Solution: Why Pricing Under $\mathbb{P}$ Admits Arbitrage

Exercise: Why Pricing Under P Admits Arbitrage

Part 1: the arbitrage portfolio

Step 1: replicate the call. Find

(\phi, \psi)

such that

\phi S_T + \psi =

call payoff in each state. With

r = 0

, the bond position

\psi

is constant.

Up state: $\phi \cdot 110 + \psi = 10$
Down state: $\phi \cdot 90 + \psi = 0$

Subtracting: $\phi \cdot 20 = 10 \Rightarrow \phi = 0.5$ . Then $\psi = -\phi \cdot 90 = -45$ .

So the replicating portfolio is: long $0.5$ shares of stock, short $45$ in the bond (i.e. borrow $45$ ). Its cost at $t = 0$ :

V_0^{\text{rep}} = 0.5 \cdot 100 + (-45) = 5

This confirms the no-arbitrage price is $5$ (matching the risk-neutral formula).

Step 2: the arbitrage strategy. The naïve trader sells a call at

7

. You:

Buy the replicating portfolio at cost $5$ .
Sell a call to the naïve trader at $7$ .

Net cashflow at $t = 0$ : $+7 - 5 = +2$ (receive $2$ today).

Step 3: verify zero liability at $T$ . At time

T

You own a portfolio worth $0.5 S_T - 45$ .
You owe the call holder $(S_T - 100)^+$ .

In the up state: portfolio value $= 0.5 \cdot 110 - 45 = 10$ ; call obligation $= 10$ . Net: $0$ .

In the down state: portfolio value $= 0.5 \cdot 90 - 45 = 0$ ; call obligation $= 0$ . Net: $0$ .

Arbitrage summary. You received

2

t = 0

and owe nothing at

T

in either state. This is a textbook arbitrage: positive initial wealth, zero terminal liability.

Part 2: per-state profit

By construction of the replicating portfolio, the portfolio-minus-obligation is exactly $0$ in both states. Your realised profit is the $\$ 2 $received at$ t = 0$ — deterministic and state-independent. That is the whole point of replication: arbitrage is risk-free.

If you had merely taken a directional view (sell call naked, hope for down state), you'd profit $\$ 7 $in the down state and lose$ $3 $in the up state. That's speculation, not arbitrage. The replicating portfolio converts the speculation into a locked-in$ $2$.

Part 3: why $p$ is irrelevant

The replicating portfolio depends only on the two possible prices of the stock and the risk-free rate. Nowhere in the calculation does the probability

p

of the up move appear — the stock-holdings

\phi

and bond-holdings

\psi

are pinned down by the two linear equations

\phi u S_0 + \psi = \text{payoff}_u

and

\phi d S_0 + \psi = \text{payoff}_d

, which involve only

u, d, S_0

, and the payoffs.

The only inputs to the no-arbitrage price are the possible values of

S_T

, the payoff function, and the risk-free rate. The real-world probability

p

does not enter. The risk-neutral probability

q

appears in the expectation-form price

V_0 = e^{-rT}\mathbb{E}^{\mathbb{Q}}[\text{payoff}]

, but

q

is itself computable from

u, d, r

alone — it is not an independent input.

Part 4: the trinomial / incomplete market

The martingale condition $\mathbb{E}^{\mathbb{Q}}[S_T] = S_0$ becomes (with $r = 0$ ):

q_u \cdot 110 + q_m \cdot 100 + q_d \cdot 90 = 100

Plus $q_u + q_m + q_d = 1$ and each $q \in (0, 1)$ for equivalence.

Rearranging: $q_u \cdot 10 + q_m \cdot 0 - q_d \cdot 10 = 0 \Rightarrow q_u = q_d$ . So any triple of the form

(q_u, q_m, q_d) = (t, 1 - 2t, t), \qquad t \in (0, 1/2)

satisfies both constraints and defines an equivalent martingale measure.

Two explicit examples. Take

t = 0.3

(q_u, q_m, q_d) = (0.3, 0.4, 0.3)

Call price: $e^{-0} \cdot (0.3 \cdot 10 + 0.4 \cdot 0 + 0.3 \cdot 0) = 3$ .

Take $t = 0.45$ : $(0.45, 0.1, 0.45)$ .

Call price: $0.45 \cdot 10 = 4.5$ .

Two different risk-neutral measures produce two different call prices — any number in

(0, 5)

is attainable. The market is incomplete: the call is not replicable by a portfolio in

(\text{stock}, \text{bond})

alone (you'd need a third instrument that resolves the three-state risk). Every risk-neutral measure is "admissible" in the no-arbitrage sense, but there is no unique no-arbitrage price.

Takeaways

Arbitrage = replication + wrong price. When a derivative is mispriced relative to its replicating portfolio, the arbitrage is automatic: buy the cheap side, sell the expensive side, hold to maturity.
The real-world probability $p$ is not a price input. Only $u, d, r$ and the payoff function determine the price. This is sharpest evidence that $\mathbb{P}$ and $\mathbb{Q}$ serve different purposes.
Completeness $\Leftrightarrow$ unique $\mathbb{Q}$ . A market with $n$ states and fewer than $n$ hedging instruments admits a multi-parameter family of risk-neutral measures — every choice is an equally valid no-arbitrage price, and the market alone cannot pick one. This is why stochastic volatility models require calibration.