Stochastic Process

HW3

1

$X_{(r)}$ 的分布.

$\set{X_{(r)}\in[x, x+\Delta x)}$ $\Delta x\to 0$ 时，该事件发生当且仅当：

$X_{(1)},\cdots,X_{(r-1)}$ $r-1$ $x$ ；
$x\leqslant X_{(r)}<x+\Delta x$ ；
$X_{(r+1)},\cdots,X_{(n)}$ $n-r$ $x+\Delta x$ .

这样的分组方式共有

(\binom{n}{1}) (\binom{n - 1}{r - 1}) = n \cdot \frac{(n - 1)!}{(r - 1)! (n - r)!} = \frac{n!}{(n - r)! (r - 1)!}

种，且对于每一组分组，其出现的概率为

\begin{aligned} P (X_{(r)} \in [x, x + Δ x)) & = P (X_{(1)} < x, \dots, X_{(r - 1)} < x, x ⩽ X_{(r)} < x + Δ x, X_{(r + 1)} ⩾ x + Δ x, \dots, X_{(n)} ⩾ x + Δ x) \\ = [P (X_{(1)} < x)]^{r - 1} \cdot P (x ⩽ X_{(r)} < x + Δ x) \cdot [P (X_{(n)} ⩾ x + Δ x)]^{n - r} \\ = F^{r - 1} (x) \cdot (F (x + Δ x) - F (x)) \cdot [1 - F (x + Δ x)]^{n - r} . \end{aligned}

$X_{(r)}$ $F_r(x)$ $f_r(x)$ $\set{X_{(r)}\in[x, x+\Delta x)}$ 发生的概率为

\begin{matrix} P (x ⩽ X_{(r)} < x + Δ x) = F_{r} (x + Δ x) - F_{r} (x) = \frac{n!}{(n - r)! (r - 1)!} \cdot F^{r - 1} (x) \cdot (F (x + Δ x) - F (x)) \cdot [1 - F (x + Δ x)]^{n - r}, \\ \frac{F_{r} (x + Δ x) - F_{r} (x)}{Δ x} = \frac{n!}{(n - r)! (r - 1)!} \cdot F^{r - 1} (x) [1 - F (x + Δ x)]^{n - r} \cdot \frac{F (x + Δ x) - F (x)}{Δ x} . \end{matrix}

所以，

\begin{aligned} f_{r} (x) & = lim_{Δ x \to 0} \frac{F_{r} (x + Δ x) - F_{r} (x)}{Δ x} \\ = \frac{n!}{(n - r)! (r - 1)!} F^{r - 1} (x) lim_{Δ x \to 0} [1 - F (x + Δ x)]^{n - r} \cdot \frac{F (x + Δ x) - F (x)}{Δ x} \\ = \frac{n!}{(n - r)! (r - 1)!} F^{r - 1} (x) [1 - F (x)]^{n - r} \cdot f (x) . \end{aligned}

2

P (N (t) - N (0) = n) = \frac{(λ t)^{n}}{n!} e^{- λ t} = \frac{(2 t)^{n}}{n!} e^{- 2 t}, n = 0, 1, 2, \dots .

(1)

P (N (3) - N (0) = 0) = {\frac{(2 t)^{n}}{n!} e^{- 2 t} |}_{t = 3, n = 0} = e^{- 6} (\approx 0.00248) .

(2)

$A=\set{N(3)-N(0)=4}$ $B=\set{N(6)-N(0)\geqslant 7}$ $P(B|A)$ .

\begin{matrix} P (N (3) - N (0) = 0) = {\frac{(2 t)^{n}}{n!} e^{- 2 t} |}_{t = 3, n = 0} = e^{- 6}, \\ P (N (3) - N (0) = 1) = {\frac{(2 t)^{n}}{n!} e^{- 2 t} |}_{t = 3, n = 1} = 6 e^{- 6}, \\ P (N (3) - N (0) = 2) = {\frac{(2 t)^{n}}{n!} e^{- 2 t} |}_{t = 3, n = 2} = 18 e^{- 6}, \\ P (N (3) - N (0) ⩽ 2) = \sum_{i = 0}^{2} P (N (3) - N (0) = i) = 25 e^{- 6} . \end{matrix}

\begin{aligned} P (B | A) & = \frac{P (A B)}{P (A)} \\ = \frac{P (N (3) - N (0) = 4, N (6) - N (0) ⩾ 7)}{P (N (3) - N (0) = 4)} \\ = \frac{P (N (3) - N (0) = 4, N (6) - N (3) ⩾ 3)}{P (N (3) - N (0) = 4)} \\ = \frac{P (N (3) - N (0) = 4) P (N (6) - N (3) ⩾ 3)}{P (N (3) - N (0) = 4)} \\ = P (N (6) - N (3) ⩾ 3) \\ = P (N (3) - N (0) ⩾ 3) \\ = 1 - P (N (3) - N (0) ⩽ 2) \\ = 1 - 25 e^{- 6} (\approx 0.938) . \end{aligned}

