What a Short Call Payoff Graph Shows
A payoff graph records the value of an option position at expiration for different possible prices of the underlying asset. For a short call, the graph is especially easy to misread: the call writer’s payoff is not the buyer’s positive payoff. It is zero when the call is not exercised, then becomes more negative as the underlying price rises above the exercise price.
The vertical axis in this tutorial is the short position’s expiration payoff per unit of the underlying. It does not include any premium received when the contract was entered into, so it is not a profit graph. The horizontal axis is the underlying price at expiration. The examples use hypothetical settlement prices and do not assume an interest rate, mortality model, or probability distribution; none is needed to calculate an expiration payoff.
Why the Payoff Has This Shape
A call gives its holder the right to buy the underlying asset for \(K\). At expiration, if \(S_T\le K\), buying at \(K\) would not be advantageous compared with the market price. The call is not exercised, and its expiration payoff is zero to both sides of the contract. At \(S_T=K\), buying for the exercise price is no better than buying at the market price, so the payoff is still zero.
If \(S_T>K\), the holder can buy for \(K\) an asset worth \(S_T\). The holder’s payoff per unit is \(S_T-K\). The writer’s payoff is the opposite amount, \(K-S_T\), because the writer is on the other side of the contract. This gives the piecewise formula above.
This equality is also a direct way to construct the graph. First locate the long call’s expiration payoff at each terminal price; then reverse the sign of every vertical value. A long call’s zero payoff stays zero, and every positive payoff becomes negative. The exercise price remains the point where the graph changes direction.
Put \(K\) on the horizontal axis. It is the boundary between the zero-payoff region and the region where the short call has a negative payoff.
For every \(S_T\le K\), plot payoff zero. This part of the graph lies on the horizontal axis.
For each price above the strike, calculate \(K-S_T\). These points lie below zero and form a straight, downward-sloping ray.
The graph meets the horizontal axis at \((K,0)\). Moving to the right from that point, the payoff must become increasingly negative.
Graph Features to Read Correctly
For nonnegative underlying prices, the graph begins at \(S_T=0\) and remains on the zero-payoff axis until \(S_T=K\). At the strike, the payoff is exactly zero, so there is no jump or gap. Above \(K\), the payoff is \(K-S_T\). For example, a price one dollar above the strike gives a payoff of negative one dollar per underlying unit; a price five dollars above gives negative five dollars per unit.
The portion above the strike is a straight line with slope \(-1\) when both axes are measured in dollars per underlying unit. The line continues downward as \(S_T\) increases; it does not turn back toward zero. The payoff is never positive for the writer under this definition. These features provide useful checks when sketching without plotting many points.
| Terminal price | Short call payoff | Graph location |
|---|---|---|
| \(S_T<K\) | 0 | On the horizontal axis, left of the strike |
| \(S_T=K\) | 0 | At the point \((K,0)\) |
| \(S_T>K\) | \(K-S_T<0\) | Below the axis, right of the strike |
Worked Example: Plotting a Short Call at Several Prices
A hypothetical short call has an exercise price of \(K=\$50\). For a graph, calculate its expiration payoff at underlying prices of \(\$42\), \(\$50\), \(\$53\), \(\$61\), and \(\$82\). Report payoff per underlying unit. The listed prices are deterministic scenarios, not a probability distribution; no interest rate or mortality assumption applies.
Method. Compare each terminal price with the strike. Use zero at or below \(K\), and use \(K-S_T\) above \(K\). This avoids assigning the long call’s positive payoff to the writer.
At \(S_T=\$42\), the price is below the strike, so the payoff is \(\$0\).
At \(S_T=\$50\), the price equals the strike, so the payoff is also \(\$0\).
At \(S_T=\$53\), the price exceeds the strike, so the payoff is \(\$50-\$53=-\$3\).
At \(S_T=\$61\), the payoff is \(\$50-\$61=-\$11\).
At \(S_T=\$82\), the payoff is \(\$50-\$82=-\$32\).
The graph coordinates are \((42,0)\), \((50,0)\), \((53,-3)\), \((61,-11)\), and \((82,-32)\). Plot the points and connect them: the graph is horizontal at zero through the strike, then slopes downward to the right. As a check, each price above \(\$50\) produces a payoff equal to the negative of its excess over \(\$50\).
Deriving the Graph from the Contract
The formula is not a separate rule to memorize for each part of the graph. It follows from the call’s exercise condition and the fact that the writer’s payoff is opposite the holder’s payoff. For a long call, the holder exercises only when doing so produces positive value. The holder’s expiration payoff is therefore the greater of \(S_T-K\) and zero.
For each terminal price, the contract’s two sides have equal and opposite expiration payoffs. Negating the holder’s formula gives the writer’s formula:
The condition \(S_T-K\le 0\) is equivalent to \(S_T\le K\), which proves the zero-payoff portion, including the strike itself. When \(S_T>K\), subtracting \(S_T\) from \(K\) gives a negative result. As \(S_T\) rises by one dollar while remaining above \(K\), \(K-S_T\) falls by one dollar. That is why the graph is a straight line sloping downward to the right.
