概率论与数理统计

HW1

T1.2

(1)

A = A_{1} {\overset{―}{A}}_{2} {\overset{―}{A}}_{3} + {\overset{―}{A}}_{1} A_{2} {\overset{―}{A}}_{3} + {\overset{―}{A}}_{1} {\overset{―}{A}}_{2} A_{3}

(2)

$A$ 是“三次均未击中目标”的对立事件，即

A = \overset{―}{{\overset{―}{A}}_{1} {\overset{―}{A}}_{2} {\overset{―}{A}}_{3}} = A_{1} \cup A_{2} \cup A_{3}

(3)

$A_1$ $A'$ ）是“第二次、第三次均未击中”的对立事件，即

A^{'} = \overset{―}{{\overset{―}{A}}_{2} {\overset{―}{A}}_{3}}

从而

A = A_{1} \cdot A^{'} = A_{1} \overset{―}{{\overset{―}{A}}_{2} {\overset{―}{A}}_{3}} = A_{1} (A_{2} \cup A_{3})

(4)

“最多击中一次”是“三次均未击中目标”和“仅有一次击中目标”的和事件，即

A = {\overset{―}{A}}_{1} {\overset{―}{A}}_{2} {\overset{―}{A}}_{3} + A_{1} {\overset{―}{A}}_{2} {\overset{―}{A}}_{3} + {\overset{―}{A}}_{1} A_{2} {\overset{―}{A}}_{3} + {\overset{―}{A}}_{1} {\overset{―}{A}}_{2} A_{3}

T1.8

$A\cup B\cup C$ . 由容斥原理（inclusive-exclusive principle），有

\begin{matrix} P (A \cup B \cup C) \\ = P (A) + P (B) + P (C) - P (A B) - P (A C) - P (B C) + P (A B C) \\ = 1 / 3 + 1 / 3 + 1 / 3 - 1 / 8 - 0 - 1 / 8 + 0 \\ = 3 / 4 \end{matrix}

T1.12

$A$ 为“甲胜利”.

$A$ $A_1$ ，则

P (A_{1}) = p^{2} + (\binom{2}{1}) p (1 - p) \cdot p = p^{2} + 2 p^{2} (1 - p) = - 2 p^{3} + 3 p^{2}

$A$ $A_2$ ，则

\begin{aligned} P (A_{2}) & = p^{3} + (\binom{3}{2}) p^{2} (1 - p) \cdot p + (\binom{4}{2}) p^{2} (1 - p)^{2} \cdot p \\ = p^{3} + 3 p^{3} (1 - p) + 6 p^{3} (p^{2} - 2 p + 1) \\ = p^{3} + 3 p^{3} - 3 p^{4} + 6 p^{5} - 12 p^{4} + 6 p^{3} \\ = 6 p^{5} - 15 p^{4} + 10 p^{3} \end{aligned}

\begin{aligned} f (p) & = P (A_{1}) - P (A_{2}) \\ = - 2 p^{3} + 3 p^{2} - 6 p^{5} + 15 p^{4} - 10 p^{3} \\ = - 6 p^{5} + 15 p^{4} - 12 p^{3} + 3 p^{2} \\ = 3 p^{2} (- 2 p^{3} + 5 p^{2} - 4 p + 1) \end{aligned}

$g(p)=-2p^3+5p^2-4p+1$ ，则

g^{'} (p) = - 6 p^{2} + 10 p - 4 = - 6 (p - \frac{2}{3}) (p - 1)

$g(p)$ $\left(\dfrac12,\dfrac23\right)$ $\left(\dfrac23,1\right)$ $g\left(\dfrac12\right)=g(1)=0$ $\left(\dfrac12,1\right)$ $g(p)<0$ $f(p)<0$ ，即

P (A_{1}) < P (A_{2})

$p>\dfrac12$ 时，五局三胜制更有利.

