# Median versus Harmonic Mean As Log Normal Data Summary

I have a set of data that follows a lognormal distribution (it is fixed-distance, variable-speed situation https://stats.stackexchange.com/a/23130/55305). I am trying to summarize the data in a single value and I would normally expect to use the harmonic mean. However, a co-worker suggested that a median might be a better measure.

I can't think of an argument for or against using the median to summarize the data ("average") versus using the harmonic mean. What are some things I should consider? Which will tell me what that the other won't?

As a note, I'm already capturing the mean and SD of the log of the values so that I can plug them into Wolfram Alpha for graphing (https://www.wolframalpha.com/input/?i=LogNormalDistribution[μ,σ]), but those data don't tell a human, at a glance, how fast or slow the system is on "average" since they are log values.

• Given $\log(Y) \sim N(\mu, \sigma^2)$, what do you want to present in terms of $\mu$ and $\sigma^2$ or the function of them? Then find the best estimate of it. – user158565 Nov 28 '18 at 17:27
• The median of a lognormal corresponds one to one with the median and the mean of the corresponding normal distribution; it is also the geometric mean of the same lognormal. If the lognormal has median $M$ then log($M$) is the centre of the corresponding normal. I haven't seen that the harmonic mean has any nice properties for the lognormal. But this is neither here nor there if interpretation of any other summary measure is more natural for your application. (I distrust "is lognormal" because at best real data follow named distributions approximately and I want to see the approximation.) – Nick Cox Nov 28 '18 at 17:34
• @Nick Agreed. Note, though, that the reciprocal of $X$ has a lognormal distribution when $X$ does and the harmonic mean of $X$ is the reciprocal of the arithmetic mean of $1/X.$ This reduces the situation to the (very) familiar one of relating the geometric and arithmetic means of lognormal variables. – whuber Nov 28 '18 at 17:39