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?


1 Answer 1


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.

enter image description here

  • 1
    $\begingroup$ how do you account for the value of a call option going to the value of the underlying under this same intuitive idea? $\endgroup$
    – will
    Dec 14, 2021 at 22:08
  • $\begingroup$ That is a very good question. I have limited intuition around $d_{+}$, but here are my 2 cents: While $d_{-}$ is $\text{Prob}(S_{T}>K)$, $d_{+}$ is $\text{Prob}(S_{T}e^{\sigma T}>K)$. $d_{+}$ obviously goes towards 1, when $\sigma\rightarrow\infty$ and I presume that this is due to the mean of the log-normal distribution increasing as $\sigma$ increases. Remember that for the digital call we only care whether it ends up in the money, but for the call we also care about the stock price at expiry. So the mean doesn't matter in the first case, but does in the last. $\endgroup$
    – mmencke
    Dec 15, 2021 at 21:52
  • $\begingroup$ I don't want to even get to the maths if possible. Ultimately I black Scholes breaks when you go to infinity vol (I challenge you to find someone who will trade options or digis anywhere near these theoretical values). For me a preferable answer would be one which is entirely handwavey and even better model free. The option value converging to the stock price makes sense to me, as the option has to increase in value as you increase vol, and there is no scenario that makes sense with it worth more than the stock... $\endgroup$
    – will
    Dec 16, 2021 at 7:51
  • $\begingroup$ ... Does that mean that all options become equal and thus all call spreads are worth zero? That's a difficult notion to swallow. Of course the idea of infinite vol is also silly, but I'd hope some intuition was still possible. $\endgroup$
    – will
    Dec 16, 2021 at 7:52
  • 1
    $\begingroup$ This answer also has charts showing the CDF and PDF of a lognormal together with the mean, median and mode for various values of sigma. It also discusses why the value of a call option is not zero (more details here as @Will asks. It is a "competition" of limits. $\endgroup$
    – AKdemy
    Feb 5 at 21:11

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.