# Behavior of AR process

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 $$[-a, a]$$?

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

## 1 Answer

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)))