Black-Scholes-Merton: I · II · III
The same formula, reached twice more — once as heat, once as a change of units — and then the assumptions, each pulled on until the formula bends or breaks.
Vol. II derived
\[C \;=\; S_t\,\Phi(d_1) - K e^{-r\tau}\,\Phi(d_2), \qquad d_{1,2} \;=\; \frac{\ln(S_t/K)+\bigl(r\pm\tfrac12\sigma^2\bigr)\tau}{\sigma\sqrt{\tau}}, \tag{1}\]by integrating the payoff against a lognormal whose drift is $r$. The rain check — $S=K=100$, $\tau=1$, $r=5\%$, $\sigma=20\%$ — costs $10.45$. This volume reaches $(1)$ by two other roads, then revisits the assumptions of Vol. I.
1. The heat equation
Vol. II never wrote a differential equation for the option price. Black and Scholes did; it was their route. It needs Itô’s lemma, Vol. II’s $(3)$: for $dX=a\,dt+b\,dW$,
\[df(t,X_t) \;=\; \Bigl(f_t + a f_x + \tfrac12 b^2 f_{xx}\Bigr)dt + b f_x\,dW_t. \tag{3}\]The hedge
Let $V(S,t)$ be the option’s price, and hold the portfolio $\Pi=-V+\Delta S$: short one option, long $\Delta$ shares. Expand $dV$ with $(3)$, using $dS=\mu S\,dt+\sigma S\,dW$:
\[d\Pi \;=\; -\Bigl(V_t+\mu S V_S+\tfrac12\sigma^2S^2V_{SS}\Bigr)dt - \sigma S V_S\,dW + \Delta\bigl(\mu S\,dt+\sigma S\,dW\bigr).\]Choose $\Delta=V_S$. The $dW$ terms cancel — this is delta-hedging — and so do the $\mu$ terms. What is left has no randomness:
\[d\Pi \;=\; -\Bigl(V_t+\tfrac12\sigma^2S^2V_{SS}\Bigr)dt.\]A riskless portfolio must earn the riskless rate, $d\Pi=r\Pi\,dt$, or there is a free lunch. Setting the two equal gives the Black–Scholes PDE:
\[V_t + rS V_S + \tfrac12\sigma^2 S^2 V_{SS} - rV \;=\; 0, \tag{17}\]with $V(S,T)=(S-K)^+$. The $dW$ cancellation eats $\mu$, just as $\Delta=2/3$ did on the two-leaf tree. For the rain-check numbers, $\Delta=V_S=\Phi(d_1)\approx 0.637$.
Heat in disguise
This is a linear parabolic PDE. You have seen one: the heat equation $u_t=u_{xx}$, which says a rod’s temperature at a point rises when the point is colder than the average of its neighbours. Let $\tau=T-t$, $x=\ln(S/K)$, $V(S,t)=U(x,\tau)$. Then $V_t=-U_\tau$, $V_S=U_x/S$, and $S^2 V_{SS}=U_{xx}-U_x$. Substitute:
\[U_\tau \;=\; \tfrac12\sigma^2 U_{xx} + \bigl(r-\tfrac12\sigma^2\bigr)U_x - r U.\]Rescale $\tilde\tau=\sigma^2\tau/2$, set $k=2r/\sigma^2$, and kill lower-order terms with $U=e^{\alpha x+\beta\tilde\tau}W$, $\alpha=(1-k)/2$, $\beta=\alpha^2+(k-1)\alpha-k$. Then
\[W_{\tilde\tau} \;=\; W_{xx},\]the heat equation, with initial temperature profile $W(x,0)=e^{-\alpha x}(Ke^{x}-K)^+$. (Rain-check numbers: $k=2.5$, $\alpha=-3/4$, $\beta=-49/16$.) Time runs backwards from expiry: the kink in the payoff at $x=0$ is a hot spot on the rod, and time to expiry lets it diffuse.
The solution is the initial profile smeared by the heat kernel,
\[W(x,\tilde\tau) \;=\; \int_{-\infty}^{\infty} G(x-y,\tilde\tau)\,W(y,0)\,dy, \qquad G(x,\tilde\tau)=\frac{1}{\sqrt{4\pi\tilde\tau}}\exp\Bigl(-\frac{x^2}{4\tilde\tau}\Bigr).\]$G$ is the density of $\mathcal{N}(0,2\tilde\tau)$, and $2\tilde\tau=\sigma^2\tau=\mathrm{Var}(\ln S_T)$ under $\mathbb{Q}$. The heat kernel is the lognormal of Vol. II, written in log-coordinates, and the integral above is Vol. II’s integral $(15)$ with the substitutions undone. Evaluating it is the same completing-the-square, and returns $(1)$. Feynman–Kac, in this instance, is that sentence: the solution of the PDE is an expectation against the process.
