Options: Payoffs, Strategies, Replication, and Risk-Neutral Pricing

An option lets its buyer choose whether to transact at an agreed price. That choice changes the shape of the payoff: the buyer can benefit from favorable outcomes while declining an unfavorable exercise. The right has a price, however, and an attractive payoff diagram is not the same as a profitable investment.

Learning goals and conventions

You should finish able to distinguish option payoffs from profits, construct common strategies, interpret equity as a contingent claim, replicate an option in a binomial tree, and explain why risk-neutral probabilities are pricing tools rather than forecasts.

Use European options, one unit of underlying per option, a non-dividend-paying stock, and frictionless trading unless specified otherwise. Real contracts have specified multipliers, exercise procedures, settlement rules, and deadlines. All strategy legs below have the same underlying and expiration. Interest rates are effective per tree period except in the continuous-time section.

Symbol	Meaning
$S_t$	Underlying asset price at date $t$
$K$	Strike or exercise price
$T$	Expiration date
$C_t$, $P_t$	Call and put values
$u$, $d$	One-period stock-price multipliers
$r$	Risk-free effective interest rate per tree step
$\Delta$	Shares held in a replicating portfolio
$B_0$	Money invested in the risk-free asset today; negative means borrowing

1. Calls, puts, and exercise styles

A call gives its holder the right to buy at $K$. A put gives the right to sell at $K$. The writer has the corresponding obligation if the option is exercised. Cash-settled options replace asset delivery with the specified cash difference.

At expiration:

$$C_T=(S_T-K)^+=\max(S_T-K,0), \qquad P_T=(K-S_T)^+.$$

The notation $x^+$ means the positive part of $x$. The holder may let an option expire; the writer cannot choose to ignore a valid exercise or assignment.

A European option permits exercise at expiration only. An American option permits exercise on allowed dates through expiration, subject to its terms. These labels describe exercise rights, not the exchange’s geographic location.

Moneyness and time value

A call is in the money when $S>K$; a put is in the money when $S<K$. Either is at the money when $S=K$. The opposite inequalities describe out-of-the-money options.

An option can be valuable before expiry even when currently out of the money, because favorable future outcomes remain possible. For an American option, immediate exercise value is available and market value cannot be lower than that value in the frictionless model. For European options, immediate exercise is unavailable; familiar “intrinsic value plus nonnegative time value” statements require care, particularly for puts when interest rates are positive.

2. Payoff, premium, and profit

Take strike $K=100$:

Terminal stock price	Long call payoff	Long put payoff	Short call payoff	Short put payoff
80	0	20	0	-20
100	0	0	0	0
120	20	0	-20	0

A long option’s payoff is nonnegative, but the premium is paid today. At a one-year effective financing rate $r$, the terminal financed profit of a long call is

$$\Pi_T=(S_T-K)^+-C_0(1+r).$$

For a call costing 8 with strike 100 and financing 5%, terminal break-even is 108.40. At $S_T=105$, the payoff is 5 but financed profit is $5-8.40=-3.40$. Ignoring financing, the commonly quoted break-even is $K+C_0=108$; state which convention you use.

A premium-paid long option can lose its entire premium. A naked short call has theoretically unbounded terminal loss as the underlying rises. A short put’s payoff loss can approach the strike when a nonnegative underlying falls to zero. Premium receipts and collateral requirements must be considered separately from those payoffs.

3. A protective put is a floor on a matched position

Holding one share plus one put gives

$$S_T+(K-S_T)^+=\max(S_T,K).$$

For an exactly matched position, this creates a terminal value floor of $K$, before premium and financing costs and assuming performance of the contract. It is not a guarantee that the investor cannot lose money relative to the initial outlay.

Worked example

Buy a share for 100 and a put with strike 95 for premium 4. Ignore interest for this strategy example. Initial cost is 104.

$S_T$	Stock value	Put payoff	Combined value	Profit versus 104 cost
60	60	35	95	-9
95	95	0	95	-9
120	120	0	120	16

The terminal value floor is 95, and maximum loss is 9 under these assumptions. Upside remains uncapped. For a nonnegative stock, the put itself pays at most $K$ if the stock goes to zero; that maximum put payout is different from the combined position’s floor. A European put protects at its expiry, not necessarily at a forced earlier sale date. An index put may not exactly protect a different portfolio.

