Kinks, Slopes, And Boundary Values

Reading the Payoff Near the Exercise Price

An expiration payoff diagram is easiest to interpret by locating the exercise price first. On either side of that price, the option payoff follows a straight line; at the exercise price, the line changes direction or becomes flat. That change is called a kink. Reading the kink correctly helps identify both the zero-payoff region and the rate at which the payoff changes as the underlying price moves.

Let $S_T$ be the underlying asset’s price at expiration and let $K>0$ be the option’s exercise price. The discussion assumes a single option on one unit of the underlying, with no transaction costs or other adjustments. It concerns expiration payoff, not the premium paid or received and therefore not profit. The underlying price satisfies $S_T\geq 0$.

Definition: The slope of a straight segment in a payoff diagram is the change in payoff divided by the change in $S_T$. A slope of $1$, for example, means the payoff increases by one dollar when the underlying’s expiration price increases by one dollar. A kink occurs where the slope changes. A payoff is continuous at a point if its limiting value from either side equals its value at that point.

Payoff Functions and Their Boundaries

A long call gives the holder the right to buy the underlying for $K$. At expiration, exercising is valuable only when $S_T>K$. A long put gives the holder the right to sell for $K$, so exercising is valuable only when $S_T<K$. The long-position payoffs are therefore:

C_T=\max(S_T-K,0)= \begin{cases} 0, & S_T\leq K,\\ S_T-K, & S_T>K, \end{cases} \qquad P_T=\max(K-S_T,0)= \begin{cases} K-S_T, & S_T<K,\\ 0, & S_T\geq K. \end{cases}

At $S_T=K$, both payoffs are zero: neither exercising nor letting the option expire produces an intrinsic payoff. This point is the boundary between the zero-payoff region and the region where the option has positive payoff. A short position has the opposite payoff of the corresponding long position:

-C_T= \begin{cases} 0, & S_T\leq K,\\ K-S_T, & S_T>K, \end{cases} \qquad -P_T= \begin{cases} S_T-K, & S_T<K,\\ 0, & S_T\geq K. \end{cases}

These boundary statements apply to payoff, not profit. If an option was bought or sold for a premium, that premium changes profit but does not change the expiration payoff functions above. In particular, a zero-payoff point is not automatically a break-even point for the position.

Slopes, Kinks, and Continuity

The slopes can be found by examining the linear expression in each region. For a long call, the payoff is flat below $K$, then rises one-for-one above $K$. For a long put, it falls one-for-one below $K$, then is flat above $K$. Short-position slopes have the opposite signs.

Position	Slope for $S_T<K$	Slope for $S_T>K$
Long call	0	1
Short call	0	-1
Long put	-1	0
Short put	1	0

The slopes in the table are the slopes of the payoff as a function of $S_T$, measured in payoff dollars per dollar of underlying price. The slope exactly at $S_T=K$ is not defined: the left-hand and right-hand slopes differ. That is the kink. It does not mean the payoff itself jumps.

Why the Payoff Is Continuous at the Strike

The continuity follows directly by checking the value at the boundary against the values approached from each side. For a long call, the payoff to the left of $K$ is zero. To the right it is $S_T-K$, which approaches zero as $S_T$ approaches $K$. At $K$, the payoff is also zero. The put has the same boundary value, although its sloping segment is on the opposite side.

\lim_{S_T\to K^-} C_T=0,\qquad \lim_{S_T\to K^+} C_T=\lim_{S_T\to K^+}(S_T-K)=0,\qquad C_T(K)=0.

For a long put, the left-hand payoff $K-S_T$ approaches zero, and the right-hand payoff is zero. Thus both one-sided limits equal the payoff at the strike. Negating either payoff to obtain the short position also preserves continuity, because the negative of a function with matching boundary values still has matching boundary values.

The slopes, however, do not match. For a long call, the slope changes from $0$ to $1$; for a long put, it changes from $-1$ to $0$. Hence the payoff is continuous at $K$, but is not differentiable there. A diagram should show the two line segments meeting at the strike, not a gap or jump.

Worked Example: Locating a Long Call’s Boundary and Kink

A hypothetical call has exercise price $K=\$48$. Determine its payoff at expiration prices of $\$43$, $\$48$, and $\$55$; state the zero-payoff region and the slopes on each side of the strike. The method is to apply the piecewise payoff rule separately in each region. This avoids treating the exercise price itself as a positive-payoff point.

At $S_T=\$43$. Since $\$43<\$48$, the call is in its zero-payoff region: $C_T=\max(43-48,0)=\max(-5,0)=\$0$.

At $S_T=\$48$. At the boundary, $C_T=\max(48-48,0)=\max(0,0)=\$0$.

At $S_T=\$55$. Since $\$55>\$48$, $C_T=\max(55-48,0)=\max(7,0)=\$7$.

For all $S_T\leq \$48$, the payoff is zero, so the slope below the strike is $0$. For $S_T>\$48$, the payoff is $S_T-48$, so the slope above the strike is $1$. The payoff from the left and at the strike is zero, and the payoff from the right approaches $48-48=\$0$; this confirms continuity. The kink is at $\$48$, where the slope changes from $0$ to $1$. Common mistake avoided: assigning a positive payoff at $S_T=K$; the intrinsic payoff there is exactly zero.