T1.13

$A_n$ $n$ $P(A) =\left(1-\dfrac12\right)^{n-1}\cdot\dfrac12=\left(\dfrac12\right)^n$ ；

$B_n$ $n$ $P(B)=\dfrac{1}{n+1}\leqslant\dfrac12\left(n\geqslant1,当且仅当\ n=1\ 时取等\right)$ ；

$C_n$ $n$ 次硬币后出现正面并胜利”，则

P (C_{n}) = P (A_{n} B_{n}) = P (A_{n}) P (B_{n}) = {(\frac{1}{2})}^{n} \cdot \frac{1}{n + 1} ⩽ {(\frac{1}{2})}^{n} \cdot \frac{1}{2} = {(\frac{1}{2})}^{n + 1}

$C$ 为“甲胜利”，则

P (C) = P (\sum_{n = 1}^{\infty} C_{n}) ⩽ \sum_{n = 1}^{\infty} P (C_{n}) ⩽ lim_{n \to \infty} (\sum_{k = 1}^{n} {(\frac{1}{2})}^{k + 1}) = lim_{n \to \infty} (\frac{\frac{1}{4} (1 - {(\frac{1}{2})}^{n + 1})}{1 - \frac{1}{2}}) = \frac{1}{2}

$P(C)<\dfrac12$ . 综上，这规则对乙更有利.

T1.17

$3:00$ $3:15$ $3:30$ $3:45$ $4:00$ .

$3:00$ $x$ $y$ $0<x,\ y\leqslant60$ ）.

$A_1$ 为“甲、乙同乘第一辆车”，则

A_{1} = {(x, y) | 0 < x, y ⩽ 15}

同理，

A_{2} = {(x, y) | 15 < x, y ⩽ 30}

A_{3} = {(x, y) | 30 < x, y ⩽ 45}

A_{4} = {(x, y) | 45 < x, y ⩽ 60}

易知

P (A_{i}) = \frac{15^{2}}{60^{2}} = \frac{1}{16}, i = 1, 2, 3, 4

$A$ 为“甲、乙同乘一辆车”，则

P (A) = P (\sum_{i = 1}^{4} A_{i}) = \sum_{i = 1}^{4} P (A_{i}) = \frac{1}{4}

T1.20

$A=\{掷出\ 2\ 的倍数\}$ $B=\{掷出\ 6\}$ $C=\{掷出\ 3\ 的倍数\}$ ，则

P (A) = \frac{1}{3}, P (B) = \frac{1}{6}, P (C) = \frac{1}{2},

P (A B) = P (B) = \frac{1}{6}, P (B C) = P (B) = \frac{1}{6}, P (A C) = \frac{1}{6},

P (A | B) = \frac{P (A B)}{P (B)} = 1, P (B | C) = \frac{P (B C)}{P (C)} = \frac{1}{3}, P (A | C) = \frac{P (A C)}{P (C)} = \frac{1}{3}

$P(A|B)>P(A)$ $P(B|C)>P(B)$ $P(A|C)=P(A)$ .

T1.23

(1)

$A_r$ 为所求事件，则

P (A_{r}) = {(\frac{n - 1}{n})}^{r}

(2)

$r$ 次，则：

$r\leqslant n$ 时，
$P (B_{r}) = \frac{n}{n} \cdot \frac{n - 1}{n} \cdot \frac{n - 2}{n} \dots \frac{n - r + 1}{n} = \frac{A_{n}^{r}}{n^{r}}$
$r > n$ $P(B_r)=0$ .

(3)

(3.1)

$A_r'$ 为所求事件，则

P (A_{r}^{'}) = {(\frac{n - 1}{n})}^{m r}

(3.2)

$r$ 次，则

$mr\leqslant n$ 时，
$\begin{matrix} \begin{aligned} P (B_{r}^{'}) & = (\frac{n}{n} \cdot \frac{n - 1}{n} \dots \frac{n - m + 1}{n}) \cdot (\frac{n - m}{n} \cdot \frac{n - m - 1}{n} \dots \frac{n - 2 m + 1}{n}) \dots (\frac{n - (r - 1) m}{n} \cdot \frac{n - (r - 1) m - 1}{n} \dots \frac{n - r m + 1}{n}) \\ = \frac{n (n - 1) \dots (n - m r + 1)}{n^{m r}} \\ = \frac{A_{n}^{m r}}{n^{m r}} \end{aligned} \end{matrix}$
$mr>n$ $P(B_r')=0$ .

T1.24

(1)

$\dfrac{10}{50}$ ；

$\dfrac{18}{30}$ .

$A=\{第一次取到的零件是一等品\}$ ，则其概率为

P (A) = \frac{1}{2} \cdot \frac{10}{50} + \frac{1}{2} \cdot \frac{18}{30} = \frac{2}{5}

(2)

$B=\{第二次取到的零件是一等品\}$ $AB=\{两次取到的零件都是一等品\}$ .

