Can the standard deviation be calculated for harmonic mean? Can the standard deviation be calculated for the harmonic mean? I understand that the standard deviation can be calculated for arithmetic mean, but if you have harmonic mean, how do you calculate the standard deviation or CV?
 A: Here is an example for Exponential r.v's.
The harmonic mean for $n$ data points is defined as
$$S = \frac{1}{\frac{1}{n} \sum_{i=1}^n X_i}$$
Suppose you have $n$ iid samples of an Exponential random variable,  $X_i \sim {\rm Exp}(\lambda)$. The sum of $n$ Exponential variables follows a Gamma distribution 
$$\sum_{i=1}^n X_i \sim {\rm Gamma}(n, \theta)$$
where $\theta = \frac{1}{\lambda}$. We also know that
$$\frac{1}{n} {\rm Gamma}(n, \theta) \sim {\rm Gamma}(n, \frac{\theta}{n})$$
The distribution of $S$ is therefore
$$S \sim {\rm InvGamma}(n, \frac{n}{\theta})$$
The variance (and standard deviation) of this r.v. are well known, see, for example here.
A: My answer to a related question points out that the harmonic mean of a set of positive data $x_i$ is a weighted least squares (WLS) estimate (with weights $1/x_i$).  You can therefore compute its standard error using WLS methods.  This has some advantages, including simplicity, generality, and interpretability as well as being automatically produced by any statistical software that allows weights in its regression calculation.
The principal disadvantage is that the calculation does not produce good confidence intervals for highly skewed underlying distributions.  That's likely to be a problem with any general-purpose method: the harmonic mean is sensitive to the presence of even a single tiny value in the dataset.
To illustrate, here are empirical distributions of $20$ independently generated samples of size $n=12$ from a Gamma(5) distribution (which is modestly skewed).  The blue lines show the true harmonic mean (equal to $4$) while the red dashed lines show the weighted least squares estimates.  The vertical gray bands around the blue lines are approximate two-sided 95% confidence intervals for the harmonic mean.  In this case, in all $20$ samples the CI covers the true harmonic mean.  Repetitions of this simulation (with random seeds) suggest the coverage is close to the intended 95% rate, even for these small datasets.

Here is the R code for the simulation and figures.
k <- 5             # Gamma parameter
n <- 12            # Sample size
hm <- k-1          # True harmonic mean
set.seed(17)

t.crit <- -qt(0.05/2, n-1)
par(mfrow=c(4, 5))
for(i in 1:20) {
  #
  # Generate a random sample.
  #
  x <- rgamma(n, k)
  #
  # Estimate the harmonic mean.
  #
  fit <- lm(x ~ 1, weights=1/x)
  beta <- coef(summary(fit))[1, ]
  message("Harmonic mean estimate is ", signif(beta["Estimate"], 3), 
          " +/- ", signif(beta["Std. Error"], 3))
  #
  # Plot the results.
  #
  covers <- abs(beta["Estimate"] - hm) <= t.crit*beta["Std. Error"]
  plot(ecdf(x), main="Empirical CDF", sub=ifelse(covers, "", "***"))
  rect(beta["Estimate"] - t.crit*beta["Std. Error"], 0, 
       beta["Estimate"] + t.crit*beta["Std. Error"], 1.25, 
       border=NA, col=gray(0.5, alpha=0.10))
  abline(v = hm, col="Blue", lwd=2)
  abline(v = beta["Estimate"], col="Red", lty=3, lwd=2)
}

A: What I would suggest is to use the following formula as a substitute for the standard deviation:
$$\sigma=\sqrt{\frac{N}{\sum_{i=1}^{N}{\left(\frac{1}{\hat{x}}-\frac{1}{x_i}\right)^2}}},$$
where $\hat{x} = \frac{N}{\sum \frac{1}{x_i}}$. The nice thing about this formula is that it is minimized when $\hat{x} = \frac{N}{\sum \frac{1}{x_i}}$, and it has the same units as the standard deviation would (which are the same units as $x$ has). 
This is in analogy to the standard deviation, which is the value that $\sqrt{\frac{1}{N}\sum(\hat{x}-x_i)^2}$ takes when it is minimized over $\hat{x}$. It is minimized when $\hat{x}$ is the mean: $\hat{x}=\mu=\frac{1}{N}\sum x_i$.
A: There is some concern that mpiktas's CLT requires a bounded variance on $1/X$.  It is true that $1/X$ has crazy tails when $X$ has positive density around zero.  However, in many applications using the harmonic mean,  $X\ge1$.  Here, $1/X$ is bounded by $1$, giving you all the moments that you want! 
A: Harmonic mean is the average amount by which all data differs from the mean. Notice: THIS IS EXACTLY THE SAME DEFINITION AS THE STANDARD DEVIATION! Therefore, no difference exists between these terms. They are the same!
