# How to calculate impulse responses for a given autoregressive process?

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.

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*}

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

• 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? Sep 9, 2015 at 9:48