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first time here, so I hope that you will could help me!

I went to simulate stationary and positive AR(1) process such as $X[t]=a*X_{t-1}+ \sqrt{1-a^2}*e_t$, where $e\sim N(n,0,1)$ and $a=0.1$. In order to get positive value, can I just use the absolute value of X? Is an absolute value of an AR(1) still an AR(1)?

Here are my simulation steps :

n=100

e=rnorm(n)

a=0.1

X=rep(0,n)

X[1]<-rnorm(1,0,1)

for(t in 2:n){X[t]=a*X[t-1]+ sqrt(1-a^2)*e[t]}

X<-abs(X)
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    Consider noting what programming language you use when you write code. It is not obvious to everyone that you use R. – 2017-01-15
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    Yes, I forget to notice that! I working under R – 2017-01-15

1 Answers 1

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I chose to answer your question in a list form since I felt your question consisted of several sub-questions.

  1. It is important to note whether the sequence $\{e_t\}$ consists of independent random variables or not.

  2. Instead of writing $X_t = aX_{t-1} + \sqrt{1-a^2}e_t$ with $e_t\sim\mathcal N(0,1)$, we rather write $X_t = aX_{t-1} + e_t$ with $e_t \sim \mathcal N(0,1-a^2)$. This is because then $X_t$ is on the standard form of an $\operatorname{AR}(1)$ process. You may say that $\{e_t\}$ is a white noise process with variance $\sigma^2 = 1-a^2$.

  3. If you take the absolute value of the process, I suppose you may still be able to find $e'_t$ such that $\lvert X_t\rvert = Y_t = c+ a'Y_{t-1} + e'_t$ is a stationary time series and thus that $Y_t$ is an $\operatorname{AR}(1)$ process for some mean-zero process $e'_t$. If $X_t$ is stationary so is $Y_t$ which is usually the important part.

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    First, thank you for your answer...Yes, the ${e_t}$ are independent and as they follow a normal distribution, it still has the same distribution with adding absolute value? and can you give me some points about how to proof that $Y_t$ is an AR(1)? – 2017-01-15