例题:时间序列

ε t {\varepsilon_t} 是一个均值为0,方差为1的独立同分布随机时间序列,定义如下随机过程:
X t = ε t 1 2 ε t 1 + 1 2 ε t 2 X_t=\varepsilon_t-\frac{1}{2}\varepsilon_{t-1}+\frac{1}{2}\varepsilon_{t-2}

  • E X t E(X_t) V a r ( X t ) {\rm Var}(X_t)
    E X t = 0 E(X_t)=0
    V a r ( X t ) = ( 1 + 1 4 + 1 4 ) V a r ( ϵ i ) = 3 2 {\rm Var}(X_t)=(1+\frac{1}{4}+\frac{1}{4}){\rm Var}(\epsilon_i)=\frac{3}{2}
  • 证明自相关函数 ρ 1 = 1 2 \rho_1=-\frac{1}{2} , ρ 2 = 1 3 \rho_2=\frac{1}{3}
    ρ 1 = C o v ( X t , X t + 1 ) V a r ( X t ) \rho_1=\frac{{\rm Cov}(X_t,X_{t+1})}{{\rm Var}(X_t)}
    其中,

C o v ( X t , X t + 1 ) = E ( X t , X t + 1 ) E ( X t ) E ( X t + 1 ) E ( X t , X t + 1 ) = E [ ( ε t 1 2 ε t 1 + 1 2 ε t 2 ) ( ε t + 1 1 2 ε t + 1 2 ε t 1 ) ] = 3 4 {\rm Cov}(X_t,X_{t+1})=E(X_t,X_{t+1})-E(X_t)E(X_{t+1}) E(X_t,X_{t+1})\\=E[(\varepsilon_t-\frac{1}{2}\varepsilon_{t-1}+\frac{1}{2}\varepsilon_{t-2})(\varepsilon_{t+1}-\frac{1}{2}\varepsilon_{t}+\frac{1}{2}\varepsilon_{t-1})]=-\frac{3}{4}
所以 ρ = 1 / 2 \rho=-1/2。
同理
ρ 2 = C o v ( X t , X t + 2 ) V a r ( X t ) = 1 / 2 3 / 2 = 1 3 \rho_2=\frac{{\rm Cov }(X_t,X_{t+2})}{{\rm Var}(X_t)}=\frac{1/2}{3/2}=\frac{1}{3}
-当k>2, ρ k \rho _k 应为多少?
当k>2, E ( X t ) ( X t + k ) E(X_t)(X_{t+k}) 都是 ε t \varepsilon_t 的交叉项,故为0.因而其协方差为零。
ρ k = C o v ( X t , X t + k ) V a r ( X t ) = 0 \rho_k=\frac{{\rm Cov}(X_t,X_{t+k})}{{\rm Var}(X_t)}=0

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转载自blog.csdn.net/weixin_39174856/article/details/104068466