\begin{aligned} Y_t &= a Y_{t-1} + e_t, \\ Z_t &= Y_t + H_t, \\ \end{aligned}
where $H_t$ is independent of $Y_t$.
I'm trying to understand what ARMA model $Z_t$ corresponds to but I'm not really sure.
Can someone provide a quick explanation?
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asked May 10, 2017 at 11:53
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$\begingroup$ @RichardHardy H_t is defined as being independent from Y_t and drawn from a normal distribution. Can you explain why it is AR(1)? Is it simply because it's only varying component is Y_t which is an AR(1) process? $\endgroup$– tryingtolearnCommented May 10, 2017 at 12:15
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$\begingroup$ Consider adding the self-study tag and reading its Wiki. Then show us what you have done already and where you got stuck. $\endgroup$– Richard HardyCommented May 10, 2017 at 13:19
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$\begingroup$ If i$H_t$ is independent of $e_t$ and has 0 mean then the process $Z_t$ it just an AR(1) process with a larger variance. But although you have said that $H_t$ is normally distributed, you have not mentioned independence and its relationship to $e_t$. $\endgroup$– Michael R. ChernickCommented May 10, 2017 at 13:30
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$\begingroup$ @MichaelChernick Hi Michael, it simply states that $H_t$ is a white noise process with variance $\sigma^2_h$ and specifies no relationship with the $e_t$. Could you clarify for me why it's an AR(1)? $\endgroup$– tryingtolearnCommented May 10, 2017 at 13:36
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$\begingroup$ If you only add two independent normal random variables each with zero mean you get another normal random variable with 0 mean and variance equal to the sum of the variances $\endgroup$– Michael R. ChernickCommented May 10, 2017 at 13:41
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2 Answers
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We can see that $Y_t$ is an AR(1) process with $a$ parameter.
We can find the autocorrelation function of $Z_t$ by first calculating its autocovariance.
$$\text{cov}(Z_t, Z_{t-k}) = \text{cov}(Y_t + h_t, Y_{t-k} + H_{t-k})$$
this gives
$$\text{cov}(Z_t, Z_{t-k}) = \text{cov}(Y_t, Y_{t-k}) + \text{cov}(Y_t, H_{t-k}) + \text{cov}(H_t, Y_{t-k}) + \text{cov}(H_t, H_{t-k})$$
therefore when k = 0, we get $$ \gamma_Z(0) = \sigma^2_Y + \sigma^2_H $$ and when k > 0, we get $$\gamma_Z(k) = \gamma_Y(k) $$
From the equation for autocovariance for AR(1) models (since $Y_t$ is and AR(1)), $$ \gamma_Y(k) = a * \gamma_Y(0) \text{ and since } \gamma_Y(0) = \sigma^2_Y$$
this gives us an autocorrelation function of
$$\rho_Z(k) = \frac{a * \sigma^2_Y}{\sigma^2_Y + \sigma^2_H}$$
this has the form $$\rho_Z(k) = A a^{k-1}$$
which is typical of ARMA$(1,1)$ models and therefore implies that $Z_t$ is an ARMA$(1,1)$ model.
answered May 11, 2017 at 12:37
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$Z_t$ is not described by an autoregressive model because
$$ Z_t = a Y_{t-1} + e_t + H_t $$
No lagged values of $Z_t$ are present in the right-hand-side, and so there is no auto-regression. The fact that the variable $Y_t$ appears lagged, makes the model dynamic, but not autoregressive also.
(I presume that $e_t$ is not assumed to be a function of lagged values of $Z$)
answered May 10, 2017 at 13:45
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$\begingroup$ Well, you can re-arrange it to $Z_t = aZ_{t-1} + H_t - a H_{t-1} + e_t$, so I wouldn't say it's "not autoregressive". It's really just an AR(1) process observed with a bit of iid noise on top of it. It's not quite an AR(1) because the impact of the innovations ($e_t$) "carry forward" but the observation errors ($H_t$) do not. $\endgroup$ Commented May 10, 2017 at 15:17
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$\begingroup$ @ChrisHaug Granted, such reformulations are usually possible with dynamic models, but I believe "losing sight" of the $Y$ variable is detrimental. $\endgroup$ Commented May 10, 2017 at 15:35
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$\begingroup$ I don't disagree. In fact, I think the original formulation in the question is actually the clearest one because it maps directly to the state space framework: the state is AR(1), and the observation is that plus some iid observation noise. $\endgroup$ Commented May 10, 2017 at 15:52
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$\begingroup$ I solved it but I didn't get the same answer. I will update my post with the solution. $\endgroup$ Commented May 11, 2017 at 12:24