Worked Example: Checking a Long Put on Both Sides of Its Strike

A hypothetical put has $K=\$72$. Evaluate its payoff when $S_T=\$65$, $\$72$, and $\$80$, and describe its zero-payoff region, slopes, and behavior at the boundary. Use the put payoff rule $P_T=\max(K-S_T,0)$, keeping the subtraction in the order $K-S_T$.

At $S_T=\$65$. The underlying is below the strike: $P_T=\max(72-65,0)=\max(7,0)=\$7$.

At $S_T=\$72$. The underlying equals the strike: $P_T=\max(72-72,0)=\$0$.

At $S_T=\$80$. The underlying is above the strike: $P_T=\max(72-80,0)=\max(-8,0)=\$0$.

The put’s payoff is zero for every $S_T\geq \$72$. Below $\$72$, its payoff is $72-S_T$, whose slope is $-1$; above $\$72$, its payoff is flat, with slope $0$. Approaching the strike from below gives $72-72=\$0$, which matches both the value at the strike and the limit from above. The graph is continuous but has a kink at $\$72$. Common mistake avoided: reversing the subtraction and calculating $S_T-K$ for a put, which would give the wrong sign below the strike.

Worked Example: Reading a Short Put’s Payoff Around the Boundary

A hypothetical investor is short one put with $K=\$36$. Find the short-position payoff for $S_T=\$30$, $\$36$, and $\$41$, then identify the two slopes. Use the short payoff as the negative of the long put payoff; this method keeps the position sign explicit and prevents confusing a short put with a long put.

At $S_T=\$30$. The long put would pay $\max(36-30,0)=\$6$. The short put payoff is its negative, $-\$6$.

At $S_T=\$36$. The long put payoff is $\max(36-36,0)=\$0$; its negative is also $\$0$.

At $S_T=\$41$. The long put payoff is $\max(36-41,0)=\max(-5,0)=\$0$, so the short put payoff is $\$0$.

For $S_T<\$36$, the short payoff is $S_T-36$, with slope $1$. For $S_T>\$36$, the payoff is zero, with slope $0$. At the boundary, the left-hand expression gives $36-36=\$0$, matching the value there and the right-hand value. The payoff is continuous and has a kink at $\$36$. Common mistake avoided: reporting the long put’s positive $\$6$ as the short position’s payoff; the short position has the opposite sign.

Common Interpretation Traps

When reading or sketching an expiration diagram, separate three questions: where the payoff is zero, what the slope is on each open interval, and what value the payoff takes at the strike. Those checks catch most boundary errors.

Confusing the kink with a jump. The slope changes at $K$, but all four positions have payoff zero at the strike, and their one-sided limits agree with that value.
Giving a slope at the kink. A single slope at $K$ is not defined because the left-hand and right-hand slopes differ. State both one-sided slopes instead.
Using profit instead of payoff. Premiums shift profit but do not alter the option’s expiration payoff or the location of its payoff kink.
Ignoring the position sign. Short call and short put payoffs are the negatives of their corresponding long payoffs, so their slopes reverse sign as well.
Extending a line beyond its applicable region. A long put does not keep rising as $S_T$ increases past $K$; its payoff is zero there. Similarly, a long call is flat below $K$.

Key takeaway: At expiration, the strike $K$ is the shared boundary where call and put payoffs are zero. Payoffs meet continuously at that boundary, while their slopes change there. Identify the applicable region first, then calculate the payoff and state the one-sided slopes.

Check Your Understanding

For each original problem, determine the requested expiration payoff features before checking the answer.

A long call has $K=\$57$. Find its payoff at $S_T=\$52$, $\$57$, and $\$63$, and state its zero-payoff region and slopes. (Answer: Payoffs are $\$0,\ \$0,\ \$6$; zero for $S_T\leq \$57$; slopes $0$ below and $1$ above.)
A long put has $K=\$44$. Find its payoff at $S_T=\$38$ and $\$49$, then state its boundary value, slopes, and whether it is continuous at $K$. (Answer: Payoffs are $\$6$ and $\$0$; value at $K$ is $\$0$; slopes $-1$ below and $0$ above; continuous with a kink.)
A short call has $K=\$83$. Calculate its payoff at $S_T=\$80$, $\$83$, and $\$91$. State the zero-payoff region and the slope above the strike. (Answer: Payoffs are $\$0,\ \$0,\ -\$8$; zero for $S_T\leq \$83$; slope above the strike is $-1$.)
A short put has $K=\$29$. Calculate its payoff at $S_T=\$24$, $\$29$, and $\$33$. Verify the boundary by comparing the left-hand and right-hand expressions. (Answer: Payoffs are $-\$5,\ \$0,\ \$0$; the left expression $S_T-29$ gives $\$0$ at the boundary, matching the right-hand value $0$.)
For a long put with $K=\$61$, a student claims the slope at the strike is $-1$ because the payoff immediately below $K$ changes one-for-one with the underlying price. Correct the claim by giving both one-sided slopes and explaining the boundary value. (Answer: The left-hand slope is $-1$, the right-hand slope is $0$, so no single slope exists at $K$; the payoff at $K$ and both one-sided limits are $\$0$.)

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