# ARMA process and Impulse response

I need to solve the following problem and I don't know where to look for relevant information. Does anyone have a good source when ARMA processes are input/output to and LTI (Linear Time Invariant)? Or perhaps some initial explanation on how to think about this problem? I do understand the relationship between autocorrelation, spectral density and impulse responses though.

Hi: The model you wrote is an ARMA(1,1) but I prefer the notation:

$$Y_t = \phi Y_{t-1} + \epsilon_t - \theta \epsilon_{t-1}$$ where $$\theta = \frac{3}{2}$$ and $$\phi = \frac{4}{25}$$.

Using the lagged operator notation, we can simplify the above: to $$Y_{t}(1-\phi L) = \epsilon_t(1 - \theta L)$$.

Therefore, $$Y_{t} = \epsilon_{t} \left[\frac{1-\theta L}{1-\phi L}\right]$$

I'm not sure what the method is for writing this is an infinite sum of $$\epsilon_t$$. If you can figure that out, the impulse response is straightforward. Unfortunately, I'm at a loss right now for how to do that. Maybe a good night's sleep will lead to a revelation. If anyone knows, feel free to finish this answer off.

Actually, I worked on it on scrap paper and I think I got it. The $$(1- \phi L)$$ in the bottom can be written as infinite series in the top. So, we get

$$Y_t = \epsilon_t (1-\theta L)(1 + \phi L + \phi^2 L^2 + \ldots +\phi^n L^n + \ldots )$$ \

$$= \epsilon_t(1 + (\phi - \theta)L + (\phi - \theta)\phi L^2 + (\phi - \theta) \phi^2 L^3 + \ldots + (\phi - \theta) \phi^{n-1}L^n$$

But the series above is an infinite series, so it can be written as:

$$= \sum_{j=0}^{\infty} \psi_{j} \epsilon_{t-j}$$

where $$\psi_0 = 1$$ and $$\psi_j = (\phi - \theta)\phi^{j-1}$$.

So, since the definition of the impulse response is the response due to the initial $$\epsilon$$ being 1.0 and then zero after that, the IRF becomes:

$$\sum_{j=0}^{\infty} \psi_{j}$$

Of course, you should put the actual numbers in for $$\psi_{j}$$.

If you have any questions, let me know. It's late and I could have made an algebra mistake somewhere but hopefully not a conceptual one.