# 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...

• You are missing the brackets around v*sqrt(T) Jul 27, 2018 at 21:02

$S_T$ 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.

• Thanks Andrew, just want to make sure I understand E[S(T)]=S=110. E is expected value which is also the mean? So while the expected value is 110, more the probability distribution at T falls below the K of 100? Jul 30, 2018 at 15:43

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.