Piecewise Profit Equations for Calls

From Call Payoff to Expiration Profit

A call’s payoff at expiration depends on whether the underlying asset price is above the strike. Its profit also depends on the premium: the buyer paid it, while the writer received it. To write a piecewise profit equation correctly, keep those two effects separate and use the same time basis for both.

Throughout this tutorial, assume a call with strike $K$ expires in $T$ years. Let $S_T$ be the asset price at expiration, and let $C_0$ be the premium per share paid by the buyer and received by the writer at time 0. Assume a constant effective annual interest rate $i$, and assume the premium could be invested or financed at that rate. Profit is measured at expiration, with no transaction costs or taxes. The call is held until expiration, when it is either exercised or expires unexercised.

Definition: On an expiration-value basis, the premium amount carried to expiration is $C_0(1+i)^T$. The buyer’s expiration profit is the call payoff minus this accumulated premium. The writer’s expiration profit is the accumulated premium received minus the call payoff.

The call payoff is $\max(S_T-K,0)$. Thus, when $S_T\leq K$, the call expires without exercise and the payoff is zero. When $S_T>K$, exercising produces a payoff of $S_T-K$. The equations below include the strike boundary in the first region; because the payoff is zero at $S_T=K$, either region’s formula gives the same value there.

The Piecewise Equations

Write $Q=C_0(1+i)^T$ for the premium carried to expiration. The purchased call has a negative premium cost in both regions. Its payoff is zero below or at the strike and $S_T-K$ above the strike:

\text{Long-call profit at expiration}= \begin{cases} -Q, & S_T\leq K,\\ S_T-K-Q, & S_T>K. \end{cases}

The written call has the opposite payoff, but the premium received is positive to the writer. Therefore:

\text{Short-call profit at expiration}= \begin{cases} Q, & S_T\leq K,\\ Q-(S_T-K)=K-S_T+Q, & S_T>K. \end{cases}

These are expiration-profit equations, not just payoff equations. If a problem explicitly asks for a simple payoff-minus-initial-premium calculation, it is using a different convention: it is not carrying the premium to expiration. Follow the stated convention, and do not subtract a time-0 amount directly from an expiration amount unless the problem defines profit that way.

Why the Equations Hold

The equations follow by comparing the position’s expiration value with the premium’s expiration value. This also explains why the buyer and writer have opposite profit signs.

Carry the premium forward.
A dollar at time 0 accumulates to $1+i$ dollars after one year. Over $T$ years, the accumulation factor is $(1+i)^T$, so the premium amount at expiration is $Q=C_0(1+i)^T$.

Determine the payoff by region.
If $S_T\leq K$, the call is not exercised and its payoff is 0. If $S_T>K$, exercising gives the asset for $K$ and produces value $S_T$, so the payoff is $S_T-K$.

Compare payoff and premium value.
The buyer’s net expiration value is payoff minus $Q$. The writer’s is $Q$ minus payoff. Substituting the payoff from each region gives the two piecewise equations.

For example, in the exercised region the buyer’s profit is $(S_T-K)-Q$, while the writer’s profit is $Q-(S_T-K)$. The expressions are negatives of each other because they describe opposite sides of the same call contract, with the same premium and expiration payoff.

Worked Example: A Purchased Call Across the Strike

A buyer purchases one call with strike $K=\$80$ for a premium of $C_0=\$6.40$. It expires in $T=2$ years. Assume a constant effective annual interest rate of $i=4\%$, and measure profit at expiration. Write the piecewise equation and find the profit if $S_T$ is $\$75$, $\$80$, $\$90$, and $\$100$.

Method and premium accumulation. Carry the time-0 premium to expiration:

Q=C_0(1+i)^T =\$6.40(1.04)^2 =\$6.40(1.0816) =\$6.92224.

The final premium value is $\$6.92$ to the nearest cent. Keep the unrounded amount for the intermediate profit calculations.

