12
$\begingroup$

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%.

$\endgroup$
  • $\begingroup$ real options... have they started to take into account that one is short one's competitor's options ? $\endgroup$ – nicolas May 22 '11 at 18:52
  • 1
    $\begingroup$ Do you mean, that it's just one of the models for the non-constant (in this case, time-dependent) volatility? Then the form of it has some logic: as you've noticed it decreases with time (if $t$ is time, not time-to-maturity) and the higher the drift, the lower the effect of the "real" vol on the diffusion coefficient $\sigma'$. $\endgroup$ – Ulysses Nov 7 '14 at 8:26
  • $\begingroup$ I read thru the original Boeing approach paper (doi.org/10.1111/j.1745-6622.2007.00140.x) and don't see the formula you ask about. Can you give a link? $\endgroup$ – q.t.f. May 21 '18 at 21:33
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.