A call option: payoff is not profit

A financial asset is a set of rights. A call option makes that idea particularly useful: the right can have positive value at expiry while the trade still loses money. To understand why, separate what the contract delivers from what you paid to acquire it.

Consider one European call on one unit of an underlying asset. Its strike is 100, and you pay a premium of 8 when you buy it. Hold it until expiry, ignore fees, taxes and financing costs, and measure the exercise outcome in cash-equivalent terms. There is no additional contract multiplier in this example.

The call gives you the right to buy the underlying for 100 at expiry. The strike is the purchase condition if you exercise; the premium is what you paid earlier to obtain that choice. Letting the right expire does not refund the premium. Keep the two dates in view: at purchase you pay 8; at expiry you receive whatever economic value the right then provides. The contract and its terms ↗

At $S_T=90$, exercise offers no positive difference. Payoff is 0 and profit is −8. At 100 the difference is still zero, so both figures stay the same. At 105, payoff is 5 but profit is −3. The option is in the money and has a positive payoff, yet the trade loses money.

At 108, payoff is 8 and profit is zero. At 120, payoff is 20 and profit is 12. Keep the two columns separate as you work through the table.

The strike of 100 marks where positive intrinsic value begins. The break-even price is 108 because $S_T-100-8=0$ there.

Pause at 105. Why take the exercise value if the trade still loses 3? Abandoning a value of 5 would leave a loss of 8 instead. The premium has already been paid; the expiry choice is between value of 5 and zero.

Spend two minutes with the controls. First reproduce 105 and 108. Then change only the premium from 8 to 12. Predict which line moves before looking at it, find the new break-even point, and restore the original parameters.

With $S_T$ on the horizontal axis, payoff stays at zero below the strike and rises with slope one above it. The profit line sits 8 lower and crosses zero at 108. The horizontal axis shows alternative expiry outcomes, not the passage of time. Moving the marker to 120 asks what happens if the underlying finishes at 120.

Suppose the option trades at 12 before expiry. Selling then produces a profit of 12−8=4 for the original buyer. The payment comes from the buyer taking over the position; the seller no longer holds its later expiry payoff.

That quote does not change the function $\max(S_T-100,0)$. A new buyer paying 12 has a different cost basis and therefore a different profit line. If you retain the position, your total profit measured from the original purchase still subtracts 8. If instead you compare holding now with selling now,12 is the sale proceeds you forgo. The starting point determines which comparison you are making.

Moving the expiry underlying price from 90 to 100 changes the call payoff by zero. Moving it from 110 to 120 changes payoff by 10. Equal changes in the underlying do not always produce equal changes in the option outcome.

Now suppose, purely for this probability example, that expiry prices of 90 and 120 are equally likely. Their average is 105. But the call payoffs are 0 and 20, whose average is 10—not the payoff of 5 obtained by inserting the average price into the formula. Averaging before applying a nonlinear function is different from averaging its outcomes.

Expected expiry profit in this example is 10−8=2. The 10 is an average future payoff under the stated probabilities;8 is the initial purchase price. The average still leaves a possible realised loss of 8 in the lower-price state.

Try the following questions without looking back at the table. Explain the calculation in words as well as numbers. We now have a payoff for each state; the next idea is conditional expectation, which asks how to average those outcomes when information changes the relevant states and their weights.