Finance-Theory · Unit 12 · Video 3 · Interactive Practice

Option Payoff Diagrams: The Kink and the Premium

IKey Formulas

FormulaNameMeaning
CT=Max[0, ST−K]C_T = \text{Max}[0,\ S_T - K]
PT=Max[0, K−ST]P_T = \text{Max}[0,\ K - S_T]		Payoff at maturityCall: flat at 0, then slope +1+1 above KK. Put: flat at 0, then slope −1-1 below KK, capped at KK
Net=Max[0, ST−K]−C\text{Net} = \text{Max}[0,\ S_T - K] - CNet payoff (call)The payoff line shifted down by the premium at every stock price (interest on the premium ignored over short horizons)
ST=K+C(call)S_T = K + C \quad\text{(call)}
ST=K−P(put)S_T = K - P \quad\text{(put)}		Break-evenWhere the net payoff line crosses zero
ST−S0S0vsMax[0, ST−K]−CC\dfrac{S_T - S_0}{S_0}\quad \text{vs}\quad \dfrac{\text{Max}[0,\ S_T - K] - C}{C}Stock return vs call returnStraight line vs a line flat at −100%-100\% that bends up at the strike

Key Insight: The kink at the strike price is the option. Stocks, bonds, futures and forwards all have straight-line payoffs in the underlying price; an option's payoff bends, so its risk and return are different in kind.

IIVisualization 1 — From Payoff to Net Payoff

The net payoff is the kinked payoff line shifted down by the premium; break-even is where it crosses zero.

💡 These are values at maturity. Before maturity a call with the stock below $20 is still worth something, because the stock may yet rise above the strike before expiry.

IIIVisualization 2 — Stock Return vs Call Return

Bought today for $20 or $2, the stock's return is a straight line; the call's bends at the strike and climbs ten times as fast.

IVQuiz Questions

Problem 1 · Net Payoff of a Call

Given (illustrative figures): a call with a strike price of $35 was bought for a premium of $2.75. At maturity the stock price is $41. Ignoring interest on the premium, find the net payoff.

✅ Correct! The payoff of $6 less the $2.75 premium nets $3.25.
❌ Close, but… $6 is the gross payoff Max[0, 41−35]\text{Max}[0,\ 41 - 35]. The net payoff subtracts the premium you paid for the call.
❌ Check the sign. The premium is a cost to the buyer, so it is subtracted from the payoff, not added.
❌ Not quite. Losing the whole premium happens only when the call finishes at or below the strike. Here the stock ends above $35, so the call is exercised.
❌ Not quite. Net payoff =Max[0, ST−K]−C= \text{Max}[0,\ S_T - K] - C.
Show solution

The call is in the money at maturity, so it is exercised:

CT=Max[0, 41−35]=6C_T = \text{Max}[0,\ 41 - 35] = 6

Subtract the premium paid for it:

Net=CT−C=6−2.75=3.25\text{Net} = C_T - C = 6 - 2.75 = 3.25

The net payoff is $3.25. Break-even for this call is K+C=35+2.75=37.75K + C = 35 + 2.75 = 37.75, and $41 is above it.

Problem 2 · Break-Even for a Put

Given (illustrative figures): a put with a strike price of $40 costs a premium of $3. Ignoring interest on the premium, find the stock price at maturity at which the put buyer exactly breaks even.

✅ Correct! At $37 the put pays $3, which exactly repays the premium.
❌ Check the direction. K+CK + C is the call's break-even. A put pays when the stock falls, so its break-even lies below the strike: K−PK - P.
❌ Not quite. At $40 the put pays Max[0, 40−40]=0\text{Max}[0,\ 40 - 40] = 0, so the buyer is down the full $3 premium.
❌ Not quite. Below $40 the put is exercised, but just below the strike its payoff is smaller than the $3 premium. The net payoff is still negative until the stock falls another $3.
❌ Not quite. Set the net payoff Max[0, K−ST]−P\text{Max}[0,\ K - S_T] - P equal to zero.
Show solution

Below the strike the put's net payoff is (K−ST)−P(K - S_T) - P. Setting it to zero:

40−ST−3=0⟹ST=K−P=40−3=3740 - S_T - 3 = 0 \quad\Longrightarrow\quad S_T = K - P = 40 - 3 = 37

Break-even is a stock price of $37. Above $37 the buyer loses part or all of the premium; below it the buyer profits, up to a maximum net payoff of K−P=37K - P = 37 dollars if the stock goes to zero.