Piecewise equation. For $S_T\leq \$80$, the call payoff is zero, so profit is $-\$6.92224$. For $S_T>\$80$, subtract the accumulated premium from the exercise payoff:

\text{Profit}= \begin{cases} -\$6.92224, & S_T\leq \$80,\\ S_T-\$80-\$6.92224, & S_T>\$80. \end{cases}

Evaluate each price. At $S_T=\$75$, the price is below the strike, so profit is $-\$6.92224$, or $-\$6.92$. At $S_T=\$80$, the call is not exercised and the profit is again $-\$6.92224$, or $-\$6.92$.

At $S_T=\$90$, the call is exercised. Its payoff is $\$90-\$80=\$10$, so profit is $\$10-\$6.92224=\$3.07776$, or $\$3.08$. At $S_T=\$100$, payoff is $\$100-\$80=\$20$, so profit is $\$20-\$6.92224=\$13.07776$, or $\$13.08$.

Check. The two prices at or below the strike produce the same loss, since neither produces an exercise payoff. Above the strike, each additional dollar in $S_T$ adds one dollar to profit. The common mistake this method avoids is subtracting the original $\$6.40$ instead of its expiration value.

Worked Example: A Written Call and the Premium Received

A writer sells a call with strike $K=\$125$ and receives a premium of $C_0=\$8$ at time 0. The call expires in $T=2$ years. Assume a constant effective annual interest rate of $i=5\%$. Write the writer’s piecewise expiration-profit equation, then calculate profit for $S_T=\$118$, $\$125$, and $\$140$.

Method and accumulated premium. The writer receives the premium at time 0 and carries it forward at the assumed rate:

Q=C_0(1+i)^T =\$8(1.05)^2 =\$8(1.1025) =\$8.82.

When $S_T\leq \$125$, the writer keeps the premium’s accumulated value and pays no call payoff. When $S_T>\$125$, the writer’s profit is the accumulated premium minus $S_T-\$125$. Therefore:

\text{Profit}= \begin{cases} \$8.82, & S_T\leq \$125,\\ \$8.82-(S_T-\$125)=\$133.82-S_T, & S_T>\$125. \end{cases}

Evaluate each price. At $S_T=\$118$, the call expires unexercised, so profit is $\$8.82$. At $S_T=\$125$, the payoff is still zero, so profit remains $\$8.82$. At $S_T=\$140$, the writer owes a payoff of $\$140-\$125=\$15$, giving profit $\$8.82-\$15=-\$6.18$.

Check. Above the strike, the writer’s profit falls dollar-for-dollar as $S_T$ rises: the equation gives $\$133.82-\$140=-\$6.18$. The common mistake this method avoids is using $S_T-K$ as the writer’s profit; that is the call payoff to the buyer, not the writer’s net result.

Reading the Regions and Checking Signs

The boundary $S_T=K$ is useful as a quick consistency check. The call payoff there is zero, so the buyer’s profit is $-Q$, and the writer’s is $+Q$. Substituting $S_T=K$ into the above-strike expressions gives those same values:

K-K-Q=-Q,\qquad Q-(K-K)=Q.

The strict inequality in the exercised region, $S_T>K$, and the inclusive inequality in the unexercised region, $S_T\leq K$, make the cases complete without assigning the boundary twice. At the boundary the payoff is zero whether or not an exercise decision is described as indifferent; the profit equations agree there.

A reliable sign check is to ask what happens when the call expires unexercised. The buyer paid the premium, so the buyer has a loss of $Q$. The writer received it, so the writer has a gain of $Q$. Above the strike, the buyer’s exercise payoff is positive before accounting for the premium, and that same payoff is a cost to the writer.

Worked Example: Compare the Buyer’s and Writer’s Profits

One call has strike $K=\$52$, an initial premium of $C_0=\$3$, and a one-year term. Assume a constant effective annual interest rate of $i=4\%$. A buyer and a writer each hold one side of this same contract. Find both expiration profits when $S_T=\$49$ and when $S_T=\$60$.