What this road buys you is numerics. When the payoff or the boundary makes the integral intractable — an American put, a barrier — the PDE is still a PDE, and finite differences on a grid solve it.
2. Change of numeraire
The two $\Phi$’s in $(1)$ are both probabilities that the call finishes in the money, $\lbrace S_T>K\rbrace$. Yet $\Phi(d_1)\approx 0.637$ and $\Phi(d_2)\approx 0.560$. Same event, two probabilities. The explanation is that they are measured in different units of account. To say that precisely we first need the tool Vol. II asserted: changing probabilities without changing the spread.
The calculation below 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.
Girsanov, from two Gaussians
On the tree, the likelihood ratio on one step is $p^{\ast}/p$ on up and $(1-p^{\ast})/(1-p)$ on down. Then $\mathbb{E}^{\ast}[X]=\mathbb{E}[X\,\xi]$. Over $n$ steps, $\xi$ is the product of those ratios. That product is Girsanov, before the limit.
One Gaussian. Under $\mathbb{P}$, $W_t\sim\mathcal{N}(0,t)$. We want $\mathbb{Q}$ under which $W_t\sim\mathcal{N}(-\theta t,\,t)$, so that $W_t+\theta t$ is standard Wiener. Divide the two densities:
\[\xi_t \;=\; \frac{\exp\bigl(-(w+\theta t)^2/2t\bigr)}{\exp\bigl(-w^2/2t\bigr)}\Bigg|_{w=W_t} \;=\; \exp\bigl(-\theta W_t - \tfrac12\theta^2 t\bigr).\]For any test function, $\mathbb{E}^{\mathbb{Q}}[f(W_t)]=\mathbb{E}^{\mathbb{P}}[f(W_t)\,\xi_t]$. Only the center moved; the variance $t$ did not. Independent increments multiply the same ratio across steps. That is Girsanov’s theorem, from the Gaussian pdf — motivated here, not proved for general processes.
Which $\theta$. Rewrite the stock $dS=\mu S\,dt+\sigma S\,dW$ as
\[dS \;=\; r S\,dt + \sigma S\Bigl(dW + \frac{\mu-r}{\sigma}\,dt\Bigr).\]The process in parentheses is Wiener plus a constant drift $\theta=(\mu-r)/\sigma$, the market price of risk: excess return per unit of volatility. Girsanov with that $\theta$ makes $W^{\mathbb{Q}}_t=W_t+\theta t$ a Wiener process under $\mathbb{Q}$, and
\[dS_t \;=\; r S_t\,dt + \sigma S_t\,dW_t^{\mathbb{Q}}. \tag{5}\]This is what Vol. II assumed by analogy with the tree: $\mu$ becomes $r$, $\sigma$ stays. Under $\mathbb{Q}$,
\[S_T \;=\; S_t\exp\Bigl(\bigl(r-\tfrac12\sigma^2\bigr)\tau + \sigma\sqrt{\tau}\,Z\Bigr), \qquad Z\sim\mathcal{N}(0,1). \tag{7}\]Two numeraires
A numeraire is the unit you quote prices in. Dollars: the money-market account $B_t=e^{rt}$. Shares: the stock $S_t$. For each choice there is a measure that makes every asset, divided by that numeraire, a martingale. $\mathbb{Q}$ is the one for dollars-in-the-bank: $S_t/B_t$ is a fair game.
| Numeraire | Measure | Martingale | exercise probability | Role in (1) |
|---|---|---|---|---|
| bond \(B_t=e^{rt}\) | \(\mathbb{Q}\) | \(S_t/B_t\) | \(\Phi(d_2)\) | multiplies the cash \(Ke^{-r\tau}\) |
| stock \(S_t\) | \(\mathbb{Q}^{S}\) | \(B_t/S_t\) | \(\Phi(d_1)\) | multiplies the share \(S_t\) |
Vol. II split the call into a cash-or-nothing (receive $K$ if $S_T>K$) and an asset-or-nothing (receive the share if $S_T>K$). Under $\mathbb{Q}$ the cash-or-nothing is worth $Ke^{-r\tau}\,\mathbb{Q}(S_T>K)=Ke^{-r\tau}\Phi(d_2)$. The asset-or-nothing pays in shares, so price it in shares.