Problem 3 · Stock Return vs Call Return

Given (illustrative figures): a stock trades at $40 today, and a call on it with a strike price of $40 costs $2.50. At maturity the stock price is $45. Ignoring interest, find the return on buying the stock and the return on buying the call.

Return on the stock?

Return on the call?

✅ Correct! A 12.5% move in the stock became a 100% return on the call.
❌ Check the units. The stock gains $5, but a return divides that gain by the $40 paid for the share.
❌ That is the call's return. The stock was bought for $40 and gains $5; divide the gain by $40.
❌ Check the denominator. 5/455/45 divides by the ending price. A return divides the gain by what you paid today, $40.
❌ Check the stock's return. Stock return =(ST−S0)/S0= (S_T - S_0)/S_0.
❌ That is the stock's return. The call was bought for $2.50 and pays $5; its gain of $2.50 is divided by the $2.50 premium.
❌ Close, but… 5/2.50=25/2.50 = 2 compares the whole payoff with the premium. The return is the gain, payoff minus premium, divided by the premium.
❌ Check the denominator. (5−2.50)/5(5 - 2.50)/5 divides by the payoff. The amount invested is the premium, $2.50.
❌ Check the call's return. Call return =(Max[0, ST−K]−C)/C= (\text{Max}[0,\ S_T - K] - C)/C.
Show solution

Stock: bought for $40, worth $45 at maturity:

ST−S0S0=45−4040=540=0.125=12.5%\frac{S_T - S_0}{S_0} = \frac{45 - 40}{40} = \frac{5}{40} = 0.125 = 12.5\%

Call: payoff Max[0, 45−40]=5\text{Max}[0,\ 45 - 40] = 5, bought for $2.50:

Max[0, ST−K]−CC=5−2.502.50=2.502.50=1=100%\frac{\text{Max}[0,\ S_T - K] - C}{C} = \frac{5 - 2.50}{2.50} = \frac{2.50}{2.50} = 1 = 100\%

The stock returns 12.5% and the call 100%. Had the stock ended at $40, the stock would have returned 0% while the call returned −100%-100\%.

Problem 4 · How Much Steeper Is the Call?

Given: the same position as Problem 3 — stock at $40 today, call with strike $40 bought for $2.50. For stock prices at maturity above the strike, each extra $1 of STS_T raises the stock's return by some number of percentage points and the call's return by another. How many times as fast does the call's return rise?

✅ Correct! 40 points per dollar for the call against 2.5 for the stock: the ratio is S0/C=40/2.50=16S_0 / C = 40 / 2.50 = 16.
❌ Not quite. Ten times was the video's case, S0/C=20/2S_0 / C = 20 / 2. With these prices the ratio is different.
❌ Close, but… Above the strike the stock and the call payoff both rise by $1 per $1, but returns divide that dollar by very different amounts invested: $40 for the stock, $2.50 for the call.
❌ Not quite. 40 is the call's gain in percentage points per $1. Compare it with the stock's gain per $1.
❌ Not quite. Compare the slopes 1/C1/C and 1/S01/S_0.
Show solution

Above the strike, one more dollar of STS_T adds one dollar to both the stock's value and the call's payoff. Divide by the amount invested in each:

stock: 140=2.5% per dollar,call: 12.50=40% per dollar\text{stock: } \frac{1}{40} = 2.5\% \text{ per dollar}, \qquad \text{call: } \frac{1}{2.50} = 40\% \text{ per dollar} 40%2.5%=S0C=402.50=16\frac{40\%}{2.5\%} = \frac{S_0}{C} = \frac{40}{2.50} = 16

The call's return rises 16 times as fast. The same bend works in reverse: at or below the strike the call's return sits at −100%-100\% however little the stock falls.

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