Graphing A Short Put At Expiration

What a Short Put Payoff Diagram Shows

A short put is the writer’s side of a put option. At expiration, the put holder may sell the underlying asset for the contract’s exercise price, even if the asset’s market price has fallen below that amount. The writer must take the other side of that transaction if the holder exercises. A payoff diagram shows the value of the option position at expiration across possible underlying prices.

The graph is determined by two inputs: the exercise price and the underlying asset’s price at expiration. Below the exercise price, the short put has a negative payoff because the writer is obliged to buy an asset for more than its market value. At or above the exercise price, the holder does not exercise, so the short put payoff is zero.

Definition: Consider one put option on one share of an underlying asset. Let $K>0$ be the exercise price and $S_T\geq 0$ the asset’s market price at expiration time $T$. The short put payoff is the writer’s terminal option payoff, excluding any premium received earlier and any other cash flows. The model assumes the holder exercises when selling for $K$ is more favorable than selling at the market price. Thus the payoff is negative below $K$, and zero at or above $K$.

The Payoff Rule and Its Derivation

The long put holder has the right to sell for $K$. If $S_T<K$, exercising gives the holder an asset worth $S_T$ in exchange for $K$, a payoff of $K-S_T$. If $S_T\geq K$, selling in the market is at least as favorable, so the holder does not exercise and the payoff is zero. The long put payoff is therefore $\max(K-S_T,0)$. The short position has the opposite option payoff: what the holder gains from exercise is what the writer loses.

\text{Short put payoff} =-\max(K-S_T,0) = \begin{cases} S_T-K, & 0\leq S_T<K,\\ 0, & S_T\geq K. \end{cases}

This formula also follows directly from the exercise transaction. When $S_T<K$, the writer pays $K$ and receives an asset worth $S_T$. The value of that transaction to the writer is $S_T-K$, which is negative. When $S_T\geq K$, the holder does not exercise, so there is no expiration option payoff for the writer.

At the boundary $S_T=K$, the two cases agree: $S_T-K=0$, and the holder is indifferent between exercising and selling at the market price. The payoff diagram therefore has no jump at the strike; it has a kink, where the line below the strike meets the horizontal line above it.

Mark the axes.
Use $S_T$, the nonnegative underlying price at expiration, on the horizontal axis and the short put payoff per share on the vertical axis.

Mark the strike.
At $S_T=K$, the payoff is zero. This is the point where the graph changes direction.

Draw the below-strike segment.
For $0\leq S_T<K$, plot $S_T-K$. It is a straight line rising from $(0,-K)$ to $(K,0)$.

Draw the at-or-above-strike segment.
For $S_T\geq K$, the payoff remains zero, so draw a horizontal line on the zero-payoff axis.

The below-strike line has slope $+1$: for every one-dollar increase in $S_T$ while it remains below $K$, the short put payoff increases by one dollar (that is, becomes less negative). Conversely, as $S_T$ falls farther below $K$, the payoff becomes increasingly negative. The line is not extended below $S_T=0$, because a market price for the asset cannot be negative under this model.

Worked Examples

Worked Example: Plotting a Short Put from Its Strike

A hypothetical put writer has sold one put on one share with exercise price $K=\$48$. Find the payoff at expiration when $S_T$ is $\$0$, $\$35$, $\$48$, and $\$61$, then specify the points needed to sketch the payoff diagram. Ignore any premium received.

Method. Apply the piecewise short put payoff rule separately at each terminal price, checking whether the price is below or at/above the $\$48$ strike.

At $S_T=\$0$, the price is below the strike, so the payoff is $S_T-K=\$0-\$48=-\$48$ per share.

At $S_T=\$35$, it is also below the strike: $\$35-\$48=-\$13$ per share.

At $S_T=\$48$, the price equals the strike, and the payoff is $\$48-\$48=\$0$.

At $S_T=\$61$, the price is above the strike, so the option is not exercised and the payoff is $\$0$.

The key graph coordinates are $(0,-48)$ and $(48,0)$, joined by a straight line; from $(48,0)$ onward the graph is horizontal at zero. The other calculated points, $(35,-13)$ and $(61,0)$, lie on those segments. Checking confirms that the payoff is never positive for this short put. The common mistake to avoid is using $K-S_T$ as the writer’s payoff; that is the holder’s positive payoff below the strike, not the writer’s.

Worked Example: Distinguishing Both Regions of the Graph

A hypothetical put has $K=\$72$. Calculate the short put payoff per share at terminal prices of $\$66$, $\$72$, $\$79$, and $\$20$. State which points define the two graph segments and explain what happens as the price falls from $\$66$ to $\$20$.

Method. Compare each $S_T$ with $K$ first; only use $S_T-K$ in the below-strike region.