4. Combining options changes the payoff shape

The following strategy examples ignore interest and trading costs to make premium-adjusted profit easy to see. Let $x^+=\max(x,0)$.

Long straddle: movement must exceed the price of protection

Buy a call and a put with the same strike and expiration:

$$\text{Payoff}=(S_T-K)^++(K-S_T)^+=|S_T-K|.$$

With $K=100$, call premium 6, and put premium 5, total cost is 11. Terminal break-evens are 89 and 111. At $S_T=100$, both options expire worthless and the loss is 11. At 80 or 120, payoff is 20 and profit is 9.

The strategy benefits at expiry from a sufficiently large move in either direction. It is not a pure wager on realized variance: terminal payoff depends on the endpoint, and a wildly fluctuating price that returns to 100 still gives zero terminal payoff. Before expiry, implied volatility, time decay, and the path of prices affect its resale value. See the Options Industry Council’s strategy explanations.

Bull call spread: limit cost and upside

Buy a call at $K_1$ and sell a call at $K_2>K_1$:

$$\text{Payoff}=(S_T-K_1)^+-(S_T-K_2)^+.$$

Buy the 100-strike call for 8 and sell the 110-strike call for 3. Net debit is 5. Payoff is zero below 100, rises dollar for dollar between 100 and 110, and is capped at 10 above 110. Maximum loss is 5, maximum profit is 5, and break-even is 105.

These are terminal results with matching legs. American exercise and early assignment can create interim positions and funding needs even if the eventual payoff diagram has limited loss.

Long call butterfly: a terminal target region

For equally spaced strikes $K_1<K_2<K_3$ with $K_2-K_1=K_3-K_2=h$, buy one low-strike call, sell two middle-strike calls, and buy one high-strike call:

$$\text{Payoff}=(S_T-K_1)^+-2(S_T-K_2)^++(S_T-K_3)^+.$$

The payoff is zero outside the wings and reaches $h$ at $K_2$. Equal spacing is essential to this symmetric shape and zero upper tail.

For strikes 90, 100, and 110 and respective call premiums 14, 7, and 3, the net debit is $14-2(7)+3=3$. Maximum loss is 3, maximum profit is $10-3=7$, and break-evens are 93 and 107.

$S_T$	Straddle payoff, $K=100$	Bull spread payoff, 100/110	Butterfly payoff, 90/100/110
80	20	0	0
90	10	0	0
95	5	0	5
100	0	0	10
105	5	5	5
110	10	10	0
120	20	10	0

The butterfly rewards a terminal price near its middle strike. Calling it simply a bet on low variance is incomplete: a calm market ending outside the wings still produces zero payoff, and pre-expiry volatility sensitivity depends on spot, strikes, and time.

5. No-arbitrage relationships before choosing a model

For European options on a non-dividend-paying stock with the same strike and maturity, compare a stock plus put with a call plus a risk-free payment of $K$. Both pay $\max(S_T,K)$ at expiry. Therefore

$$C_0-P_0=S_0-PV_0(K).$$

With a flat annual effective rate and maturity $T$ years:

$$C_0-P_0=S_0-\frac{K}{(1+r)^T}.$$

This is put–call parity. It is a price relationship from replication, not a forecast of which option will finish in the money. Dividends change the stock-carry component; American exercise rights require different qualifications.

For nonnegative rates in this basic European model, useful bounds include

$$\max(0,S_0-PV_0(K))\le C_0\le S_0,$$ $$\max(0,PV_0(K)-S_0)\le P_0\le PV_0(K).$$

Bounds help detect inconsistent quotes or calculations before using a detailed pricing model.

6. Corporate equity as an option: the Merton interpretation

Consider a simplified firm whose only liability is a zero-coupon debt payment of face value $L$ at date $T$. Let $A_T$ be asset value then. Assume no earlier payouts, no bankruptcy costs, and limited liability.

Shareholders receive

$$E_T=\max(A_T-L,0),$$

the payoff of a call on assets with strike $L$. Debt holders receive

$$D_T=\min(A_T,L)=A_T-\max(A_T-L,0) =L-\max(L-A_T,0).$$

Thus debt is either assets minus the equity call, or default-free face value minus a put on assets. Assets plus a written put is not the correct debt decomposition. The two correct forms are assets minus a call, and a risk-free bond minus a put.

