# Superposition of random walk and autoregressive process

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 $$\sigma^{2}_{w}$$?

$$y_{t+1} = c_{t+1} + \alpha y_{t} + v_{t+1}.$$ 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).
• 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)? – Richard Hardy Nov 1 '19 at 14:40
• $v_{t}$ and $w_{t}$ are independent. The last equality is in distribution. I have edited. – ABK Nov 1 '19 at 14:42