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Correcting a Black–Scholes Volatility Transformation for Lognormal Moments

Article Quant Q&A · Author: Corn
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Summary

The note questions a proposed transformation for substituting volatility into the Black–Scholes formula when valuing real options. The proposed expression uses the mean and standard deviation of simulated payoffs, and the author observes that its behavior does not match an expected volatility level as the project approaches execution.

The response suggests a likely formula error: placing the squared ratio of standard deviation to mean inside the logarithm yields the familiar relationship between the first two moments of a lognormal variable. This connection can help translate arithmetic mean and variance into lognormal volatility. The exchange offers a conceptual correction rather than a derivation or numerical validation, and it does not establish whether the corrected transformation is appropriate for the specific real-options model or its payoff simulation.

Key ideas

  • The proposed volatility transformation uses simulated payoff mean and standard deviation.
  • The response identifies a likely missing square in the ratio inside the logarithm.
  • The corrected expression corresponds to the relationship between the first two moments of a lognormal variable.
  • A moment match alone does not establish that the transformation fits a particular real-options model.

Tags

OptionsVolatilityDerivatives pricingStatistics
Full text
# Transformation of Volatility - BS


# Transformation of Volatility - BS












I have recently seen a paper about the Boeing approach that replaces the "normal" Stdev in the BS formula with the Stdev

\begin{equation} \sigma'=\sqrt{\frac{ln(1+\frac{\sigma}{\mu})^{2}}{t}} \end{equation}

$\sigma$ and $\mu$ being the "normal" Stdev and Mean, respectively. (Both in absolute values, resulting from a simulation of the pay-offs.)

Since it is about real options, it sounds reasonable to have the volatility decrease approaching the execution date of a project, but why design the volatility like this? I have plotted the function here via Wolframalpha.com. Even though the volatility should be somewhere around 10% in this example, it never assumes that value. Why does that make sense?

I've run a simulation and compared the values. Since the volatility changes significantly, the option value changes, of course, are significant.

Here some equivalent expressions. Maybe it reminds somebody of something that might help?

$\Longleftrightarrow t\sigma'^{2}=ln(1+\frac{\sigma}{\mu})^{2}$

$\Longleftrightarrow\sqrt{exp(t\sigma'^{2})}-1=\frac{\sigma}{\mu}$

$\Longleftrightarrow\sigma=\mu\left[\sqrt{exp(t\sigma'^{2})}-1\right]$

It somehow looks similar to the arithmetic moments of the log-normal distribution, but it does not fit 100%.

## Answer by Ezy (score 1)

https://quant.stackexchange.com/a/42191

looks like a typo maybe, just like you said if

\begin{equation} \sigma'=\sqrt{\frac{ln\left(1+(\frac{\sigma}{\mu})^{2}\right)}{t}} \end{equation}

then this would match exactly the relationship between the first 2 moments of a lognormal variable.

Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)

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