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I have to derive analytic expressions for the autocorrelation function of this ARMA($2$,$1$) process:

$$y_{t} = \varphi_{1}y_{t-1} + \varphi_{2}y_{t-2} + \varepsilon_{t} + \theta\varepsilon_{t-1}$$

in function of

$$\varphi_{1}, \varphi_{2}, \theta, \sigma_{\varepsilon}^{2}.$$

I somehow managed to find the autocovariance function, but I cannot figure out how to make it into a function of only those $4$ parameters.

$$\gamma(0) = \varphi_{1}\gamma(1) + \varphi_{2}\gamma(2) + \sigma_{\varepsilon}^{2} + (\varphi_{1} + \theta)\theta * \sigma_{\varepsilon}^{2}$$

$$\gamma(1) = \varphi_{1}\gamma(0) + \varphi_{2}\gamma(1) + \theta * \sigma_{\varepsilon}^{2}$$

$$\gamma(2) = \varphi_{1}\gamma(1) + \varphi_{2}\gamma(0)$$

For $k > 1$,

$$\gamma(k) = \varphi_{1}\gamma(k-1) + \varphi_{2}\gamma(k-2).$$

I tried to change it into the autocorrelation function in order to solve it, but I have a problem with the $\rho(1)$ term which I cannot seem to find how to simplify.

$$\rho(h) = \frac{\gamma_{k}}{\gamma_{0}}$$

$$\rho(0) = \frac{\gamma_{0}}{\gamma_{0}} = 1$$

$$\rho(1) = \frac{\gamma_{1}}{\gamma_{0}} = \frac{\varphi_{1}\gamma(0) + \varphi_{2}\gamma(1) + \theta\sigma_{\varepsilon}^{2}}{\gamma(0)} = \frac{\varphi_{1}\gamma(0)}{\gamma(0)} + \frac{\varphi_{2}\gamma(1)}{\gamma(0)} + \color{red}{\frac{\theta\sigma_{\varepsilon}^{2}}{\gamma(0)}}$$

$$\rho(1) = \varphi_{1}\rho(0) + \varphi_{2}\rho(1) + \frac{\theta\sigma_{\varepsilon}^{2}}{\gamma(0)}$$

$$\rho(2) = \frac{\gamma_{2}}{\gamma_{0}} = \frac{\varphi_{1}\gamma(1) + \varphi_{2}\gamma(0)}{\gamma(0)} = \frac{\varphi_{1}\gamma(1)}{\gamma(0)} + \frac{\varphi_{2}\gamma(0)}{\gamma(0)}$$

$$\rho(2) = \varphi_{1}\rho(1) + \varphi_{2}\rho(0)$$

For $h>1$,

$$\rho(h) = \frac{\gamma_{h}}{\gamma_{0}} = \frac{\varphi_{1}\gamma(h-1) + \varphi_{2}\gamma(h-2)}{\gamma(0)} = \frac{\varphi_{1}\gamma(h-1)}{\gamma(0)} + \frac{\varphi_{2}\gamma(h-2)}{\gamma(0)}$$

$$\rho(h) = \varphi_{1}\rho(h-1) + \varphi_{2}\rho(h-2)$$

How can I do that part?

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  • $\begingroup$ en.m.wikipedia.org/wiki/Autoregressive–moving-average_model $\endgroup$ – cactus314 Mar 29 '18 at 23:50
  • $\begingroup$ See here: stats.stackexchange.com/questions/68644/… $\endgroup$ – Math1000 Mar 30 '18 at 1:40
  • $\begingroup$ I had actually looked at those two links, which is what helped get to the point I'm at right now. Unfortunately, I'm stuck, and the links don't seem to explain where to go from here, unless I am misunderstanding something. Would either of you care to point out and explain in an answer the part you believe to be relevant to the step I am stuck on? $\endgroup$ – Kaito Kid Mar 31 '18 at 14:12
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Because$$ \left\{\begin{array}{rcl} γ(0) - φ_1 γ(1) - φ_2 γ(2) &=& (1 + φ_1 θ + θ^2) σ_ε^2\\ γ(1) - φ_1 γ(0) - φ_2 γ(1) &=& θ σ_ε^2\\ γ(2) - φ_1 γ(1) - φ_2 γ(0) &=& 0 \end{array}\right. $$ solving this system of linear equations,$$ \left\{\begin{array}{l} γ(0) = \dfrac{(1 - φ_2)(1 + θ^2) + 2φ_1 θ}{(1 + φ_2)(φ_1^2 + (1 - φ_2)^2)} σ_ε^2\\ γ(1) = \dfrac{φ_1(1 + θ^2) + (1 + φ_1^2 - φ_2^2)θ}{(1 + φ_2)(φ_1^2 + (1 - φ_2)^2)} σ_ε^2\\ γ(2) = \dfrac{(φ_1^2 + φ_2 - φ_2^2 + φ_1(1 + φ_1^2 + 2φ_2 - φ_2^2) θ}{(1 + φ_2)(φ_1^2 + (1 - φ_2)^2)} σ_ε^2 \end{array}\right. $$ Using the linear recurrence relation$$ γ(n) = φ_1 γ(n - 1) + φ_2 γ(n - 2), \quad \forall n \geqslant 2 $$ all $γ(n)$ can be found.

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  • $\begingroup$ Thank you for the details, but even with this, I'm not getting there. i.imgur.com/DtdPSwl.jpg I'm not getting to zero even using this last result $\endgroup$ – Kaito Kid Apr 3 '18 at 20:15
  • $\begingroup$ @KaitoKid I don't undrrstand how you derived the identity in the photo. Could you explain? $\endgroup$ – Saad Apr 4 '18 at 0:59
  • $\begingroup$ There's probably a mistake somewhere in what I did because your result for gamma 0 is the same as other people who seem to have understood it well. I'll try harder to understand your solution. I'll award you the bounty even if I fail because it's my fault at this point though $\endgroup$ – Kaito Kid Apr 4 '18 at 2:27

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