Exercise Price Effects on Calls and Puts

Why the Exercise Price Matters

An option’s exercise price, also called its strike price, determines the price at which its holder may buy or sell the underlying asset. When comparing options, the central question is what happens if every feature except the exercise price is held fixed. For calls, a higher exercise price makes the right to buy less favorable. For puts, it makes the right to sell more favorable.

The comparison concerns option value, not just the payoff in one selected scenario. The key is to compare expiration payoffs at every possible terminal stock price. If one option’s payoff is at least as large in every state, and greater in at least one state that can occur, the option with the larger payoff cannot have a lower no-arbitrage value.

Definition: For a European call and put on the same underlying asset with the same expiration date, let $K$ be the exercise price and $S_T$ the stock price at expiration. Their expiration payoffs per share are $C_T(K)=\max(S_T-K,0)$ and $P_T(K)=\max(K-S_T,0)$. The maximum with zero means the holder does not exercise when doing so would produce a negative payoff.

The comparison assumes the options have the same underlying, expiration, settlement terms, and other contract features. For the pricing examples, assume a frictionless market, a constant risk-free effective annual rate $i$, and a risk-neutral probability distribution $Q$ under which discounted expected payoffs give European option prices. With $T$ measured in years, $v=1/(1+i)$, and this constant annual rate, the pricing rule is:

C_0(K)=v^T E_Q[\max(S_T-K,0)], \qquad P_0(K)=v^T E_Q[\max(K-S_T,0)].

This pricing representation is convenient for calculation, but the direction of the strike effect does not depend on a particular probability distribution. It follows first from comparing payoffs state by state.

Payoff Comparison and the Direction of the Price Change

Consider two exercise prices $K_1<K_2$. For a call, lowering the exercise price cannot reduce the payoff: the holder can buy the asset for less, or choose not to exercise. For a put, raising the exercise price cannot reduce the payoff: the holder can sell the asset for more, or choose not to exercise.

Theorem: For otherwise identical European options with $K_1<K_2$, the lower-strike call is worth at least as much as the higher-strike call, $C_0(K_1)\ge C_0(K_2)$. The higher-strike put is worth at least as much as the lower-strike put, $P_0(K_2)\ge P_0(K_1)$. The inequalities are strict when the risk-neutral pricing model assigns positive probability to a state where the corresponding payoff difference is positive.

The state-by-state differences show both why the inequalities hold and when the changes are strict. For calls, compare the lower-strike payoff minus the higher-strike payoff. For puts, compare the higher-strike payoff minus the lower-strike payoff:

\max(S_T-K_1,0)-\max(S_T-K_2,0) = \begin{cases} 0, & S_T\le K_1,\\ S_T-K_1, & K_1<S_T\le K_2,\\ K_2-K_1, & S_T>K_2, \end{cases}

\max(K_2-S_T,0)-\max(K_1-S_T,0) = \begin{cases} K_2-K_1, & S_T<K_1,\\ K_2-S_T, & K_1\le S_T<K_2,\\ 0, & S_T\ge K_2. \end{cases}

Every listed amount is nonnegative. In the call comparison, the difference is positive exactly when $S_T>K_1$. In the put comparison, it is positive exactly when $S_T<K_2$. Taking risk-neutral expectations preserves these inequalities because an expectation of a nonnegative quantity cannot be negative. Multiplying by the positive discount factor $v^T$ also preserves them. This justifies the price comparison:

C_0(K_1)-C_0(K_2) = v^T E_Q\!\left[\max(S_T-K_1,0)-\max(S_T-K_2,0)\right]\ge 0,

P_0(K_2)-P_0(K_1) = v^T E_Q\!\left[\max(K_2-S_T,0)-\max(K_1-S_T,0)\right]\ge 0.

The theorem gives a direction, not a fixed dollar change. For example, increasing a call’s strike by $10 does not necessarily reduce its value by $10. The call’s payoff difference depends on the terminal stock price: it is zero below the lower strike, increases between the two strikes, and reaches $10 above the higher strike. The put difference has a different state pattern. The distribution of possible terminal prices and the discounting therefore matter to the size of each price change.

