Is it optional to know what an option is? No.

This note prices a European call. We will not write the famous formula until we have the tools to read it. First: what the contract is. Then a detour to Japan, where Kiyosi Itô repaired the chain rule for paths that jitter like a stock. Then, and only then, the price.

If you have Calc 1–2, a first course in probability (bell curves, means, the substitution $y=\ln x$), and have never seen a derivative in the Wall Street sense, you are the intended reader. Every term is defined when it appears.

• What is an option?
• How do we price one of these things?
• A detour to Japan
• Brownian motion
• The stock: geometric Brownian motion
• Martingales
• Itô’s lemma
• Assumptions
• The Black–Scholes–Merton formula
• Risk-neutral pricing
• Toolkit: the lognormal
• Route I: two lognormal integrals
• Route II: change of numeraire
• The PDE, briefly
• Limitations
• References

The change-of-numeraire calculation follows Fabrice Douglas Rouah, Four Derivations of the Black-Scholes Formula. The original host is gone. A copy is here; the Wayback capture of 19 July 2024 is the provenance.

What is an option?

In ordinary English, optional means you do not have to. In finance that is almost the definition.

An option is a contract that gives its owner the right, and not the obligation, to buy or sell something at a pre-agreed price. The something is the underlying — a share of stock, a bushel of wheat. The pre-agreed price is the strike, written $K$. The deadline is expiry, written $T$. For that right you pay money up front, the premium. The rest of this note is the question: what must the premium be?

You already know the everyday version. A rain check that lets you buy a TV at today’s price next month is a call option on the TV. If the store drops the price, you ignore the rain check and buy cheaper. If the store raises the price, you use the rain check. You will only exercise when it helps you. That one-sidedness is the whole point.

Calls and puts: bets in favor and bets against

Options come in two flavors.

A call is the right to buy the underlying at the strike. It is a bet the underlying will finish above $K$. A put is the right to sell the underlying at the strike. It is a bet the underlying will finish below $K$.

Every contract has two sides. The buyer (the holder, long the option) pays the premium today and owns the right. The seller (the writer, short the option) captures that premium today and takes on the matching obligation. Whatever the buyer can choose to do, the writer can be forced to do. The buyer’s profit is the writer’s loss, dollar for dollar. An option is a zero-sum bet between those two people.

To see the cash actually move, fix a working premium of 10 dollars. We do not yet know whether 10 is the right price — that is the rest of the note — but we need a number to subtract. A stock trades at 100 dollars today. Strike (K = 100), expiry in one year.

The call buyer: right to buy at 100

You pay 10 dollars today. One year later the stock is at (S_T), and you look at the ticket.

• If (S_T = 130), you exercise. The writer must sell you the share at 100. You now hold something worth 130 that you paid 100 for, so the contract itself paid 30. Subtract the 10 you already spent: net profit (+20). Same thing in two steps if you do not want to keep the share: buy at 100 via the call, immediately sell in the open market at 130.
• If (S_T = 80), exercising would mean paying 100 for a share worth 80. You are not required to be that foolish. You let the ticket expire. The 10 is gone. Net profit (-10).
• If (S_T = 100), exercising gains you nothing. You walk away. Net profit (-10).

The buyer of a call has unlimited upside (the stock can in principle go to the moon) and limited downside (the worst case is losing the premium). That is why people buy calls.

The call writer: capturing the premium

Someone had to take the other side. That person received your 10 dollars today. In exchange they promised: “if you choose to buy at 100, I must sell to you at 100.”

• If (S_T = 80) or (S_T = 100), the buyer walks away. You keep the 10. You do nothing. Net profit (+10). This is the writer’s dream: collect premium, option expires worthless.
• If (S_T = 130), the buyer knocks. You must sell a 130-dollar share for 100.
  • If you already owned the share (covered call): you hand it over at 100 instead of selling it in the market at 130. You missed out on 30 of upside, but you had pocketed 10, so net (-20).
  • If you did not own it (naked call): you buy the share in the market at 130, sell it to the buyer at 100, and lose 30 on that round trip, again offset by the 10 you captured. Net (-20).

The writer of a call has limited upside (the premium is the most they can make) and unlimited downside (the stock can in principle go to the moon, and they still have to sell at 100). That is why writing naked calls is a dangerous way to “collect premium,” and why a covered call is the version a junior should picture first: you already own the stock, you are willing to let it get called away, and the premium is extra income if it does not.

Break-even for both sides of this call is (S_T = 110): the stock has to rally 10 just to recoup the 10 you paid (or, for the writer, that is where the obligation starts to eat the premium).

Each row sums to zero. That is the bet.