For $L=100$, asset values 70 and 130 produce equity payoffs 0 and 30, and debt payoffs 70 and 100. Adding debt and equity always recovers asset value in this simplified structure. Real firms have multiple liabilities, covenants, earlier default possibilities, and recovery costs, so this is a model rather than a literal description of every capital structure.

7. One-step binomial replication: keep bond units consistent

A stock costs $S_0$ and moves to $uS_0$ or $dS_0$ after one period. An option pays $C_u$ or $C_d$. Invest $B_0$ in a risk-free asset today; it becomes $B_0(1+r)$ at the next date.

Matching both outcomes requires

$$\Delta uS_0+B_0(1+r)=C_u, \qquad \Delta dS_0+B_0(1+r)=C_d.$$

Solving gives

$$\Delta=\frac{C_u-C_d}{(u-d)S_0}, \qquad B_0=\frac{uC_d-dC_u}{(u-d)(1+r)}.$$

The cost today, and hence the option value, is

$$C_0=\Delta S_0+B_0.$$

Do not discount $B_0$ again. The supplied outline mixed today’s bond investment with its maturity value. If instead $B_1=(uC_d-dC_u)/(u-d)$ denotes the terminal bond payoff, then $C_0=\Delta S_0+B_1/(1+r)$. Either notation works when used consistently.

Worked example with both-state verification

Let $S_0=100$, $u=1.2$, $d=0.8$, $r=5\%$, and call strike 100. The stock becomes 120 or 80; the call pays 20 or zero.

$$\Delta=\frac{20}{40}=0.5,\qquad B_0=\frac{1.2(0)-0.8(20)}{(1.2-0.8)(1.05)} =-38.095238\ldots$$

Borrow 38.0952 and buy half a share. The cost is $50-38.0952=11.9048$.

State	Half-share value	Debt repayment	Replicating payoff	Call payoff
Up	60	-40	20	20
Down	40	-40	0	0

The identical state-by-state payoffs establish the price under the model’s trading assumptions. An option mispriced relative to this attainable portfolio would permit an offsetting arbitrage.

8. Risk-neutral valuation and a two-step tree

Define

$$q=\frac{(1+r)-d}{u-d}.$$

The no-arbitrage condition $d<1+r<u$ ensures $0<q<1$. Substituting the replication solution gives

$$C_0=\frac{qC_u+(1-q)C_d}{1+r}.$$

For the example, $q=(1.05-0.8)/0.4=0.625$, so $C_0=0.625(20)/1.05=11.9048$.

The subjective up probability $p$ does not enter this replication price. This does not mean real probabilities or risk preferences never matter: they influence the stock’s market price and investment decisions. It means that, conditional on the tree, stock price, financing opportunity, and complete replication, the derivative price follows without another subjective probability input. In incomplete markets, replication may not determine one unique price.

Backward induction over two periods

Use the same stock multipliers and 5% rate per period, with a call expiring after two periods. Terminal stock prices are 144, 96, and 64; call payoffs for strike 100 are 44, zero, and zero.

At the first up node:

$$C_u=\frac{0.625(44)+0.375(0)}{1.05}\approx26.1905.$$

At the first down node $C_d=0$. Working back again:

$$C_0=\frac{0.625(26.190476\ldots)}{1.05}\approx15.5896.$$

The hedge must change after the first step: delta at the up node is $44/(144-96)=0.9167$, while at the down node it is zero. At the root it is approximately $26.1905/(120-80)=0.6548$. Replication is a dynamic trading strategy, not generally a fixed stock holding.

American options add an exercise comparison

At each node, an American option’s value is the larger of immediate exercise value and discounted continuation value. In a non-dividend-paying stock model with nonnegative interest rates, early exercise of an American call is not optimal under the standard assumptions. Puts can be worth exercising early; dividends can change the call conclusion.

