In the case where  are i.i.d. random variables, then

Now, what if  are identically distributed, but no longer independent. What if we have an autoregressive process? Assume that

Then

can be written

Here, we will express the variance as a function of  and , but it is possible to use also , since, in the context of an ,

Now, since  we get

which can be simplified, since

i.e.

So, the variance of the mean can be writen as

Observe that if  is large enough,

This asymptotic relationship is well known actually. A simple way to get it is the following. One can can write

or equivalently

But actually, the first relationship is probably more interesting to get an asymptotic approximation,

In the context of an  process, this can be writen

Thus, we get the following well-known relationship

In the case where  is an i.i.d. sequence, i.e. , then we get the relationship mentioned initially. And in the case of a random walk… unfortunately, we cannot use that relationship. But observe that

i.e.

which can be written

If we compare the true value and the approximation, we get the following graph,

> V=function(phi,s2=1,n=100){
+ g0=s2/(1-phi^2)
+ if(phi<1){
+ if(phi==0){v1=g0/n}
+ if(phi>0){v1=g0/n^2*(n+2*((n-1)*
+ phi^(-1)-n+phi^(n-1))/(phi^(-1)-1)^2)}
+ v2=g0/n*(1+phi)/(1-phi)
+ }
+ if(phi==1){
+ v1=(2*n+1)*(n+1)*s2/(6*n)
+ v2=NA
+ }
+ return(c(v1,v2))}
> 
> Vphi=function(phi) V(phi,1,100)
> x=seq(.01,1,by=.02)
> M=matrix(unlist(lapply(x,V)),nrow=2)
> plot(x,M[1,],type="l",col="red",log="y",
+ ylab="Variance of the average (log scale)",
+ xlab="Autoregressive coefficient")
> lines(x,M[2,],col="blue")

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