(3)

$Y\sim Exp(0.1)$ ，则

f_{Y} (y) = λ e^{- λ y} = 0.1 e^{- 0.1 y}, y > 0.

$X$ $X\sim P(12)$ ，即

P (X = x) = P (N (6) - N (0) = x) = {\frac{(2 t)^{n}}{n!} e^{- 2 t} |}_{t = 6, n = x} = \frac{12^{x}}{x!} e^{- 12}, x = 0, 1, 2, \dots .

$E(X)=12,E(Y)=10$ .

$XY$ $X,Y$ 是相互独立的随机变量，

E (X Y) = E (X) E (Y) = 120,

即到 6 月底为止因交通事故而导致的总损失的期望值为 120 万元.

3

内含瞎蒙，谨慎参考.

$N_1(t),N_2(t),N_3(t)$ $N(t)$ $N^*(t)$ $1,3,5,9,4$ 的 Poisson 过程.

$T$ ，则

P (T > t) = P (N (t) - N (0) = 0) = e^{- λ t} = e^{- 9 t}, t > 0,

$T\sim Exp(9)$ $E(T_1)=1/9$ $10/9$ 分钟.

$X=W_k^{(1)}$ $k$ $Y=W_1^{(2)}$ $Z=W_{k+1}^{(1)}$ $k+1$ 辆非白车的到达时间，则所求可以表示为

P (X < Y < Z) = 1 - P (Y ⩽ X) - P (Y ⩾ Z)

$X\sim \Gamma(k,4),Y\sim\Gamma(1,5)=Exp(5),Z\sim \Gamma(k+1,4)$ ，从而

\begin{matrix} f_{X} (x) = 4 e^{- 4 x} \frac{(4 x)^{k - 1}}{(k - 1)!}, x > 0, \\ f_{Y} (y) = 5 e^{- 5 y}, y > 0, \\ f_{Z} (z) = 4 e^{- 4 z} \frac{(4 z)^{k}}{k!}, z > 0. \end{matrix}

\begin{aligned} P (Y ⩽ X) & = \iint_{0 < y ⩽ x} f (x, y) d x d y \\ = \int_{0}^{+ \infty} d y \int_{y}^{+ \infty} f_{X} (x) f_{Y} (y) d x \\ = \int_{0}^{+ \infty} 5 e^{- 5 y} d y \int_{y}^{+ \infty} 4 e^{- 4 x} \frac{(4 x)^{k - 1}}{(k - 1)!} d x \\ = \int_{0}^{+ \infty} 5 e^{- 5 y} \cdot e^{- 4 y} \sum_{j = 0}^{k - 1} \frac{(4 y)^{j}}{j!} d y \\ = 5 \int_{0}^{+ \infty} e^{- 9 y} \sum_{j = 0}^{k - 1} \frac{(4 y)^{j}}{j!} d y \\ = 5 \sum_{j = 0}^{k - 1} \int_{0}^{+ \infty} e^{- 9 y} \frac{(4 y)^{j}}{j!} d y \\ = 5 \sum_{j = 0}^{k - 1} \frac{1}{9} {(\frac{4}{9})}^{j} \\ = \frac{5}{9} \sum_{j = 0}^{k - 1} {(\frac{4}{9})}^{j} \\ = 1 - {(\frac{4}{9})}^{k} . \end{aligned}

