Why Black–Scholes N(d₂) Is Not the Same as Option Value
Summary
The document addresses a common confusion in the Black–Scholes model: interpreting N(d₂) as a risk-neutral probability that a call finishes in the money, and wondering why that probability can approach zero as volatility becomes extremely large while the call value approaches the discounted forward value. It points to an explanatory paper on the roles of N(d₁) and N(d₂), then uses the lognormal distribution’s density and cumulative distribution to describe the apparent paradox.
As volatility rises, the probability mass shifts toward very low terminal prices, reducing the chance of finishing above a fixed strike. At the same time, the increasingly rare outcomes above the strike can be sufficiently large to raise the expected payoff. Thus the probability of exercise and the option’s expected value need not move together. The discussion concerns a limiting case in the Black–Scholes framework and flags that very long horizons or extreme volatility can produce counterintuitive model outputs. It offers conceptual interpretation rather than empirical evidence or guidance for estimating real-world exercise probabilities.
Key ideas
- N(d₂) is associated with risk-neutral exercise probability in the Black–Scholes framework, while N(d₁) has a different role.
- Increasing volatility can make a call less likely to finish above a fixed strike even as its value rises.
- Rare high terminal prices can contribute more to expected payoff as their probability falls.
- The discussion is about model-based risk-neutral quantities, not direct estimates of real-world probabilities.
- Extreme volatility and long horizons can expose counterintuitive behavior in the model’s limits.
Tags
Full text
# probablity expiring in the money ..basic question
# probablity expiring in the money ..basic question
Everyone says $N(d_2)$ is the probability of the option being exercised but stocks that have really high volatility have really expensive options indicating a high likelihood of expiring in the money. If volatility is very high we see that s $N(d_2)$ goes to 0, so doesn't that contradict our first statement? Isn't $N(d_1)$ the true probability?
## Answer by vonjd (score 4)
https://quant.stackexchange.com/a/10135
This is indeed a very basic question. Please have a look at the following paper which will answer all your questions in a very intuitive and step-by-step fashion:
Understanding $N(d_1)$ and $N(d_2)$: Risk-Adjusted Probabilities in the Black-Scholes Model by Lars Tyge Nielsen
## Answer by AKdemy (score 0)
https://quant.stackexchange.com/a/73636
I agree with @vonjd that the distinction between $N(d1)$ and $N(d2)$ is indeed a very basic question, which is why I upvoted his answer a long time ago. That said, reconciling why the probability of exercise for call options goes to zero when IVOL $\rightarrow \infty$, while at the same time you pay the discounted value of the forward $F=S_0e^{(r−q)T}$, which when the dividend yield is zero, corresponds to the current value of the stock price $S_0$, irrespective of strike, is not so trivial in my opinion.
The probabilities with increasing IVOL (3 = 300%):
Option values:
This indeed is looking a bit odd, and in my opinion, is best answered by looking at the probability density function (PDF) and cumulative density function (CDF).
The higher σ, the more the global maximum of the probability density function (the mode) shifts towards the lower bound of the lognormal distribution. On the other hand, as mentioned for example by @Jesper Tidblom,
> More and more contribution of the expected value come from values of $S_T$ when $S_T \geq K$ as $\sigma$ grows, although the probability for those values to occur goes to zero. So we have a sort of competition of limits here where the values of $S_T$ above $K$ increases faster than their probability to occur goes to zero, so to speak.
Also, expected returns decrease. Closely following the paper of Lars Tyge Nielsen as referenced by @vonjd:
In `Black Scholes`, stock prices $S_t$ at time t follow a lognormal distribution. At time 0, $$log(S_T) \sim \mathcal{N}(log(S) +(\mu -\sigma^2/2)t, \sigma^2t)$$ To be precise about $\mu$ and $\sigma^2$ we need to make a few observations about the rate of return of the stock. The continuously compounded rate of return over an interval $[0,t]$ is $$\frac{log(S_t)-log(S)}{t}$$ Given the current stock price $S$, this rate follows the normal distribution $$\mathcal{N}((\mu -\sigma^2/2),\sigma^2/t) $$ In plain English, its logarithm is normally distributed with mean $(\mu -\sigma^2/2)$ and variance $\sigma^2/t$. As $t$ grows, variance decreases towards zero, whereas the mean of the rate of return does not depend on time $t$. However, the mean depends on volatility. The chart below shows this relationship for a unit interval ($t=1$).
This can also be demonstrated by plotting the PDF of the normal distirbution.
Last but not least, the cumulated distribution function (CDF) shows the increase in the probability of $S_T$ being very small.
Therefore, the probability of exercise for a call eventually becomes zero. At the same time, the expected value of $S_T$ being above K increases faster, making the option approaching its maximum value.
Ultimately, limits can be a strange thing. Warren Buffett explained his take on the Black Scholes formula for long-dated options (which has a similar effect as large IVOL) in his 2008 letter to the Shareholders of Berkshire Hathaway.
> The Black-Scholes formula has approached the status of holy writ in finance, and we use it when valuing our equity put options for financial statement purposes. ... If the formula is applied to extended time periods, however, it can produce absurd results. In fairness, Black and Scholes almost certainly understood this point well.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.