The put buyer: right to sell at 100

A put is the other flavor: you have paid a premium (call it 8 dollars for this paragraph) for the right to sell the share at 100, even if the market has collapsed.

The clean way to see the cash is: buy the stock low in the open market, then sell it high to the put writer at the strike.

• If (S_T = 70), the share is cheap. You buy it in the market for 70, then exercise the put: the writer is forced to buy it from you at 100. You have bought low and sold high, pocket 30, minus the 8 you paid for the ticket. Net (+22).
• If (S_T = 130), nobody will let you sell a 130-dollar share to them at 100. You throw the put away. Net (-8).

So a put is not a mystical “bet against.” It is a coupon that says: if this thing gets cheap, I may purchase it cheaply in the market and put it to you at the old high price.

The put writer: capturing that premium

The put writer collected the 8 dollars and promised: “if you choose to sell to me at 100, I must buy.”

• If (S_T = 130), the put expires. Writer keeps the 8. Net (+8).
• If (S_T = 70), the buyer puts the share to them. Writer pays 100 for a share worth 70, loses 30, offset by the 8. Net (-22).

The put writer’s upside is the premium; their downside is that the stock can go to zero, in which case they pay 100 for something worthless.

Payoff versus profit

Two different words, please keep them apart.

The payoff is what the contract hands you at expiry, before subtracting the premium. For the call buyer that is (\max(S_T-K,\,0)); for the put buyer, (\max(K-S_T,\,0)). The little plus means “take this if it is positive, otherwise take zero.”

The profit is payoff minus premium for the buyer, and premium minus payoff for the writer.

When (S_T > K) people say the call is in the money (the put is out). When (S_T < K) the call is out of the money (the put is in). When (S_T = K) both are at the money. Those are nicknames for the rows. The rest of this note prices the call; the put will fall out later from an accounting identity.

Europeans and Americans

Two nationalities, and this is the only distinction that matters here.

• A European option may be exercised only at the single instant $T$. You wait until expiry, then decide.
• An American option may be exercised on any day up to and including $T$.

The American extra right sounds valuable, and for a put it is. For a call on a stock that pays no dividend, it is a theorem of Merton’s that you should never exercise early — so the American call and the European call have the same price. We price the European call. The European put will fall out of an accounting identity later, not a second integral.

How do we price one of these things?

At expiry the payoff is arithmetic: count on your fingers. The hard question is today’s premium. What should you pay now for a lottery that, one year from now, pays $\max(S_T - 100,\, 0)$?

The honest answer depends on how $S$ is going to wander between now and $T$. If the stock is almost surely going to sit at 100, the lottery is almost surely worth nothing and you should pay almost nothing. If the stock is liable to be at 200 or at 20, the lottery is juicy — you capture the 200 and walk away from the 20 — and you should pay more.

So we need a model of the wander. Stock prices are not smooth curves. They jitter. The chain rule you learned in Calculus 1 assumes a tangent line; a stock path does not offer one. In 1942 the Japanese mathematician Kiyosi Itô found the corrected chain rule for this kind of path. We cannot write down a price until we have that correction. Detour.

A detour to Japan

Itô’s observation, in one sentence: if the input to a function is a Brownian scribble, the $\tfrac12 f’‘(x)\,(dx)^2$ term in Taylor’s formula is the same size as the $f’(x)\,dx$ term, and you cannot throw it away. That leftover is the entire difference between a world you can hedge and a world you cannot.

We need four things from his world, in order.

1. Brownian motion $W_t$ — a stand-in for the market’s noise. (Also called a Wiener process. Same object, two names.)
2. Geometric Brownian motion — the stock, built from $W_t$ so the price stays positive.
3. A martingale — a fair game. This is how we will mean “no free lunch.”
4. Itô’s lemma — the corrected chain rule.

Then we can write the formula.

Brownian motion

A Brownian motion — also called a Wiener process, written $W_t$ — is the model of pure noise used in this note. Picture a pollen grain on a water surface, or a walker who at every instant takes a tiny random step up or down. The walker’s height at time $t$ is $W_t$.

It is not a stock price. It starts at $0$, it is as likely to be negative as positive, and its typical size at time $t$ is $\sqrt{t}$, not $t$. The stock will be built from it in the next section.

A coin-flip construction

Fix a time horizon $T$ and chop it into $n$ equal pieces of length $\Delta t = T/n$. Flip a fair coin at each tick. On heads walk up $\sqrt{\Delta t}$; on tails walk down $\sqrt{\Delta t}$. After time $t = k\,\Delta t$ the position is

Each step has mean $0$ and variance $\Delta t$, so the sum of $k$ steps has mean $0$ and variance $k\,\Delta t = t$. The central limit theorem — a sum of many small independent kicks becomes a bell curve — says that for large $n$ the position at a fixed $t$ is approximately $\mathcal{N}(0,t)$. Send $n\to\infty$ and the staircase becomes a continuous scribble. That scribble is Brownian motion.