Worked Example: Reversing a Long Call Payoff Schedule
A hypothetical call has exercise price \(K=\$36.50\). A schedule for the long call gives payoffs at terminal prices of \(\$30.00\), \(\$36.50\), \(\$40.00\), and \(\$48.75\). Find the corresponding short call payoffs and the graph points. Each number is a hypothetical expiration price; no probability distribution, interest rate, or mortality assumption is used.
Method. Calculate the long call payoff, then multiply each value by \(-1\). This method directly reflects the two opposite positions and helps prevent mixing up which side has a positive payoff.
At \(\$30.00\), the long call payoff is \(\max(\$30.00-\$36.50,0)=\max(-\$6.50,0)=\$0\). The short call payoff is \(-\$0=\$0\).
At \(\$36.50\), the long call payoff is \(\max(\$36.50-\$36.50,0)=\$0\). The short call payoff is also \(\$0\).
At \(\$40.00\), the long call payoff is \(\max(\$40.00-\$36.50,0)=\$3.50\). The short call payoff is \(-\$3.50\).
At \(\$48.75\), the long call payoff is \(\max(\$48.75-\$36.50,0)=\$12.25\). The short call payoff is \(-\$12.25\).
The short call graph points are \((30.00,0)\), \((36.50,0)\), \((40.00,-3.50)\), and \((48.75,-12.25)\), with all values in dollars per underlying unit. Substitution into \(K-S_T\) checks the last two: \(\$36.50-\$40.00=-\$3.50\) and \(\$36.50-\$48.75=-\$12.25\).
Worked Example: Using the Strike Boundary to Complete a Graph
A hypothetical short call’s graph is known to lie on the horizontal axis through an exercise price of \(K=\$72.00\). A candidate point at a terminal price of \(\$79.25\) is needed, as is a point at \(\$101.60\). Calculate both payoffs and state how they should be placed on the graph. These are specified scenarios only; no rate, mortality assumption, or probability distribution is involved.
Method. Use the strike to decide which formula applies, then calculate the vertical coordinate. The common mistake is to plot the excess over the strike as positive rather than using the writer’s negative payoff.
Both prices exceed \(\$72.00\), so use \(K-S_T\). At \(\$79.25\), the payoff is \(\$72.00-\$79.25=-\$7.25\). The point is \((79.25,-7.25)\), below the horizontal axis.
At \(\$101.60\), the payoff is \(\$72.00-\$101.60=-\$29.60\). The point is \((101.60,-29.60)\), farther to the right and lower on the graph.
The calculations check because the price exceeds the strike by \(\$7.25\) and \(\$29.60\), respectively, and the writer’s payoff is the negative of each excess. The graph therefore proceeds downward as the terminal price increases.
Common Mistakes and a Fast Check
The most frequent sign error is to calculate \(S_T-K\) above the strike and plot that positive amount for the short call. That is the long call’s payoff. For the short call, either negate the long call result or calculate \(K-S_T\) directly. For example, with \(K=\$40\) and \(S_T=\$47\), the incorrect calculation \(47-40=+\$7\) gives the long call payoff; the correct short call payoff is \(40-47=-\$7\).
A second error is to treat the strike as part of the negative-payoff region. At exactly \(S_T=K\), the difference is zero, not negative. A third is to draw the line above the horizontal axis to the right of the strike. A quick check is to select any price above the strike: the short call’s graph must show a negative vertical coordinate, and a still higher price must have a more negative coordinate.
Check Your Understanding
For each problem, find the short call’s expiration payoff per underlying unit and use the signs and values to check the graph’s shape.
- A hypothetical call has \(K=\$45\). Find the payoff at \(S_T=\$39\), \(\$45\), and \(\$51.75\). (Answer: \(\$0,\ \$0,\ -\$6.75\), respectively.)
- A short call has \(K=\$128\). A sketch needs points at \(S_T=\$119\), \(\$128\), and \(\$140.50\). Give the three vertical coordinates and identify the point where the graph leaves the zero-payoff axis. (Answer: \(\$0,\ \$0,\ -\$12.50\); it leaves the axis to the right of \((\$128,\$0)\).)
- A short call graph is zero through \(S_T=\$62\) and has payoff \(-\$8\) at \(S_T=\$70\). Find the exercise price and the payoff at \(S_T=\$77\). (Answer: \(K=\$62\); payoff at \(\$77\) is \(-\$15\).)
- For a short call with \(K=\$250\), calculate the graph points at \(S_T=\$200\), \(\$250\), \(\$275\), and \(\$340\). (Answer: \((200,0)\), \((250,0)\), \((275,-25)\), and \((340,-90)\), in dollars per underlying unit.)
- A candidate uses \(S_T-K\) to plot a short call with \(K=\$80\) at \(S_T=\$91\). State the candidate’s plotted value, the correct value, and the direction the point belongs on the graph. (Answer: the candidate plots \(+\$11\); the correct payoff is \(-\$11\), below the horizontal axis.)