Stochastic Process

HW8

如有错误，烦请指正.

(1)

$Z_1,Z_2\sim U(-1,1)$ ，所以

f_{Z_{1}} (z) = f_{Z_{2}} (z) = f_{Z} (z) = \frac{1}{2}, - 1 < x < 1,

$E(Z_1)=E(Z_2)=0,\mathrm{Var}(Z_1)=\mathrm{Var}(Z_2)=1/3$ .

所以

\begin{matrix} E (Z_{1}^{2}) = Var (Z_{1}) - (E (Z_{1}))^{2} = \frac{1}{3} = E (Z_{2}^{2}), \end{matrix}

$Z_1,Z_2$ $E(Z_1Z_2)=E(Z_1)E(Z_2)=0$ .

所以，

E (X (t)) = E (Z_{1} \cos λ t + Z_{2} \sin λ t) = \cos λ t E (Z_{1}) + \sin λ t E (Z_{2}) = 0

为常数；

协方差函数

\begin{aligned} R (t, t + s) & = E [(X (t) - E X (t)) (X (t + s) - E X (t + s))] \\ = E [X (t) X (t + s)] \\ = E [(Z_{1} \cos λ t + Z_{2} \sin λ t) (Z_{1} \cos λ (t + s) + Z_{2} \sin λ (t + s))] \\ = E {Z_{1}^{2} \cos λ t \cos λ (t + s) + Z_{2}^{2} \sin λ t \sin λ (t + s) + Z_{1} Z_{2} [\cos λ t \sin λ (t + s) + \sin λ t \cos λ (t + s)]} \\ = E (Z_{1}^{2}) \cos λ t \cos λ (t + s) + E (Z_{2}^{2}) \sin λ t \sin λ (t + s) + E (Z_{1} Z_{2}) \sin λ (2 t + s) \\ = \frac{1}{3} (\cos λ t \cos λ (t + s) + \sin λ t \sin λ (t + s)) \\ = \frac{1}{3} \cos λ s \end{aligned}

$s$ 有关，记为

R (τ) = \frac{1}{3} \cos λ τ .

特别，

\begin{matrix} Var (X (t)) = R (0) = \frac{1}{3}, \\ E (X^{2} (t)) = Var (X (t)) + (E (X (t)))^{2} = \frac{1}{3} < \infty . \end{matrix}

$\set{X(t)}$ 为宽平稳的.

(2)

$X(t)$ $t$ 有关，所以必然不是严平稳的.

(3)

考虑极限

lim_{T \to \infty} \frac{1}{T} \int_{0}^{2 T} (1 - \frac{τ}{2 T}) R (τ) d τ .

因为

(1 - \frac{τ}{2 T}) R (τ) = \frac{1}{3} \cos λ τ - \frac{1}{6 T} τ \cos λ τ,

所以

\begin{aligned} \int_{0}^{2 T} (1 - \frac{τ}{2 T}) R (τ) d τ & = \frac{1}{3} \int_{0}^{2 T} \cos λ τ d τ - \frac{1}{6 T} \int_{0}^{2 T} τ \cos λ τ d τ \\ = {\frac{1}{3 λ} \sin λ τ |}_{0}^{2 T} - \frac{1}{6 λ T} \int_{0}^{2 T} τ d \sin λ τ \\ = \frac{1}{3 λ} \sin 2 λ T - \frac{1}{6 λ T} (τ \sin λ τ |_{0}^{2 T} - \int_{0}^{2 T} \sin λ τ d τ) \\ = \frac{1}{3 λ} \sin 2 λ T - \frac{1}{6 λ T} (2 T \sin 2 λ T + {\frac{\cos λ τ}{λ} |}_{0}^{2 T}) \\ = \frac{1}{3 λ} \sin 2 λ T - \frac{1}{6 λ T} (2 T \sin 2 λ T + \frac{\cos 2 λ T - 1}{λ}) \\ = \frac{1}{3 λ} \sin 2 λ T - \frac{1}{3 λ} \sin 2 λ T - \frac{\cos 2 λ T - 1}{6 λ^{2} T} \\ = \frac{1 - \cos 2 λ T}{6 λ^{2} T} \end{aligned}

所以，

\frac{1}{T} \int_{0}^{2 T} (1 - \frac{τ}{2 T}) R (τ) d τ = \frac{1 - \cos 2 λ T}{6 λ^{2} T^{2}} \to 0, T \to \infty .

$\set{X(t)}$ 的均值遍历性成立.

