This questions is two part:

1) What happens when you apply the backshift operator to a constant? For example, if I have the AR process $$(1-\phi B)(y_t-\mu)=\epsilon_t$$ does that equal $$y_t-\mu-\phi By_t-\phi B\mu = \epsilon_t$$ which (I believe reduces to) $$y_t-\mu-\phi y_{t-1}-\phi \mu = \epsilon_t\longrightarrow y_t=\mu+\phi y_{t-1}+\phi \mu+\epsilon_t $$

So am I correct in assuming that the backshift of a constant (in my example $\mu$) is just the constant?

2) If I assume that $\epsilon_t\sim N(0,v)$, then what is the likelihood of the above AR process in 1?


The Backshift operator operates normally on a constant as on every other symbol. So it shifts the constant one period back -where we find that the constant has the same value as in the current period, since this is what the essence of a constant is.

For the likelihood of an AR(1) process, in this answer there is the likelihood for the case without the constant -but from there it is just a small step to here.

The chain rule will be the same, but the conditional density will be

$$Y_i | Y_{i-1},\dots,Y_0 \sim \mathcal{N}\left((1+\phi) \mu+\phi Y_{i-1},v\right) $$

You need to specify what the distribution of $Y_0$ will be (will it contain the unknown parameters $\phi$, $v$? If not, it doesn't really matter.

  • $\begingroup$ I am not sure I follow how to do this. Do you think you could just show the last step of what the likelihood should be? $\endgroup$ – user30490 Oct 16 '13 at 18:10

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