Stochastic Process

HW2

P (X (s + t) - X (s) = n) = \frac{(λ t)^{n}}{n!} e^{- λ t} = \frac{(3 t)^{n}}{n!} e^{- 3 t}, n = 0, 1, 2, \dots .

以早上 8 点作为时间零点.

(a)

$s=0,t=2,n=0$ ，则

P (X (2) - X (0) = 0) = e^{- 6} \approx 0.00248 .

(b)

$T$ $\set{T>t}$ $[2,t+2]$ 上没有客户到达，所以

\begin{matrix} P (T > t) = P (X (2 + t) - X (2) = 0) = e^{- 3 t}, \\ P (T ⩽ t) = 1 - e^{- 3 t}, \end{matrix}

$t>0$ $T\sim Exp(3)$ .

P (X (s + t) - X (s) = n) = \frac{(λ t)^{n}}{n!} e^{- λ t} = \frac{(4 t)^{n}}{n!} e^{- 4 t}, n = 0, 1, 2, \dots .

(a)

\begin{matrix} P (X (1) - X (0) = 0) = e^{- 4}, \\ P (X (1) - X (0) = 1) = 4 e^{- 4}, \\ P (X (1) - X (0) < 2) = P (X (1) - X (0) = 0) + P (X (1) - X (0) = 1) = 5 e^{- 4} \approx 0.0916 . \end{matrix}

(b)

$A=\set{X(1)-X(0)=6}$ $B=\set{X(3)-X(2)<2}$ $P(B|A)$ .

$X$ $A,B$ 相互独立，故

\begin{aligned} P (B | A) & = P (B) \\ = P (X (3) - X (2) < 2) \\ = P (X (1) - X (0) < 2) \\ = 5 e^{- 4} \\ \approx 0.0916 . \end{aligned}

计算正态分布的矩母函数.

$X\sim N(\mu,\sigma^2)$ $X$ 的矩母函数为

\begin{aligned} g (t) & = E [e^{t X}] \\ = \int_{- \infty}^{+ \infty} e^{t x} \cdot \frac{1}{\sqrt{2 π} σ} e^{- \frac{(x - μ)^{2}}{2 σ^{2}}} d x . \end{aligned}

$y=(x-\mu)/\sigma$ $x=\mu+\sigma y$ $\mathrm dx=\sigma\mathrm dy$ .

\begin{aligned} g (t) & = \int_{- \infty}^{+ \infty} e^{t x} \cdot \frac{1}{\sqrt{2 π} σ} e^{- \frac{(x - μ)^{2}}{2 σ^{2}}} d x \\ = \int_{- \infty}^{+ \infty} e^{t (μ + σ y)} \cdot \frac{1}{\sqrt{2 π} σ} e^{- \frac{y^{2}}{2}} σ d y \\ = \frac{1}{\sqrt{2 π}} e^{t μ} \int_{- \infty}^{+ \infty} e^{- \frac{y^{2}}{2} + σ t y} d y \\ = \frac{1}{\sqrt{2 π}} e^{\frac{σ^{2}}{2} t^{2} + μ t} \int_{- \infty}^{+ \infty} e^{- \frac{1}{2} (y - σ t)^{2}} d y \\ = e^{\frac{1}{2} σ^{2} t^{2} + μ t} \int_{- \infty}^{+ \infty} \frac{1}{\sqrt{2 π}} e^{- \frac{z^{2}}{2}} d z & (z = y - σ t) \\ = e^{\frac{1}{2} σ^{2} t^{2} + μ t} . \end{aligned}