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?
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.