The Black–Scholes Formula

The binomial tree was a staircase: chop time into steps, price the option at each node, fold back. Add more steps and the staircase smooths into a ramp. In 1973, Fischer Black, Myron Scholes, and Robert Merton wrote down where that ramp lands — a single closed-form equation that gives a European option’s fair price with no tree to fold at all. It won a Nobel Prize, it underpins a multi-trillion-dollar market, and it fits on one line. Let’s earn that line.

Before we do, a gut check.

The headline result

For a European call on a non-dividend-paying stock, the Black–Scholes price is:

$C = S_0\,N(d_1) - K e^{-rT}\,N(d_2)$

where

$d_1 = \frac{\ln(S_0/K) + \left(r + \tfrac{1}{2}\sigma^2\right)T}{\sigma\sqrt{T}}, \qquad d_2 = d_1 - \sigma\sqrt{T}.$

Here $N(\cdot)$ is the standard normal cumulative distribution function — for any number $z$, $N(z)$ is the probability that a draw from the bell curve lands below $z$. It runs from $0$ (far left) through $0.5$ (at zero) up toward $1$ (far right).

The put doesn’t need its own derivation — it falls straight out of put–call parity:

$P = K e^{-rT}\,N(-d_2) - S_0\,N(-d_1).$

That’s the whole machine. Six inputs go in, one price comes out. The rest of this lesson is about why it has this exact shape, because once you can read the formula, the Greeks in the next lesson are just its derivatives.

The six inputs — five you can see, one you can’t

Remember the six drivers of an option’s premium from the previous topic? Black–Scholes is exactly those six drivers turned into arithmetic. Five of them you can read off a screen this instant; the sixth you have to forecast.

That asymmetry matters enormously. Plug five knowns and one guess into the formula and you get a price. But run it backwards — feed in a market price and solve for the $\sigma$ that reproduces it — and you’ve extracted the market’s volatility forecast. That number has a name, implied volatility, and it’s the subject of two lessons from now.

The assumptions — and what each one costs you

The formula is exact, but only inside a particular model of the world. Six assumptions buy you the clean closed form; each one is also where reality eventually pokes a hole.

Assumption (a) is the keystone. “Geometric Brownian motion” is a fancy way of saying the stock’s percentage returns are random and normally distributed, so the log of the price is normal — and therefore the price itself is lognormal: it can’t go negative, and it’s skewed with a long right tail. Hold that picture; the entire meaning of the formula lives in it.

What the formula actually means

This is the payoff of the lesson. The equation isn’t a magic incantation — every piece has a plain-English job. Read it as (present value of what you receive) minus (present value of what you pay), each weighted by the right probability.

$N(d_2)$ — the risk-neutral probability of finishing in the money

Under the model’s risk-neutral world, $N(d_2)$ is the probability that the call finishes in the money, i.e. $S_T > K$ at expiry. Picture the lognormal distribution of the terminal price: $N(d_2)$ is the shaded mass sitting above the strike — the slice of futures where exercising actually pays off.

Crank up volatility or time and that right tail fattens, pushing more probability above $K$. Crank them down and the distribution tightens around today’s price. Play with it directly — drag σ and T and watch the in-the-money mass (which is $N(d_2)$) swell and shrink:

$N(d_1)$ — the delta, the hedge ratio

$N(d_1)$ is the call’s delta: how much the option’s price moves for a $1 move in the stock, and the share-fraction you must hold to hedge it. The term $S_0\,N(d_1)$ is the present value of receiving the stock, conditional on actually exercising — the expected value of the share you’ll collect, weighted toward the cases where you do collect it. (Note $N(d_1) \geq N(d_2)$ always, since $d_1 > d_2$: the share’s value gets more weight than the raw exercise probability, because when you do exercise you tend to be deep in the money.)

$K e^{-rT}\,N(d_2)$ — the present value of paying the strike

The second term is what you hand over. You pay the strike $K$ only if you exercise, which happens with probability $N(d_2)$; you pay it at expiry, so discount it back with $e^{-rT}$. That product — strike, discounted, times the chance you actually pay it — is the present value of the cash you’ll fork over.

So the call price is simply:

$\underbrace{S_0\,N(d_1)}_{\text{PV of the stock you get}} \;-\; \underbrace{K e^{-rT}\,N(d_2)}_{\text{PV of the cash you pay}}.$

Get this and the formula stops being symbols. It’s a probability-weighted “what you get minus what you give.”

A full worked example

Let’s price an at-the-money one-year call with clean numbers, showing every step. Inputs:

• Spot $S_0 = 100$
• Strike $K = 100$ (at the money)
• Risk-free rate $r = 0.05$ (5%)
• Volatility $\sigma = 0.20$ (20%)
• Time $T = 1$ year

Step 1 — the spread term.

$\sigma\sqrt{T} = 0.20 \times \sqrt{1} = 0.20.$

Step 2 — compute $d_1$. Since $S_0 = K$, we have $\ln(S_0/K) = \ln(1) = 0$, so the numerator is just the drift term:

$d_1 = \frac{0 + \left(0.05 + \tfrac{1}{2}\cdot 0.20^2\right)\cdot 1}{0.20} = \frac{0.05 + 0.02}{0.20} = \frac{0.07}{0.20} = 0.35.$

Step 3 — compute $d_2$.

$d_2 = d_1 - \sigma\sqrt{T} = 0.35 - 0.20 = 0.15.$

Step 4 — look up the normal CDF. Use this small reference table.

So $N(d_1) = N(0.35) \approx 0.6368$ and $N(d_2) = N(0.15) \approx 0.5596$.

Read $N(d_2)$ as a sentence: there’s about a 56% risk-neutral chance this call finishes in the money.

Step 5 — assemble the price. First the discount factor: $e^{-rT} = e^{-0.05} \approx 0.9512$. Then:

$C = S_0\,N(d_1) - K e^{-rT}\,N(d_2) = 100\cdot 0.6368 - 100\cdot 0.9512\cdot 0.5596.$

$C = 63.68 - 53.23 \approx 10.45.$

The fair value of this one-year at-the-money call is about $10.45 per share — roughly 10.5% of the stock price, a useful rule of thumb for a 20%-vol, one-year, at-the-money option.

Pricing the matching put

We don’t re-derive anything — put–call parity does the work. With the same inputs, the put price uses the complementary probabilities $N(-d_1)$ and $N(-d_2)$. Try it before peeking.

Putting it together

Black–Scholes turns the binomial staircase into one closed-form line. Feed it six inputs — five observable, only $\sigma$ a forecast — assume a lognormal terminal price with constant vol and rate, and out comes a European option’s fair value as the present value of the stock you’d receive minus the present value of the strike you’d pay, each weighted by the right probability. $N(d_2)$ is the chance you finish in the money; $N(d_1)$ is your delta.

Next up — The Greeks — we stop treating the price as a number and start differentiating it. Delta you’ve already met (it’s $N(d_1)$); now meet its siblings — gamma, theta, vega, and rho — each one a partial derivative of this very formula, measuring how the price reacts as one input twitches.