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Problem: we have two processes $x_{t}$ and $y_{t}$ as follows:

$x_{t} = y_{t} + w_{t}$

$y_{t} = \phi_{0} + \phi_{1}y_{t-1} + u_{t}$

$w_{t}, u_{t}$ iid with var: $\sigma_{w}^{2}, \sigma_{u}^{2}$, mean of $w_{t}$ is 0 (mean of $u_{t}$ not mentioned)

$\phi_{1} \neq 0$

then: $x_{t}$ is ARMA(1,1) with:

(i) AR part = $\phi_{1}$

(ii) MA part:

$\theta_{1} = \frac{-\sigma_{u}^{2}-(1+\phi_{1}^2)\sigma_{w}^{2}+\sqrt{\sigma_{u}^{4}+2(1+\phi_{1}^{2})\sigma_{w}^{2} \sigma_{u}^{2}+(1-\phi_{1}^{2})^{2} \sigma_{w}^{4}}}{2\phi_{1}\sigma_{w}^{2}}$

My approach: inserting the definition of $y_{t}$ on the RHS of $x_{t}$ and then replace $y_{t-1}$ gives:

$x_{t} = \phi_{0} + \phi_{1} x_{t-1} + u_{t} + w_{t} - \phi_{1} w_{t-1}$

which proves (i) and suggests $\theta_{1} = -\phi_{1}$. How can I derive result (ii) ?

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  • $\begingroup$ Is this a homework problem? Can you edit the post to 1. tag it as self-study, 2. designate the problem statement and your approach in separate sections? $\endgroup$
    – AdamO
    Commented Jan 17, 2018 at 15:41

1 Answer 1

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The key idea is to rewrite $u_t + w_t - \phi_1 w_{t-1}$ as an MA(1) process, that is, find an $\epsilon_t$ and $\theta_1$ such that:

$$u_t + w_t - \phi_1 w_{t-1} = \epsilon_t + \theta_1 \epsilon_{t-1}$$

More details

You're correct in obtaining: $$ x_t = \phi_0 + \phi_1 x_{t-1} + u_t + w_t - \phi_1 w_{t-1}$$

Let $\zeta_t = u_t + w_t - \phi_1 w_{t-1}$. Based on the Wold representation, my answer here shows, you can write $\zeta_t$ as following an MA(1) structure $\zeta_t = \epsilon_t + \theta_1 \epsilon_{t-1}$ with:

$$ \frac{1}{\theta_1} + \theta_1 = -\frac{1}{\phi_1}\left(\frac{\sigma^2_u}{\sigma^2_w} + 1 \right) - \phi_1$$ Now it's just algebra. Rewrite as a quadratic equation in $\theta_1$: $$ \phi_1 \sigma^2_w\theta^2_1 + \left[\sigma^2_u+( 1 + \phi_1^2) \sigma^2_w\right] \theta_1 + \phi_1\sigma^2_w = 0 $$ Applying solution to the quadratic equation: $$ \theta_1 = \frac{ - \left[\sigma^2_u+( 1 + \phi_1^2) \sigma^2_w \right] \pm \sqrt{\sigma^4_u+(1 + \phi^2_1)^2\sigma^4_w+ 2\sigma^2_u(1 + \phi_1^2)\sigma^2_w - 4\phi^2_1\sigma^4_w }}{2\phi_1\sigma^2_w}$$ A little more algebra that $(1 + \phi^2_1)^2\sigma^4_w -4\phi^2_1\sigma^4_w = (1 - 2 \phi_1^2 + \phi_1^4)\sigma^4_w = (1 - \phi^2_1)^2\sigma^4_w$ $$ \theta_1 = \frac{ - \left[\sigma^2_u+( 1 + \phi_1^2) \sigma^2_w \right] \pm \sqrt{\sigma^4_u+ 2\sigma^2_u(1 + \phi_1^2)\sigma^2_w + (1 - \phi^2_1)^2\sigma^4_w }}{2\phi_1\sigma^2_w}$$ And there you are (the argument also needs $|\theta_1| < 1$ which I believe will rule out one of the solutions). There may be a simpler, slicker way to do this, but this gets there.

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  • $\begingroup$ @ Matthew Gunn: Thanks a lot, highly appreciate your answer! Recalculated it on paper; think there is a typo in your answer on the 2nd last and last line: $\sigma_{w}^{4}$ instead of $\sigma_{w}^{2}$ (last term on 2nd last line and last term in sqrt ... $\endgroup$
    – DH1288Z3
    Commented Jan 17, 2018 at 19:15
  • $\begingroup$ invertibility requires $|\theta_{1}| < 1$ and this rules out $- \sqrt{\cdot}$ in the numerator. Is it possibly to show analytically that using the solution with $- \sqrt{\cdot}$ in the numerator, $\theta_{1}$ is smaller than -1, i.e in absolute value larger than 1 (with only assuming the $\sigma$s are positive and $\phi_{1} \neq 0$)? $\endgroup$
    – DH1288Z3
    Commented Jan 17, 2018 at 19:19
  • $\begingroup$ @DanHubert Thanks. I think I fixed it. There are so many $\sigma^2_w$ and $\sigma^4_w$ terms! $\endgroup$ Commented Jan 17, 2018 at 19:22
  • $\begingroup$ @DanHubert There's often a slick way to show something is $<1$ using Cauchy-Schwartz but I've reached my tolerance limit for doing messy algebra :) $\endgroup$ Commented Jan 17, 2018 at 19:27
  • $\begingroup$ @ Matthew Gunn: thanks to hint at Cauchy-Schwartz ... $\endgroup$
    – DH1288Z3
    Commented Jan 17, 2018 at 19:36

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