Worked Examples

Worked Example: Comparing Three Strikes in a Two-State Model

A hypothetical stock is priced at $80 today. One year from now its price will be either $100 or $60. Under the risk-neutral distribution, the probabilities are $Q(S_T=100)=0.6$ and $Q(S_T=60)=0.4$. The effective annual risk-free rate is $i=5\%$, so $v=1/1.05$. Compare European call and put values at strikes $50, $70, and $90.

Method. Calculate each expiration payoff in each state, multiply by its risk-neutral probability, sum, and discount by $1.05$. The model is internally consistent because $E_Q[S_T]=0.6(100)+0.4(60)=84=80(1.05)$.

Calls. At $K=50$, the up-state payoff is $\max(100-50,0)=50$, and the down-state payoff is $\max(60-50,0)=10$. The expected payoff is $0.6(50)+0.4(10)=30+4=34$, so $C_0(50)=34/1.05=\$32.38$.

At $K=70$, the up-state payoff is $\max(100-70,0)=30$. In the down state, the raw expression is $60-70=-10$, so the actual call payoff is $\max(-10,0)=0$, not $-10$. The expected payoff is $0.6(30)+0.4(0)=18$, so $C_0(70)=18/1.05=\$17.14$.

At $K=90$, the up-state payoff is $\max(100-90,0)=10$, and the down-state payoff is $\max(60-90,0)=0$. The expected payoff is $0.6(10)+0.4(0)=6$, so $C_0(90)=6/1.05=\$5.71$.

Puts. At $K=50$, the two payoffs are $\max(50-100,0)=0$ and $\max(50-60,0)=0$, so $P_0(50)=0$. At $K=70$, they are $\max(70-100,0)=0$ and $\max(70-60,0)=10$. The expected payoff is $0.6(0)+0.4(10)=4$, giving $P_0(70)=4/1.05=\$3.81$. At $K=90$, they are $\max(90-100,0)=0$ and $\max(90-60,0)=30$. The expected payoff is $0.4(30)=12$, giving $P_0(90)=12/1.05=\$11.43$.

The call values fall from $32.38 to $17.14 to $5.71 as the strike rises; the put values rise from $0 to $3.81 to $11.43. All prices are rounded to the nearest cent at the final step. Check: in each state, raising the strike never raises a call payoff and never lowers a put payoff. The mistake to avoid is treating the raw down-state call amount $-10$ at $K=70$ as a realized option payoff; the payoff is floored at zero.

Worked Example: The Price Change Need Not Equal the Strike Change

A hypothetical stock is $100 today. In one year, its terminal price is $70, $100, or $130, with risk-neutral probabilities $0.2$, $0.5$, and $0.3$, respectively. The effective annual risk-free rate is $3\%$. Compare calls and puts with strikes $K_1=90$ and $K_2=110$.

Method. Use state-by-state expected payoffs and discount by $1.03$. The expected terminal stock price is $0.2(70)+0.5(100)+0.3(130)=14+50+39=103$, which equals $100(1.03)$, as required for this risk-neutral model.

For the call at $K=90$, the payoffs at terminal prices 70, 100, and 130 are $0,10,40$. The expected payoff is $0.2(0)+0.5(10)+0.3(40)=0+5+12=17$, so $C_0(90)=17/1.03=\$16.50$. At $K=110$, the payoffs are $0,0,20$; the expected payoff is $0.2(0)+0.5(0)+0.3(20)=6$, so $C_0(110)=6/1.03=\$5.83$. The call’s value decreases by $(17-6)/1.03=\$10.68$.

For the put at $K=90$, the payoffs are $20,0,0$. The expected payoff is $0.2(20)=4$, so $P_0(90)=4/1.03=\$3.88$. At $K=110$, the payoffs are $40,10,0$. The expected payoff is $0.2(40)+0.5(10)+0.3(0)=8+5=13$, so $P_0(110)=13/1.03=\$12.62$. The put’s value increases by $(13-4)/1.03=\$8.74$.

The $20 strike increase does not imply a $20 change in either option price. The option payoff differences vary by state, and their discounted expectations determine the price changes. Values and changes are rounded to the nearest cent at the final step. Check: the call values decrease and the put values increase. The common mistake is to equate the strike difference directly with a premium difference.

