Stochastic Process

HW1

1

由方差的定义，有

Var (\sum_{i = 1}^{n} b_{i} X (t_{i})) ⩾ 0,

也就是说

Cov (\sum_{i = 1}^{n} b_{i} X (t_{i}), \sum_{i = 1}^{n} b_{i} X (t_{i})) ⩾ 0,

不妨将其改写为

Cov (\sum_{i = 1}^{n} b_{i} X (t_{i}), \sum_{j = 1}^{n} b_{j} X (t_{j})) ⩾ 0.

展开即得

\begin{matrix} E (\sum_{i = 1}^{n} b_{i} X (t_{i}) \cdot \sum_{j = 1}^{n} b_{j} X (t_{j})) - E (\sum_{i = 1}^{n} b_{i} X (t_{i})) E (\sum_{j = 1}^{n} b_{j} X (t_{j})) ⩾ 0, \\ E (\sum_{i = 1}^{n} \sum_{j = 1}^{n} b_{i} b_{j} X (t_{i}) X (t_{j})) - \sum_{i = 1}^{n} b_{i} E (X (t_{i})) \cdot \sum_{j = 1}^{n} b_{j} E (X (t_{j})) ⩾ 0, \\ \sum_{i = 1}^{n} \sum_{j = 1}^{n} b_{i} b_{j} E (X (t_{i}) X (t_{j})) - \sum_{i = 1}^{n} \sum_{j = 1}^{n} b_{i} b_{j} E (X (t_{i})) E (X (t_{j})) ⩾ 0, \\ \sum_{i = 1}^{n} \sum_{j = 1}^{n} b_{i} b_{j} (E (X (t_{i}) X (t_{j})) - E (X (t_{i})) E (X (t_{j}))) ⩾ 0, \\ \sum_{i = 1}^{n} \sum_{j = 1}^{n} b_{i} b_{j} Cov (X (t_{i}), X (t_{j})) ⩾ 0, \\ \sum_{i = 1}^{n} \sum_{j = 1}^{n} b_{i} b_{j} R_{X} (t_{i}, t_{j}) ⩾ 0. \end{matrix}

由二阶原点矩的定义，

\begin{matrix} E [{(\sum_{i = 1}^{n} b_{i} X (t_{i}))}^{2}] ⩾ 0, \\ E [\sum_{i = 1}^{n} b_{i} X (t_{i}) \cdot \sum_{i = 1}^{n} b_{i} X (t_{i})] ⩾ 0, \\ E [\sum_{i = 1}^{n} b_{i} X (t_{i}) \cdot \sum_{j = 1}^{n} b_{j} X (t_{j})] ⩾ 0, \\ E [\sum_{i = 1}^{n} \sum_{j = 1}^{n} b_{i} b_{j} X (t_{i}) X (t_{j})] ⩾ 0, \\ \sum_{i = 1}^{n} \sum_{j = 1}^{n} b_{i} b_{j} E [X (t_{i}) X (t_{j})] ⩾ 0, \\ \sum_{i = 1}^{n} \sum_{j = 1}^{n} b_{i} b_{j} r_{X} (t_{i}, t_{j}) ⩾ 0. \end{matrix}

2

条件：

$\mu_X(t)=E(X(t))=m$ $m$ 是一个常数；
$R_X(t,s)=R_X(t-s)$ $t-s$ 有关；
$E(X^2(t))<\infty$ .

$X(t)=\sin^2(t+\varTheta),-\infty<t<+\infty$ $\varTheta$ $[0,\pi]$ 上的均匀分布，则

\begin{matrix} f_{Θ} (θ) = \frac{1}{π} I_{[0, π]} (θ) . \end{matrix}

\begin{aligned} E (X (t)) & = E (\sin^{2} (t + Θ)) \\ = E (\frac{1}{2} - \frac{1}{2} \cos 2 (t + Θ)) \\ = \frac{1}{2} - \frac{1}{2} E (\cos (2 t + 2 Θ)) \\ = \frac{1}{2} - \frac{1}{2} (\cos 2 t \cdot E (\cos 2 Θ) - \sin 2 t \cdot E (\sin 2 Θ)) \\ = \frac{1}{2} - \frac{1}{2} \cos 2 t \cdot E (\cos 2 Θ) + \frac{1}{2} \sin 2 t \cdot E (\sin 2 Θ), \end{aligned}

其中

\begin{matrix} E (\cos 2 Θ) = \int_{0}^{π} \cos 2 θ \cdot \frac{1}{π} d θ = 0, \\ E (\sin 2 Θ) = \int_{0}^{π} \sin 2 θ \cdot \frac{1}{π} d θ = 0, \end{matrix}

$\mu_X(t)=E(X(t))=1/2$ .

协方差函数

\begin{aligned} R_{X} (t, s) & = Cov (X (t), X (s)) \\ = E (X (t) X (s)) - E (X (t)) E (X (s)) \\ = E (\sin^{2} (t + Θ) \cdot \sin^{2} (s + Θ)) - \frac{1}{4} \\ = E [(\sin (t + Θ) \sin (s + Θ))^{2}] - \frac{1}{4} \\ = E [{(- \frac{1}{2} \cos (t + s + 2 Θ) + \frac{1}{2} \cos (t - s))}^{2}] - \frac{1}{4} \\ = \frac{1}{4} E [\cos^{2} (t + s + 2 Θ)] - \frac{1}{2} \cos (t - s) E [\cos (t + s + 2 Θ)] + \frac{1}{4} \cos^{2} (t - s) - \frac{1}{4}, \end{aligned}

其中

\begin{matrix} E [\cos (t + s + 2 Θ)] = \cos (t + s) E [\cos 2 Θ] - \sin (t + s) E [\sin 2 Θ] = 0, \\ \begin{aligned} E [\cos^{2} (t + s + 2 Θ)] & = \frac{1}{2} + \frac{1}{2} E [\cos (2 t + 2 s + 4 Θ)] \\ = \frac{1}{2} + \frac{1}{2} \int_{0}^{π} \cos (2 t + 2 s + 4 θ) \cdot \frac{1}{π} d θ \\ = \frac{1}{2}, \end{aligned} \end{matrix}

所以

R_{X} (t, s) = \frac{1}{4} \cos^{2} (t - s) - \frac{1}{8} .

$s=t$ ，则

\begin{matrix} R_{X} (t, t) = Var (X (t)) = E (X^{2} (t)) - (E (X (t)))^{2} = E (X^{2} (t)) - \frac{1}{4} = \frac{1}{8}, \\ E (X^{2} (t)) = \frac{3}{8} . \end{matrix}

$X(t)$ 为宽平稳过程.