9. From the tree to Black–Scholes–Merton

Convergence requires a suitable shrinking-step construction, not merely adding arbitrary periods. In the standard Cox–Ross–Rubinstein approximation with continuous annual rate $r_f$, volatility $\sigma$, and step length $\Delta t$:

$$u=e^{\sigma\sqrt{\Delta t}},\qquad d=e^{-\sigma\sqrt{\Delta t}},\qquad q=\frac{e^{r_f\Delta t}-d}{u-d}.$$

As $\Delta t\to0$ with fixed maturity, the stock process and appropriate European option prices converge to their Black–Scholes limits under the model assumptions. The continuous-time replication argument yields

$$\frac{\partial V}{\partial t} +\frac12\sigma^2S^2\frac{\partial^2V}{\partial S^2} +r_fS\frac{\partial V}{\partial S}-r_fV=0.$$

For a European call on a non-dividend-paying stock, with $\tau=T-t$:

$$C=SN(d_1)-Ke^{-r_f\tau}N(d_2),$$ $$d_1=\frac{\ln(S/K)+(r_f+\sigma^2/2)\tau}{\sigma\sqrt{\tau}}, \qquad d_2=d_1-\sigma\sqrt{\tau}.$$

$N$ is the standard normal cumulative distribution function. Delta is $N(d_1)$ in this model. The corresponding put follows from parity.

For $S=K=100$, $r_f=5\%$ continuous, $\sigma=20\%$, and one year, $d_1=0.35$, $d_2=0.15$, and the call is approximately 10.4506. This is not the same model as a single step with fixed 20% up/down moves, so its price need not equal the one-step example.

Constant volatility, continuous diffusion paths, frictionless continuous trading, and feasible financing are strong assumptions. Jumps, stochastic volatility, discrete hedging, and trading costs can leave material hedge error. Delta removes the modeled local stock-price exposure; it does not eliminate every possible market or company-specific risk.

10. Lecture 12 bridge: mean, variance, and covariance

Replication prices a claim under a model. Portfolio analysis asks how uncertain returns behave together. For discrete return outcomes $R_i$ with real-world probabilities $p_i$:

$$\mu=\mathbb{E}[R]=\sum_i p_iR_i, \qquad \operatorname{Var}(R)=\sum_i p_i(R_i-\mu)^2, \qquad \sigma_R=\sqrt{\operatorname{Var}(R)}.$$

For two returns:

$$\operatorname{Cov}(R_A,R_B)=\mathbb{E}[(R_A-\mu_A)(R_B-\mu_B)], \qquad \rho_{AB}=\frac{\operatorname{Cov}(R_A,R_B)}{\sigma_A\sigma_B}.$$

With portfolio weights $w$ and $1-w$:

$$\sigma_P^2=w^2\sigma_A^2+(1-w)^2\sigma_B^2 +2w(1-w)\operatorname{Cov}(R_A,R_B).$$

For equally likely returns of 20% and -10%, expected return is 5%, variance is 0.0225, and standard deviation is 15%. These are population moments for the stated probabilities. A sample variance estimated from observations usually uses a different denominator correction.

Two equally weighted assets with 20% volatility each and zero correlation produce portfolio volatility $\sqrt{0.5^2(0.2)^2+0.5^2(0.2)^2}\approx14.14\%$. Their risks do not add directly because covariance matters.

Historical comparisons of stocks and bonds require consistent horizons, reinvested distributions, inflation treatment, and arithmetic versus geometric averaging. Long-run historical average returns are estimates, not promised future returns. Options also have asymmetric returns, so variance alone can miss important tail and liquidity exposures.

11. Historical return patterns: hypotheses to test

The course’s Risk and Return lectures introduce empirical patterns. They are historical findings to investigate, not dependable trading rules:

Pattern	Historical hypothesis or observation
Size	Smaller firms have outperformed larger firms in some samples
January	Returns have shown calendar seasonality, particularly among some small-stock samples
Value	High book-to-market, equivalently low price-to-book, portfolios have outperformed in some samples
Momentum	Recent winners have tended to outperform recent losers over intermediate horizons
Accruals	High-accrual firms have shown lower subsequent returns relative to low-accrual firms in studied samples
Issuance	Some equity-issuing firms have shown subsequent underperformance relative to selected benchmarks

Momentum research often ranks prior months 2–12, skipping the most recent month, rather than mechanically using every version of a trailing 12-month return. The Kenneth French Data Library provides research portfolio definitions and data for investigating size, value, and momentum.

