Автоковариантность процесса ARMA (2,1) - вывод аналитической модели для

Мне нужно вывести аналитические выражения для автоковариантной функции $\gamma\left(k\right)$ процесса ARMA (2,1), обозначенного как:

$y_t=\phi_1y_{t-1}+\phi_2y_{t-2}+\theta_1\epsilon_{t-1}+\epsilon_t$

Итак, я знаю, что:

$\gamma\left(k\right) = \mathrm{E}\left[y_t,y_{t-k}\right]$

так что я могу написать:

$\gamma\left(k\right) = \phi_1 \mathrm{E}\left[y_{t-1}y_{t-k}\right]+\phi_2 \mathrm{E}\left[y_{t-2}y_{t-k}\right]+\theta_1 \mathrm{E}\left[\epsilon_{t-1}y_{t-k}\right]+\mathrm{E}\left[\epsilon_{t}y_{t-k}\right]$

затем, чтобы вывести аналитическую версию автоковариантной функции, мне нужно подставить значения - 0, 1, 2 ..., пока я не получу рекурсию, которая действительна для всех k, превышающих некоторое целое число.$k$$k$

Поэтому я подставляю и прорабатываю это, чтобы получить:$k=0$

$$\gamma \left(0\right) = \mathrm{E}\left[y_t,y_t\right] = \phi_1 \mathrm{E}\left[y_{t-1}y_t\right] + \phi_2 \mathrm{E}\left[y_{t-2}y_t\right]+\theta_1 \mathrm{E}\left[\epsilon_{t-1}y_t\right]+\mathrm{E}\left[\epsilon_ty_t\right]\\ $$

Теперь я могу упростить первые два из этих терминов, а затем заменить как и раньше:$y_t$

$$\gamma\left(0\right) = \phi_1 \gamma\left(1\right) + \phi_2 \gamma\left(2\right)\\ + \theta_1 \mathrm{E}\left[\epsilon_{t-1} \left(\phi_1 y_{t-1} +\phi_2 y_{t-2} +\theta_1 \epsilon_{t-1} + \epsilon_t \right)\right]\\ + \mathrm{E}\left[\epsilon_t \left(\phi_1 y_{t-1} +\phi_2 y_{t-2} +\theta_1 \epsilon_{t-1} + \epsilon_t \right)\right]$$

Затем я умножаю восемь слагаемых:

$$+\theta_1\phi_1\mathrm{E}\left[\epsilon_{t-1}y_{t-1}\right]\\ +\theta_1\phi_2\mathrm{E}\left[\epsilon_{t-1}y_{t-2}\right]\\ +\theta_1^2\mathrm{E}\left[\left(\epsilon_{t-1}\right)^2\right]=\theta_1^2\sigma_{\epsilon}^2\\ +\theta_1\mathrm{E}\left[\epsilon_{t-1}\epsilon_{t}\right]=\theta_1\mathrm{E}\left[\epsilon_{t-1}\right]\mathrm{E}\left[\epsilon_{t}\right]=0\\ +\phi_1\mathrm{E}\left[\epsilon_{t}y_{t-1}\right]\\ +\phi_2\mathrm{E}\left[\epsilon_{t}y_{t-2}\right]\\ +\theta_1\mathrm{E}\left[\epsilon_t\epsilon_{t-1}\right] = \theta_1\mathrm{E}\left[\epsilon_{t}\right]\mathrm{E}\left[\epsilon_{t-1}\right]=0 \\ +\mathrm{E}\left[\left(\epsilon_t\right)^2\right] = \sigma_{\epsilon}^2$$

So, I am left needing to resolve the four remaining terms. I want to use the same logic for lines 1, 2, 5 and 6 as I used on lines 4 and 7 - for example for line 1:

$\theta_1\phi_1\mathrm{E}\left[\epsilon_{t-1}y_{t-1}\right] = \theta_1\phi_1\mathrm{E}\left[\epsilon_{t-1}\right]\mathrm{E}\left[y_{t-1}\right] = 0$ because $\mathrm{E}\left[\epsilon_{t-1}\right]=0$.

Similarly for lines 2, 5 and 6. But I have a model solution that suggests the expression for $\gamma\left(0\right)$ simplifies to:

$\gamma\left(0\right) = \phi_1\gamma\left(1\right)+\phi_2\gamma\left(2\right) +\theta_1\left(\phi_1+\theta_1\right)\sigma_{\epsilon}^2+\sigma_{\epsilon}^2$

This suggests my simplification as described above would miss the term with the coefficient $\phi_1$ - which under my logic should be 0. Is my logic at fault, or is the model solution I found incorrect?

