Why High Volatility Can Put a Binary Call’s Exercise Probability Below 50%
Summary
The document explains why a Black–Scholes cash-or-nothing call can be priced below 50% even when the underlying is above the strike. Its central point is that the model uses a lognormal distribution for the future price, whose positive skew means the expected price does not determine the probability of finishing above a threshold. Under the risk-neutral measure, the digital call price is the discounted exercise probability, represented by the cumulative normal value associated with d2.
The discussion uses the stated example and describes how increasing volatility shifts the lognormal density’s mode toward lower prices, while leaving a long upper tail. It notes that the call’s exercise probability can tend toward zero as implied volatility becomes very large. This is an explanation of the model’s probability calculation, not a forecast of real-world odds: the pricing measure, assumptions about lognormal returns, and discounting matter, and the example alone does not establish whether the option is fairly valued.
Key ideas
- A lognormal future-price distribution is positively skewed, so its mean does not determine the chance of exceeding the strike.
- A digital call’s Black–Scholes value is the discounted risk-neutral probability of exercise.
- Higher volatility can lower that probability even when spot is above the strike.
- In the model, the call exercise probability approaches zero as volatility becomes extremely large.
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Full text
# Black-Scholes pricing of binary options
# Black-Scholes pricing of binary options
I'm trying understand something basic about Black-Scholes pricing of binary options. In my example above, the current price is over the strike price. The volatility is extreme but I'm still having trouble understanding why the price of the binary option (which I'm interpreting as the probability of expiring in the money) would be below 50 (50% odds). Assuming a random walk from the current price, isn't it more likely that it would expire above the strike?
Black-Scholes gives an implied price of ~ 0.390. Which I interpret as a 39% chance of expirying in the money? Why wouldn't' it be more then 50%?
```
S = 110 #current_price
K = 100 #ATM strike
v = 1.20 #annualized volatility
r = 0.00 #interest rate
T = 0.44 #days remaining (annualized)
d2 = (log(S/K) + (r - 0.5 * v**2) * T) / (v*sqrt(T))
print exp(-r * T) * norm.cdf(d2)
```
> 0.390...
## Answer by Andrew (score 7, accepted)
https://quant.stackexchange.com/a/40982
$S_T$ is log-normal distributed and therefore skewed. In particular $E[S_T]=S=110$ (no drift), but $Q(S_T>S)<Q(S_T<S)$. For example if S=K=100 you don't get a value of 0.5 as you might expect, but a lower value since you have norm.cdf($-0.5\sigma\sqrt{T}$)$<$ norm.cdf($0$)=0.5.
## Answer by AKdemy (score 6)
https://quant.stackexchange.com/a/74541
Just adding a few graphical explanations to Andrew's answer because I think thez could help in understanding the result.
The Black Scholes price of a Digital corresponds to the discounted probability of exercise,($N(d2)$), which for call options goes to zero when IVOL $\rightarrow \infty$. The figure below shows the probabilities with increasing IVOL (3 = 300%):
To understand this result, it is helpful to look at the probability density function (PDF) and cumulative density function (CDF) of the lognormal distribution.
The higher σ, the more the global maximum of the probability density function (the mode) shifts towards the lower bound of the lognormal distribution.
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 if implied vol is sufficiently large.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.