Risk-Neutral Pricing

Walk onto any derivatives desk in the world and ask how they price an option — any option, on anything — and you will get one recipe back:

V_0 = e^{-rT}\,\mathbb{E}_{\mathbb{Q}}\!\big[\,\text{payoff}\,\big].

Discount the expected payoff. Reasonable enough — except the expectation is taken under probabilities that nobody believes. The measure \mathbb{Q} describes a fictional world in which every asset, however risky, drifts at the risk-free rate r. Stocks that everyone expects to beat the bank account are assumed not to. And yet the number that falls out is the true, arbitrage-free price — the same price a hedger would charge. This page is about why a deliberately wrong-looking probability gives an exactly right answer.

The clue is already in hand. The Black–Scholes PDE gave us a price in which the stock's real-world drift \mu had mysteriously vanished. Risk-neutral pricing is a second, probabilistic road to the same price that makes the disappearance of \mu not a surprise but the whole point. Under the real-world measure \mathbb{P} the stock obeys dS = \mu S\,dt + \sigma S\,dW. By Girsanov's theorem we can switch to an equivalent risk-neutral measure \mathbb{Q} under which the Brownian motion \tilde W absorbs the drift change, and the stock becomes a geometric Brownian motion with drift r:

dS_t = r S_t\,dt + \sigma S_t\,d\tilde W_t.

The volatility \sigma is untouched — Girsanov reweights probabilities, it cannot change the size of the wiggle — and \mu is gone for good. But before we ride the continuous-time machinery, let us watch the whole trick happen in a world with exactly two outcomes, where every number can be checked by hand.

The trick in miniature: one step, two roads

Take the simplest market that can carry an option. A stock is at S_0 = 100. In one period it will be worth either 120 (up) or 90 (down) — nothing else. A bank account turns 1 into 1.05 over the period. We want today's price of a European call with strike K = 100, which will pay 20 in the up state and 0 in the down state. Note what we have not been told: the probability of the up-move. We will price the option anyway — twice.

Road 1 — replication. Following the logic of replication, build a portfolio of \Delta shares and B in the bank that matches the option's payoff in both states:

\begin{aligned} 120\,\Delta + 1.05\,B &= 20 \quad &\text{(up)}\\ 90\,\Delta + 1.05\,B &= 0 \quad &\text{(down)} \end{aligned}

Subtracting: 30\,\Delta = 20, so \Delta = \tfrac{2}{3}, and then B = -60/1.05 = -57.14 (borrow). Two states, two instruments, two equations — the payoff is hit exactly. By no-arbitrage the option must cost what the portfolio costs today:

V_0 = \Delta S_0 + B = \tfrac{2}{3}\cdot 100 - 57.14 = 9.52.

No probability was used anywhere. The price is forced by hedging alone.

Road 2 — risk-neutral expectation. Now ask a strange question: what probability q of the up-move would make the stock, on average, grow at exactly the bank's rate? Solve

100 = \frac{q \cdot 120 + (1-q)\cdot 90}{1.05} \;\Longrightarrow\; q = \frac{1.05\cdot 100 - 90}{120 - 90} = \tfrac{1}{2}.

Then price the option as its discounted expected payoff under q:

V_0 = \frac{q\cdot 20 + (1-q)\cdot 0}{1.05} = \frac{10}{1.05} = 9.52.

The same number — and it always will be, because road 2 is road 1 in disguise: solve the two replication equations symbolically and the answer rearranges algebraically into a discounted expectation with weight q = \dfrac{(1+r) - d}{u - d} (here u = 1.2, d = 0.9). Here is the punchline to carry through the rest of the page: q came from no-arbitrage, not from anyone's beliefs. The market's real probability of the up-move — call it p — never appeared in either calculation. Whether investors think the stock is a rocket or a dud, the call costs 9.52, because at any other price the hedger eats a free lunch. Step through the figure to watch both roads meet.

Scaling up: backward induction on a two-step tree

Add a second period, with the same up/down factors u = 1.2, d = 0.9 and rate 5\% per step. From 100 the stock reaches 144, 108 or 81, and the call pays 44, 8 or 0. Because u, d and r are the same at every node, so is q = \tfrac{1}{2} — and the pricing rule simply repeats itself, one step at a time, from the leaves backward:

V_u = \frac{\tfrac12\cdot 44 + \tfrac12\cdot 8}{1.05} = 24.76, \qquad V_d = \frac{\tfrac12\cdot 8 + \tfrac12\cdot 0}{1.05} = 3.81, V_0 = \frac{\tfrac12\cdot 24.76 + \tfrac12\cdot 3.81}{1.05} = 13.61.

