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Use the Black-Scholes-Merton model to value a European option on a futures contract, a foreign currency or a stock that pays dividends.

FRM® Part I · Valuation and Risk Models · The Black-Scholes-Merton Model · FRM1.VR.15.g

The idea

The BSM formula extends to other underlyings by treating them as assets with a continuous yield q. For a stock with yield q, the call value is c = S0 × e^(−qT) × N(d1) − K × e^(−rT) × N(d2). The first input is d1 = [ln(S0 ÷ K) + (r − q + σ²÷2) × T] ÷ (σ × √T). The second input is d2 = d1 − σ × √T. For known discrete dividends during a European option's life, subtract their present value from S0 and use the ordinary formula. For a currency option, q is the foreign risk-free rate. For an option on a futures price F0, Black's model gives c = e^(−rT) × [F0 × N(d1) − K × N(d2)]. In that formula d1 = [ln(F0 ÷ K) + σ² × T ÷ 2] ÷ (σ × √T). Then d2 = d1 − σ × √T as before. This is the case q = r. Typical errors are using spot in the futures formula and swapping the domestic and foreign rates.

Formulas

BSM value of a European call on an asset with a continuous yield

c = S0 × e^(−q × T) × N(d1) − K × e^(−r × T) × N(d2)

S0
the price of the underlying today
q
the continuous yield: dividend yield, or the foreign risk-free rate for a currency
T
the time to expiry in years
N
the standard normal cumulative distribution function
d1
[ln(S0 ÷ K) + (r − q + σ² ÷ 2) × T] ÷ (σ × √T)
d2
d1 − σ × √T
K
the strike price
r
the continuously compounded domestic risk-free rate

The yield enters both d1 and the stock leg. Swapping domestic and foreign rates for a currency option is the common error.

Black's model for a European call on a futures price

c = e^(−r × T) × [F0 × N(d1) − K × N(d2)]

r
the continuously compounded risk-free rate
T
the time to expiry in years
F0
the futures price today
N
the standard normal cumulative distribution function
d1
[ln(F0 ÷ K) + σ² × T ÷ 2] ÷ (σ × √T)
d2
d1 − σ × √T
K
the strike price

The whole bracket is discounted, futures price included, because a futures price needs no carry. Using the spot formula or the spot price is the trap.

A practice question

From the question bank, with the reason each wrong option is wrong.

Under Black's model, the value of a European call on a futures price F_0 with strike K, maturity T, risk-free rate r and volatility sigma is:

  1. F_0 x N(d1) - K x e^(-rT) x N(d2)Why not: Applies the spot-price formula to the futures price, discounting only the strike.
  2. e^(-rT) x [K x N(d2) - F_0 x N(d1)]Why not: Reverses the terms, giving the negative of the call value.
  3. e^(rT) x [F_0 x N(d1) - K x N(d2)]Why not: Compounds at the risk-free rate instead of discounting.
  4. e^(-rT) x [F_0 x N(d1) - K x N(d2)] - correct

In Black's model the futures price needs no cost of carry, and the whole payoff expectation, F_0 N(d1) - K N(d2), is discounted at the risk-free rate.

A harder one

A European call on a futures price has F_0 = 62, K = 60, maturity six months, r = 5% (continuous) and volatility 25%. Cumulative normal values: N(0.274) = 0.6079, N(0.097) = 0.5387. The value of the call under Black's model is closest to:

  1. 5.37Why not: Omits the overall discount factor e^(-rT).
  2. 5.24 - correct
  3. 5.50Why not: Multiplies by e^(rT) instead of discounting.
  4. 6.17Why not: Uses the spot-style formula, discounting only the strike and leaving the futures price undiscounted.

d1 = [ln(62/60) + 0.25² x 0.5/2]/(0.25 x sqrt(0.5)) = 0.274, d2 = 0.097. c = e^(-0.025) x (62 x 0.6079 - 60 x 0.5387) = 0.9753 x 5.368 = 5.24.

The hardest tier

A stock trading at 80 pays a continuous dividend yield of 3%. A European put has strike 75, six months to maturity, r = 5% (continuous) and volatility 30%. Cumulative normal values: N(-0.457) = 0.3237, N(-0.245) = 0.4031. The value of the put is closest to:

  1. 3.61Why not: Uses the no-dividend BSM formula, ignoring the yield in both the d terms and the stock leg.
  2. 3.98 - correct
  3. 9.64Why not: Values the call, S e^(-qT) N(d1) - K e^(-rT) N(d2), rather than the put.
  4. 3.20Why not: Multiplies the stock price by e^(+qT) instead of e^(-qT).

d1 = 0.457 and d2 = 0.245 include (r - q). p = K e^(-rT) N(-d2) - S e^(-qT) N(-d1) = 75 x e^(-0.025) x 0.4031 - 80 x e^(-0.015) x 0.3237 = 29.49 - 25.51 = 3.98.

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