The density ratio that changes $\mathbb{Q}$ into $\mathbb{Q}^{S}$ is the terminal value of the new numeraire relative to the old, rebased to $1$ today:
\[\frac{d\mathbb{Q}^{S}}{d\mathbb{Q}} \;=\; \frac{S_T/S_t}{B_T/B_t} \;=\; \frac{S_T}{S_t e^{r\tau}}.\]Under $\mathbb{Q}$, from $(7)$,
\[\frac{S_T}{S_t e^{r\tau}} \;=\; \exp\bigl(-\tfrac12\sigma^2\tau + \sigma\sqrt{\tau}\,Z\bigr).\]That is the Girsanov density $\xi=\exp(-\theta W-\tfrac12\theta^2 t)$ from above, with $\theta=-\sigma$ and $W_{\tau}=\sqrt{\tau}\,Z$. The tilt adds $\sigma\sqrt{\tau}$ to $Z$. The event $\lbrace S_T>K\rbrace$ is $\lbrace Z>-d_2\rbrace$ under $\mathbb{Q}$; after the tilt it is $\lbrace Z+\sigma\sqrt{\tau}>-d_2\rbrace=\lbrace Z>-d_1\rbrace$, so
\[\mathbb{Q}^{S}(S_T>K) \;=\; \Phi(d_1).\]The asset-or-nothing is therefore $S_t\,\Phi(d_1)$: today’s share, times the exercise probability in share units. The cash-or-nothing is $Ke^{-r\tau}\,\Phi(d_2)$: discounted strike, times the exercise probability in dollar units. That is $(1)$, with both $\Phi$’s now ordinary probabilities, each belonging to the numeraire that multiplies it. No integral was completed; the square was completed once, inside Girsanov.
Why is the share-measure probability higher? Under $\mathbb{Q}^S$, outcomes are weighted by how many dollars the share is worth in them. The in-the-money outcomes are exactly the ones where the share is worth a lot, so they get more weight. $\Phi(d_1)-\Phi(d_2)$ is that reweighting.
The same trick is the workhorse whenever $r$ is not constant: price a payoff in units of a zero-coupon bond maturing at $T$, and the discount factor comes out of the expectation even when rates are random. That is the $T$-forward measure, and §3 uses it.
3. The assumptions, revisited
Vol. I listed the assumptions before they were used. Now we know exactly where each was spent, so we can pull on each one and watch. The two-$\Phi$ shape is not a law of nature. It is a receipt for a short list of hypotheses. Perturb one and the formula either morphs (same skeleton, different inputs or an extra factor) or breaks (the completing-the-square step no longer lands on $\Phi$).
The risk-free rate
Where it was spent: $r$ appears twice. It discounts, $e^{-r\tau}$; and it is the drift of the stock under $\mathbb{Q}$.
Known, but not constant. If $r(t)$ is a deterministic function of time, replace $r\tau$ everywhere by $\int_t^T r(u)\,du$. The formula survives intact; in practice you read that integral off today’s yield curve, as the yield on a zero-coupon bond maturing at $T$.
Random. If $r_t$ is itself random and correlated with $S$, then $\mathbb{E}^{\mathbb{Q}}\bigl[e^{-\int r}(S_T-K)^+\bigr]$ no longer splits: the discount factor does not come out in front. The fix is §2’s: use the $T$-maturity bond $P(t,T)$ as the numeraire. Under that measure the forward $F=S_t/P(t,T)$ is a martingale, and if $F$ is lognormal with total variance $v^2$, the call is
\[C \;=\; P(t,T)\bigl[F\,\Phi(d_1) - K\,\Phi(d_2)\bigr], \qquad d_{1,2}=\frac{\ln(F/K)\pm\tfrac12 v^2}{v}.\]Two $\Phi$’s again, but $v^2$ now includes the bond’s volatility and its correlation with the stock. For short-dated equity options this is a rounding error. For long-dated ones, or options on bonds, it is the whole story.
Borrowing is not lending. The hedge in §1 borrows to buy shares. If you borrow at $r_b$ and lend at $r_\ell<r_b$, a call-buyer’s replication and a call-writer’s replication cost different amounts, and no-arbitrage pins the price only to an interval, not a point. Black–Scholes is the midpoint of a band whose width is the spread.
