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How to show that $y_t=x_t+\nu_t$, where $x_t$ is an $AR(p)$ process and $\nu_t$ is a white noise, follows an ARMA(p,p) process?

Say $x_t=\phi x_t + \epsilon_t$. Then replacing $x_t=y_t-\nu_t$, we get $y_t=\phi y_{t-1}+\nu_t-\phi \nu_{t-1} + \epsilon_t$. Doing this with an AR(p) process always yields the $\epsilon_t$ adding the MA(p) process.

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  • 2
    $\begingroup$ Can you show your attempt? Otherwise this will be closed as off-topic. $\endgroup$ – luchonacho Mar 24 '17 at 14:46
  • $\begingroup$ Please correct the math typos in your question (indices and signs). $\endgroup$ – Alecos Papadopoulos Mar 24 '17 at 19:39
  • $\begingroup$ Do you know beforehand that this can be shown? $\endgroup$ – luchonacho Mar 25 '17 at 10:02
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You have $$y_t = x_t + v_t \tag{1} $$ and $$ \phi(B)x_t = e_t. $$ Applying $\phi(B)$ to both sides of (1) yields \begin{align} \phi(B)y_t &= \phi(B)x_t + \phi(B) v_t \\ &= e_t + \phi(B) v_t. \tag{2} \end{align} Consider the right hand side of (2). This is clearly a covariance stationary process. By the Wold decomposition theorem it must have a moving average representation. Since the autocovariance function cuts off for lags $k>p$ it must be a $MA(p)$ process, say $(1-\theta_1B-\dots-\theta_p B^p) u_t$. Hence, $y_t$ must be a $ARMA(p,p)$ process.

From the left hand side of (2), it is clear that its autoregressive parameters are equal to those of $x_t$. The moving average parameters $\theta_1,\theta_2,\dots,\theta_p$ and the white noise variance $\sigma_u^2$ of this $ARMA(p,p)$ process can be found by equating the autocovariance function of the right hand side of (2) with that of $\theta(B) u_t$ for lags $k=0,1,\dots,p$ and solving the $p+1$ resulting non-linear equations \begin{align} (1+\theta_1^2+\dots+\theta_p^2)\sigma_u^2 &= \sigma_e^2 + (1+\phi_1^2 +\dots +\phi_p^2)\sigma_v^2\\ (-\theta_1 + \theta_1\theta_2 +\dots+\theta_{p-1}\theta_p)\sigma_u^2 &= (-\phi_1 + \phi_1\phi_2 +\dots+\phi_{p-1}\phi_p)\sigma_v^2\\ &\vdots \tag{3} \\ (-\theta_{p-1} + \theta_1\theta_p)\sigma_u^2 &= (-\phi_{p-1} + \phi_1\phi_p)\sigma_v^2 \\ \theta_p \sigma_u^2&= \phi_p\sigma_v^2. \end{align}

Here is a R-function that solves these equations and returns the parameters of the $ARMA(p,p)$-model.

arplusnoise2arma <- function(phi,se = 1,sv) {
  p <- length(phi) # order of process
  # autocovariance of right hand side
  gamma0 <- ltsa:::tacvfARMA(theta=phi, maxLag = p,sigma2 = sv)
  gamma0[1] <- gamma0[1] + se
  # non-linear equations to solve resulting from equating autocov functions
  f <- function(par) { 
    gamma1 <- ltsa::tacvfARMA(theta=par[1:p], maxLag = p, sigma2 = exp(par[p+1]))
    gamma0-gamma1
  }
  # solve the non-linear system
  fit <- rootSolve:::multiroot(f, c(phi,1), maxiter=1000, rtol=1e-12)
  # parameters of the new ARMA, possibly non-invertible
  theta <- fit$root[1:p]
  sigma2 <- exp(fit$root[p+1])
  # reparameterize the MA-part to make it invertible by moving roots outside unit circle
  r <- 1/polyroot(c(1,-theta))
  for (i in 1:p) {
    if (Mod(r[i])>1) {
      sigma2 <- sigma2*r[i]^2
      r[i] <- 1/r[i]
    }
  }
  sigma2 <- Re(sigma2)
  # compute the new coefficients of the MA-polynomial
  polycoef <- 1
  for (i in 1:p)
    polycoef <- c(polycoef,0) - r[i]*c(0,polycoef)
  theta <- Re(-polycoef[-1])
  # return the invertible ARMA(p,p) model
  list(model=list(phi=phi,theta=theta,sigma2=sigma2),estim.precis=fit$estim.precis)
}

The following example checks that the autocovariance functions indeed are the same for a simple stationary AR(3) model and the computed ARMA(3,3) model:

> phi <- c(.2, -.1, .2) 
> Mod(polyroot(c(1,-phi)))
[1] 1.678659 1.725853 1.725853
> result <- arplusnoise2arma(phi,1,.5)
> result
$model
$model$phi
[1]  0.2 -0.1  0.2

$model$theta
[1]  0.07286795 -0.04104890  0.06545496

$model$sigma2
[1] 1.527768


$estim.precis
[1] 4.176867e-14

> do.call(ltsa:::tacvfARMA, c(result$model, maxLag=10))
 [1]  1.5793650794  0.1904761905 -0.0317460317  0.1904761905  0.0793650794 -0.0095238095
 [7]  0.0282539683  0.0224761905 -0.0002349206  0.0033561905  0.0051899683
> ltsa:::tacvfARMA(phi=phi,theta=NULL,maxLag=10)
 [1]  1.0793650794  0.1904761905 -0.0317460317  0.1904761905  0.0793650794 -0.0095238095
 [7]  0.0282539683  0.0224761905 -0.0002349206  0.0033561905  0.0051899683
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  • $\begingroup$ Two questions: How did you derive the autocovariance functions? Why the lag[0] items don't match in the simulation, 1.57 vs. 1.07? $\endgroup$ – Cagdas Ozgenc Jul 26 '19 at 6:18
  • $\begingroup$ @CowboyTrader The expressions for the autocovariance functions on the left and right hand sides of (3) follows from en.wikipedia.org/wiki/…. The last two lines of code compares the autocovariance function of the resulting ARMA$(p,p)$ model with that of $x_t$. The autocovariance function of $y_t$ is that of $x_t$ plus $\mbox{Var}(v_t)=0.5$ at lag 0 so this explains the difference at lag 0. Again, this follows from the same rule for the covariance of linear combinations. $\endgroup$ – Jarle Tufto Jul 26 '19 at 9:33

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