Worked Example: Finding When the Strike Effect Is Strict

A hypothetical stock has terminal prices of $50, $65, and $90 after one year, with risk-neutral probabilities $0.25$, $0.25$, and $0.50$. The effective annual risk-free rate is $0\%$. Compare strikes $K_1=60$ and $K_2=70$.

Method. Since $i=0\%$, the discount factor is $v=1$. The distribution is consistent with a current stock price of $0.25(50)+0.25(65)+0.50(90)=12.5+16.25+45=73.75$.

For the call at $K=60$, payoffs at terminal prices 50, 65, and 90 are $0,5,30$. Its expected payoff and price are $0.25(0)+0.25(5)+0.50(30)=0+1.25+15=\$16.25$. At $K=70$, the payoffs are $0,0,20$, giving $0.25(0)+0.25(0)+0.50(20)=\$10.00$. The lower-strike call is worth $6.25 more.

For the put at $K=60$, the payoffs are $10,0,0$, giving $0.25(10)=\$2.50$. At $K=70$, the payoffs are $20,5,0$, giving $0.25(20)+0.25(5)=5+1.25=\$6.25$. The higher-strike put is worth $3.75 more.

The call price change is strict because the model assigns positive probability to prices above $60, where the lower-strike call has a greater payoff. The put price change is strict because there is positive probability of a terminal price below $70. Check: the call difference is $6.25 and the put difference is $3.75; both are nonnegative, as the theorem requires. The mistake to avoid is assuming every strike change produces a strict price change: if all relevant terminal prices lie in the region where the compared payoffs are equal, the price change can be zero.

Common Traps and Practical Interpretation

A reliable comparison begins by holding all non-strike terms fixed and writing the payoff at each relevant terminal price. This separates what follows mechanically from the contract from what depends on the assumed distribution and interest rate. In particular:

Do not reverse the direction. A higher call strike is less favorable to the buyer, while a higher put strike is more favorable to the buyer.
Apply the zero floor. A negative expression such as $S_T-K$ for a call is not a negative payoff. The call payoff is $\max(S_T-K,0)$.
Do not infer the price change from the strike change alone. The payoff difference is state-dependent; discounting and the probabilities of the states affect its value.
Distinguish weak from strict inequalities. Payoff dominance implies a weak price ordering. A strict price difference requires positive probability on states where the payoff difference is positive under the pricing model.
Compare like with like. Changing the strike while also changing the expiration or underlying does not isolate the exercise-price effect.

Key takeaway: For otherwise identical European options, raising the exercise price cannot increase call value and cannot decrease put value. The exact change is the discounted value of the state-by-state payoff difference, not simply the change in the strike.

Check Your Understanding

For each question, compare otherwise identical European options and state the relevant price effect. Round dollar values to the nearest cent where needed.

A call’s strike rises from $40 to $55. In every possible state, the terminal stock price is at most $40. What is the change in the call payoff and value? (Answer: Both remain unchanged; both calls expire worthless.)
A put’s strike rises from $75 to $85. The terminal stock price is $80 with certainty, the expiration is one year, and the effective annual rate is $4\%$. Calculate the increase in put value. (Answer: The payoff rises from $0 to $5, so value increases by $5/1.04=\$4.81$.)
A stock has one-year risk-neutral terminal prices $50 and $90 with probabilities $0.4$ and $0.6$. The effective annual rate is $5\%$. Find the call values at strikes $60 and $80. (Answer: $C_0(60)= [0.4(0)+0.6(30)]/1.05=\$17.14$; $C_0(80)=[0.4(0)+0.6(10)]/1.05=\$5.71$.)
Using the same terminal prices, probabilities, and rate as in Question 3, find the put values at strikes $60 and $80. (Answer: $P_0(60)=[0.4(10)+0.6(0)]/1.05=\$3.81$; $P_0(80)=[0.4(30)+0.6(0)]/1.05=\$11.43$.)
A call’s strike rises from $100 to $120. The terminal stock price is $90 with probability $0.2$, $110 with probability $0.5$, and $140 with probability $0.3$. The effective annual rate is $0\%$. By how much does the call value fall? (Answer: The statewise payoff reductions are $0, $10, and $20; the expected reduction is $0.2(0)+0.5(10)+0.3(20)=\$11$, so the call value falls by $11.)

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