The three panels are the same sequence of coin flips, grouped into fewer, larger steps on the left and drawn almost continuously on the right. Brownian motion is the right-hand picture, taken as a mathematical limit.

The four properties, in symbols

A standard Brownian motion is a random function $t\mapsto W_t$ satisfying:

1. Starts at zero. $W_0 = 0$.
2. Independent increments. For $t>s$, the future step $W_t-W_s$ does not depend on the path before time $s$. The next wiggle does not remember how it got here.
3. Gaussian increments. $W_t - W_s \sim \mathcal{N}(0,\, t-s)$. In words: over an interval of length $\Delta t$ the step is drawn from a bell curve with mean $0$ and variance $\Delta t$. In particular the position itself is \(W_t \sim \mathcal{N}(0,t), \qquad \mathbb{E}[W_t] = 0, \qquad \mathrm{Var}(W_t) = t.\) The typical size of $W_t$ is therefore the standard deviation $\sqrt{t}$, not $t$. Over a short interval the typical move is $\sqrt{\Delta t}$, which is much larger than $\Delta t$ itself when $\Delta t$ is small.
4. Continuous paths. The graph of $t\mapsto W_t$ has no jumps. It is a jagged scribble with no breaks.

Normal reminder. We write $Z\sim\mathcal{N}(m,s^2)$ to mean that $Z$ has the bell-curve (Gaussian, normal) density

The number $m$ is the mean, $s^2$ is the variance. The standard normal is the special case $\mathcal{N}(0,1)$. Later we will need the cumulative distribution function (cdf) of this law: the area under the bell curve to the left of a point $x$, written

If that integral looks unfriendly, just remember: $\Phi(0)=1/2$ (half the bell is to the left of zero), $\Phi$ increases from $0$ to $1$, and $\Phi(-x)=1-\Phi(x)$. We will only ever evaluate $\Phi$, never compute the integral by hand.

On the left, every path starts at $0$ and wanders equally above and below the axis — several paths are negative, which a stock price cannot be. The shaded trumpet is the band $\pm 2\sqrt{t}$; at each fixed $t$ about 95% of paths sit inside it, because a normal random variable is within two standard deviations of its mean about 95% of the time. On the right, many independent runs are stopped at $t=1$ and histogrammed. The histogram is the $\mathcal{N}(0,1)$ density, which is property 3 at $t=1$.

The increment $dW_t$, and why ordinary calculus fails

Write $dW_t$ for the increment of $W$ over an instant of length $dt$. Property 3 says $dW_t$ is approximately $\mathcal{N}(0,dt)$, so $(dW_t)^2$ is typically of size $dt$, not $(dt)^2$. That is the multiplication table we will use:

The third rule has a name, quadratic variation. Chop $[0,t]$ into $n$ steps of length $\Delta t=t/n$. Each increment $\Delta W_i$ satisfies $\mathbb{E}[(\Delta W_i)^2]=\Delta t$, so the sum of the squares has expectation $t$. The variance of that sum shrinks like $1/n$ and vanishes as $n\to\infty$. The sum of squared wiggles converges to $t$: in the infinitesimal shorthand, $(dW_t)^2=dt$. Ordinary calculus throws $(dx)^2$ away. Here it is the same size as $dt$, so it must be kept. That is why, when we later differentiate functions of $S_t$, we will need Itô’s lemma rather than the chain rule from Calculus 1.

The paths are so jagged that, with probability $1$, they have no tangent line anywhere. You cannot write $dW_t/dt$. You can only write $dW_t$.

The stock: geometric Brownian motion

$W_t$ is a bad model for a stock. It can be negative, and a $100 stock and a $10 stock should not make dollar moves of the same typical size. The model used here is geometric Brownian motion: the percentage move is Brownian.

Read the equation as a recipe for a short interval $dt$:

• the stock grows by a deterministic fraction $\mu\,dt$ (the drift $\mu$ is the expected rate of return);
• plus a random fraction $\sigma\,dW_t$ (the volatility $\sigma$ scales the Brownian increment).

Because the noise is multiplied by the current price $S_t$, a $200 stock wiggles twice as many dollars as a $100 stock, and $S_t$ stays positive. The $d$ on the left is the same kind of increment as $dW_t$: it is not a derivative, it is a small change.