$\dfrac{\mathrm{C}_{10}^2}{\mathrm{C}_{50}^2}=\dfrac{9}{245}$ ；

$\dfrac{\mathrm{C}_{18}^2}{\mathrm{C}_{30}^2}=\dfrac{51}{145}$ .

从而，

P (A B) = \frac{1}{2} \cdot \frac{9}{245} + \frac{1}{2} \cdot \frac{51}{145} = \frac{276}{1421}

所以有

P (B | A) = \frac{P (A B)}{P (A)} = \frac{276}{1421} \cdot \frac{5}{2} = \frac{690}{1421} \approx 0.486

T1.30

$B$ $\overline B$ $\varOmega$ $P(A)>0$ ，则
$P (B | A) = \frac{P (A B)}{P (A)} = \frac{P (A | B) P (B)}{P (A | B) P (B) + P (A | \overset{―}{B}) P (\overset{―}{B})}$

$A=\{测出阳性\}$ $B=\{带菌\}$ $\overline B=\{不带菌\}$ $P(B)=0.1$ $P(\overline B)=0.9$ $P(A|B)=0.95$ $P(A|\overline B)=0.01$ .

(1)

P (B | A) = \frac{P (A | B) P (B)}{P (A | B) P (B) + P (A | \overset{―}{B}) P (\overset{―}{B})} = \frac{0.95 \times 0.1}{0.95 \times 0.1 + 0.01 \times 0.9} = \frac{95}{104} \approx 0.913

(2)

$A_i=\{第\ i\ 次测出阳性\},\ i=1,\ 2.$ $A'=A_1A_2=\{两次均测出阳性\}$ $P(A_i|B)=0.95,P(A_i|\overline B)=0.01,\ i=1,\ 2$ .

$P(A'|B)=P(A_1A_2|B)=P(A_1|B)P(A_2|B)=0.95^2$ $P(A'|\overline B)=P(A_1A_2|\overline B)=P(A_1|\overline B)P(A_2|\overline B)=0.01^2$ .

P (B | A^{'}) = \frac{P (A^{'} | B) P (B)}{P (A^{'} | B) P (B) + P (A^{'} | \overset{―}{B}) P (\overset{―}{B})} = \frac{{0.95}^{2} \times 0.1}{{0.95}^{2} \times 0.1 + {0.01}^{2} \times 0.9} = \frac{9025}{9034} \approx 0.999

T1.33

$A_i=\{从甲袋中取\ i\ 个白球和 (2-i)个黑球放入乙袋\},\ i=0,\ 1,\ 2$ $B=\{从乙袋中取出白球\}$ $A_i$ $\varOmega$ 的一个划分.

$P(A_0)=\dfrac{\mathrm{C}_2^2}{\mathrm{C}_7^2}=\dfrac{1}{21}$ $P(A_1)=\dfrac{\mathrm{C}_5^1\mathrm{C}_2^1}{\mathrm{C}_7^2}=\dfrac{10}{21}$ $P(A_2)=\dfrac{\mathrm{C}_5^2}{\mathrm{C}_7^2}=\dfrac{10}{21}$ ；

$P(B|A_0)=\dfrac{4}{11}$ $P(B|A_1)=\dfrac{5}{11}$ $P(B|A_2)=\dfrac{6}{11}$ .

(1)

考察全概率公式.

P (B) = \sum_{i = 0}^{2} P (B | A_{i}) P (A_{i}) = \frac{4}{11} \cdot \frac{1}{21} + \frac{5}{11} \cdot \frac{10}{21} + \frac{6}{11} \cdot \frac{10}{21} = \frac{38}{77}

(2)

考察 Bayes 公式.

$P(A_0|B)$ .

P (A_{0} | B) = \frac{P (B A_{0})}{P (B)} = \frac{P (B | A_{0}) P (A_{0})}{\sum_{i = 0}^{2} P (B | A_{i}) P (A_{i})} = \frac{\frac{4}{11} \cdot \frac{1}{21}}{\frac{38}{77}} = \frac{2}{57}

所求概率为其对立事件的概率，则

P = P ({\overset{―}{A}}_{0} | B) = 1 - P (A_{0} | B) = 1 - \frac{2}{57} = \frac{55}{57}

HW2

摆了

HW3

T2.22

P (X ⩽ x) = \frac{\int_{0}^{x} y (t) d t}{\int_{0}^{2} y (t) d t} = \frac{- \frac{x^{3}}{3} + x^{2}}{\frac{4}{3}} = - \frac{x^{3}}{4} + \frac{3}{4} x^{2} .

\begin{matrix} F (x) = - \frac{x^{3}}{4} + \frac{3}{4} x^{2}, 0 ⩽ x ⩽ 2, \\ f (x) = F^{'} (x) = - \frac{3}{4} x^{2} + \frac{3}{2} x, 0 ⩽ x < 2. \end{matrix}

T2.27

$a$ $b=a+t$ $0\leqslant a <1,\ 0<t<1,\ b=a+t\leqslant1$ .

