Case 098Option pricing and arbitrage checksCore

A stock is at Rs 250. Price a six-month 260 call by Black-Scholes with 30% volatility and a 6% rate: compute d1, d2, the two probabilities and the call, then the put by parity.

The situation

Rushikulya Salt trades at Rs 250 and pays no dividend over the next six months. An interviewer hands you a normal table and asks you to price a European call struck at Rs 260, expiring in six months, with volatility of 30% a year and a continuously compounded rate of 6%.

You are to show d1 and d2, read the two probabilities, compute the call, then get the put on the same strike from put-call parity rather than from the formula, and say what each piece of the formula means.

Your task

What are d1, d2, N(d1), N(d2), the call price and the put price, and what does each piece mean?

Quick check

Before computing: the call is 10 rupees out of the money with six months to go. Roughly what is it worth?

Worked solution

Try it on paper, then open one step at a time.

30-second answerThe answer to give first

d1 is 0.0626, d2 is -0.1495, N(d1) is 0.5250, N(d2) is 0.4406, the call is about Rs 20.08 and the put about Rs 22.39. The call is 250 times N(d1), Rs 131.24, less the strike discounted to 252.32 times N(d2), Rs 111.16. Parity gives the put as call minus stock plus discounted strike. N(d2) is the risk-neutral chance the call finishes in the money; N(d1) is its delta.

Step 1How do you get d1 and d2 without slipping?

Build them from three pieces and write each down. The log of stock over strike, ln(250/260), is -0.0392: the call starts 3.9% out of the money. The drift term, (r plus half of vol squared) times T, is (0.06 plus 0.045) times 0.5, 0.0525. Volatility times root T is 0.30 times 0.7071, 0.2121. So d1 is (-0.0392 plus 0.0525) over 0.2121, which is 0.0626, and d2 is d1 minus 0.2121, which is -0.1495. A careful candidate says the three numbers out loud before dividing; most mistakes are a dropped minus sign on the log or forgetting the half in front of vol squared.

The relationship
d1=ln⁡(S/K)+(r+12σ2)TσTd2=d1−σTC=S N(d1)−Ke−rTN(d2)d_1 = \frac{\ln(S/K) + (r + \tfrac{1}{2}\sigma^2)T}{\sigma\sqrt{T}} \qquad d_2 = d_1 - \sigma\sqrt{T} \qquad C = S\,N(d_1) - K e^{-rT} N(d_2)
S, Kstock price 250 and strike 260
\sigma\sqrt{T}volatility over the option's life, 0.2121
N(\cdot)the standard normal cumulative probability
K e^{-rT}the strike discounted six months at 6%, 252.32
What it says in wordsThe call is the stock weighted by one probability minus the discounted strike weighted by another.
Step 2What do the two probabilities mean?

Think of a raffle ticket that pays a prize only if a number is drawn above 260. Its value is the chance you win times what you collect, less the chance you must pay times what you pay. N(d2), 0.4406, is the risk-neutral probability that the stock ends above 260, so the call holder pays the strike; N(d1), 0.5250, is the option's deltaHow much the option price moves for a one rupee move in the stock. and weights the stock the holder receives, which is larger because the stock is worth more in exactly the states where the option is exercised. The call is 250 times 0.5250, Rs 131.24, less 252.32 times 0.4406, Rs 111.16: Rs 20.08.

Black-Scholes in five steps, each feeding the next1. InputsS 250, K 260T 0.5, vol 30%r 6%2. d1 and d2d1 = 0.0626d2 = -0.1495gap = 0.21213. ProbabilitiesN(d1) = 0.5250N(d2) = 0.4406from the normal table4. Call131.24 - 111.16= 20.08Rs a share5. Put by parity20.08 - 250+ 252.32= 22.39The call is what you expect to receive in stock, less what you expect to pay in strike:S x N(d1)131.24PV(K) x N(d2)111.16Call20.08
Inputs give d1 of 0.0626 and d2 of -0.1495, which give N(d1) of 0.5250 and N(d2) of 0.4406; the call is Rs 131.24 of expected stock less Rs 111.16 of expected discounted strike, Rs 20.08, and parity gives the put at Rs 22.39.
Step 3How does parity give the put without a second formula?

A call minus a put on the same strike and expiry behaves exactly like owning the stock and owing the strike at expiry, so C minus P equals S minus the discounted strike. Rearranged, P is C minus S plus K times e to the minus rT: 20.08 minus 250 plus 252.32, which is Rs 22.39. The put is worth more than the call here because it is 10 rupees in the money, partly offset by the rate, which lifts the forward to about 257.61 and favours the call. Computing the put directly from the formula gives the same Rs 22.39, which is a check worth doing out loud.

QuantityValueWhat it is
ln(S/K)-0.0392how far out of the money, in log terms
(r + vol sq. / 2) T0.0525drift over six months
vol x root T0.2121one standard deviation over the life
d1, d20.0626, -0.1495standardised distances
N(d1), N(d2)0.5250, 0.4406delta, and chance of exercise
Call, put20.08, 22.39Rs a share
Each intermediate number in the Black-Scholes calculation for the 260 call, ending at a call of Rs 20.08 and a put of Rs 22.39.

Close with the limits, briefly, because a desk interviewer will ask. The model assumes a constant volatility, a lognormal stock with no jumps, no dividends and European exercise. Real markets price different volatilities for different strikes, so the 30% here would itself be read off a skew. The formula is a translation device between a price and a volatility, and its value is that everyone uses the same translation.

Where candidates lose it

The common loss is using the strike instead of the discounted strike in the second term, which understates the call by about seven rupees times N(d2). The strike is paid at expiry and must be discounted.

The second is reading N(d1) as the probability of exercise. That is N(d2); N(d1) is the delta, larger because it weights the stock by the states where it is high.

What the interviewer asks next

  • Raise volatility to 35%. Roughly how much does the call price change, using vega?
  • The stock will pay a Rs 5 dividend in three months. How do you adjust the inputs?
  • Why is N(d1) larger than N(d2), and when are they nearly equal?

Company names and figures are illustrative.