Is there a possibility to use a recurrent equation to calculate the impulse responses for an AR(p) process?

$$Y_t = \rho_1Y_{t-1} + \ldots + \rho_pY_{t-p} + e_t $$

It is quite a theoretical question, but I do not know how to start with the AR(1) process.


2 Answers 2


Elaborating on Martin's answer, you will want to compare coefficients in the general AR(p) case.

First, write the AR(p) process in compact lag operator notation, using $\rho(L)=1-\rho_1L-\ldots-\rho_pL^p$.

We have from $\rho(L)Y_t=\epsilon_t$ and the lag operator statement of an $MA(\infty)$ process, $Y_t=\psi(L)\epsilon_{t}$, that

$$\rho(L)Y_t=\rho(L)\psi(L)\epsilon_t=\epsilon_t$$ Hence, $\rho(L)\psi(L)=1$. Now the two polynomials $\rho(L)\psi(L)$ and 1 (the latter of order 0) are identical if and only if the coefficients of each power are identical.

Example for an $AR(2)$:

We obtain $$ (1-\rho_1L-\rho_2L^2)(\psi_0+\psi_1L+\psi_2L^2+\psi_3L^3+\ldots)=1 $$ Matching powers of $L$ yields \begin{align*} \psi_0&=1\\ \psi_1-\rho_1\psi_0&=0\Rightarrow\psi_1=\rho_1\\ -\rho_2\psi_0-\rho_1\psi_1+\psi_2&=0\Rightarrow\psi_2=\rho_2+\rho_1^2\\ \ldots& \end{align*}

  • $\begingroup$ Does the require that the lag polynomial is invertible? $\endgroup$
    – 123
    Commented Sep 6, 2018 at 21:28
  • $\begingroup$ How do we handle this if we have known coefficients on LHS attached to our lag operator that are larger than 1 $\endgroup$
    – 123
    Commented Sep 6, 2018 at 22:32

Not sure if your equation is correct. I guess you meant

$$ y_t = \rho_1 y_{t-1} + \dots + \rho_p y_{t-p} + \epsilon_t $$

In case of a AR(1) process you have to cast it into its MA($\infty$) (or 'covariance stationary') representation by reinserting the past observations $y_{t-j}$ where $j=1,\dots,\infty$:

$$ y_t = c + \rho_1 y_{t-1} + \epsilon_t . $$

With $y_{t-1} = c + \rho_1 y_{t-2} + \epsilon_{t-1}$ it follows that

$$ y_t = c + \rho_1 (c + \rho_1 y_{t-2} + \epsilon_{t-1}) + \epsilon_t. $$

If you do that infinitely you end up with

$$ y_t = \frac{c}{1-\rho} + \sum_{j=0}^{\infty} \psi_j \epsilon_{t-j} \quad \text{with} \quad \psi_j = \rho^j $$ where we used that $\sum_{j=0}^{\infty}\rho^j c = \frac{c}{1-\rho}$ and the condition that $\rho < 1$.

The impsule-responses can now be calculated by $$ \frac{\partial y_{t+j}}{\partial \epsilon_t} = \psi_j. $$

You basically have to do the same for higher order AR(p) processes. There you can compute the $\psi_j$'s by 'comparing coefficients'. An important condition for an $AR(p)$ process to have a covariance stationary representation is that its roots all lie outside the unit circle (thats why we assumed that $\rho < 1$).

I hope this rudimentary info helps you...

  • $\begingroup$ Thank you very much for your extensive answer! It really helped me. Do you also know if there is a possibility to display the AR(p) process with reference to an AR(1) model? $\endgroup$
    – OST_EE
    Commented Sep 9, 2015 at 9:48

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