## MC exercise

In this Monte Carlo exercise, we empirically assess the sampling distribution of $$\hat{\phi}$$, the OLS estimate of the AR(1) coefficient $$\phi$$: $y_t = \phi y_{t-1} + \epsilon_t \qquad \epsilon_t \sim N(0,1).$ When $$|\phi|<1$$, it is well known that $\hat{\phi}|\phi \sim N\left(\phi,\frac{1}{\sum_{t=1}^n y_{t-1}^2}\right) \ \ \ \mbox{and} \ \ \ t = \sqrt{\sum_{t=1}^n y_{t-1}^2}(\hat{\phi}-\phi) \sim N(0,1)$ Let us generate $$S=100$$ samples of size $$n$$ from the above AR(1) model and compute the $$t$$ statistic for different samples sizes $$n$$ and difference persistence parameter $$\phi$$. The following plot contrasts the standard normal against the sampling distribution of the $$t$$ test above. Notice that when $$\phi=0.5$$ the two densities are almost identical. On the other hand, when $$\phi=1$$ (unit root) the sampling densities is shifted to the left of the standard normal density.

set.seed(31416)
S  = 1000
par(mfrow=c(2,3))
for (phi in c(0.5,1.0)){
for (n in c(100,200,1000)){
tt = rep(0,S)
for (s in 1:S){
y = rep(0,n)
for (t in 2:n){
y[t] = rnorm(1,phi*y[t-1],1)
}
xtx     = crossprod(y[1:(n-1)])
xty     = crossprod(y[2:n],y[1:(n-1)])
phi.hat = xty/xtx
tt[s]    = sqrt(xtx)*(phi.hat-phi)
}
L = round(quantile(tt,0.05),3)
plot(density(tt),main="",xlab="")
title(paste("phi=",phi," - n=",n,sep=""))
legend("topleft",legend=c(paste("q(5%)=",L,sep="")))
xxx = seq(-5,5,length=200)
lines(xxx,dnorm(xxx),col=2)
}
}