At $\$66$, $S_T<K$, so the payoff is $\$66-\$72=-\$6$.

At $\$72$, the price equals $K$, giving a payoff of $\$0$. At $\$79$, the price exceeds $K$, so the payoff remains $\$0$.

At $\$20$, the price is below $K$, and the payoff is $\$20-\$72=-\$52$.

The line segment below the strike runs from $(0,-72)$ to $(72,0)$; the horizontal segment begins at $(72,0)$. As the price falls from $\$66$ to $\$20$, the payoff changes from $-\$6$ to $-\$52$, a decrease of $\$46$, matching the $\$46$ fall in the asset price. The common mistake is to treat the payoff as zero just because the option is out of the money for the writer; below the strike, the writer’s payoff is negative.

Worked Example: Inferring the Strike from Payoff Observations

A hypothetical short put on one share has payoff $-\$9$ when $S_T=\$29$, and payoff $-\$2$ when $S_T=\$36$. At $S_T=\$40$, its payoff is observed to be zero. Infer the strike and identify the graph’s kink and its below-strike intercept.

Method. A negative short put payoff means the observation is below the strike and satisfies $S_T-K$. Use a negative-payoff observation to solve for $K$, then check the other observations.

From the first observation, $\$29-K=-\$9$, so $K=\$38$. Check the second observation: $\$36-\$38=-\$2$, which agrees with the stated payoff.

The observed zero payoff at $\$40$ is consistent because $\$40\geq \$38$, placing it in the at-or-above-strike region. The kink is $(38,0)$. Since the price of the underlying cannot fall below zero, the below-strike segment begins at $(0,-38)$. The complete sketch has a rising straight segment from $(0,-38)$ to $(38,0)$, followed by a horizontal zero-payoff segment.

The check is important: a zero payoff at $\$40$ alone would only imply that the strike is no greater than $\$40$; it would not identify the strike exactly. The negative observations supply the information needed to solve for $K$. The common mistake is to infer $K$ from the zero observation alone.

Reading the Diagram Correctly

The shape is simple, but the sign and boundary conventions matter. The vertical coordinate is the short position’s option payoff, not the value of the underlying asset and not the premium received when the option was written. A negative coordinate indicates a loss on the option payoff at expiration. A zero coordinate means the put has no expiration payoff for the writer, not that the underlying asset is worth zero.

The strike is the point where the two pieces meet. The short put payoff is continuous there: approaching $K$ from below gives $S_T-K$ approaching zero, and at or above $K$ the payoff is exactly zero. The graph is flat to the right of the strike because every higher terminal price leaves the holder with no reason to sell at $K$.

Do not confuse a short put with a short call. Below the put’s strike, a declining underlying price makes the short put payoff more negative; above the strike, the payoff stays at zero. Also keep payoff separate from profit: the diagram here excludes any premium and other cash flows, so it cannot by itself show the writer’s total profit.

Key takeaway: For one short put, the expiration payoff is $S_T-K$ when $S_T<K$, and zero when $S_T\geq K$. Plot the line from $(0,-K)$ to $(K,0)$, then continue horizontally at zero. Lower prices below the strike mean more negative payoff.

Check Your Understanding

For each problem, use the short put payoff per share at expiration, excluding any premium. Draw or describe the graph where requested.

A short put has $K=\$31$. Find its payoffs at $S_T=\$24$, $\$31$, and $\$39$. (Answer: $-\$7,\ \$0,\ \$0$ per share.)
A short put has $K=\$62$. Give the graph coordinates at $S_T=0$, $S_T=\$47$, and $S_T=\$70$, and state the horizontal portion’s payoff. (Answer: $(0,-\$62)$, $(\$47,-\$15)$, and $(\$70,\$0)$; the graph is horizontal at zero for $S_T\geq \$62$.)
A short put’s payoff is $-\$8$ at $S_T=\$26$, and its payoff is zero at $S_T=\$35$. Infer the strike and calculate the payoff at $S_T=0$, $\$30$, and $\$41$. (Answer: $K=\$34$; payoffs are $-\$34$, $-\$4$, and $\$0$ per share.)
A short put has $K=\$50$. The underlying price falls from $\$58$ to $\$43$ by expiration. What is the change in the short put payoff between those terminal prices? (Answer: The payoff changes from $\$0$ to $-\$7$, a decrease of $\$7$ per share.)
A hypothetical short put has $K=\$44$. Give the piecewise payoff rule and evaluate it at $S_T=\$35$, $\$44$, and $\$51$. State the graph’s intercept at $S_T=0$. (Answer: $S_T-\$44$ for $0\leq S_T<\$44$, and $\$0$ for $S_T\geq \$44$; the three payoffs are $-\$9$, $\$0$, and $\$0$, and the vertical intercept is $(0,-\$44)$.)

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