Fundamental Theorem Of Asset Pricing

Two questions sit underneath every pricing model ever written. First: given a screenful of prices, how would you know whether they are coherent, whether some clever combination of them hands out free money? Second: why is it legitimate to price a derivative by inventing probabilities that nobody believes and taking an average with them? These look unrelated. The fundamental theorem of asset pricing says they are the same question.

Start at the bookmaker

A bookmaker prices a three-horse race. Convert his odds into implied chances and you get 40 percent, 35 percent and 20 percent. Add them: 95 percent. Something is missing, and you can go and collect it. Stake $40 on the first horse, $35 on the second, $20 on the third, $95 in total. One of them must win, and whichever does pays you $100. You made $5 with certainty, without knowing anything about horses.

Real bookmakers never let that happen. They set odds whose implied chances sum to more than one, which is their margin. The rule they are obeying is simple and it is the whole theorem in miniature: a price list is safe from free money exactly when you can read it as a genuine set of probabilities, all positive, adding to one. The prices come first; the probabilities are what you back out of them.

Markets are that race with more horses. The horses are the possible future states of the world, the bets are the traded assets, and the implied chances have a name.

The first theorem

Write $S_t$ for the price of a traded asset at time $t$, and $B_t$ for the value of a bank account that started at $1 and earns interest. Write $\mathcal{F}_t$ for everything you know at time $t$. Then:

FTAP 1. There is no arbitrage if and only if there exists a probability measure $Q$, equivalent to the real one, under which every asset divided by the bank account is a martingale.

In symbols, that last clause is

$$\frac{S_t}{B_t} \;=\; \mathbb{E}^{Q}\!\left[\, \frac{S_T}{B_T} \;\middle|\; \mathcal{F}_t \,\right].$$

In plain English: measure everything in units of the bank account instead of dollars, and under the made-up probabilities $Q$ every asset becomes a fair game, its best guess for tomorrow is exactly its value today. Three words in that sentence carry weight. Measure means $Q$ is a full assignment of probabilities to outcomes. Equivalent means $Q$ and reality agree on what is possible, and disagree only on how likely; nothing real is assigned zero chance and nothing impossible gets a positive one. Martingale means fair game, no drift.

Turn it around and you have the pricing rule that every desk runs:

$$V_0 \;=\; e^{-rT}\,\mathbb{E}^{Q}\!\left[\, \text{payoff at } T \,\right].$$

In plain English: the price today is the average payoff under the made-up probabilities, discounted back at the risk-free rate $r$. The theorem is what licenses that line.

Worked example 1: finding Q, then checking it against a hedge

A stock trades at $100. In one period it is worth either $120 or $90. Rates are zero, so $1 in the bank stays $1.

Step 1, find the weights. We need a number $q$ between 0 and 1 with $100 = q(120) + (1-q)(90)$. Expand: $100 = 90 + 30q$, so $30q = 10$ and $q = 1/3$. The real-world chance of an up move might be 60 percent, or 10 percent; it plays no part.

Step 2, price a call. Take a call struck at $105. It pays $120 - 105 = 15$ in the up state and nothing in the down state. So

$$C \;=\; \tfrac{1}{3}(15) + \tfrac{2}{3}(0) \;=\; 5.$$

In plain English: weight each payoff by its $Q$ chance and add. The call is worth $5.

Step 3, prove it without any probability at all. Buy $\Delta$ shares and put $B$ in the bank so the pair matches the call in both states. Matching the difference between states fixes the shares: $\Delta = (15 - 0)/(120 - 90) = 0.5$. Matching the down state fixes the cash: $0.5(90) + B = 0$, so $B = -45$, meaning you borrow $45. The cost today is $0.5(100) - 45 = 5$.

Both routes give $5. That is not a coincidence, it is the theorem: the artificial average and the hedging cost are the same number, always.

Step 4, break it. Now suppose the stock were quoted at $130 with the same two outcomes. Solve again: $130 = 90 + 30q$ gives $q = 4/3$, bigger than one. No probability exists, so FTAP 1 promises an arbitrage, and it is easy to find: sell the stock at $130 today, buy it back for at most $120 later, pocket at least $10 for certain.

The second theorem: is Q the only one?

FTAP 2. An arbitrage-free market is complete if and only if the measure $Q$ is unique. Complete means every payoff can be replicated by trading the assets you have.

In plain English: one set of weights means one price for everything and a hedge for everything. Many sets of weights means many defensible prices, and payoffs you simply cannot manufacture.

Worked example 2: three outcomes, two assets, no unique price

Same stock at $100, zero rates, but now three possible outcomes: $120, $100 or $90. Call the weights $q_1, q_2, q_3$. They must satisfy

• $120q_1 + 100q_2 + 90q_3 = 100$ (today's price is the average),
• $q_1 + q_2 + q_3 = 1$ (they are probabilities),
• all three strictly positive (equivalence).

Two equations, three unknowns, so there is a whole family. Eliminating $q_2$ gives $20q_1 = 10q_3$, that is $q_3 = 2q_1$, and then $q_2 = 1 - 3q_1$. Any $q_1$ between 0 and $1/3$ works:

$q_1$	weights $(q_1, q_2, q_3)$	price of the 105 call
0.20	(0.20, 0.40, 0.40)	$3.00
0.25	(0.25, 0.25, 0.50)	$3.75
0.30	(0.30, 0.10, 0.60)	$4.50

The call pays $15 in the top state only, so its price is simply $15q_1$. Every value from just above $0 to just below $5 is arbitrage-free. There is no single right answer, because with one stock and one bond you cannot build a portfolio that pays 15, 0, 0 across three states. This is an incomplete market, and the endpoints $0 and $5 are precisely the no-arbitrage bounds you would get by pure logic.

Changing the measure changes the tilt, not the shakiness

The explorer below draws sample paths of a stock. Drag the drift slider and watch the whole bundle tilt up or down; drag volatility and watch it fan out. Moving from the real measure to $Q$ is exactly the first slider and never the second: the pricing measure changes the expected growth rate to $r$ and leaves the amount of shaking untouched. That is why volatility is the input you must estimate honestly and expected return is the input that cancels out.

What this means in practice

Every pricing library on a derivatives desk is FTAP written in code. A Monte Carlo engine simulates under $Q$, averages the payoffs, and discounts, which is the pricing rule verbatim. A tree does the same with explicit weights. Calibration is the act of choosing $Q$ so that the model reprints the market's quoted vanillas, and then using that same $Q$ on the exotic you actually have to price.

FTAP 2 is where model risk enters. Real markets are incomplete: volatility moves on its own, prices jump, credit defaults arrive out of nowhere. The moment you add a risk you cannot hedge with traded instruments, $Q$ stops being unique, and two respectable models calibrated to the same vanilla quotes will disagree on the exotic. That disagreement is not a bug in either model; the theorem says it is unavoidable, and the honest response is a reserve and a range, not a single number.

Key terms

• Arbitrage — a position costing nothing today that can never lose and might gain.
• Equivalent measure — a reweighting of probabilities that agrees on what is possible.
• Martingale — a process whose expected future value equals its value now.
• Numeraire — the asset you measure everything in; here, the bank account.
• Complete market — every payoff is replicable, equivalently $Q$ is unique.