# Variance of the Average of a Sequence

In the case where $\{Y_1,\cdots,Y_n\}$ are i.i.d. random variables, then

$\text{Var}\left(\frac{1}{n}\sum_{t=1}^n Y_t\right)=\frac{\text{Var}(Y_t)}{n}$

Now, what if $\{Y_1,\cdots,Y_n\}$ are identically distributed, but no longer independent. What if we have an autoregressive process? Assume that

$Y_t=\phi Y_{t-1}+\varepsilon_t$

Then

$\text{Var}\left(\frac{1}{n}\sum_{t=1}^n Y_t\right) = \frac{1}{n^2}\left(\sum_{t=1}^n \text{Var}(Y_t)+\sum_{s\neq t}\text{Cov}(Y_s,Y_t) \right)$

can be written

$\text{Var}\left(\frac{1}{n}\sum_{t=1}^n Y_t\right) = \frac{1}{n^2}\left(n\gamma(0) + \sum_{h=1}^{n-1} 2 (n-h) \gamma(h)\right)$

Here, we will express the variance as a function of $\gamma(0)$ and $\phi$, but it is possible to use also $\sigma^2$, since, in the context of an $AR(1)$,

$\gamma(0)=\frac{\sigma^2}{1-\phi^2}$

Now, since $\gamma(h)=\phi^h \gamma(0)$ we get

$\text{Var}\left(\frac{1}{n}\sum_{t=1}^n Y_t\right) = \frac{\gamma(0)}{n^2}\left(n\gamma(0) + \sum_{h=1}^{n-1} 2 (n-h) \phi^h\right)$

which can be simplified, since

$\sum_{h=1}^{n-1} 2(n-h) \phi^h=2\phi^{n-1} \sum_{h=1}^{n-1} (n-h) \left(\frac 1\phi\right)^{n-h-1}=2\phi^{n-1} \frac {\partial}{\partial x}\left. \sum_{i=1}^{n} x^{n-i} \right\vert_{x=1/\phi}$

i.e.

$\sum_{h=1}^{n-1} 2(n-h) \phi^h=2\phi^{n-1} \frac{(n-1)\phi^{-(n+1)} - n\phi^{-n} + \phi^{-1}}{\phi^{-1}(\phi^{-1}-1)^2}=2 \frac{(n-1)\phi^{-1} - n + \phi^{n-1}}{(\phi^{-1}-1)^2}$

So, the variance of the mean can be writen as

$V=\mathrm{Var}\left(\frac{1}{n}\sum_{t=1}^n Y_t\right)=\frac{\gamma(0)}{n^2}\left[n + 2 \frac{(n-1)\phi^{-1} - n + \phi^{n-1}}{(\phi^{-1}-1)^2}\right]$

Observe that if $n$ is large enough,

$V=\frac{\gamma(0)}{n^2}\left[n + 2 \frac{(n-1)\phi^{-1} - n + \phi^{n-1}}{(\phi^{-1}-1)^2}\right]\sim \frac{\gamma(0)}{n}\frac{1+\phi}{1-\phi}$

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

$V=\mathrm{Var}\left(\frac{1}{n}\sum_{t=1}^n Y_t\right)=\frac{\gamma(0)}{n}\left[ \sum_{h=-n+1}^{n-1}\left(1-\frac{\vert h\vert}{n}\right)\rho(h) \right]$

or equivalently

$V=\mathrm{Var}\left(\frac{1}{n}\sum_{t=1}^n Y_t\right)=\frac{\gamma(0)}{n}\left[ 1+2\sum_{h=1}^{n-1}\left(1-\frac{h}{n}\right)\rho(h) \right]$

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

$V=\mathrm{Var}\left(\frac{1}{n}\sum_{t=1}^n Y_t\right)\sim\frac{\gamma(0)}{n}\left[ \sum_{h=-\infty}^{\infty}\rho(h) \right]$

In the context of an $AR(1)$ process, this can be writen

$\sum_{h=-\infty}^{\infty}\rho(h) = \frac{1+\phi}{1-\phi}$

Thus, we get the following well-known relationship

$V=\frac{\gamma(0)}{n}\cdot \frac{1+\phi}{1-\phi}$

In the case where $\{Y_1,\cdots,Y_n\}$ is an i.i.d. sequence, i.e. $\phi=0$, then we get the relationship mentioned initially. And in the case of a random walk… unfortunately, we cannot use that relationship. But observe that

$V=\frac{1}{n}\text{Var}\left(\sum_{t=1}^n\sum_{h=1}^t\varepsilon_h\right)$

i.e.

$V=\frac{1}{n}\text{Var}\left(n\varepsilon_1+(n-1)\varepsilon_2+\cdots+\varepsilon_n\right)$which can be written

$V=\frac{\text{Var}(\varepsilon)}{n}\sum_{h=1}^n h^2=\frac{(2n+1)(n+1)}{6n}\text{Var}(\varepsilon)$

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")

This site uses Akismet to reduce spam. Learn how your comment data is processed.