It is known that variance of AR(1) $y_t=\phi y_{t-1}+\varepsilon_{t}$ is $$\text{Var}(y)=\frac{\sigma_\varepsilon^2}{1-\phi^2}$$ Let $\varepsilon_{t}$ has continuous uniform distribution on $[-1, 1]$ and $y_0 = 0$
How can we estimate how often a process will go beyond given interval $[-a, a]$?

  • 2
    $\begingroup$ Is the question about the proportion of time spent outside the interval or about the number of crossings of the boundary? $\endgroup$
    – Yves
    Nov 14, 2022 at 13:07
  • $\begingroup$ In some interpretations the answer is always. Please explain what it means by "process go[ing] beyond [a] given interval." $\endgroup$
    – whuber
    Jan 31 at 15:51

1 Answer 1


By the $MA(\infty)$ representation, we may write $y_t$ as a linear combination of the uniform $\varepsilon_t$. This weighted sum should then, for the process running for sufficiently many periods $n$, behave approximately normally (the exact distribution of a linear combination of uniforms will be difficult to handle), so that it will exceed the normal $\alpha/2$ and $1-\alpha/2$ quantiles times the standard deviation of $y$ approximately a fraction $\alpha$ of observations.

Here is an illustration showing that the 5 and 95% quantiles are exceeded 10% of the times on average. (In practice, you could estimate $\phi$ and $\sigma^2_\varepsilon$ from the series.) For other values of $a$ you'd back out which quantile they correspond to. (Or you resort to the simulation approach for a given value of $a$ directly.)

sigma.eps <- 2^2/12 # the variance of a uniform with support of width two
phi <- 0.5

v.y <- 2^2/12/(1-phi^2)

alpha <- 0.10
a <- qnorm(1-alpha/2)
boundary <- a*sqrt(v.y)

n <- 2000

# a single run
y <- arima.sim(list(ar=phi), n, innov = runif(n, -1, 1))
mean(abs(y) > boundary)

# a little "simulation study"
mean(replicate(10000, mean(abs(arima.sim(list(ar=phi), n, innov = runif(n, -1, 1))) > boundary)))

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