Economic Modelling · Principles of option pricing

No-Arbitrage Principle and Option Price Bounds

Updated 11 October 2026 · Fact-checked

Arbitrage is a trade that costs nothing, can never lose and may gain. Assuming none exists, you compare two portfolios with ordered payoffs at expiry. The cheaper one must cost no more today. This gives bounds such as max(S₀ − K·e^(−rT), 0) ≤ c ≤ S₀ for a European call on a non-dividend share.

Understand No-Arbitrage Principle and Option Bounds

An arbitrage is a strategy that needs no net investment, has no chance of a loss, and has a positive chance of a gain. Markets are assumed to have no such opportunities. If one appeared, traders would exploit it until prices moved and it vanished.

The key idea is the law of one price. If two portfolios give the same payoff in every state at time T, they must have the same value today. A stronger form is dominance: if portfolio A pays at least as much as portfolio B in every state, and more in some state, then A must cost more than B today.

To get a bound, build two portfolios. Choose them so that one always pays at least as much as the other at expiry. Then the value ordering today follows. To prove it, assume it fails, and show the trade that makes a risk-free profit: buy the cheaper portfolio, sell the dearer one.

The standard setting is a European option on a share paying no dividends, with a constant continuously compounded risk-free rate r and maturity T. Let S₀ be the share price, K the strike, c the European call price and p the European put price. Money at time T is discounted by e^(−rT).

These bounds are model-free. They use no assumption about how the share price moves. That is why they are tested: they are the base for put-call parity and for the binomial and Black-Scholes models.

Key rules to remember

Call upper bound: c ≤ S₀; A call gives the right to buy one share, so it cannot be worth more than the share. Also holds for American calls (C ≤ S₀).
Put upper bound (European): p ≤ K·e^(−rT); The best payoff of a put at T is K, so its value cannot exceed the present value of K.
Put upper bound (American): P ≤ K; An American put can be exercised at once for at most K.
European call lower bound (no dividends): c ≥ max(S₀ − K·e^(−rT), 0); Needs a non-dividend-paying share. With a known dividend present value D, use S₀ − D − K·e^(−rT).
European put lower bound (no dividends): p ≥ max(K·e^(−rT) − S₀, 0); Follows from put-call parity and c ≥ 0, or from a direct portfolio argument.
Put-call parity (European, no dividends): c + K·e^(−rT) = p + S₀; Holds exactly under no-arbitrage. Use it to move between call and put bounds.
American call, no dividends: C = c; It is never optimal to exercise early, so the American call equals the European call.
American option bounds: S₀ − K ≤ C − P ≤ S₀ − K·e^(−rT); Holds for non-dividend shares. Note the inequality, since parity is not exact for American options.

How to solve No-Arbitrage Principle and Option Bounds questions

Use this routine for any 'prove the bound' or 'find the arbitrage' question.

1State the assumptions: no arbitrage, no dividends (unless given), no transaction costs, constant risk-free rate r, option type (European or American).
2Write down the bound you want to prove, with the inequality direction clearly.
3Build two portfolios, A and B. Put the one you expect to be worth more as A.
4Compute the payoff of each portfolio at time T (or at exercise time) in every relevant case, for example S_T > K and S_T ≤ K.
5Show that A pays at least as much as B in all cases. Then state that A must be worth at least as much as B today, otherwise arbitrage exists.
6Say how the arbitrage works if the bound fails: buy the cheaper portfolio, sell the dearer, bank the difference, and the payoffs at T cannot lose.
7For a numerical question, plug in values, compare the market price with the bound, and name the trade: which side to buy, which to sell, and the riskless profit at T.

Quickest way: Compare payoffs, then compare prices

When to use it: Use this for numerical 'is there an arbitrage?' questions and for MCQs on bounds.

Compute the lower bound max(S₀ − K·e^(−rT), 0) for a call, or max(K·e^(−rT) − S₀, 0) for a put.
Compute the upper bound: S₀ for a call, K·e^(−rT) for a European put.
If the market price lies outside the range, an arbitrage exists.
If the price is below the lower bound, buy the option and take the opposite position in the other leg. If above the upper bound, sell the option.
For put and call together, test put-call parity: c + K·e^(−rT) against p + S₀. Buy the cheaper side and sell the dearer.

