The purpose of this short note is to walk through some of the equations used for the "Black Swan Ponds"in a recent visualization I have been playing with. This was done to double-check some things, and serve as a point of reference in case someone finds an issue with the equations that are being used or their implementation.

The visualization is available here: https://learnforeverlearn.com/blackscholes/?tab=options&strategy=flock-of-christmas-trees

In this case, using 3 standard deviations as boundary,

with σ=0.5 and t=0.083

What I am doing is highlighting the region where the stock price after time $t$ is at least $N$ standard deviations from the mean - following "standard" assumptions regarding the stock price' behavior being "geometric Brownian motion".

I don't think that there is a definitive definition of the value for $N$ which constitutes a Black Swan. Just that it is "extremely unlikely", relative to standard methods of quantifying probability. I don't think that one wants to necessarily predict/estimate probabilities of them, rather than simply seeing where this region might lie relative to "normal" assumptions - "normal" in both the statistical and non-statistical senses.

Assuming that the stock price follows a geometric Brownian motion process, ultimately this means that the distribution of stock price follows a lognormal distribution with arithmetic(!) mean and variance given by (see, e.g., wikipedia),

$$ \begin{eqnarray} E(S_t) &=& \text{arithmetic mean} = S_0 * e^{\mu t} \nonumber \\ Var(S_t) &=& \text{arithmetic variance} = S_0^2 * e^{2 \mu t} \left({ e^{ \sigma^2 t} - 1 }\right) \nonumber \end{eqnarray} $$ where $$ \begin{eqnarray} S_0 &=& \text{initial stock price} \nonumber \\ t &=& \text{time to expiration} \nonumber \\ \mu &=& \text{drift} \nonumber \\ \sigma &=& \text{volatility} \nonumber \\ \end{eqnarray} $$In the case at hand, $\mu=0$, so this ultimately reduces to $$ \begin{eqnarray} E(S_t) &=& \text{arithmetic mean} = \text{m} = S_0 \nonumber \\ Var(S_t) &=& \text{arithmetic variance} = \nu^2 = S_0^2 \left({ e^{ \sigma^2 t} - 1 }\right) \nonumber \end{eqnarray} $$ where $$ \begin{eqnarray} S_0 &=& \text{initial stock price} \nonumber \\ t &=& \text{time to expiration} \nonumber \\ \sigma &=& \text{volatility} \nonumber \\ \end{eqnarray} $$ Then, if, $N$ is how many standard deviations constitute "far enough from the mean" to "be" a Black Swan, then the boundaries of the "Black Swan Ponds" are given by

These are the bounds I am currently using in the visualization, with $N=3$ based on an informal review of a few things on the internet (feel free to suggest a better value).

One thing that has struck me here is that the Black Swan Ponds can visually appear to be a bit "closer" than one might expect when using only three standard deviations. Thus, while this might need to be tweaked a bit, the primary purpose of the ponds is to simply try to integrate into the viz information on the potential impacts of the tails that are implied by "normal" assumptions.

## No comments:

## Post a Comment