Method and premium accumulation. First calculate the shared premium value:

Q=C_0(1+i)^T =\$3(1.04)^1 =\$3.12.

The buyer’s piecewise profit is $-\$3.12$ for $S_T\leq \$52$, and $S_T-\$52-\$3.12$ for $S_T>\$52$. The writer’s profit is $+\$3.12$ for $S_T\leq \$52$, and $\$3.12-(S_T-\$52)$ for $S_T>\$52$.

At $S_T=\$49$. Since $\$49\leq \$52$, the call expires unexercised. The buyer’s profit is $-\$3.12$, and the writer’s is $+\$3.12$.

At $S_T=\$60$. Since $\$60>\$52$, the payoff is $\$60-\$52=\$8$. The buyer’s profit is $\$8-\$3.12=\$4.88$; the writer’s is $\$3.12-\$8=-\$4.88$.

Check. In each case, the two profits sum to zero: $-\$3.12+\$3.12=0$ below the strike, and $\$4.88-\$4.88=0$ above it. This is expected for opposite sides of the same contract when both profits use the same premium accumulation convention. The sign check catches the common error of assigning the writer the buyer’s profit.

Common Exam Traps

The piecewise structure is simple, but errors often come from mixing up the payoff, the premium, or the position. Check the following before finalizing an equation:

Subtracting the wrong premium amount. On an expiration-value basis, use $C_0(1+i)^T$, not $C_0$. The initial premium and expiration payoff are at different times.
Giving the writer the buyer’s sign. The buyer’s profit is payoff minus accumulated premium; the writer’s is accumulated premium minus payoff.
Using the exercise payoff below the strike. When $S_T\leq K$, the call payoff is zero, not $S_T-K$. Applying the latter expression below the strike would give a negative payoff rather than the call’s actual payoff.
Omitting the premium in one region. A call that expires unexercised still has a profit or loss: $-Q$ for the buyer and $+Q$ for the writer.
Leaving the boundary unclear. State the cases as $S_T\leq K$ and $S_T>K$, then verify that the formulas agree at $S_T=K$.

Key takeaway: First write the call payoff by region, then carry the premium to expiration, and finally apply the position sign. A purchased call has payoff minus accumulated premium; a written call has accumulated premium minus payoff.

Check Your Understanding

For each problem, use expiration-value profit and retain unrounded values until the final dollar amount.

A buyer purchases a call with $K=\$64$ for $C_0=\$3.50$. It expires in two years at a constant effective annual rate of $i=4\%$. Find the buyer’s profit at $S_T=\$60$ and $S_T=\$70$. (Answer: $-\$3.79$ and $+\$2.21$, respectively.)
A writer receives $C_0=\$5$ for a one-year call with $K=\$95$. Assume $i=6\%$. Find the writer’s profit at $S_T=\$90$ and $S_T=\$103$. (Answer: $+\$5.30$ and $-\$2.70$, respectively.)
A call has $K=\$40$, $C_0=\$8$, and $T=1$ year. Assume $i=0\%$. Write the buyer’s piecewise expiration-profit equation, identify the stock price where profit is zero, and find profit at $S_T=\$50$. (Answer: $-\$8$ for $S_T\leq \$40$, and $S_T-\$48$ for $S_T>\$40$; zero profit at $S_T=\$48$; profit at $\$50$ is $+\$2$.)
A buyer and writer take opposite sides of a one-year call with $K=\$150$ and $C_0=\$12$. Assume $i=5\%$. Find each side’s profit when $S_T=\$145$ and $S_T=\$158$. (Answer: at $\$145$, buyer $-\$12.60$ and writer $+\$12.60$; at $\$158$, buyer $-\$4.60$ and writer $+\$4.60$.)
A writer sells a call with $K=\$72$ and $C_0=\$4$, expiring in one year at $i=5\%$. If the writer’s expiration profit is $-\$7.80$, find $S_T$. (Answer: $S_T=\$84$; the negative profit places the price above the strike, so $\$4(1.05)-(\$84-\$72)=-\$7.80$.)

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