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The arithmetic mean of a lognormal distribution is greater than the geometric mean (which equals the median). So its percentile is greater than 50%. But how much greater? It depends on the geometric standard deviation, which defines the asymmetry of the distribution.

What function defines the percentile of the arithmetic mean (of a lognormal distribution) from the geometric standard deviation?

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It's easiest to work on the log-scale, with a normal distribution.

The log of the mean of a lognormal in the usual parameterization is $\mu+\frac12\sigma^2$. Its z-value is therefore $\frac12\sigma$, so I believe this quantile should be at the $\Phi(\frac12\sigma)$ point of the cdf.

A little more formally, let $Y\sim \text{logN}(\mu,\sigma^{2})$, then:

$$\begin{align}P(Y\leq E[Y]) &= P(Y\leq e^{\mu+\frac12 \sigma^2})\\ &= P(\log(Y)\leq\mu+\frac12 \sigma^2)\\ &= P(\frac{\log(Y)-\mu}{\sigma}\leq\frac12 \sigma)\\ &= P(Z\leq\frac12 \sigma)\\ &= \Phi(\frac12 \sigma)\end{align}$$

e.g. at $\sigma=0.2$, the mean is at about the 54th percentile.

Here's a quick simulation in R as a check on my work -

 nsim = 1000000
 mu = 0
 sig = 0.2
 mnln = exp(mu+sig^2/2)
 y = rlnorm(nsim,mu,sig)
 c(sim = mean( y < mnln ), exact = pnorm( sig/2 ) )
       sim     exact 
 0.5392810 0.5398278 

Note, however, that $\sigma$ here is the standard deviation of the natural logs of the lognormal random variable. The geometric standard deviation equals $\exp(\sigma)$ (https://en.wikipedia.org/wiki/Geometric_standard_deviation#Relationship_to_log-normal_distribution).

So now if $\sigma_g = \exp(\sigma)$ then $\Phi(\frac12 \sigma)=\Phi(\frac12 \log(\sigma_g))$, where "log" is the natural logarithm. This equation computes the quantile; multiply by 100 to obtain the percentile, graphed below.

enter image description here

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  • $\begingroup$ #Glen_b thanks for the instant solution. $\endgroup$ Commented Aug 20, 2022 at 17:22
  • $\begingroup$ Thanks Harvey, I appreciate the addition. I actually generated a similar plot for myself (with $\sigma$ on the x axis) - and computed the $\sigma$ where the mean was at the 75th and 95th percentiles respectively - but hesitated to include it - it's sometimes hard to judge when such things would be welcome and when I might be labouring the point. $\endgroup$
    – Glen_b
    Commented Aug 21, 2022 at 1:02

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