Opposite Payoffs For Long And Short Positions

Introduction

A long option and a short position in that same option have opposite expiration payoffs. This is a precise relationship: at every possible underlying price at expiration, the amount received by the long position from the option is the amount paid by the short position, and vice versa. It holds for both calls and puts, including when the option expires exactly at its strike price.

The relationship is useful for checking formulas and graphs. If the short-position graph is not the long-position graph reflected across the horizontal axis, at least one payoff has been assigned the wrong sign. The discussion here concerns payoff at expiration, before any premium, transaction cost, or other cash flow is included.

Definition: Assume one standard option on one underlying asset, with strike price $K$, and let $S_T$ be the asset price at expiration. A long call's expiration payoff per unit of underlying is $\max(S_T-K,0)$; a short call's is the negative of that amount. A long put's payoff is $\max(K-S_T,0)$; a short put's is the negative of that amount. These formulas omit premiums and transaction costs.

The Algebra Behind Opposite Payoffs

The notation $\max(x,0)$ means the greater of $x$ and zero. A call gives its holder the right to buy at $K$. At expiration, that right has positive payoff only when buying at $K$ is preferable to buying at the market price $S_T$: that occurs when $S_T>K$, and the payoff is $S_T-K$. Otherwise its payoff is zero. A put gives its holder the right to sell at $K$, so its payoff is positive only when $S_T<K$, in which case it is $K-S_T$.

A short position has the opposite contractual payoff to the corresponding long position. The short party owes the amount the long party receives under the option at expiration. Algebraically, multiplying the long payoff by $-1$ gives the short payoff:

\begin{aligned} \text{Long call: } C_L(S_T)&=\max(S_T-K,0),\\ \text{Short call: } C_S(S_T)&=-\max(S_T-K,0),\\ \text{Long put: } P_L(S_T)&=\max(K-S_T,0),\\ \text{Short put: } P_S(S_T)&=-\max(K-S_T,0). \end{aligned}

The piecewise forms make the signs explicit. For a call, when $S_T\le K$, neither side has an exercise payoff. When $S_T>K$, the long call receives $S_T-K$ and the short call pays $S_T-K$. For a put, when $S_T<K$, the long put receives $K-S_T$ and the short put pays that amount. When $S_T\ge K$, neither has an exercise payoff.

\begin{aligned} C_S(S_T)&= \begin{cases} 0, & S_T\le K,\\ K-S_T, & S_T>K, \end{cases} & P_S(S_T)&= \begin{cases} S_T-K, & S_T<K,\\ 0, & S_T\ge K. \end{cases} \end{aligned}

Derivation: For every real number $x$, $-\max(x,0)=\min(-x,0)$. If $x\ge 0$, both sides equal $-x$; if $x<0$, both sides equal $0$. Substituting $x=S_T-K$ gives the short-call formula, while substituting $x=K-S_T$ gives the short-put formula. Thus, at each $S_T$, the short payoff is exactly the negative of the matching long payoff—not merely an approximation based on whether the option is exercised.

Reading the Graphs as Reflections

Put $S_T$ on the horizontal axis and payoff on the vertical axis. To obtain the short-position graph from the matching long-position graph, keep each horizontal coordinate unchanged and reverse the sign of its vertical coordinate. In other words, reflect each point $(S_T,y)$ to $(S_T,-y)$ across the horizontal axis.

For a call, both long and short payoffs are zero at every $S_T\le K$, so their graphs coincide with the horizontal axis on that interval. Above $K$, the long-call payoff increases with slope $+1$; the short-call payoff decreases with slope $-1$. Both pass through $(K,0)$. For a put, the long-put payoff decreases with slope $-1$ as $S_T$ rises below $K$, while the short-put payoff increases with slope $+1$ on that interval. At and above $K$, both payoffs are zero.

These slopes are per unit of underlying and assume the standard payoff formulas above. A graph for a position in multiple contracts scales the vertical payoff values by the number of contracts and any contract multiplier, but the sign-reversal relationship remains. The strike is an important check: all four payoffs equal zero at $S_T=K$.

Worked Example: Reflecting a Call Payoff

A hypothetical investor holds one call with strike $K=\$54$. For expiration prices of $\$39$, $\$54$, $\$58$, and $\$71$, calculate the long-call and short-call payoffs per share. Then identify the graph coordinates for each position.

Method. Apply $\max(S_T-K,0)$ at each price, then negate each long payoff to obtain the short payoff. This method avoids assigning a negative payoff to a short call when the call expires out of the money.

At $S_T=\$39$, $S_T-K=39-54=-15$, so the long-call payoff is $\max(-15,0)=\$0$ and the short-call payoff is $-\$0=\$0$.

At $S_T=\$54$, $S_T-K=54-54=0$. Both payoffs are $\$0$.

At $S_T=\$58$, $S_T-K=58-54=4$. The long-call payoff is $\max(4,0)=\$4$, and the short-call payoff is $-\$4$.

At $S_T=\$71$, $S_T-K=71-54=17$. The long-call payoff is $\max(17,0)=\$17$, and the short-call payoff is $-\$17$.

The long-call coordinates are $(39,0)$, $(54,0)$, $(58,4)$, and $(71,17)$. The corresponding short-call coordinates are $(39,0)$, $(54,0)$, $(58,-4)$, and $(71,-17)$, with dollar amounts on the vertical axis. Each short coordinate keeps the same $S_T$ and reverses the long payoff's sign. The check is that the sum of the two payoffs is zero at every listed price.