Each path is one draw of $W$, turned into a price by $(2)$. Compare with the previous figure: these paths cannot cross zero, and they spread in proportion to the level of $S$. A European call with the dashed strike pays the excess over $100$ if the path ends above the line, and zero otherwise. Pricing the call is averaging that payoff over every such path — under a fair-game probability we will define after the assumptions, not under the real-world drift $\mu$.

Assumptions

Four assumptions do the mathematics. A fifth is the trading plumbing that lets the mathematics be a price.

1. The stock follows the geometric Brownian motion $(2)$, with $\mu$ and $\sigma$ constant. One source of randomness, namely the single Brownian motion $W_t$. No sudden jumps, no volatility that itself wiggles. This is what makes $S_T$ lognormal.

2. The interest rate $r$ is a known constant. You may borrow or lend any amount at this rate. A dollar deposited in the money-market account (a riskless savings account) grows to $B_t = e^{rt}$. The exponential is continuous compounding: interest is added every instant, so the growth factor over time $t$ is $e^{rt}$ rather than $(1+r)^t$. Merton later let $r$ wander; the two-$\Phi$ formula does not survive that generalization in this form.

3. No dividends during the life of the option. A cash dividend drops $S$ by the amount paid and would change both the hedge and the law of $S_T$. The repair, recorded at the end, is one letter: a continuous yield $q$ replaces $S_t$ by $S_t e^{-q\tau}$ and $r$ by $r-q$ inside $d_1,d_2$.

4. No arbitrage, continuous trading, a complete market. An arbitrage is a riskless profit — a free lunch. Continuous trading means you may rebalance at every instant. A market is complete when every payoff you can write down can be manufactured by trading the stock and the money-market account. Here there is one source of noise and two traded assets, so the market is complete. The manufactured (replicating) portfolio’s value is the price, and that is why the physical drift $\mu$ will cancel: the hedge does not contain it.

Plumbing, said once: you may sell a stock you do not own (short selling) and use the cash, securities may be held in any fractional amount, and there are no transaction costs. None of that enters an integral.

The gap between this universe and listed options is the subject of Derman and Miller’s The Volatility Smile. The market still quotes in the language of the formula we are about to write.

Martingales

A stochastic process is a family of random variables indexed by time — a random path. A process $M_t$ is a martingale if

In words: a martingale is a fair game. Given what you know now, the expected future value is the present value. You should not expect to win or to lose. The notation $\mathcal{F}_t$ is the usual name for “the information available at time $t$,” so the same sentence is written $\mathbb{E}[M_T\mid\mathcal{F}_t]=M_t$.

Brownian motion itself is a martingale: $\mathbb{E}[W_T\mid W_t]=W_t$, because the remaining increment $W_T-W_t$ has mean $0$ and is independent of the past. The physical stock $S_t$ is not a martingale — it has drift $\mu$, which is why anyone bothers to own it.

Derivative pricing lives on a different assignment of probabilities, written $\mathbb{Q}$ and called the risk-neutral measure. A probability measure is just a consistent way of assigning probabilities to outcomes. The real-world measure is written $\mathbb{P}$. We say $\mathbb{Q}$ is equivalent to $\mathbb{P}$ when the two agree on which events are impossible (probability zero); they may disagree on how likely the possible events are. Under $\mathbb{Q}$ the discounted stock $S_t/B_t$ — the stock measured in time-$0$ dollars, by dividing out the growth of the money-market account — is a martingale: every traded asset earns $r$ on average, so risk is not paid extra. The call price is then an expectation under this fair-game measure, multiplied by $e^{-r\tau}$ to bring the future payoff back to today.

Two stand-ins, for the rest of the note:

• Brownian motion $W_t$ stands in for the market’s noise.
• A martingale stands in for the market’s fairness, once the probabilities have been changed so that risk is not paid extra.

Itô’s lemma

You already know the chain rule. If $f$ is a smooth function of a smooth path $x(t)$, then $df = f’(x)\,dx$. Taylor’s formula actually produces a second term, $\tfrac12 f’‘(x)\,(dx)^2$, which Calculus 1 throws away because $(dx)^2$ is negligible compared to $dx$.

For Brownian motion that discard is illegal: $(dW_t)^2 = dt$, same size as the terms you kept. Itô’s lemma is the chain rule with that leftover put back.

Let $X_t$ satisfy $dX_t = a\,dt + b\,dW_t$, with $a$ and $b$ allowed to depend on $X$ and $t$. For a function $f(t,x)$ that is twice differentiable in $x$ and once in $t$ (the usual meaning of smooth here),

The first two terms in the $dt$ coefficient are the ordinary chain rule. The third is Itô’s correction. It is present because Taylor’s formula produces a $\tfrac12 f_{xx}(dX)^2$ term and $(dX)^2 = b^2\,dt$ is of order $dt$, not smaller.