(1)

$\Theta$ ，

\begin{matrix} f_{Θ} (θ) = \frac{1}{2 π}, 0 < θ < 2 π, \\ E (Θ) = π, Var (Θ) = \frac{π^{2}}{3}, E (Θ^{2}) = Var (Θ) + E^{2} (Θ) = \frac{4 π^{2}}{3}, \end{matrix}

\begin{matrix} E (\cos (ω t + Θ)) = \int_{0}^{2 π} \cos (ω t + θ) f_{Θ} (θ) d θ = \frac{1}{2 π} \int_{0}^{2 π} \cos (ω t + θ) d θ = 0, \\ E (\cos (ω (2 t + s) + Θ)) = 0. \end{matrix}

$A$ ，

\begin{matrix} f (x) = \frac{x}{σ^{2}} e^{- \frac{x^{2}}{2 σ^{2}}}, x > 0, \end{matrix}

\begin{aligned} E (A) & = \int_{0}^{+ \infty} x f (x) d x \\ = \int_{0}^{+ \infty} \frac{x^{2}}{σ^{2}} e^{- \frac{x^{2}}{2 σ^{2}}} d x \\ = \int_{0}^{+ \infty} y^{2} e^{- \frac{1}{2} y^{2}} \cdot σ d y & (y = x / σ) \\ = σ {\int_{0}^{+ \infty} y^{α} e^{- β y^{2}} d y |}_{α = 2, β = 1 / 2} \\ = σ {\frac{Γ ((α + 1) / 2)}{2 β^{(α + 1) / 2}} |}_{α = 2, β = 1 / 2} \\ = σ \frac{Γ (\frac{3}{2})}{2 \times {(\frac{1}{2})}^{3 / 2}} \\ = \sqrt{2} σ Γ (1 + \frac{1}{2}) \\ = \frac{\sqrt{2}}{2} σ Γ (1 / 2) \\ = \frac{\sqrt{2 π}}{2} σ, \end{aligned}

\begin{aligned} E (A^{2}) & = \int_{0}^{+ \infty} x^{2} f (x) d x \\ = \int_{0}^{+ \infty} \frac{x^{3}}{σ^{2}} e^{- \frac{x^{2}}{2 σ^{2}}} d x \\ = σ^{2} \int_{0}^{+ \infty} y^{3} e^{- \frac{1}{2} y^{2}} d y & (y = x / σ) \\ = σ^{2} {\frac{Γ ((α + 1) / 2)}{2 β^{(α + 1) / 2}} |}_{α = 3, β = 1 / 2} \\ = σ^{2} \frac{Γ (2)}{2 \times {(\frac{1}{2})}^{2}} \\ = σ^{2} \frac{1}{1 / 2} \\ = 2 σ^{2}, \end{aligned}

Var (A) = E (A^{2}) - E^{2} (A) = 2 σ^{2} - \frac{π}{2} σ^{2} = (2 - \frac{π}{2}) σ^{2} .

$X(t)=A\cos(\omega t+\Theta)$ .

$A,\Theta$ $A$ $\Theta$ $\cos(\omega t+\Theta)$ 也相互独立，所以

E (X (t)) = E [A \cos (ω t + Θ)] = E (A) E (\cos (ω t + Θ)) = 0,

为常数；

协方差函数

\begin{aligned} R (t, t + s) & = E [(X (t) - E X (t)) (X (t + s) - E X (t + s))] \\ = E [X (t) X (t + s)] \\ = E [A^{2} \cos (ω t + Θ) \cos (ω (t + s) + Θ)] \\ = E [\frac{A^{2}}{2} (\cos ω s + \cos (ω (2 t + s) + Θ))] \\ = \frac{1}{2} \cos ω s E (A^{2}) + \frac{1}{2} E [A^{2} \cos (ω (2 t + s) + Θ)] \\ = σ^{2} \cos ω s + \frac{1}{2} E (A^{2}) E (\cos (ω (2 t + s) + Θ)) & (A^{2}, \cos (ω (2 t + s) + Θ) 相 互 独 立) \\ = σ^{2} \cos ω s \end{aligned}

$s$ 有关，记为

R (τ) = σ^{2} \cos ω τ .

特别，

\begin{matrix} Var (X (t)) = R (0) = σ^{2}, \\ E (X^{2} (t)) = Var (X (t)) + E^{2} (X (t)) = σ^{2} < \infty . \end{matrix}

$\set{X(t),t\in\R}$ 为（宽）平稳过程.

(2)

\begin{aligned} lim_{T \to \infty} \frac{1}{T} \int_{0}^{2 T} (1 - \frac{τ}{2 T}) R (τ) d τ \\ = & σ^{2} lim_{T \to \infty} \frac{1}{T} \int_{0}^{2 T} (1 - \frac{τ}{2 T}) \cos ω τ d τ \\ = & σ^{2} lim_{T \to \infty} \frac{1}{T} (\int_{0}^{2 T} \cos ω τ d τ - \frac{1}{2 T} \int_{0}^{2 T} τ \cos ω τ d τ) \\ = & σ^{2} lim_{T \to \infty} \frac{1}{T} (\frac{1}{ω} \sin 2 ω T - \frac{1}{2 T} \frac{2 ω T \sin 2 ω T + \cos 2 ω T - 1}{ω^{2}}) \\ = & σ^{2} lim_{T \to \infty} \frac{1}{T} (\frac{1}{ω} \sin 2 ω T - \frac{\sin 2 ω T}{ω} + \frac{1 - \cos 2 ω T}{2 ω^{2} T}) \\ = & σ^{2} lim_{T \to \infty} \frac{1 - \cos 2 ω T}{2 ω^{2} T^{2}} \\ = & 0. \end{aligned}

$\set{X(t),t\in \R}$ 的均值具有遍历性.