Worked Example: Reflecting a Put Payoff

A hypothetical investor compares a long put and a short position in that same put, with strike $K=\$72$. Calculate their per-share expiration payoffs at $S_T=\$49$, $\$72$, and $\$86$, and describe the graph segments.

Method. First calculate the long put as $\max(K-S_T,0)$, then take its negative for the short put. This avoids using the call expression $S_T-K$ for a put.

At $S_T=\$49$, $K-S_T=72-49=23$. The long-put payoff is $\max(23,0)=\$23$, so the short-put payoff is $-\$23$.

At $S_T=\$72$, $K-S_T=72-72=0$. The long and short payoffs are both $\$0$.

At $S_T=\$86$, $K-S_T=72-86=-14$. The long-put payoff is $\max(-14,0)=\$0$, and the short-put payoff is also $\$0$.

The long-put coordinates are $(49,23)$, $(72,0)$, and $(86,0)$; the short-put coordinates are $(49,-23)$, $(72,0)$, and $(86,0)$. Below the strike, the long put slopes down toward zero as $S_T$ increases, and the short put slopes up toward zero. At and above the strike, both graphs lie on the horizontal axis. The listed short coordinates are exactly the reflections of the corresponding long coordinates.

Using a Short Payoff to Recover a Price

The same piecewise equations can be rearranged when a payoff is known and the expiration price is unknown. The key is to check that the answer satisfies the region assumed by the relevant branch. This check prevents a valid-looking algebraic answer from being used on the wrong side of the strike.

Worked Example: Inferring the Expiration Price From a Short Call

A hypothetical short call has strike $K=\$52$. At expiration its payoff per share is $-\$9$. Find $S_T$, verify the applicable branch, and give three points that show how its graph relates to the long-call graph.

Method. Because the payoff is negative, the short call must be in the $S_T>K$ branch, where $C_S(S_T)=K-S_T$. This method avoids treating the stated negative payoff as a loss including a premium; it is the option payoff alone.

Set the short-call formula equal to the given payoff: $52-S_T=-9$. Rearranging gives $-S_T=-61$, hence $S_T=\$61$. The branch condition is satisfied because $61>52$. As a direct check, the long-call payoff is $\max(61-52,0)=\$9$, and the short-call payoff is $-\$9$.

At $S_T=\$40$, both call payoffs are zero. At $S_T=\$52$, both are zero at the strike. At $S_T=\$61$, the long-call point is $(61,9)$, and the short-call point is $(61,-9)$. These points show a shared zero-payoff portion through the strike and opposite payoffs above it.

Common Mistakes and Why the Distinction Matters

The opposite-payoff identity is a useful consistency check when translating option terms into equations or sketching a payoff diagram. It does not mean that a short position has the same cash flows as a long position, nor does it describe profit after accounting for the price paid or received to enter the position.

Reversing the sign in the wrong region. A short call has zero payoff when $S_T\le K$, not a positive payoff. Above $K$, its payoff is $K-S_T$, which is negative.
Using the call expression for a put. The long put begins with $K-S_T$, not $S_T-K$. The short put's nonzero branch is $S_T-K$ for $S_T<K$.
Changing the horizontal coordinate when reflecting. Reflection across the horizontal axis changes the payoff sign, not the underlying price. For example, $(58,4)$ reflects to $(58,-4)$, not to $(54,-4)$.
Confusing payoff with profit. Payoff here is determined only by the option's expiration settlement. Profit also depends on the premium and potentially other cash flows. Those amounts are not part of these equations.
Forgetting the boundary. At $S_T=K$, the call and put payoff formulas both give zero. This shared boundary should appear on both long and short graphs.

Key takeaway: For the same contract and expiration price, short payoff equals negative long payoff. On a payoff graph, the short position is the reflection of the corresponding long position across the horizontal axis. This treatment is aligned with the SOA VEE Accounting and Finance syllabus.

Check Your Understanding

For each problem, use expiration payoff only; exclude premiums and transaction costs.

A call has strike $K=\$36$. Find the short-call payoffs per share at $S_T=\$30$, $\$36$, and $\$44$. (Answer: $\$0,\ \$0,\ -\$8$, respectively.)
A put has strike $K=\$95$. Find the short-put payoffs per share at $S_T=\$80$, $\$95$, and $\$109$. (Answer: $-\$15,\ \$0,\ \$0$, respectively.)
A long put with strike $K=\$58$ has payoff coordinates $(40,18)$, $(58,0)$, and $(75,0)$. Give the corresponding short-put coordinates and state the reflection rule. (Answer: $(40,-18)$, $(58,0)$, and $(75,0)$; retain each $S_T$ coordinate and negate its payoff.)
A short call with strike $K=\$73$ has expiration payoff $-\$12$ per share. Find $S_T$ and verify the branch. (Answer: $S_T=\$85$, which satisfies $S_T>K$; the matching long-call payoff is $\$12$.)
An investor is short four calls, each covering one share, with strike $K=\$41$. Find the aggregate option payoff at $S_T=\$48$ and at $S_T=\$39$. (Answer: $-\$28$ at $\$48$, since $4(41-48)=-28$; $\$0$ at $\$39$, since each call expires without an exercise payoff.)

What You'll Learn