Apply this to $f(s)=\ln s$ and to the stock $(2)$. Then $f_s = 1/s$, $f_{ss} = -1/s^2$, and

\begin{align} d\ln S_t &= \frac{1}{S_t}\,dS_t + \tfrac12\Bigl(-\frac{1}{S_t^2}\Bigr)(\sigma S_t\,dW_t)^2
&= \mu\,dt + \sigma\,dW_t - \tfrac12\sigma^2\,dt
&= \bigl(\mu - \tfrac12\sigma^2\bigr)dt + \sigma\,dW_t. \end{align}

The correction lowers the drift of the logarithm by $\tfrac12\sigma^2$. Integrate from $t$ to $T$ and use $W_T-W_t \stackrel{d}{=} \sqrt{\tau}\,Z$ with $Z\sim\mathcal{N}(0,1)$:

Thus $\ln S_T$ is Gaussian, so $S_T$ is lognormal, with

The $\tfrac12\sigma^2$ that left the log-drift returns in the mean of the lognormal: $\mathbb{E}[S_T\mid S_t] = S_t e^{\mu\tau}$. Nothing is lost. It is booked in a different ledger. Under the fair-game probabilities of the next section the same calculation will hold with $\mu$ replaced by $r$, and that lognormal is the one we will integrate against.

We now have the four tools. Time to write the price.

The Black–Scholes–Merton formula

Fischer Black and Myron Scholes published the following in 1973; Robert Merton derived the same price from a more general theory the same year. The spot $S_t$ is the stock price right now. The volatility $\sigma$ is how wildly the logarithm of the stock wiggles, measured as a standard deviation per square-root year. The time remaining is $\tau = T-t$. The risk-free interest rate $r$ is constant. Then the European call is worth

In English, before we derive it: the call is a package of $\Phi(d_1)$ shares, minus a package of $K$ dollars delivered at expiry with probability $\Phi(d_2)$, brought back to today by the discount $e^{-r\tau}$. The two numbers $d_1$ and $d_2$ are just places on the bell curve. We will see exactly which places.

The brown line is the expiry payoff you already understand. The blue curve is $(1)$, drawn as a function of today’s spot, with one year left, $r=5\%$, $\sigma=20\%$. The shaded gap is time value: extra worth from still having time left for the stock to wander. The rest of the note is why the blue curve is that shape.

A numerical check

Take the call we walked through by hand: spot $S = 100$, strike $K = 100$, one year left, $r = 5\%$, $\sigma = 20\%$. Then

$\Phi(0.35) \approx 0.6368$, $\Phi(0.15) \approx 0.5596$, and

So the rain check on a 100-dollar stock, struck at 100, one year out, is worth about ten dollars and forty-five cents. That is the number the algebra below is trying to explain.

The same arithmetic in Python:

from math import log, exp, sqrt, erf

def Phi(x):
    return 0.5 * (1 + erf(x / sqrt(2)))

S, K, r, sig, tau = 100, 100, 0.05, 0.20, 1.0
d1 = (log(S / K) + (r + 0.5 * sig**2) * tau) / (sig * sqrt(tau))
d2 = d1 - sig * sqrt(tau)
C = S * Phi(d1) - K * exp(-r * tau) * Phi(d2)
print(C)  # 10.450583572185565

Risk-neutral pricing

Equation $(1)$ does not contain $\mu$, the stock’s expected return. That is the first surprise, and it is the reason a market can exist: two people who disagree about whether the stock will go up can still agree on the call price, provided they agree on $\sigma$.

The mechanism is a change of probabilities. We already named the fair-game measure $\mathbb{Q}$ in the martingales section. Here is how you get there from the real world.

Girsanov’s theorem is the fact that you may add a drift to a Brownian motion and still have a Brownian motion, provided you change which paths you treat as likely. The tilt we want is the market price of risk $\theta = (\mu-r)/\sigma$: excess return on the stock per unit of volatility. Set

Then $W^{\mathbb{Q}}$ is Brownian motion under the risk-neutral measure $\mathbb{Q}$ (the notation $\mathbb{Q}\sim\mathbb{P}$ means the two measures are equivalent: they agree on which events are impossible). Substitute $dW = dW^{\mathbb{Q}} - \theta\,dt$ into $(2)$:

\begin{align} dS_t &= \mu S_t\,dt + \sigma S_t\bigl(dW_t^{\mathbb{Q}} - \theta\,dt\bigr)
&= r S_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}}. \tag{5} \end{align}

The physical drift $\mu$ has been replaced by $r$. That is the content of “risk-neutral.”

