x = 2 x=2 x = 2 1 < x < 3 1<x<3 1 < x < 3
f ( 2 ) = 4 7 > 2 7 = f ( 1 ) , f(2)=\frac47>\frac27=f(1), f ( 2 ) = 7 4 > 7 2 = f ( 1 ) , and the density decreases from x = 2 x=2 x = 2 x = 3 x=3 x = 3
2 . \boxed{2}. 2 . 