$F(a+t)-F(a)=g(t)$ ，则

\frac{F (a + t) - F (a)}{t} = \frac{g (t)}{t}

$t\to0$ ，则

lim_{t \to 0} \frac{g (t)}{t} = F^{'} (a) = f (a) .

$\displaystyle\lim_{t\to0}g(t)=0$ $\displaystyle\lim_{t\to0}\dfrac{g(t)}t=c$ .

$a$ 的取值无关，故

F^{'} (a) = f (a) = c, 0 < a < 1.

又因为

\int_{0}^{1} f (a) d a = 1,

$c=1$ $f(x)=1,0<x<1$ . 同时，

F (x) = \int_{0}^{x} f (a) d a = x, 0 < x < 1.

$X$ $(0,1)$ 上的均匀分布.

T2.37

(3)

$Y=1/X$ $g(x)=1/x$ $(-\infty,0),(0,+\infty)$ 分别单调减.

$I_1=(-\infty,0)$ $g(x)=1/x$ $h_1(y)=1/y$ $h'_1(y)=-1/y^2$ 连续，则

f (h_{1} (y)) | h_{1}^{'} (y) | I_{1} = \frac{1}{π (1 + (1 / y)^{2})} \cdot | - \frac{1}{y^{2}} | \cdot I_{(- \infty, 0)} = \frac{1}{π (1 + y^{2})} I_{(- \infty, 0)};

$I_2=(0,+\infty)$ 上，同理可得

f (h_{2} (y)) | h_{2}^{'} (y) | I_{2} = \frac{1}{π (1 + y^{2})} I_{(0, + \infty)} .

此时由密度变换公式，

f_{1} (y) = \sum_{j = 1}^{2} f (h_{j} (y)) | h_{j}^{'} (y) | I_{j} = \frac{1}{π (1 + y^{2})}, y \neq 0.

$f_1(0)=\dfrac1\pi$ ，这对分布函数没有影响. 此时，

f_{1} (y) = \frac{1}{π (1 + y^{2})}, - \infty < y < + \infty .

$X$ $1/X$ 具有相同的分布.

HW4

T2.40

$X\sim U(0,1)$ $f(x)=1,0<x<1$ ，其分布函数为

F (x) = \int_{0}^{x} f (t) d t = x .

(1)

$g(x)=e^x$ $(0,1)$ $h(y)=\ln y$ $|h'(y)|=h'(y)=1/y$ . 故概率密度函数为

f_{1} (y) = f (h (y)) | h^{'} (y) | = \frac{1}{y}, 1 < y < e .

(2)

$g(x)=1/x$ $(0,1)$ $h(y)=1/y$ $h(y)=-1/y^2,|h(y)|=1/y^2$ . 故概率密度函数为

f_{1} (y) = f (h (y)) | h^{'} (y) | = \frac{1}{y^{2}}, 1 < y < + \infty .

(3)

$g(x)=-\dfrac{\ln x}{\lambda}$ $(0,1)$ $h(y)=e^{-\lambda y}$ $h'(y)=-\lambda e^{-\lambda y},|h'(y)|=\lambda e^{-\lambda y}$ . 故概率密度函数为

f_{1} (y) = f (h (y)) | h^{'} (y) | = λ e^{- λ y}, 0 < y < + \infty .

T2.42

$g(x)=F(x)=\displaystyle\int_{-\infty}^x f(t)\ \mathrm dt$ $F(x)$ 为分布函数，一定为非减函数，从而一定是增函数.

$Y=F(X)$ 的分布函数为

F_{1} (y) = P (Y ⩽ y) = P (F (X) ⩽ y) = \int_{F (x) ⩽ y} f (x) d x .

$F(x)$ $F(x)\leqslant y\iff x\leqslant h(y)$ ，则

F_{1} (y) = \int_{F (x) ⩽ y} f (x) d x = \int_{- \infty}^{h (y)} f (x) d x = F (h (y)) = y .