Whether a pattern challenges market efficiency depends on the expected-return benchmark, trading costs, risk exposures, information availability, and robustness. Risk compensation, behavioral effects, data mining, and implementation frictions are competing explanations. Effects can weaken, reverse, or vary across markets and periods. Avoid presenting historical outperformance as a current universal fact.

12. Practice questions

Ignore interest in strategy-profit questions unless a financing rate is explicitly given.

1. A call has strike 50 and premium 4. What are payoff and profit at terminal stock prices 45 and 60? If the one-year financing rate is 5%, what is financed break-even?
2. Buy stock at 80 and a put at strike 75 for premium 3. What are the combined terminal value floor and maximum loss?
3. A straddle at strike 100 costs 12 in total. Find the terminal break-evens and profit if the stock finishes at 108.
4. A bull call spread uses strikes 90 and 105 with net debit 6. Find maximum payoff, maximum profit, maximum loss, and break-even.
5. An equally spaced butterfly uses strikes 40, 50, and 60 with net debit 2. Find maximum profit and break-evens.
6. A firm has a single debt payment of 100. Find debt and equity payoffs when terminal asset value is 65 and when it is 140.
7. A stock costs 50 and becomes 60 or 40 next period. The risk-free rate is 10%. A call has strike 50. Find delta, today’s bond investment, risk-neutral probability, and call value.
8. In question 7, suppose someone discounts the already calculated $B_0$ again when computing the portfolio cost. Why is that wrong?
9. Two equally likely returns are 12% and -4%. Find expected return and population standard deviation.

13. Answers and reasoning

1. At 45, payoff is zero and profit is -4. At 60, payoff is 10 and profit is 6. Financed break-even is $50+4(1.05)=54.20$.
2. The terminal value floor is 75. Initial cost is 83, so maximum loss is 8 under the stated matched-position assumptions.
3. Break-evens are 88 and 112. At 108, payoff is 8 and profit is -4: a move in the right direction can still be too small.
4. Maximum payoff is 15, maximum profit is 9, maximum loss is 6, and break-even is 96.
5. Maximum payoff is 10 and maximum profit is 8. Break-evens are 42 and 58.
6. At assets 65, debt gets 65 and equity zero. At assets 140, debt gets 100 and equity 40. Debt plus equity equals assets in both cases.
7. Call payoffs are 10 and zero. $\Delta=10/20=0.5$. Debt repayment is 20, so $B_0=-20/1.10\approx-18.1818$. The price is $25-18.1818=6.8182$. Also $q=(1.10-0.8)/(1.2-0.8)=0.75$, giving $0.75(10)/1.10=6.8182$.
8. $B_0$ is already the present value of the bond position. The correct cost is $\Delta S_0+B_0$, or equivalently $\Delta S_0+B_1/(1+r)$ when using terminal bond payoff $B_1$.
9. Mean is 4%. Deviations are +8 and -8 percentage points, so variance is 0.0064 and population standard deviation is 8%.

Further reading

• Andrew W. Lo, MIT OpenCourseWare, Options: course materials on payoffs, strategies, and replication.
• Options Industry Council, Options Strategies and Strategy Questions: premium costs and strategy structures.
• MIT OpenCourseWare, Risk and Return: statistical foundations and historical return patterns.
• Kenneth R. French, Data Library: research factors, portfolio construction, and downloadable historical data.

Feynman study lab

Explain it simply

An option buys a right to choose later; its payoff differs from its profit after paying the premium.

Concrete example. A call struck at 100 pays 20 when the asset ends at 120. With premium 8, profit is 12 before financing and fees.

Teach it back without looking

Close the lesson and explain the idea aloud to a beginner in two minutes. Define each technical term you use, walk through the example, and state one assumption or limitation. Then answer:

Why is a call’s expiry payoff never negative while the buyer can still lose money?

Check your explanation after you try

The buyer can let it expire, but the premium has already been paid. Payoff excludes that initial cost.

Find the gap and try again

If you hesitated, relied on a formula you could not explain, or missed the self-check, write down the exact gap. Revisit the relevant section of this lesson and work through its example. Close the notes, explain it again in simpler language, then change one number or assumption and predict what changes. Recalculate or check the lesson to test your prediction. Try once more tomorrow from memory.