If a variable X runs in AR(1) process X = alpha*X(-1) + e,

where absolute value of alpha < 1, X(-1) is a lagged variable and e is an error term.

What happens to X in steady state. Does X equal to 1 in steady state? I know this is a stupid question. Can anyone please give me an intuition of why X is equal to 1?

Hi, here you have a simple first order auto-regressive process:

x_t = \alpha x_{t-1}+\varepsilon_t

Since you want to compute the deterministic steady state we can drop the zero mean disturbance:

x_t = \alpha x_{t-1}

So the steady state x^{\star} must satisfy:

x^{\star} = \alpha x^{\star}

and the only possible value for the steady state is x^{\star}=0 (certainly not 1). If you want to have a non zero mean steady state with an auto-regressive process you need to add a constant:

x_t = c + \alpha x_{t-1}+\varepsilon_t

and you can check, following the same line of reasoning, that the steady state is then x^{\star}=\frac{c}{1-\alpha}, so the steady state will be equal to one iff c-1-\alpha.

Best,
Stéphane.

1 Like

Hi Stéphane,

Thank you so much. I really appreciate the knowledge.

Sincerely,

Eric