Given is the AR(1) model: $y_t=\rho y_{t-1}+\epsilon_t$ with $\epsilon_t $ i.i.d. $N(0, \sigma^2), t=1,...,n$ and $y_0=0$ with $\epsilon_0\sim N(0,\frac{\sigma^2}{1-\rho^2}),\lvert\rho\rvert<1;$

Why is $\epsilon_0$ there and why its variance is $(\frac{\sigma^2}{1-\rho^2})$? What is the reason behind setting up distribution for $\epsilon_0$ equal to long term variance of $y_t$?


1 Answer 1

  1. the error term and it's variance are usually assumed to be constant (even though it's subscripted by $t$). The expectation is usually assumed to be zero and the variance is assumed to be the long term variance of the model. ( see answer in 3) for details ).

  2. You need a value for starting the process so, a most of time, the convention is to set the initial value, $y_0$ to zero. Assuming the series is long enough, it really won't make a difference what you set it to, during estimation.

  3. To calculate the long term variance of the model, you can rewrite the model (using the lag operator and the fact that $t$ starts at 1) as $y_t = \sum_{i=0}^{\infty} \rho^{i} \epsilon_{t-i+1}$.

Then calculate the variance of the RHS using the infinite series formula. You'll get what you have stated in your question.

EDIT: I CALCULATE THE LONG TERM VARIANCE BELOW #===================================================================

VAR $(y_t) = $ VAR $\left(\sum_{i=0}^{\infty} \rho^{i} \epsilon_{t-i+1}\right)$ $ = \left(\frac{\sigma^2}{1-\rho^2}\right) $

Notice that the long term variance of the model is also equal to the variance of $\epsilon_{0}$.

  • $\begingroup$ with step 3 i get the variance of $y_t$ or according to my question $\epsilon_t$. why the initial noise term $\eta_0$ has the same distribution as var($\epsilon_t$). $\endgroup$ Jan 21, 2023 at 13:19
  • $\begingroup$ Hi: you shouldn't use $\epsilon_t$ as the response because it's normally used as the error term. So I replaced your $\epsilon_t$ with $y_t$ and your $\eta_t$ with $\epsilon_t$. My apologies for confusion but my notation is more standard. $\endgroup$
    – mlofton
    Jan 22, 2023 at 0:09
  • $\begingroup$ eitherway your step 3 only gives me the variance result interms of $y_t$ where as i need an expression for variance of $\epsilon_0$ $\endgroup$ Jan 22, 2023 at 0:13
  • $\begingroup$ i corrected the terms in your favor. would u please again have a look at my question? $\endgroup$ Jan 22, 2023 at 0:20
  • $\begingroup$ at time $t$, the only random component is $\epsilon_t$ so what's happening is that they are setting the variance of $\epsilon_0$ to the long term variance of the model. The long term variance of the model is VAR $(\sum_{i=0}^{\infty} \rho^{i} \epsilon_{t-i})$. Let me explain this in the answer itself. $\endgroup$
    – mlofton
    Jan 22, 2023 at 0:29

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