Why Extreme Volatility Drives Black–Scholes Digital Prices Apart
Summary
The document examines the Black–Scholes prices of digital calls and puts as volatility approaches infinity, assuming a zero interest rate. From the model’s formulas, it identifies limiting prices of zero for the call and one for the put, independent of moneyness. It asks why this result differs from the intuition that an at-the-money option should have equal chances of finishing in or out of the money.
The answer attributes the asymmetry to the lognormal distribution used in Black–Scholes, which is not symmetric: as volatility rises, probability mass is pushed toward zero. The response frames zero as an absorbing boundary and argues that the chance of reaching it during the option’s life tends toward one in the limit. This is a model-based limiting argument, not a description of ordinary market conditions. The document provides an intuitive explanation but does not discuss dividends, nonzero rates, or alternative price processes.
Key ideas
- In the stated Black–Scholes setup, digital call and put prices approach opposite limits as volatility grows without bound.
- The limiting prices do not depend on moneyness under the formula given.
- The equal-probability intuition assumes symmetry, but the Black–Scholes lognormal distribution is asymmetric.
- The answer explains the put’s limiting value by the increasing probability mass near the zero boundary.
- The result is an extreme-volatility model limit, not a general prediction for traded options.
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Full text
# Limit of digital call and put price when volatility goes to infinity
# Limit of digital call and put price when volatility goes to infinity
The price a digital call and put in the Black-Scholes model is given by $$c^d = \Phi (d_-), \qquad p^d = \Phi (-d_-), \qquad \text{with} \qquad d_- = \dfrac{\log S_t / K}{\sigma \sqrt{T}} - \dfrac{1}{2}\sigma \sqrt{T}.$$
I am assuming $r = 0$, since interest rates are unrelated to the question.
It is easy to see that, as the volatility goes to infinity, the digital call price will go to zero whereas the price for the digital put will tend to one. Moreover, the price is independent of the moneyness. Taking the example of the digital call, one could argue that this limit makes sense as one could understand the value of a digital call as the limit of a infinitely-narrow call-spread. When volatility increases, both prices approach each other and therefore the difference goes to zero. However, we can see that this exercise only works for the digital call, and fails for the digital put.
Intuitively, and lets consider an ATM case for simplicity, I would argue that as $\sigma$ increases, the distribution flattens-out, and therefore there is a 50-50 chance that the option finishes OTM and ITM. So, naively, I would price both the digital call and put at 0.5. But apparently this is not the case, as stated at the beginning.
So the question is, what does fail in the reasoning here in the last paragraph?
## Answer by mmencke (score 2, accepted)
https://quant.stackexchange.com/a/69106
Your intuition is correct for a symmetric distribution. However, the log-normal distribution, as is assumed in the Black-Scholes model, is an asymmetric distribution. I have illustrated the effect of increasing the volatility, while holding everything else equal. The probability mass is squeezed towards zero.
The way I think about this in terms of intuition is that zero is an absorbing barrier for the process. This means that when the process hits zero, then the process "dies". If volatility goes towards infinity then the probability of hitting that absorbing barrier in the life of the option goes towards 1. Hence the price of a put option should be 1 and the price of a call should be zero.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.