The worked solution also suggest that "analogously" $\gamma\left(1\right)$ can be found as:

$\gamma\left(1\right) = \phi_1\gamma\left(0\right)+\phi_2\gamma\left(1\right) + \theta_1\sigma_{\epsilon}^2$

and for $k>1$:

$\gamma\left(k\right) = \phi_1\gamma\left(k-1\right)+\phi_2\left(k-2\right)$

I hope the question is clear. Any assistance will be much appreciated. Thank you in advance.

This is a question related to my research, and is not in preparation for any exam or coursework.

If the ARMA process is causal there is a general formula that provides the autocovariance coefficients.

Consider the causal $\text{ARMA}(p,q)$ process

$$y_t = \sum_{i = 1}^p \phi_i y_{t-1} + \sum_{j = 1}^q \theta_j \epsilon_{t - j} + \epsilon_t,$$ where $\epsilon_t$ is a white noise with mean zero and variance $\sigma_\epsilon^2$. By the causality property, the process can be written as $$y_t = \sum_{j = 0}^\infty \psi_j \epsilon_{t - j},$$ where $\psi_j$ denotes the $\psi$-weights.

The general homogeneous equation for the autocovariance coefficients of a causal $\text{ARMA}(p,q)$ process is

$$\gamma (k) - \phi_1 \gamma (k-1) - \cdots - \phi_p \gamma (k-p) = 0, \quad k \geq \max (p, q+1),$$ with initial conditions $$\gamma (k) - \sum_{j = 1}^p \phi_j \gamma (k-j) = \sigma_\epsilon^2 \sum_{j = k}^q \theta_j \psi_{j - k}, \quad 0 \leq k < \max (p, q+1).$$

Your calculation mistake in your original question lies in

$$\theta_1\phi_1\mathrm{E}\left[\epsilon_{t-1}y_{t-1}\right] = \theta_1\phi_1\mathrm{E}\left[\epsilon_{t-1}\right]\mathrm{E}\left[y_{t-1}\right] = 0 \qquad \text{(mistaken)}$$

You cannot separate the expectation $\mathrm{E}\left[\epsilon_{t-1}y_{t-1}\right]$ - $\epsilon_{t-1}$ and $y_{t-1}$ are not independent.

OK. So the process of writing the post actually pointed me to the solution.

Consider the Expectation terms 1, 2, 5 and 6 from above that I thought should be 0.

Immediately for terms 5 - $\mathrm{E}\left[\epsilon_ty_{t-1}\right]$ - and 6 - $\mathrm{E}\left[\epsilon_ty_{t-2}\right]$: these terms are definitely zero, because $y_{t-1}$ and $y_{t-2}$ are independent of $\epsilon_t$ and $\mathrm{E}\left[\epsilon_t\right] = 0$.

However, terms 1 and 2 look as though the Expectation is of two correlated variables. So, consider the expressions for $y_{t-1}$ and $y_{t-2}$ thus:

$$y_{t-1} = \phi_1y_{t-2}+\phi_2y_{t-3}+\theta_1\epsilon_{t-2}+\epsilon_{t-1}\\ y_{t-2} = \phi_1y_{t-3}+\phi_2y_{t-4}+\theta_1\epsilon_{t-3}+\epsilon_{t-2}$$

And recall term 1 - $\phi_1\theta_1\mathrm{E}\left[\epsilon_{t-1}y_{t-1}\right]$. If we multiply both sides of the expression for $y_{t-1}$ by $\epsilon_{t-1}$ and then take Expectations, it is clear that all terms on the right hand side except the last become zero (because the values of $y_{t-2}$, $y_{t-3}$, and $\epsilon_{t-2}$ are independent of $\epsilon_{t-1}$ and $\mathrm{E}\left[\epsilon_{t-1}\right]=0$) to give:

$$\mathrm{E}\left[\epsilon_{t-1}y_{t-1}\right] = \mathrm{E}\left[\left(\epsilon_{t-1}\right)^2\right] = \sigma_{\epsilon}^2$$

So term 1 becomes $+\phi_1\theta_1\sigma_{\epsilon}^2$. For term 2, it should be clear that, by the same logic, all terms are zero.

Hence the original model answer was correct.

However, if anyone can suggest an alternative way to obtain a general (even if messy) solution, I would be very pleased to hear it!