Let $\tilde S_t = S_t e^{-rt}$ be the stock in time-$0$ dollars. The ordinary product rule, plus Itô (the mixed increment $dS\cdot d(e^{-rt})$ is a $dW\cdot dt$ term, which the multiplication table sends to $0$), gives, under $\mathbb{Q}$,

No $dt$ term: $\tilde S$ is a $\mathbb{Q}$-martingale, so $\mathbb{E}^{\mathbb{Q}}[\tilde S_T\mid\mathcal{F}_t] = \tilde S_t$. In a complete, arbitrage-free market the time-$t$ price of a payoff $h(S_T)$ is the discounted risk-neutral expectation

For the call, $h(x)=(x-K)^+$:

Under $\mathbb{Q}$, $(4)$ holds with $\mu=r$, so $S_T$ is lognormal with parameters

Equation $(6)$ is now an integral against that lognormal. Two traders who disagree about $\mu$ still agree on $C$ if they agree on $\sigma$: $\mu$ is not in $(7)$.

Toolkit: the lognormal

This is the engine. Let $Y\sim\mathcal{N}(m,s^2)$ and $X=e^Y$, so $X$ is lognormal and $X>0$ with probability $1$. Completing the square in the Gaussian integral for $e^Y$ gives the mean; a second moment computation gives the variance:

The density of $X$ follows from the change of variable $y=\ln x$, $dy = dx/x$:

The extra $1/x$ is the stretching factor from the substitution (sometimes called the Jacobian): $dy=dx/x$. The cdf is the Gaussian cdf evaluated at $\ln x$:

The call does not need $\mathbb{E}[X]$ and does not need $\mathbb{E}[X\mid X>K]$. It needs the contribution of the upper tail, the partial expectation

The symbol $1{{X>K}}$ is the indicator of the event ${X>K}$: it equals $1$ when $X>K$ and $0$ otherwise, so $X\,1{{X>K}}$ keeps $X$ on the upper tail and throws the rest away.

This is not the conditional tail expectation $\mathbb{E}[X\mid X>K]$, which would divide $(11)$ by $1-F_X(K)$. (That mix-up appears in some notes, including Rouah’s; the integral they write is $(11)$, and $(11)$ is the object that enters the call.)

Substitute $(9)$. The $x$ in the integrand cancels the $1/x$ in the density:

Change variable $y=\ln x$, so $x=e^y$ and $dx=e^y\,dy$. The lower limit becomes $\ln K$:

The exponent in the integrand is $y - (y-m)^2/(2s^2)$. Expand the square and collect terms over a common denominator:

\begin{align} y - \frac{(y-m)^2}{2s^2} &= \frac{2s^2 y - (y^2 - 2my + m^2)}{2s^2} = \frac{-y^2 + 2(m+s^2)y - m^2}{2s^2}. \end{align}

Rewrite the quadratic in $y$ by completing the square. The identity $y^2 - 2(m+s^2)y = \bigl(y-(m+s^2)\bigr)^2 - (m+s^2)^2$ turns the numerator into

Divide by $2s^2$:

The constant $m+s^2/2$ comes out of the integral, and what remains is a normal density with mean $m+s^2$ and variance $s^2$:

The integral is $\mathbb{P}\bigl(\mathcal{N}(m+s^2,\,s^2)>\ln K\bigr)$, which in standard units is $\Phi\bigl((-\ln K+m+s^2)/s\bigr)$. Therefore

That is the identity Route I spends. Completing the square shifted the mean of $Y$ from $m$ to $m+s^2$. In the call, that shift is exactly the gap between $d_2$ and $d_1$.

Route I: two lognormal integrals

Start from $(6)$ and split the payoff on the event ${S_T>K}$. Write $F$ for the cdf of $S_T$ under $\mathbb{Q}$, so $\int g\,dF$ means the expected value of $g(S_T)$ — if $F$ has density $f$, this is the ordinary integral $\int g(x)\,f(x)\,dx$:

\begin{align} C &= e^{-r\tau}\int_K^{\infty}(S_T-K)\,dF(S_T)
&= e^{-r\tau}\int_K^{\infty} S_T\,dF(S_T) \;-\; e^{-r\tau}K\int_K^{\infty} dF(S_T). \tag{15} \end{align}