This is backward induction: at every node the value is the one-step discounted q-expectation of the two values ahead of it. Equivalently, compound the steps and read the price in one shot as a discounted expectation over the leaves,

V_0 = \frac{q^2\cdot 44 + 2q(1-q)\cdot 8 + (1-q)^2\cdot 0}{1.05^2} = \frac{11 + 4}{1.1025} = 13.61,

where q^2,\; 2q(1-q),\; (1-q)^2 are the risk-neutral probabilities of the three leaves. Let the number of steps grow and the time step shrink, and this leaf-weighted sum converges to an integral against a lognormal density — the tree's q becomes the continuous-time measure \mathbb{Q}. That is the world we now return to.

The pricing formula, line by line

The engine is a single structural fact. Under \mathbb{Q} the discounted stock price e^{-rt}S_t is a martingale — a fair game — because choosing the drift to be r exactly offsets the discount factor's decay. (In the tree this was precisely the defining equation for q: today's price equals the discounted expected tomorrow.) The same holds for the discounted option value: it is built by trading the stock and the (riskless) bank account, so it inherits the martingale property. We ride that fact straight to the price.

Step 1 — the discounted value is a \mathbb{Q} -martingale. Write V_t for the option's value at time t. Then e^{-rt}V_t is a martingale under \mathbb{Q}, so its best forecast equals its current value:

e^{-rt}V_t = \mathbb{E}_{\mathbb{Q}}\!\big[\,e^{-rT}V_T \,\big|\, \mathcal{F}_t\,\big].

Step 2 — solve for V_t. Multiply both sides by e^{rt}, pulling the time-t discount factor inside (it is \mathcal{F}_t-measurable, a constant to the expectation):

V_t = e^{rt}\,\mathbb{E}_{\mathbb{Q}}\!\big[\,e^{-rT}V_T \,\big|\, \mathcal{F}_t\,\big] = e^{-r(T - t)}\,\mathbb{E}_{\mathbb{Q}}\!\big[\,V_T \,\big|\, \mathcal{F}_t\,\big].

Step 3 — substitute the terminal payoff. At expiry the option is its payoff, V_T = \text{payoff}(S_T). For a European call this is (S_T - K)^+:

V_t = e^{-r(T - t)}\,\mathbb{E}_{\mathbb{Q}}\!\big[\,(S_T - K)^+ \,\big|\, \mathcal{F}_t\,\big].

Step 4 — read it at t = 0. With nothing yet observed, the conditional expectation is an ordinary one:

V_0 = e^{-rT}\,\mathbb{E}_{\mathbb{Q}}\!\big[\,(S_T - K)^+\,\big].

The recipe is now a slogan: price = discount × expected payoff, with the expectation taken in the world where the stock drifts at r. No \mu in sight — and no PDE either, though the two answers agree (the Feynman–Kac formula is the bridge).

Let \mathbb{Q} be the risk-neutral measure, under which S is a geometric Brownian motion of drift r, dS_t = r S_t\,dt + \sigma S_t\,d\tilde W_t. Then for any European derivative with payoff g(S_T):

The name is honest. A genuinely risk-neutral investor demands no extra reward for bearing uncertainty, so in their world every asset — risky stock and safe bond alike — drifts at the single rate r, and future cashflows are discounted at that same r. We are not claiming investors are risk-neutral; we are exploiting that the option's price, being perfectly hedgeable, is the same in that fictitious world as in the real one.

That is why \mu may be replaced by r for free. The replication argument behind the PDE and this change of measure are two faces of one coin: hedging removes the directional bet, so the market's view of the stock's direction (its \mu, its risk premium) is irrelevant to the price. Only the spread \sigma matters. Reassuringly, this expectation gives the same number the PDE does.

The continuous version: the integral Black–Scholes computes

The formula asks for an expectation, so let us set it up as an honest integral. Under \mathbb{Q}, solving the GBM with drift r gives the terminal price in closed form,

S_T = S_0\,\exp\!\Big(\big(r - \tfrac{1}{2}\sigma^2\big)T + \sigma\sqrt{T}\,Z\Big), \qquad Z \sim N(0,1) \text{ under } \mathbb{Q},

so S_T is lognormal with the risk-neutral drift baked in. The call price is then a one-dimensional integral against the standard normal density \varphi:

V_0 = e^{-rT}\int_{-\infty}^{\infty} \Big(S_0\,e^{(r - \sigma^2/2)T + \sigma\sqrt{T}\,z} - K\Big)^{\!+}\,\varphi(z)\,dz.