Common mistakes in No-Arbitrage Principle and Option Bounds

Using K instead of K·e^(−rT) in the European lower bound.
Students remember the payoff S_T − K and forget to discount the strike.
Fix: The strike is paid at T, so it must be discounted. Only American exercise-now arguments use plain K.
Forgetting the max with zero in the lower bound.
S₀ − K·e^(−rT) can be negative for out-of-the-money options.
Fix: An option price cannot be negative. Always write max(…, 0).
Applying the no-dividend result when dividends are paid.
The formulas are memorised without their conditions.
Fix: Check the question for dividends. If there is a known dividend, subtract its present value from S₀ in the bounds and parity.
Describing the arbitrage trade in the wrong direction.
Students mix up which portfolio is cheap.
Fix: Always buy the cheaper portfolio and sell the dearer one. Then check that the net cash at time 0 is positive and the net payoff at T is at least zero.
Treating put-call parity as exact for American options.
Parity is learned in one form and used everywhere.
Fix: Exact parity is for European options. For American options only the inequality from the bounds list applies.
Proving a bound with only one case for S_T.
Time pressure leads to checking only the in-the-money case.
Fix: Show the payoffs for both S_T > K and S_T ≤ K in a small list, so the dominance is clear.

Worked examples

Example 1

A non-dividend-paying share trades at ₹100. A 1-year European call with strike ₹90 trades at ₹8. The continuously compounded risk-free rate is 5% per year. Show that an arbitrage exists and describe it.

Show the solution

Lower bound: c ≥ S₀ − K·e^(−rT) = 100 − 90·e^(−0.05).
e^(−0.05) = 0.951229, so 90 × 0.951229 = 85.61. The bound is 100 − 85.61 = 14.39.
The market price ₹8 is below ₹14.39, so the call is too cheap.
Trade at time 0: buy the call (−₹8), short sell the share (+₹100), and lend ₹85.61 at the risk-free rate (−₹85.61). Net cash at time 0: 100 − 8 − 85.61 = +₹6.39.
At T = 1 the loan returns ₹90. If S_T > 90, exercise the call: pay ₹90, receive the share, close the short. Net payoff 90 − 90 = 0.
If S_T ≤ 90, let the call lapse and buy the share at S_T to close the short. Net payoff 90 − S_T ≥ 0.
So the payoff is never negative and positive when S_T < 90, with ₹6.39 received up front. This is an arbitrage.

Answer: The call is priced below its lower bound of about ₹14.39. Buy the call, short the share and lend ₹85.61 for a risk-free gain of ₹6.39 now with no loss at expiry.

Example 2

Prove that for a European put on a non-dividend-paying share, p ≥ K·e^(−rT) − S₀.

Show the solution

Assume no arbitrage. Build Portfolio A: one European put plus one share. Build Portfolio B: cash K·e^(−rT) invested at rate r.
Payoff of A at T if S_T < K: the put pays K − S_T and the share is worth S_T. Total K.
Payoff of A at T if S_T ≥ K: the put is worth 0 and the share is S_T. Total S_T ≥ K.
Payoff of B at T: K·e^(−rT) × e^(rT) = K in every case.
So A pays at least K, which is at least what B pays, in every state.
By no-arbitrage, A is worth at least as much as B today: p + S₀ ≥ K·e^(−rT).
Rearranging gives p ≥ K·e^(−rT) − S₀. Since p ≥ 0 as well, p ≥ max(K·e^(−rT) − S₀, 0).
If it failed, buy A, sell B by borrowing K·e^(−rT). Cash now would be positive and the payoff at T never negative.

Answer: p ≥ max(K·e^(−rT) − S₀, 0), shown by dominance of Portfolio A (put plus share) over Portfolio B (cash K·e^(−rT)).

Exam tips

Write the assumptions first: no arbitrage, no dividends, no costs, constant r. Examiners award marks for stating them.
In proofs, tabulate payoffs for S_T > K and S_T ≤ K. A clear table-like list earns the dominance marks.
In numerical questions, give the trade, the time-0 cash flow and the payoff at T. Do not stop at saying an arbitrage exists.
Use continuous discounting e^(−rT) unless the question gives an annual effective rate. Then discount with (1 + i)^(−T).
In MCQs, test each option against the bounds quickly. Options that break the upper or lower bound can be removed at once.

Practice questions from Principles of option pricing

No-Arbitrage Principle and Option Bounds in other exams

The same ground in other exams, if you are preparing for more than one or want another angle on it.

No-Arbitrage Principle and Option Bounds: frequently asked questions

What is arbitrage in option pricing with an example?

Arbitrage is a way to make a risk-free profit with no net outlay. For example, if a call trades below S₀ − K·e^(−rT), you buy the call, short the share and lend the present value of the strike. You receive cash now and cannot lose at expiry.

What are the upper and lower bounds for a European call option?

For a non-dividend-paying share, max(S₀ − K·e^(−rT), 0) ≤ c ≤ S₀. With known dividends of present value D, the lower bound becomes max(S₀ − D − K·e^(−rT), 0).

Why is an American call on a non-dividend share never exercised early?

The call is worth at least S₀ − K·e^(−rT), which exceeds S₀ − K when r > 0 and T > 0. Selling the option beats exercising it. So the American call has the same value as the European call.

Do these bounds depend on a model for the share price?

No. They use only the no-arbitrage assumption and payoff comparisons. This is why they hold under binomial, Black-Scholes or any other consistent model.