Wednesday, December 11, 2019

option pricing - Probability of exercise in the Black-Scholes Model


What's the intuition behind the fact that the limit of $\mathcal{N}(d_2)$, i.e. the (risk-neutral) probability of exercise, in the Black-Scholes Model tends to $0$ when the volatility tends to infinity?



Answer




As a random variable, the terminal asset price has a semi-infinite support, bounded at zero. Intuitively, this means that when increasing volatility while keeping all other parameters unchanged (same mean, here reflected by the forward price which is vol independent), the distribution tries to extend itself on both side of the definition domain but hits a boundary at zero, where probability accumulates (probability mass).


This is not very rigourous but I hope you get the idea, you could try plotting the lognormal pdf as $\sigma \to \infty$ for visual inspection.


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