其中，积分
$\begin{aligned} I_{k - 1} (y) & = \int_{y}^{+ \infty} 4 e^{- 4 x} \frac{(4 x)^{k - 1}}{(k - 1)!} d x \\ = - \int_{y}^{+ \infty} \frac{(4 x)^{k - 1}}{(k - 1)!} d e^{- 4 x} \\ = - ({\frac{(4 x)^{k - 1}}{(k - 1)!} e^{- 4 x} |}_{y}^{+ \infty} - \int_{y}^{+ \infty} e^{- 4 x} d \frac{(4 x)^{k - 1}}{(k - 1)!}) \\ = \frac{(4 y)^{k - 1}}{(k - 1)!} + \int_{y}^{+ \infty} e^{- 4 x} \cdot 4 (k - 1) \frac{(4 x)^{k - 2}}{(k - 1)!} d x \\ = \frac{(4 y)^{k - 1}}{(k - 1)!} + \int_{y}^{+ \infty} 4 e^{- 4 x} \frac{(4 x)^{k - 2}}{(k - 2)!} d x \\ = \frac{(4 y)^{k - 1}}{(k - 1)!} e^{- 4 y} + I_{k - 2} (y) . \end{aligned}$
$I_0(y)=\mathrm e^{-4y}$ . 所以，
$I_{k - 1} = e^{- 4 y} \sum_{j = 0}^{k - 1} \frac{(4 y)^{j}}{j!} .$
积分
$\begin{aligned} J_{j} & = \int_{0}^{+ \infty} e^{- 9 y} \frac{(4 y)^{j}}{j!} d y \\ = - \frac{1}{9} \int_{0}^{+ \infty} \frac{(4 y)^{j}}{j!} d e^{- 9 y} \\ = - \frac{1}{9} ({\frac{(4 y)^{j}}{j!} e^{- 9 y} |}_{0}^{+ \infty} - \int_{0}^{+ \infty} 4 e^{- 9 y} \frac{(4 y)^{j - 1}}{(j - 1)!} d y) \\ = \frac{4}{9} \int_{0}^{+ \infty} e^{- 9 y} \frac{(4 y)^{j - 1}}{(j - 1)!} d y \\ = \frac{4}{9} J_{j - 1} . \end{aligned}$
$J_0=1/9$ . 所以，
$J_{j} = \frac{1}{9} {(\frac{4}{9})}^{j} .$

\begin{aligned} P (Y ⩾ Z) = 1 - P (Y < Z) & = 1 - \iint_{0 < y < z} g (y, z) d y d z \\ = 1 - \int_{0}^{+ \infty} d y \int_{y}^{+ \infty} f_{Y} (y) f_{Z} (z) d z \\ = 1 - \int_{0}^{+ \infty} 5 e^{- 5 y} d y \int_{y}^{+ \infty} 4 e^{- 4 z} \frac{(4 z)^{k}}{k!} d z \\ = 1 - \int_{0}^{+ \infty} 5 e^{- 5 y} \cdot e^{- 4 y} \sum_{j = 0}^{k} \frac{(4 y)^{j}}{j!} d y \\ = 1 - 5 \int_{0}^{+ \infty} e^{- 9 y} \sum_{j = 0}^{k} \frac{(4 y)^{j}}{j!} d y \\ = 1 - 5 \sum_{j = 0}^{k} \int_{0}^{+ \infty} e^{- 9 y} \frac{(4 y)^{j}}{j!} d y \\ = 1 - 5 \sum_{j = 0}^{k} \frac{1}{9} {(\frac{4}{9})}^{j} \\ = 1 - \frac{5}{9} \sum_{j = 0}^{k} {(\frac{4}{9})}^{j} \\ = {(\frac{4}{9})}^{k + 1} . \end{aligned}

所以，

P (X < Y < Z) = 1 - P (Y ⩽ X) - P (Y ⩾ Z) = {(\frac{4}{9})}^{k} - {(\frac{4}{9})}^{k + 1} = \frac{5}{9} {(\frac{4}{9})}^{k} .

（如果这么做是对的的话）用结论得了，现场推要出人命的.
、

\begin{matrix} P (W_{k}^{(1)} < W_{1}^{(2)}) = P (X < Y) = {(\frac{λ_{1}}{λ_{1} + λ_{2}})}^{k} = {(\frac{4}{4 + 5})}^{k} = {(\frac{4}{9})}^{k}, \\ P (Y ⩽ X) = 1 - {(\frac{4}{9})}^{k} . \\ P (W_{k + 1}^{(1)} < W_{1}^{(2)}) = P (Z < Y) = {(\frac{λ_{1}}{λ_{1} + λ_{2}})}^{k + 1} = {(\frac{4}{9})}^{k + 1}, \\ P (X < Y < Z) = 1 - P (Y ⩽ X) - P (Y ⩾ Z) = 1 - P (Y ⩽ X) - P (Y > Z) = {(\frac{4}{9})}^{k} - {(\frac{4}{9})}^{k + 1} = \frac{5}{9} {(\frac{4}{9})}^{k} . \end{matrix}