Under $\mathbb{Q}$, $S_T$ is lognormal with the parameters $(7)$. The first integral is the partial expectation $L_{S_T}(K)$. Feed $(7)$ into $(14)$:

\begin{align} e^{m+s^2/2} &= \exp\Bigl(\ln S_t + \bigl(r-\tfrac12\sigma^2\bigr)\tau + \tfrac12\sigma^2\tau\Bigr) = S_t e^{r\tau}, \end{align}

and the argument of $\Phi$ is

\begin{align} \frac{-\ln K + m + s^2}{s} &= \frac{-\ln K + \ln S_t + \bigl(r-\tfrac12\sigma^2\bigr)\tau + \sigma^2\tau}{\sigma\sqrt{\tau}} = \frac{\ln(S_t/K)+\bigl(r+\tfrac12\sigma^2\bigr)\tau}{\sigma\sqrt{\tau}} = d_1. \end{align}

So $L_{S_T}(K) = S_t e^{r\tau}\,\Phi(d_1)$, and the first term of $(15)$ is $S_t\Phi(d_1)$.

The second integral is the upper tail of the cdf. By $(10)$,

The argument is

and $1-\Phi(-d_2)=\Phi(d_2)$. The second term of $(15)$ is therefore $K e^{-r\tau}\Phi(d_2)$. Combine:

which is $(1)$.

Reading the two terms. $\Phi(d_2)=\mathbb{Q}(S_T>K)$ is the risk-neutral probability of exercise, so the second term is a cash-or-nothing contract: $K$ dollars paid at $T$ if and only if the call finishes in the money ($S_T>K$), and nothing otherwise. (A claim to one dollar at $T$ is a zero-coupon bond, worth $e^{-r\tau}$ today, which is why $K$ is multiplied by $e^{-r\tau}$.) The first term is a share-or-nothing: you receive one share if $S_T>K$, and nothing otherwise. $\Phi(d_1)$ is the same exercise probability after the mean shift $m\mapsto m+s^2$ that $(14)$ performed — equivalently, as Route II will show, the exercise probability when the unit of account is the stock itself.

For the at-the-money numbers above (spot equals strike), $d_1$ and $d_2$ sit $\sigma\sqrt{\tau}=0.20$ apart. That gap is the Itô correction, visible on the page.

Put-call parity. A European call minus a European put with the same strike and expiry is a forward — a contract that obliges you to buy the stock at $K$ at time $T$, with no choice. A forward on a non-dividend stock is worth $S_t - K e^{-r\tau}$:

This is an accounting identity, not a pricing model. Substitute $(1)$:

For those same numbers, $P\approx 5.57$. The call is worth more than the put because the forward is already profitable once $r>0$.

Remark (standard-normal coordinates). Writing $S_T = S_t\exp\bigl((r-\sigma^2/2)\tau + \sigma\sqrt{\tau}\,Z\bigr)$ with $Z\sim\mathcal{N}(0,1)$, the event ${S_T>K}$ becomes ${Z>-d_2}$, and the partial expectation $\mathbb{E}[S_T 1_{{Z>-d_2}}]$ is the same completed square as $(12)$–$(14)$, now in the variable $Z$. Same algebra, different name for the integration variable.

Route II: change of numeraire

A numeraire is the asset used as the unit of account. Quote everything in dollars and the numeraire is the money-market account $B_t=e^{rt}$. Quote everything in shares and the numeraire is $S_t$. For a numeraire $N$ there is a measure $\mathbb{N}$ that makes every asset, divided by $N$, a martingale, and the price of a payoff $V(S_T,T)$ is

For $N=B$ this is $(6)$ and $\mathbb{N}=\mathbb{Q}$. Take $N=S$ instead. The call, measured in shares, is $(1-KZ_T)^+$ with $Z=1/S$, and $\Phi(d_1)$ will appear as an ordinary probability under the associated measure $\mathbb{Q}^{S}$.

Start from $(5)$. The relative bond $\tilde B_t = B_t/S_t$ is not a $\mathbb{Q}$-martingale: Itô produces leftover drift $\sigma^2$. Kill it by Girsanov in the other direction,

so $d\tilde B_t = -\sigma\tilde B_t\,dW_t^{\mathbb{Q}^{S}}$ is a $\mathbb{Q}^{S}$-martingale. Relative to the stock, the bond is now a fair game.

For $Z_t=1/S_t$ the derivatives are $Z_S=-1/S^2$ and $Z_{SS}=2/S^3$. Itô plus $(5)$, then the change of Brownian motion, gives

One more Itô, on $Y=\ln Z$:

So $\ln Z_T$ is normal with mean and variance

and $Z_T$ is lognormal. In particular $\mathbb{E}^{\mathbb{Q}^{S}}[Z_T]=e^{u+v/2}=e^{-r\tau}/S_t$, which is the martingale property $\mathbb{E}^{\mathbb{Q}^{S}}[B_T/S_T]=B_t/S_t$.

