The Black-Scholes-Merton formula
1. The idea
Level 4 valued an option on a tree whose time to expiry \(T\) is cut into \(n\) time steps of \(\Delta t = T/n\). With more steps, the tree follows the share price more finely, and its value settles at one number. For a European option on a share without dividends, Fischer Black, Myron Scholes and Robert Merton found that number as a formula, the Black-Scholes-Merton formula:
\[ c = S_0\,N(d_1) - K e^{-rT} N(d_2), \qquad p = K e^{-rT} N(-d_2) - S_0\,N(-d_1), \]
\[ d_1 = \frac{\ln(S_0/K) + (r + \sigma^2/2)\,T}{\sigma\sqrt{T}}, \qquad d_2 = d_1 - \sigma\sqrt{T}, \]
where \(c\) and \(p\) are the prices of a European call and put, \(S_0\) is the share price today, \(K\) the strike price, \(T\) the time to expiry in years, \(r\) the risk-free rate with continuous compounding and \(\sigma\) the share’s volatility, the standard deviation of its yearly return with continuous compounding (Level 4), both as decimals, and \(\ln\) the natural logarithm. \(N(x)\) is the cumulative standard normal distribution: the probability that a normally distributed variable with mean 0 and standard deviation 1 is below \(x\). A normal table, a spreadsheet’s NORM.S.DIST(x, TRUE) or the box in section 3 gives it, and \(N(-x) = 1 - N(x)\).
The formula is the tree’s method in the limit: the present value of the option’s expected payoff under the risk-neutral probabilities of Level 4. \(N(d_2)\) is the risk-neutral probability that a call ends in the money; \(S_0 N(d_1)\) is the present value of receiving the share when it does, and \(Ke^{-rT}N(d_2)\) the present value of paying the strike price then.
An example. A share trades at $50, and a six-month European call on it has a strike price of $52; the risk-free rate is 4% and the volatility 30%. Then \(\sigma\sqrt{T} = 0.3 \times \sqrt{0.5} = 0.212132\),
\[ d_1 = \frac{\ln(50/52) + (0.04 + 0.3^2/2) \times 0.5}{0.212132} = 0.0155, \qquad d_2 = 0.0155 - 0.212132 = -0.1966, \]
so \(N(d_1) = 0.5062\), \(N(d_2) = 0.4221\) and \(c = 50 \times 0.5062 - 52\,e^{-0.04 \times 0.5} \times 0.4221 = \$3.80\). By put-call parity (Level 5), the put with the same terms costs \(p = c + Ke^{-rT} - S_0 = 3.80 + 50.97 - 50 = \$4.77\), the price the formula gives too.
The formula assumes that the share’s returns with continuous compounding are normally distributed with a constant volatility, that the risk-free rate is constant, that the share pays no dividends before expiry (the next lesson adds them) and that trading has no costs.
2. Try it
If an answer is wrong, we will see what went wrong and how to fix it.
3. From trees to the formula
First, the time step of a tree for this call.
Now move the number of steps. Each tree is valued backwards node by node, as in Level 4, with the time step \(\Delta t = T/n\); only the number of steps changes. The tree’s value swings around the formula’s value and settles at it, although one more step does not always bring it closer.
\(N(x)\) for any \(x\), as a normal table gives it:
Then the price on a chart. First, the call’s lowest price from Level 5.
Now move today’s share price. The formula’s price stays above the lowest price. While the lowest price is above zero, the gap is the price of the put with the same terms, by put-call parity: the value of the chance that the call ends below the strike price, where it is not exercised. Where the lowest price is zero, the gap is the call’s whole price.
4. One worked example, then your turn
In the worked example, we price a put.
Now a call with hints, one step at a time.
And one on your own: a put.
5. Practice
Each question has new numbers. You master the skill when you get three right in a row, with at least one right answer for a call and one for a put. You can always open the solution, but then that question does not count.
6. Coming back to it
When you have mastered a skill, it comes back on the start page for a short review after 1 day. After each right answer the next review comes later: after 3, 7, 14 and then 30 days. After a wrong one, the reviews start again at 1 day.
This skill: