94 Quiz: Properties of AR processes

Caution

Section omitted to comply with the Honor Code

--- date: 2024-11-05 title: "Quiz: Properties of AR processes" subtitle: Time Series Analysis description: "The AR(P) process, its state-space representation, the characteristic polynomial, and the forecast function" categories: - Bayesian Time Series keywords: - time series - AR(p) process - reciprocal roots - characteristic polynomial - forecast function - autocorrelation function - R code jupyter: python3 --- ::::: {.content-visible unless-profile="HC"} ::: {.callout-caution} Section omitted to comply with the Honor Code ::: ::::: ::::: {.content-hidden unless-profile="HC"} ::: {#exr-1 } Consider the following $AR(2)$ process, $Y_t = 0.5 Y_{t-1} + 0.24 Y_{t-2} + \epsilon_t \qquad \epsilon_t \sim \mathcal{N}(0,v).$ Give the value of one of the reciprocal roots of this process. ::: ::: {.solution} we first need to find the roots of the characteristic polynomial $$ 1 - 0.5 z - 0.24 z^2 = 0 $$ {#eq-char-poly-q1} ```{python} import numpy as np # AR coefficients: Y_t = φ1 * Y_{t-1} + φ2 * Y_{t-2} + ε_t phi = np.array([0.5, 0.24]) # Characteristic polynomial: z^2 - φ1*z - φ2 = 0 roots = np.roots([1, -phi[0], -phi[1]]) print(roots) ``` One of the reciprocal roots of the $AR(2)$ process is $0.8$. ::: ::: {#exr-2 } Assume the reciprocal roots of an $AR(2)$ process are $0.7$ and $-0.2$. Which is the corresponding form of the autocorrelation function $\rho(h)$ of this process? ::: ::: {.solution} - [ ] $\rho(h) = (a+bh)0.7^h, \quad h > 0$ where $a$ and $b$ are constants. - [ ] $\rho(h) = (a+bh)0.3^h, \quad h > 0$ where $a$ and $b$ are constants. - [ ] $\rho(h) = (a+bh)(0.3^h + 0.7^h), \quad h > 0$ where $a$ and $b$ are constants. - [x] $\rho(h) = a(0.7)^h + b(-0.3)^h, \quad h > 0$ where $a$ and $b$ are constants. ::: ::: {#exr-3 } Assume that an $AR(2)$ process has a pair of complex reciprocal roots with modulus $r=0.95$ and period $\lambda=7.1$. Which following options corresponds to the correct form of its autocorrelation function, $\rho(h)$? ::: ::: {.solution} - [ ] $\rho(h) = a(0.95)^h \cos(7.1h + b), \quad h > 0$ where $a$ and $b$ are constants. - [x] $\rho(h) = a(0.95)^h \cos(2\pi h / 7.1 + b), \quad h > 0$ where $a$ and $b$ are constants. - [ ] $\rho(h) = a(0.95)^h, \quad h > 0$ where $a$ and $b$ are constants. - [ ] $\rho(h) = (a + bh)(0.95)^h, \quad h > 0$ where $a$ and $b$ are constants. ::: ::: {#exr-4 } Given the following $AR(2)$ process, $Y_t = 0.5Y_{t-1} + 0.36Y_{t-2} + \epsilon_t, \quad \epsilon_t \sim \mathcal{N}(0, v)$ The $h=3$ steps-ahead forecast function $f_t(3)$ has the following form: ::: ::: {.solution} - [X] $f_t(3) = c_{1t}(0.9)^3 + c_{2t}(-0.4)^3$ for $c_{1t}$ and $c_{2t}$ constants. - [ ] $f_t(3)= c_{1t} (3)^{0.9} + c_{2t} (3)^{-0.4}$ for $c_{1t}$ and $c_{2t}$ constants. - [ ] $f_t(3)= c_{1t} (1.1)^3 + c_{2t} (-2.5)^3$ for $c_{1t}$ and $c_{2t}$ constants. - [ ] $f_t(3)= (0.9)^3 (c_{1t}+ c_{2t}3)$ for $c_{1t}$ and $c_{2t}$ constants. we have the following characteristic polynomial $1 - 0.5z - 0.36z^2 = 0$ ```{python} import numpy as np # AR coefficients: Y_t = φ1 * Y_{t-1} + φ2 * Y_{t-2} + ε_t phi = np.array([0.5, 0.36]) # Characteristic polynomial: z^2 - φ1*z - φ2 = 0 roots = np.roots([1, -phi[0], -phi[1]]) print(roots) #reciprocals = 1 / roots #print(reciprocals) ``` i.e. the reciprocal roots are approximately $1.11$ and $-2.5$. these give us a forecast function of the form $$ f_t(h) = c_{1t} (1.11)^h + c_{2t} (-2.5)^h $$ ::: :::::