The payoff is zero until the exponential climbs past K — i.e. for z above a threshold z^{*} found by solving S_T = K. Above it the integrand splits into two pieces: a K-piece, which is just K times a normal tail probability, and an S_T-piece, where the exponential of z can be absorbed into the Gaussian by completing the square. Both pieces come out in terms of the normal CDF \Phi — the calculation closes, and the two \Phi(d_1), \Phi(d_2) terms of the celebrated Black–Scholes formula are exactly these two pieces. We save the algebra for that page; here the point is structural: the Black–Scholes formula is nothing more than this risk-neutral expectation evaluated in closed form.

And look once more at what the integral contains: S_0, K, r, \sigma, T. The real-world drift \mu — the one parameter every investor argues about, the hardest of all to estimate — is simply not an input. Hedging removed it: the replicating portfolio neutralises the directional bet continuously, just as \Delta = \tfrac{2}{3} neutralised it in one stroke on the tree, so the price cannot depend on which way anyone thinks the stock is headed. That is the deepest practical consequence of risk-neutral pricing: it turns an unknowable forecasting problem into a hedging problem with observable inputs.

The single most common misreading of this subject: treating risk-neutral probabilities as predictions. They are not. A \mathbb{Q}-probability is a price in disguise — in the one-step tree, q/1.05 is literally today's price of a security paying 1 in the up state and nothing otherwise (a state price), and q is that state price scaled so the two states sum to one. Because investors pay a premium for payoffs that arrive in bad times, \mathbb{Q} systematically overweights bad states relative to real frequencies. Reading a "risk-neutral default probability" off credit spreads as a forecast of actual default rates — a category error analysts genuinely commit — will overstate the true frequency, sometimes severalfold, because the gap between \mathbb{Q} and \mathbb{P} is the risk premium.

Second trap: the discounting and the measure come as a pair. The theorem says the \mathbb{Q}-expectation of the discounted payoff is the price — V_0 = \mathbb{E}_{\mathbb{Q}}[e^{-rT}g(S_T)]. Take the expectation under \mathbb{Q} but forget the discount, or discount a \mathbb{P}-expectation at r, and the number is wrong in both cases. \mathbb{Q} is defined relative to the bank account as the yardstick (the "numéraire"); change the yardstick and the measure changes with it.

Here is the fun inversion. This page computes prices from \mathbb{Q}; a trader can run it backwards and compute \mathbb{Q} from prices. Differentiate the call price twice with respect to the strike and (Breeden–Litzenberger, 1978) out pops the risk-neutral density itself: \frac{\partial^2 C}{\partial K^2} = e^{-rT}\,q_{\mathbb{Q}}(K). Since real markets quote calls at dozens of strikes, desks reconstruct the whole market-implied distribution of S_T every day.

The picture it reveals changed in a single afternoon. Before the crash of October 1987, the extracted density looked comfortably lognormal. Ever since, equity-index densities have worn a fat left tail — out-of-the-money puts trade rich, the "volatility smile" — because the market permanently re-priced the possibility of a crash. And the idea has a noble pedigree: the state prices behind q are the Arrow–Debreu securities of 1950s general-equilibrium theory — one abstract claim per state of the world — dreamt up to prove theorems about economies, and now, half a century on, extracted daily from the option board as a working trading-desk tool.

Pricing by simulation

The formula V_0 = e^{-rT}\mathbb{E}_{\mathbb{Q}}[(S_T - K)^+] is also a recipe: simulate many risk-neutral paths of the stock, collect each path's discounted payoff, and average. This is Monte-Carlo pricing, and for payoffs too tangled for a closed form — path-dependent exotics, baskets on twenty underlyings — it is how banks actually compute the expectation. Below are several sample paths of dS = rS\,dt + \sigma S\,d\tilde W from S_0 = 100; the dashed line is the strike K, and the figure prints the Monte-Carlo estimate \hat V_0 from this batch. Refresh for a fresh draw — each run prints the \sigma it used.

Paths finishing above K pay S_T - K; those below pay nothing. With only a handful of paths the estimate is noisy — hit Refresh a few times and watch \hat V_0 bounce around. The law of large numbers tames it: the standard error shrinks like 1/\sqrt{N}, so a production run uses tens of thousands of paths (plus variance-reduction tricks) to pin the price down. In the limit the average is the Black–Scholes price — the same number the PDE produces, which the Feynman–Kac formula guarantees. Notice, one last time, which drift the simulated paths use: r, not \mu. Simulate under \mathbb{P} and average, and you get a forecast, not a price.