Negative, or near zero. The lognormal needs a positive underlying. When the underlying is itself a rate, and rates can go through zero — euro rates after 2014 — the log in $d_{1,2}$ is undefined. Go back to an additive Gaussian, $S_T=S+\sigma\sqrt{\tau}\,Z$ (Bachelier, 1900), and the same integral gives a single $\Phi$ of $(S-K)/(\sigma\sqrt{\tau})$ plus a density term, with no logarithms. The log in $d_{1,2}$ is not decoration. It is the Jacobian of Vol. II §2, still visible in the answer.
Dividends
Where it was spent: under $\mathbb{Q}$ the stock’s drift had to be $r$, otherwise you could pocket the excess over the bond by holding it. And in Vol. I, Merton’s no-early-exercise argument for calls.
If the stock pays a continuous yield $q$ — a cash stream $qS\,dt$ while you hold it — the capital-gain drift under $\mathbb{Q}$ must be $r-q$. Total return, gain plus yield, equals $r$. Any more is a free lunch.
The prepaid forward is then $S_t e^{-q\tau}$ instead of $S_t$. Vol. II’s integral still closes. The share piece picks up $e^{-q\tau}$, and $r$ is replaced by $r-q$ inside $d_1$ and $d_2$:
\[C \;=\; S_t e^{-q\tau}\,\Phi(d_1) - K e^{-r\tau}\,\Phi(d_2), \qquad d_{1,2} \;=\; \frac{\ln(S_t/K)+\bigl(r-q\pm\tfrac12\sigma^2\bigr)\tau}{\sigma\sqrt{\tau}}.\]Give the rain-check stock a $2\%$ yield. Then $d_1=0.25$, $d_2=0.05$, and $C\approx 9.23$, cheaper than $10.45$: the stock itself now pays you, so the call is less of a reason to hold.
The same skeleton prices three cousins, once you name what $q$ is.
- A call on an index. The stocks in the basket pay dividends, so the index has a yield $q$.
- A call on a foreign currency (Garman–Kohlhagen). Holding euros earns the euro interest rate $r_f$; that rate is $q$.
- A call on a futures (Black 1976). You put up no cash to hold the futures, so you save the financing $r$. That saving is a yield $q=r$, and the formula collapses to $e^{-r\tau}\bigl[F\Phi(d_1)-K\Phi(d_2)\bigr]$ — the bond-numeraire formula above, with a constant rate.
A discrete cash dividend of size $D$ just before $T$ is a different animal: $S_T$ is then a lognormal minus $D$, which is not lognormal, and $(1)$ as written does not apply. The common patch is to subtract the present value of $D$ from $S_t$ and apply $(1)$ to what is left. And a discrete dividend revives early exercise for American calls: just before the stock goes ex-dividend, the call-holder may do better to exercise and collect $D$ than to watch the share price drop by $D$.
The smile
Where it was spent: one $\sigma$, constant, for every strike and every path. That is what made $\ln S_T$ a single Gaussian.
Implied volatility is the one input you cannot read off a screen, so the market reads it backwards. Given a quoted price, $\sigma_{\mathrm{imp}}$ is the number you feed into $(1)$ to recover that quote. Our formula at $20\%$ says $10.45$; if the market shows $12$, implied vol is some number above $20\%$. Traders quote “vol $22$” rather than a dollar price, because that number compares across strikes and tenors.
If the model were right, $\sigma_{\mathrm{imp}}$ would be the same for every strike. It is not. Before October 1987, equity implied vols were roughly flat. After the crash, out-of-the-money puts — insurance against another one — traded at prices no single $\sigma$ could fit. Plot $\sigma_{\mathrm{imp}}$ against strike and you get a U in currencies (a smile) and a downward slope in equities (a smirk: crash insurance is dear). Traders using $(1)$ this way are using it as a language. They do not believe $\ln S_T$ is Gaussian.
The smile is the market telling you which assumption to drop. Each fix corresponds to a line of Vol. II’s derivation.
- Fat tails. If $\ln S_T$ is Student-$t$ rather than Gaussian, completing the square fails: that move is a property of $\exp(-\tfrac12 z^2)$, not of densities in general. No closed form; integrate numerically. Fatter tails than the lognormal make far-out-of-the-money options dearer, which is a smile.
