Equivalent Martingale Measures

A bank account growing at the riskless rate r turns 1 today into e^{rt} at time t. To compare a risky price S_t against that benchmark we strip out the guaranteed growth — we discount — giving the discounted price

\tilde{S}_t = e^{-rt} S_t.

Under the real-world measure \mathbb{P} a stock typically drifts upward faster than the bank (that excess is its risk premium), so \tilde{S}_t drifts up too — it is not fair. The central trick of arbitrage pricing is to change measure: find a new probability measure \mathbb{Q}, equivalent to \mathbb{P}, under which the discounted price is a fair game.

An equivalent martingale measure (EMM), also called a risk-neutral measure, is a measure \mathbb{Q} \sim \mathbb{P} under which the discounted price is a martingale:

\mathbb{E}_{\mathbb{Q}}\big[\tilde{S}_t \mid \mathcal{F}_s\big] = \tilde{S}_s \qquad \text{for all } s \le t.

The word equivalent is doing real work: by the Radon–Nikodym relation, \mathbb{Q} and \mathbb{P} must agree on what is possible — we are allowed to reweight the odds, never to conjure or forbid an outcome.

Pricing by discounted expectation, derived line by line

Here is the payoff for all that machinery. Consider a claim that pays H_T at maturity T (a call option pays (S_T - K)^+, say). Let V_t be its arbitrage-free price at time t. We will show

V_0 = \mathbb{E}_{\mathbb{Q}}\big[e^{-rT} H_T\big]

— today's price is just the discounted expected payoff under \mathbb{Q}.

Step 1 — discount the value process. Define the discounted price of the claim itself, \tilde{V}_t = e^{-rt} V_t. The no-arbitrage principle says any tradeable asset must be priced consistently with the stock, so — exactly like \tilde{S}_t — the discounted claim price is a \mathbb{Q}-martingale:

\mathbb{E}_{\mathbb{Q}}\big[\tilde{V}_t \mid \mathcal{F}_s\big] = \tilde{V}_s.

Step 2 — apply the martingale property from 0 to T. Take s = 0 and t = T. Conditioning on \mathcal{F}_0 (which knows nothing but constants) is just an ordinary expectation:

\tilde{V}_0 = \mathbb{E}_{\mathbb{Q}}\big[\tilde{V}_T \mid \mathcal{F}_0\big] = \mathbb{E}_{\mathbb{Q}}\big[\tilde{V}_T\big].

Step 3 — substitute the discount factors. At time 0 there is nothing to discount, so \tilde{V}_0 = e^{0}V_0 = V_0. At maturity the claim is worth its payoff, V_T = H_T, so \tilde{V}_T = e^{-rT} H_T. Plugging both in,

V_0 = \mathbb{E}_{\mathbb{Q}}\big[e^{-rT} H_T\big] = e^{-rT}\,\mathbb{E}_{\mathbb{Q}}[H_T],

the last equality pulling out the constant discount factor. Price a claim by taking its expected payoff under \mathbb{Q} and discounting. The whole subject of derivative pricing is the project of computing this one expectation for ever more elaborate H_T.

For a market with riskless rate r:

Discounting by e^{-rt} is a choice of numéraire — the yardstick we measure every price in. Here the yardstick is the bank account B_t = e^{rt}, and \tilde{S}_t = S_t / B_t is the stock priced in units of the bank account. A martingale measure is one in which prices-in-units-of-the-numéraire are fair; change the numéraire (to the stock itself, or to a bond) and you change which measure does the job — a flexibility that pays off in the forward-measure tricks of interest-rate modelling.

The name risk-neutral comes from a striking fact. Rearranging the martingale condition \mathbb{E}_{\mathbb{Q}}[\tilde{S}_t] = \tilde{S}_0 back into undiscounted terms gives \mathbb{E}_{\mathbb{Q}}[S_t] = e^{rt} S_0, so under \mathbb{Q} the stock is expected to grow at exactly the riskless rate r — the same as the bank. Every asset earns r on average; the risk premium has been reweighted away. It is not that investors are truly indifferent to risk, but that the pricing measure behaves as if they were. \mathbb{Q} is a computational device, not a forecast — real-world frequencies still live under \mathbb{P}.

Two measures, one path

The same chance outcome \omega tells two stories. Under \mathbb{P} the discounted price \tilde{S}_t carries the stock's excess drift \mu - r and climbs on average; under the equivalent measure \mathbb{Q} that drift is reweighted to zero and \tilde{S}_t merely wanders — a martingale around its start. Step through to see the upward-drifting \mathbb{P} path and the level \mathbb{Q} path drawn from the same Brownian shocks, and Refresh for a fresh \omega.