Consider the ARMA(1,1) process \begin{align} y_t=a_1y_{t-1}+b_1\epsilon_{t-1}+\epsilon_t, \end{align} and assume $\epsilon_t$~$N(\mu_t,\sigma_t)$. And $\mu_t$ and $\sigma_t$ are all known.

$f(y_t|y_{t-1})$ and $f(y_t>c|y_{t-1})$ are easy to calculate. For instance $f(y_t|y_{t-1})$ is the Normal density with mean $a_1y_{t-1}+b_1\epsilon_{t-1}$ and variance $\sigma^2_t$.

But how do I calculate $f(y_t|y_{t-1}>c)$ and $f(y_t>c|y_{t-1}>c)$?

Many thanks!


1 Answer 1


Assuming $y_t$ is stationary and $|a_1|<1$.

First calculate the joint distribution of $f(y_t,y_{t-1})$. This you can calculate using the conditions

$f(y_t,y_{t-1}) = f(y_t|y_{t-1}) f(y_{t-1})$

You already know about this part $f(y_t|y_{t-1})$ and $f(y_{t-1})$ is the unconditional density function (The distribution will be normal and you can calculate the mean and variance).

From the joint distribution (bivariate normal) you can calculate the probabilities.

Eg: $f(y_t | y_{t-1}>c) = \dfrac{\int_{c}^{\infty} f(y_t, y_{t-1})dy_{t-1} } {\int_{c}^{\infty} f(y_{t-1})dy_{t-1}}$

I think these are good hints to proceed your problem.

  • $\begingroup$ Thanks so much for the post! I was aware of the steps but I didn't know how to calculate the unconditional density, i.e. $f(y_{t-1})$. That was where I got stuck..... $\endgroup$
    – dynamic89
    Commented Feb 11, 2014 at 21:46
  • $\begingroup$ @dynamic89, If you are assuming $\epsilon_t$ are iid, by repeated substitution you can write $y_t$ as sum of $\epsilon_t$. Sum of independent normal rvs-> normal $\endgroup$
    – vinux
    Commented Feb 12, 2014 at 2:51
  • $\begingroup$ $\epsilon_t$ is not iid though $\endgroup$
    – dynamic89
    Commented Feb 12, 2014 at 7:42
  • $\begingroup$ to be precise, $\epsilon_t$ are independent, not identically distributed $\endgroup$
    – dynamic89
    Commented Feb 12, 2014 at 7:43
  • $\begingroup$ That means, the series is not stationary. But still you get $y_t$ as sum of independent normal random variables (Hence $y_t$ is normally distributed). $\endgroup$
    – vinux
    Commented Feb 12, 2014 at 7:59

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