-
Wandering volatility. If $\sigma_t$ is itself random (Heston), then $s$ is random, and conditional on $s$ the stock is lognormal. The call is an average of Black–Scholes prices, one per realised $s$:
\[\mathbb{E}[(S_T-K)^+] \;=\; \mathbb{E}\bigl[e^{m+s^2/2}\Phi(d_{+}(s)) - K\Phi(d_{-}(s))\bigr].\]Averaging over $s$ fattens both tails. Correlate $\sigma$ with $S$ negatively — vol rises when the market falls — and the left tail fattens more: a smirk.
- Jumps. Wiener paths are continuous; a crash is a gap. Merton added Poisson jumps on top of the Wiener process. Conditional on the number of jumps, $\ln S_T$ is Gaussian, so the price is a Poisson-weighted sum of $\Phi$-pairs. You cannot hedge a jump with $\Delta$ shares, so the market is no longer complete, and the price is no longer unique: it depends on what the market charges for jump risk.
- Local volatility. Or let $\sigma=\sigma(S,t)$, fitted so that every quoted strike is recovered exactly (Dupire, 1994). One function reproduces the whole smile today. But $S_T$ is then typically not lognormal, and the two-$\Phi$ formula is being used as a quoting dictionary, not as a model.
A deterministic but time-varying $\sigma(t)$, by contrast, changes nothing: $s^2=\int_t^T\sigma(u)^2\,du$ is still one number, and $(1)$ survives with that $s$. That is the term structure of volatility, and it is the one perturbation in this section that the formula absorbs.
Everything else
European exercise. Vol. II integrated over the law of $S_T$ only; the path was invisible. An American put can be exercised at any time, so its value depends on the whole path of opportunities. There is a free boundary — a critical stock price below which you exercise — and no two-$\Phi$ formula. The honest computation is the tree of Vol. II with one change: at each node, take the maximum of exercising now and the discounted risk-neutral value of holding. Or the PDE of §1, with the constraint $V\ge (K-S)^+$ everywhere. Barrier options (knocked out if $S$ touches a level) and Asian options (paying on the average of $S$) break the formula for the same reason: the payoff sees the path.
Frictionless and continuous. The hedge in §1 rebalances continuously. With transaction costs, continuous rebalancing costs infinitely much, so you rebalance discretely and the hedge leaks. The price becomes a band again, widening with costs and with how often you trade.
| Assumption | Perturb | Two \(\Phi\)'s? |
|---|---|---|
| \(\ln S_T\) Gaussian | Student-\(t\); mixture (Heston); jumps (Merton) | breaks; average of \(\Phi\)'s; sum of \(\Phi\)'s |
| one \(\sigma\) | \(\sigma(t)\) deterministic; \(\sigma(S,t)\); smile \(\sigma(K)\) | survives as \(s^2=\int\sigma^2\); breaks; quoting dictionary |
| payoff \((S_T-K)^+\) | American; barrier; Asian; cash-or-nothing | breaks; breaks; breaks; is the second \(\Phi\) |
| \(r\) constant | \(r(t)\) deterministic; random \(r_t\) | survives; morphs (bond numeraire) |
| one rate for borrowing and lending | \(r_b>r_\ell\) | a band, not a price |
| pricing law is \(\mathbb{Q}\) | physical \(\mu\) instead of \(r\) | shape lives, number is wrong |
| multiplicative, \(S_T>0\) | additive Gaussian (Bachelier) | different \(\Phi\), no \(\ln\) |
| no dividends | yield \(q\); discrete \(D\) | morphs (\(e^{-q\tau}\)); patched |
| frictionless, continuous | transaction costs | a band, not a price |
The assumptions are doing a lot of work. They are why a rain check on a stock has a two-line formula and a rain check on an average, or on a stock that can jump, does not.
What was proved
The theorem we proved is narrower than the word “option.” It prices a European call on a stock that pays no dividend, driven by one Wiener process with a constant $\sigma$, in a market where you can trade continuously and without cost at a known rate $r$. In that complete market, the unique no-arbitrage price of $(S_T-K)^+$ is $(1)$. We reached it three ways: an integral against a lognormal, the heat equation, and a change of unit. Using that theorem as a price, as a quoting convention, or as the first term of an approximation is a separate decision.
Three volumes: the contract and its assumptions; the noise, the lognormal, and the integral; heat, numeraires, and nuance. The rain check is $10.45$.
Cheers.
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).
- J. C. Cox, S. A. Ross, and M. Rubinstein, Option Pricing: A Simplified Approach, J. Financial Economics 7 (1979). The $n$-step tree.
- 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.