1
$\begingroup$

Let us consider the following model:

$$ y_{t} = c_{t} + \alpha y_{t-1} + v_{t} \\ c_{t+1} = c_{t} + w_{t} $$ where $v_{t} \in \mathcal{N}(0, \sigma^{2}_{v})$ and $w_{t} \in \mathcal{N}(0, \sigma^{2}_{w})$ are independent.

The model above is a superposition of random walk and autoregressive process.

Is there a common approach to estimate $\alpha$, $\sigma^{2}_{v}$ and $\sigma^{2}_{w}$?

$\endgroup$
1
$\begingroup$

I have the following idea, please, criticise!

$$ y_{t+1} = c_{t+1} + \alpha y_{t} + v_{t+1}. $$ Therefore, $$ y_{t+1} - y_{t} = c_{t+1} - c_{t} + \alpha (y_{t} - y_{t-1}) + v_{t+1} - v_{t}. $$ Next, note that $c_{t+1} - c_{t} = w_{t}$. Then $$ y_{t+1} - y_{t} = \alpha (y_{t} - y_{t-1}) + w_{t} - v_{t} + v_{t+1}, $$ which is equivalent (equivalent in distribution) to $$ y_{t+1} - y_{t} = \alpha (y_{t} - y_{t-1}) + \frac{\sqrt{(\sigma^{2}_{v} + \sigma^{2}_{w})}}{\sigma_{v}}v_{t} + v_{t+1}. $$

Therefore, $(y_{t+1} - y_{t})$ is ARMA(1,1) process and the parameters can be estimated using ARMA(1,1).

$\endgroup$
  • $\begingroup$ All looks good to me, except: how did you obtain that $w_t-v_t$ is a constant times $v_t$ (the result of moving from the penultimate to the last equation)? $\endgroup$ – Richard Hardy Nov 1 '19 at 14:40
  • $\begingroup$ $v_{t}$ and $w_{t}$ are independent. The last equality is in distribution. I have edited. $\endgroup$ – ABK Nov 1 '19 at 14:42
  • $\begingroup$ I do not think equivalence in distribution is enough to declare the model ARMA(1,1). I mean, the last equation is ARMA(1,1), but the penultimate equation is not. I do not have an immediate proof available, but I guess a counterexample could be illustrated by simulations (given sufficient estimation precision). $\endgroup$ – Richard Hardy Nov 1 '19 at 14:47
  • 1
    $\begingroup$ It looks like they are doing the thing we are discussing here. However, that does not seem to change anything of what we discussed. The conditional distributions are not equal, unless I am wrong (which is possible; if so, I would like to get corrected). If this does not matter for parameter estimation and if parameter estimation is your goal, then you are good to go. The question that is not obvious for me is, does this really not matter for parameter estimation? $\endgroup$ – Richard Hardy Nov 1 '19 at 15:20
  • 1
    $\begingroup$ @Richard Hardy: I definitely don't speak from authority and am always trying to learn from the gurus on this list. I've just seen that technique so many times because it's often it's used to show equivalences. For example, take the random walk plus noise and the ARIMA(0,1,1), You can show their equivalence using the same technique. If I can find a proof of the statement, I'll let you know. I've looked before and wasn't successful. $\endgroup$ – mlofton Nov 3 '19 at 3:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.