Why the Normal Distribution’s 10th–90th Percentile Range Scales with Volatility

Summary

The document proves that for a normally distributed variable with fixed mean and positive standard deviation, the difference between its 90th and 10th percentiles divided by the standard deviation is constant. The quantile function expresses each percentile as the mean plus the standard deviation times the corresponding standard normal quantile. Subtracting the two expressions cancels the mean and leaves a fixed difference of standard normal quantiles.

The answer gives the resulting ratio as approximately 2.56. This is a direct consequence of the location-scale form of the normal distribution, rather than a special property of a particular dataset. The result applies to the stated normal model; it does not establish the same relationship for other distributions or for empirical data that depart from normality.

Key ideas

• A normal distribution’s quantile at a fixed probability equals its mean plus its standard deviation times a standard normal quantile.
• Subtracting two normal quantiles removes the mean, leaving a difference proportional to standard deviation.
• The 90th-to-10th percentile range divided by standard deviation is constant across positive standard deviations.
• The stated constant is approximately 2.56 for the normal distribution.
• The result depends on the normal distribution assumption and need not hold for other distributions.

Tags

# Can someone prove (or disprove) this assertion about the normal distribution?


# Can someone prove (or disprove) this assertion about the normal distribution?












Let $X$ be distributed as a $Normal (\mu, \sigma^2)$. Then for a fixed $\mu$ it is always the case that:

\begin{equation} \frac{90th quantile-10th quantile}{\sigma}=constant \quad \forall \sigma>0 \end{equation}

Thanks in advance!

## Answer by Kevin (score 2, accepted)

https://quant.stackexchange.com/a/51263

Let $p\in(0,1)$. The corresponding quantile function of $X\sim N(\mu,\sigma^2)$ is given by $$F_X^{-1}(p)=\mu+\sigma\Phi^{-1}(p)=\mu+\sqrt{2}\sigma\mathrm{erf}^{-1}(2p-1),$$ where $\Phi^{-1}$ is the inverse of the cumulative distribution function of a standard normally distributed random variable and $\mathrm{erf}^{-1}$ is the inverted error function.

Thus, \begin{align} \frac{\mathrm{Quantile}(0.9)-\mathrm{Quantile}(0.1)}{\sigma}&=\frac{\mu+\sigma\Phi^{-1}(0.9)-(\mu+\sigma\Phi^{-1}(0.1))}{\sigma} \\ &=\Phi^{-1}(0.9)-\Phi^{-1}(0.1) \\ &\approx 2.56. \end{align}

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.