The share-denominated call is nonzero when $Z_T<1/K$. Equation $(16)$ becomes

By the lognormal cdf $(10)$,

$\Phi(d_1)$ is the probability of exercise when the unit of account is the stock.

The second piece is $K$ times a lower-tail partial expectation, which is the full mean minus the upper-tail identity $(14)$ applied to $Z$ at the barrier $1/K$. The $\Phi$ argument of that upper tail is $-d_2$, so

Therefore

Same formula. The two $\Phi$ terms are now both probabilities, each belonging to the numeraire that multiplies it.

The PDE, briefly

Black and Scholes did not start from $(6)$. They started from a portfolio $\Pi=-V+\Delta S$ (short one option, long $\Delta$ shares). Expand $dV$ with Itô’s lemma, choose $\Delta=V_S$ so the $dW$ terms cancel — that choice is delta-hedging, and $V_S$ is the delta of the option — and require the now-riskless $\Pi$ to earn $r$, or else there is an arbitrage against the money-market account. The result is the Black–Scholes partial differential equation (PDE): an equation relating the partial derivatives of $V$ with respect to $t$ and $S$,

with $V(S,T)=(S-K)^+$. The drift $\mu$ cancelled with the $dW$ terms. The second derivative $V_{SS}$ is the gamma; $\tfrac12\sigma^2 S^2 V_{SS}$ is Itô’s correction, now sitting in a PDE.

The Feynman–Kac theorem is the dictionary between $(17)$ and $(6)$: a PDE of this shape, with drift coefficient $rS$ and discount $r$, is solved by $e^{-r\tau}\mathbb{E}^{\mathbb{Q}}[h(S_T)\mid S_t]$ along the stock $(5)$. The hedging argument produces the PDE; Feynman–Kac translates it into the integral Route I already evaluated.

A further change of variables — $x=\ln(S/K)$, time-to-go rescaled by $\sigma^2/2$, an exponential prefactor to absorb discounting — turns $(17)$ into the heat equation $W_{\tilde\tau}=W_{xx}$, the same PDE that describes temperature spreading along a rod. The solution that starts as a spike at the origin and then smears out is a Gaussian (the heat kernel). Integrate a Gaussian and you get $\Phi$. The algebra is written out in Rouah §7 of the hosted note. The bell curve in $(1)$ is both the law of where geometric Brownian motion ends up and the shape of spreading heat: the same assumption, two costumes.

Limitations

Dividends. A continuous yield $q$ changes the risk-neutral drift of $S$ from $r$ to $r-q$. The same Route I, with $m=\ln S_t+(r-q-\sigma^2/2)\tau$, produces Merton’s formula

$r$ replaced by $r-q$ inside $d_1$ and $d_2$ as well. Index options live here ($q$ is the basket yield), FX options live here ($q$ is the foreign rate), and futures options live here ($q=r$: Black 1976).

The smile. After 1987 the market stopped believing in a single $\sigma$. Implied volatility is the number $\sigma$ you must feed into $(1)$ to recover the price the market is actually quoting. Plot that number against strike and you get a smile, or in equities a smirk. Traders still quote in Black–Scholes implied vol. They do not believe the lognormal assumption that produced it.

Jumps, stochastic vol, Americans. Merton added sudden jumps (a Poisson process: events that arrive at random times). Heston let $\sigma_t$ itself wander. Dupire showed that a whole surface of implied vols is equivalent to a local-vol model — $\sigma$ depends on $S$ and $t$ — that refits every ordinary call and put. None of these has a two-$\Phi$ formula of the same shape. The American put has no closed form of this kind; a binomial tree (the coin-flip walk, used as a calculator) is the honest computation.

What has been proved is a theorem about a complete market driven by one Brownian motion: the unique no-arbitrage price of $(S_T-K)^+$ is $(1)$. Using that theorem as a price, as a quoting convention, or as the first term of an approximation is a separate decision.

References

• Fabrice Douglas Rouah, Four Derivations of the Black-Scholes Formula. Local copy. Wayback, 19 July 2024.
• F. Black and M. Scholes, The Pricing of Options and Corporate Liabilities, JPE 81 (1973).
• R. C. Merton, Theory of Rational Option Pricing, Bell Journal 4 (1973).
• K. Itô, On Stochastic Differential Equations, Mem. AMS (1951).
• J. Hull, Options, Futures, and Other Derivatives.
• S. Shreve, Stochastic Calculus for Finance II.
• E. Derman and M. B. Miller, The Volatility Smile.