$[0,1]$ $0\leqslant y\leqslant 1$ .

$f_1(y)=\dfrac{\mathrm dF_1(y)}{\mathrm dy}=1,0\leqslant y\leqslant 1$ $Y=F(X)$ $(0,1)$ 上的均匀分布.

T2.44

$Y=g(X)$ $1$ 的指数分布，则

f_{1} (y) = λ e^{- λ y} = e^{- y}, 0 < y < + \infty .

其分布函数为

F_{1} (y) = P (Y ⩽ y) = P (g (X) ⩽ y) = \int_{g (x) ⩽ y} f (x) d x .

$y=g(x)$ $x=h(y)$ $g(x)\leqslant y\iff x\leqslant h(y)$ ，则

F_{1} (y) = \int_{g (x) ⩽ y} f (x) d x = \int_{0}^{h (y)} f (x) d x .

两边求导得

\begin{matrix} f_{1} (y) = f (h (y)) h^{'} (y), \\ e^{- y} = 2 (1 - x) \frac{d x}{d y}, \\ e^{- y} d y = 2 (1 - x) d x, \\ \int e^{- y} d y = \int 2 (1 - x) d x, \\ \int e^{- y} d (- y) = \int d (1 - x)^{2}, \\ e^{- y} + C = (1 - x)^{2} . \end{matrix}

$C=0$ ，得

y = g (x) = - 2 \ln (1 - x), 0 < x < 1.

且

g^{'} (x) = \frac{2}{1 - x} > 0,

$g(x)$ $(0,1)$ 上的单调增函数.

T2.48

$\displaystyle\int_0^3f(x)\ \mathrm dx=1$ 得

\frac{1}{a} \int_{0}^{3} x^{2} d x = {\frac{x^{3}}{3 a} |}_{0}^{3} = \frac{27}{3 a} = 1, a = 9.

从而

f (x) = \frac{x^{2}}{9}, F (x) = \int_{0}^{x} f (t) d t = \frac{x^{3}}{27}, 0 < x < 3.

(1)

$(1,2)$ $Y$ 的分布函数为

F_{1} (y) = P (1 < Y ⩽ y) = P (g (X) ⩽ y) = \int_{1 < g (x) ⩽ y} f (x) d x = \int_{1}^{y} \frac{x^{2}}{9} d x = {\frac{x^{3}}{27} |}_{1}^{y} = \frac{y^{3} - 1}{27} .

P (Y = 2) = P (X ⩽ 1) = F (1) = \frac{1}{27}, P (Y = 1) = P (X > 2) = 1 - P (X ⩽ 2) = 1 - F (2) = 1 - \frac{8}{27} = \frac{19}{27} .

从而，

P (1 ⩽ Y ⩽ y) = P (Y = 1) + P (1 < Y ⩽ y) = \frac{19}{27} + \frac{y^{3} - 1}{27} = \frac{y^{3} + 18}{27} .

$Y$ 的分布函数为

\begin{matrix} F (y) = {\begin{cases} 0 & , y ⩽ 1, \\ \frac{y^{3} + 18}{27} & , 1 < y < 2, \\ 1 & , y ⩾ 2. \end{cases} \end{matrix}

(2)

$B_1=\{X|g(X)=1\}=\{X|2\leqslant X<3\}$ $B_2=\{X|g(X)=X\}=\{X|1<X<2\}$ $B_3=\{X|g(X)=2\}=\{X|0<X\leqslant1\}$ $\varOmega=\{X|0<X<3\}$ $B_1,B_2,B_3$ $\varOmega$ 的一个划分.

\begin{matrix} P (B_{1}) = P (2 ⩽ X < 3) = 1 - P (X < 2) = 1 - F (2) = \frac{19}{27}, \\ P (B_{2}) = P (1 < X < 2) = F (0 < X < 2) - P (0 < X ⩽ 1) = F (2) - F (1) = \frac{7}{27}, \\ P (B_{3}) = P (0 < X ⩽ 1) = F (1) = \frac{1}{27} . \end{matrix}

由全概率公式，

\begin{aligned} P (X ⩽ Y) & = P (X ⩽ g (X)) \\ = P (X ⩽ 1 | g (X) = 1) P (g (X) = 1) + P (X ⩽ X | g (X) = X) P (g (X) = X) + P (X ⩽ 2 | g (X) = 2) P (g (X) = 2) \\ = P (X ⩽ 1 | 2 ⩽ X ⩽ 3) P (B_{1}) + P (X ⩽ X | 1 < X < 2) P (B_{2}) + P (X ⩽ 2 | 0 < X ⩽ 1) P (B_{3}) \\ = 0 + 1 \times \frac{7}{27} + 1 \times \frac{1}{27} \\ = \frac{8}{27} . \end{aligned}

T2.49

$X\sim U(0,1)$ $f(x)=1,0<x<1$ ，其分布函数为

F (x) = \int_{0}^{x} f (t) d t = x .

