Consider stationary autoregression AR(1): $$u_t=\beta u_{t-1}+ \varepsilon_t, \quad t \in \mathbb{Z}.$$
$\{\varepsilon_t\}$ - i.i.d. $N(0,\sigma^2)$ random variables.
I know that $\mathbf{E}u_t =0$ and $\mathbf{E}u_t^2=\sigma^2/(1-\beta^2).$
The question is: what is the distribution of $u_t$? What is its CDF or PDF?
Thanks in advance.
On
That is one of the objectives of the whole modeling exercise.
We desire to obtain the distribution of the solution to a time series. Thus, we need to obtain the distribution of the error terms since the AR(1) equation is determined simply by past observations of the random variable X and the error terms. So, if we knew the distribution of the error terms, we would have the distribution of the AR(1) equation. In addition, we would need to know the value of ρ , which we find an estimator for using the method of least squares.
First, we observe a plot of the data to get an idea of what trends there might be. We also create a histogram of the observed data to begin to get an idea of what distribution we could use to test for goodness-of-fit. It may be necessary to “play around” with the parameters of the histogram by decreasing or increasing the interval length for the data to be binned. This is done in order to try to fit the shape of the histogram as closely to a distribution curve of a specific probability distribution. Now we calculute the estimators for ρ and our distribution function and use the chi-squared goodness-of-fit test to determing whether the observed data could be of the hypothesized distribution.
For full details, please read from the source: http://www.math.utah.edu/~zhorvath/ar1.pdf
Iterating the recursion, one sees that, for every $t$, $$u_t=\sum_{s\geqslant0}\beta^s\varepsilon_{t-s}, $$ where the family $(\varepsilon_s)$ is i.i.d. normal $(0,\sigma^2)$, hence each $u_t$ is normal $(0,\tau^2)$, where $\tau^2$ is indeed $$ \tau^2=\sum_{s\geqslant0}\beta^{2s}\sigma^2=\frac{\sigma^2}{1-\beta^2}. $$