Is it possible to graph this? Is it possible to graph this?  I tried doing it on a calculator but didn't have any success.  If someone can and post the graph I would greatly appreciate it.
$$\frac{1}{\bar{m}_{T}}\sqrt{\frac{1}{12} \sum_{h=T-11}^{T} (m_{h} - \bar{m}_{T})^2}.$$
If the equation is hard to read, it is just the standard deviation of 12 observations $(m_{h})$, $h=T-11, T-10, \ldots, T$, over the mean of the same 12 observations. The standard deviation is the population standard deviation and the mean $\bar{m}_{T}$ is simple average.
 A: I have no idea if this is what the poster is looking for, but I was curious so started exploring and thought I'd share.  Using more traditional notation, let $X_1, ..., X_{n}$ be i.i.d. with $X_i \sim N(\mu, \sigma^2)$.  Then $E(\bar X | X_1)$ is a linear function of $X_1$ and $E(\sum(X_i-\bar X)^2/n | X_1)$ is a quadratic function of $X_1$; so if the mean is always positive, the expected value of the variance/mean is quadratic also.  Not too hard to check if you're patient with your algebra; I'm not, so will demonstrate by simulation. Whether the sd/mean is quadratic or not is left to the reader, who is again hopefully more patient with algebra than I. 
An intuitive explanation is that when one of the random variables is above or below the true mean, the sample mean is also more likely to be above or below the true mean.  But the sample standard deviation is more likely to be large when one of the random variables is far from the true mean (either above or below) and more likely to be small when it's near the true mean.
 
set.seed(5)
N <- 1000
n <- 12

d <- matrix(rnorm(n*N,mean=8),ncol=n)
m <- rowMeans(d)
s <- apply(d, 1, sd)

par(mfrow=c(1,3))
plot(d[,1],m, col="#88888888", xlab=expression(m[1]), ylab="mean", main="mean")
lines(lowess(d[,1], m))
plot(d[,1],s, col="#88888888", xlab=expression(m[1]), ylab="sd", main="sd")
lines(lowess(d[,1], s))
plot(d[,1],s/m, col="#88888888", xlab=expression(m[1]), ylab="sd/mean", main="sd/mean")
lines(lowess(d[,1], s/m))

A: Typically, when data are serially indexed like this, they form a series (usually a time series): the formula asks for the coefficient of variation within a twelve-period moving window (as if the data were monthly and the moving window were a year, for instance).  If this is the case, graphing takes two steps:


*

*For each index $T$ of 12 or greater, compute the CV for the window spanning indexes $T-11$ through $T$.  This gives a derived series $CV_T$, also indexed by $T$, for all $T \ge 12$.

*Plot the pairs $(T, CV_T)$.
For example, the data (shown for $T=1, 2, \ldots, 100$) might look like

and the derived series--the "moving CV"--would then look like