(1)

$g(x)=\dfrac{x}{1-x}$ $Y=g(X)$ 的分布函数为

F_{1} (y) = P (Y ⩽ y) = P (\frac{X}{1 - X} ⩽ y) = \int_{\frac{x}{1 - x} ⩽ y} f (x) d x = \int_{0}^{y / (1 + y)} 1 d x = \frac{y}{1 + y}, 0 < y < + \infty .

其概率密度函数为

f_{1} (y) = \frac{d F_{1} (y)}{d y} = \frac{1}{(1 + y)^{2}}, 0 < y < + \infty .

(2)

\begin{matrix} g (x) = {\begin{cases} x & , a < x < 1, \\ 0 & , 0 < x ⩽ a . \end{cases} \end{matrix}

则

$z<0$ $P(Z\leqslant z)=0$ ；
$0\leqslant z\leqslant a$ $P(Z\leqslant z)=P(Z=0)=P(0<X\leqslant a)=F(a)=a$ ；
$a<z< 1$ $P(Z\leqslant z)=P(0<X\leqslant a)+P(a<X\leqslant z)=P(0<X\leqslant z)=F(z)=z$ ；
$z\geqslant1$ $P(Z\leqslant z)=1$ .

所以，分布函数为

\begin{matrix} F_{1} (z) = {\begin{cases} 0 & , z < 0, \\ a & , 0 ⩽ z ⩽ a, \\ z & , a < z < 1, \\ 1 & , z ⩾ 1. \end{cases} \end{matrix}

(3)

\begin{matrix} g (x) = {\begin{cases} x^{2} + x & , 0 < x ⩽ b, \\ x^{2} & , b < x < 1. \end{cases} \end{matrix}

$g(x)$ $(0,b],(b,1)$ 上分别严格增.

$I_1=(0,b]$ $g(x)=x^2+x$ $h_1(w)=\sqrt{w+\dfrac14}-\dfrac12$ ，其导函数
$h_{1}^{'} (w) = \frac{1}{2 \sqrt{w + \frac{1}{4}}}$
连续，则
$f (h_{1} (w)) | h_{1}^{'} (w) | = \frac{1}{2 \sqrt{w + \frac{1}{4}}}, 0 < w ⩽ b^{2} + b .$
$I_2=(b,1)$ $g(x)=x^2$ $h_2(w)=\sqrt w$ $h_2'(w)=\dfrac1{2\sqrt w}$ 连续，则
$f (h_{2} (w)) | h_{2}^{'} (w) | = \frac{1}{2 \sqrt{w}}, b^{2} < w < 1.$

$0<b<\dfrac{\sqrt 5-1}2$ $b^2+b<1$ . 由密度变换公式，

\begin{matrix} f_{1} (w) = \sum_{j = 1}^{2} f (h_{j} (w)) | h_{j}^{'} (w) | I_{j} = {\begin{cases} \frac{1}{2 \sqrt{w + 1 / 4}} & , 0 < w ⩽ b^{2}, \\ \frac{1}{2 \sqrt{w + 1 / 4}} + \frac{1}{2 \sqrt{w}} & , b^{2} < w < b^{2} + b, \\ \frac{1}{2 \sqrt{w}} & , b^{2} + b ⩽ w < 1; \end{cases} \end{matrix}

$\dfrac{\sqrt5-1}{2}\leqslant b<1$ $b^2+b\geqslant1$ . 由密度变换公式，

\begin{matrix} f_{1} (w) = \sum_{j = 1}^{2} f (h_{j} (w)) | h_{j}^{'} (w) | I_{j} = {\begin{cases} \frac{1}{2 \sqrt{w + 1 / 4}} & , 0 < w ⩽ b^{2}, \\ \frac{1}{2 \sqrt{w + 1 / 4}} + \frac{1}{2 \sqrt{w}} & , b^{2} < w < 1. \end{cases} \end{matrix}