题目
Problem
The continuous random variable X X X
f ( x ) = { 3 28 ( e 1 2 x + 4 e − 1 2 x ) , 0 ≤ x ≤ 2 ln 3 , 0 , otherwise. f(x) =
\begin{cases}
\dfrac{3}{28}\big(e^{\frac{1}{2}x} + 4e^{-\frac{1}{2}x}\big), & 0 \leq x \leq 2\ln 3,\\
0, & \text{otherwise.}
\end{cases} f ( x ) = ⎩ ⎨ ⎧ 28 3 ( e 2 1 x + 4 e − 2 1 x ) , 0 , 0 ≤ x ≤ 2 ln 3 , otherwise. (a) Find the cumulative distribution function of X X X
[3]
The random variable Y Y Y Y = e 1 2 X Y = e^{\frac{1}{2}X} Y = e 2 1 X
(b) Find the probability density function of Y Y Y
[3]
(c) Find the 30th percentile of Y Y Y
[3]
(d) Find E ( Y 4 ) \mathrm{E}(Y^4) E ( Y 4 )
[2]
题目中文翻译
连续随机变量 X X X
f ( x ) = { 3 28 ( e 1 2 x + 4 e − 1 2 x ) , 0 ≤ x ≤ 2 ln 3 , 0 , otherwise. f(x) =
\begin{cases}
\dfrac{3}{28}\big(e^{\frac{1}{2}x} + 4e^{-\frac{1}{2}x}\big), & 0 \leq x \leq 2\ln 3,\\
0, & \text{otherwise.}
\end{cases} f ( x ) = ⎩ ⎨ ⎧ 28 3 ( e 2 1 x + 4 e − 2 1 x ) , 0 , 0 ≤ x ≤ 2 ln 3 , otherwise. (a) 求 X X X
(3)
随机变量 Y Y Y Y = e 1 2 X Y = e^{\frac{1}{2}X} Y = e 2 1 X
(b) 求 Y Y Y
(3)
(c) 求 Y Y Y
(3)
(d) 求 E ( Y 4 ) \mathrm{E}(Y^4) E ( Y 4 )
(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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在支撑区间上从 0 0 0
答题过程
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For 0 ≤ x ≤ 2 ln 3 0\le x\le2\ln3 0 ≤ x ≤ 2 ln 3
F X ( x ) = 3 28 ∫ 0 x ( e u / 2 + 4 e − u / 2 ) d u = 3 14 [ e u / 2 − 4 e − u / 2 ] 0 x = 3 14 ( e x / 2 − 4 e − x / 2 + 3 ) . \begin{align*}
&\,F_X(x)\\
=&\,\frac3{28}\int_0^x\bigl(\mathrm e^{u/2}+4\mathrm e^{-u/2}\bigr)\,\mathrm{d}u\\
=&\,\frac3{14}\big[\mathrm e^{u/2}-4\mathrm e^{-u/2}\big]_0^x\\
=&\,\frac3{14}\bigl(\mathrm e^{x/2}-4\mathrm e^{-x/2}+3\bigr).
\end{align*} = = = F X ( x ) 28 3 ∫ 0 x ( e u /2 + 4 e − u /2 ) d u 14 3 [ e u /2 − 4 e − u /2 ] 0 x 14 3 ( e x /2 − 4 e − x /2 + 3 ) . The CDF is zero for x < 0 x<0 x < 0 0 ≤ x ≤ 2 ln 3 0\le x\le2\ln3 0 ≤ x ≤ 2 ln 3 x > 2 ln 3 x>2\ln3 x > 2 ln 3
(b)
解法一
思路
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指数变换递增。把 Y ≤ y Y\le y Y ≤ y X ≤ 2 ln y X\le2\ln y X ≤ 2 ln y
答题过程
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The support is [ 1 , 3 ] [1,3] [ 1 , 3 ] 1 ≤ y ≤ 3 1\le y\le3 1 ≤ y ≤ 3
F Y ( y ) = F X ( 2 ln y ) = 3 14 ( y − 4 y + 3 ) . F_Y(y)=F_X(2\ln y)=\frac3{14}\Big(y-\frac4y+3\Big). F Y ( y ) = F X ( 2 ln y ) = 14 3 ( y − y 4 + 3 ) . The CDF is constant outside this interval. Therefore a density is
f Y ( y ) = { 3 14 ( 1 + 4 / y 2 ) , 1 ≤ y ≤ 3 , 0 , otherwise . f_Y(y)=
\begin{cases}
\frac3{14}(1+4/y^2), & 1\le y\le3,\\
0, & \text{otherwise}.
\end{cases} f Y ( y ) = { 14 3 ( 1 + 4/ y 2 ) , 0 , 1 ≤ y ≤ 3 , otherwise .
(c)
解法一
思路
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令新变量的累积概率为 0.3 0.3 0.3 y y y
答题过程
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The percentile q q q 3 14 ( q − 4 / q + 3 ) = 0.3 \frac3{14}(q-4/q+3)=0.3 14 3 ( q − 4/ q + 3 ) = 0.3 5 q 2 + 8 q − 20 = 0 5q^2+8q-20=0 5 q 2 + 8 q − 20 = 0 [ 1 , 3 ] [1,3] [ 1 , 3 ]
q = − 4 + 2 29 5 ≈ 1.35. q=\frac{-4+2\sqrt{29}}{5}\approx1.35. q = 5 − 4 + 2 29 ≈ 1.35.
(d)
解法一
思路
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用 y 4 y^4 y 4 [ 1 , 3 ] [1,3] [ 1 , 3 ]
答题过程
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The fourth moment is
E [ Y 4 ] = 3 14 ∫ 1 3 ( y 4 + 4 y 2 ) d y = 3 14 [ y 5 5 + 4 y 3 3 ] 1 3 = 89 5 = 17.8. \begin{align*}
&\,\operatorname E[Y^4]\\
=&\,\frac3{14}\int_1^3(y^4+4y^2)\,\mathrm{d}y\\
=&\,\frac3{14}\Big[\frac{y^5}{5}+\frac{4y^3}{3}\Big]_1^3\\
=&\,\frac{89}{5}=17.8.
\end{align*} = = = E [ Y 4 ] 14 3 ∫ 1 3 ( y 4 + 4 y 2 ) d y 14 3 [ 5 y 5 + 3 4 y 3 ] 1 3 5 89 = 17.8. 