La opción de 4,76 $ que todos creen sin valor

Una acción con volatilidad cero, un call at the money, y la respuesta refleja — cero — que suspende entrevistas reales de trading. Un solo argumento de arbitraje la valora de tres maneras, y Black–Scholes lo confirma al final.

Lambdia
22 de julio de 2026 · 8 min de lectura

Here is a question that real trading desks ask in interviews, usually early, because it sorts candidates so fast. A stock trades at $S = \$100$ and its volatility is exactly zero: the price carries no randomness at all. Interest rates sit at $5\%$ per year. What is a fair price for a European call option on this stock with strike $K = \$100$ , expiring in one year?

The answer most people give, instantly, is zero. The stock cannot move, a call pays $\max(S_T - K, 0)$ , and if $S_T$ stays at 100 the payoff is nil. Quick, clean, and wrong. The correct value, to the cent, is

c \;=\; \frac{5}{1.05} \;=\; \frac{100}{21} \;\approx\; \$4.76.

This is question 2.1 in Timothy Crack's Heard on the Street, a collection of problems actually used in Wall Street interviews. What follows is the full resolution, three times over: by arbitrage, by replication, and by taking a limit inside Black–Scholes.

Where the instant answer goes wrong

The tempting chain runs: zero volatility $\Rightarrow$ the price never moves $\Rightarrow$ $S_T = 100$ $\Rightarrow$ payoff zero $\Rightarrow$ value zero. The flaw sits in the very first arrow. Volatility measures the random part of returns. Setting $\sigma = 0$ kills the randomness, not the motion: it makes the future price certain, and certain is not the same thing as constant.

A riskless stock must earn the riskless rate

Compare two ways of parking $100 for a year. The bank turns it into $105, guaranteed. The zero-volatility stock also turns it into some number known today; call it $S_T$ . Suppose the stock pays no dividends, and suppose $S_T = 103$ . Then sell the stock short, deposit the $100 proceeds, and unwind in a year: collect 105, buy the share back for 103, keep $2. No risk, no capital, free money. If instead $S_T = 107$ , run it backwards: borrow 100, buy the share, sell at 107, repay 105, keep $2 again.

No-arbitrage fact

Under frictionless markets (short selling allowed, borrowing and lending at the same rate, no dividends), any asset whose future value is certain must grow at exactly the risk-free rate. Anything else hands out riskless profit, and prices adjust until it stops.

So the stock's only consistent path is the ramp: from 100 today to $100 \times 1.05 = 105$ in a year, with certainty. The option lets you buy at 100 something that will be worth 105. It finishes in the money for sure, with payoff

\max(S_T - K,\, 0) \;=\; \max(105 - 100,\, 0) \;=\; 5.

Fig. 1 — Zero volatility is a ramp, not a flat line. The certain payoff is the $5 gap at maturity.

A guaranteed $5 arriving in one year is not worth $5 today: money in hand could sit at the bank earning 5%. Its present value is $5/1.05 = 100/21 = 4.7619\ldots$ — the $4.76 from the board above.

Build the option out of ordinary parts

If the discounting argument feels too slick, here is the version with no trust required. Buy one share for 100 and simultaneously borrow $100/1.05 \approx 95.24$ from the bank. Your net outlay today is $100 - 95.24 = 4.76$ . A year later the share is worth 105 with certainty and the debt has grown to exactly 100, so settling leaves you holding $105 - 100 = 5$ — precisely the option's payoff, in every possible future.

Fig. 2 — Static replication: one share minus one loan pays exactly what the option pays.

Two positions with identical payoffs in every state must trade at the same price — the law of one price. The option is therefore pinned at $100 - 100/1.05 = 100/21 \approx \$4.76$ , and the hedge is as simple as hedges get: hold one share, i.e. $\Delta = 1$ . (In general, delta is 1 whenever the forward price exceeds the strike and $\sigma = 0$ ; it would be 0 in the opposite case.)

Think in forwards, and let Black–Scholes agree

The clean way to carry this lesson around: never compare the strike with where the stock is; compare it with where the stock is going. The one-year forward price here is $F = 100 \times 1.05 = 105 > K$ , and for a zero-volatility asset the call is just the discounted forward-minus-strike:

c \;=\; \frac{\max(F - K,\, 0)}{1+r} \qquad (\sigma = 0).

(1)

The famous formula concurs. Write the Black–Scholes price with $S = K = 100$ , $T = 1$ and the continuous rate chosen so that $e^{rT} = 1.05$ :

c \;=\; S\,N(d_1) - K e^{-rT} N(d_2), \qquad d_{1,2} \;=\; \frac{\ln(S/K) + \left(r \pm \tfrac{\sigma^2}{2}\right)T}{\sigma\sqrt{T}}.

As $\sigma \to 0^{+}$ , the numerators tend to $rT = \ln 1.05 > 0$ while the denominator vanishes, so $d_1, d_2 \to +\infty$ and both normal terms lock onto 1. The whole machine collapses to

c \;\longrightarrow\; S - K e^{-rT} \;=\; 100 - \frac{100}{1.05} \;=\; \frac{100}{21} \;\approx\; \$4.76,

the same number as the arbitrage argument and the same number as the ledger. (We checked the limit symbolically with a computer algebra system while producing the companion video; the convergence is monotone, with $\sigma = 0.001$ already giving 4.7619047…)

The moral the interviewers are fishing for: volatility is not the only engine inside an option. Time and interest keep running even when all the randomness is switched off. Zero volatility never meant frozen — it meant predictable, and predictable things can still grow.

Sources and further reading

Timothy Falcon Crack, Heard on the Street: Quantitative Questions from Wall Street Job Interviews— the original problem (Q2.1) and its answer.
John C. Hull, Options, Futures, and Other Derivatives— forward pricing, no-arbitrage bounds, and the Black–Scholes model.
Wikipedia: Black–Scholes model and Rational pricing (law of one price, replication).

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