CM2 · explained

The Black-Scholes formula and put-call parity, read term by term

The Black-Scholes formula looks like something to be memorised and is in fact two recognisable pieces. This page explains what each factor means, what d1 and d2 are doing, how put-call parity gives the put for free, and prices both options with the two routes agreeing exactly.

CM2 · Financial Engineering & Loss Reserving · 12 min read

What the formula is

For a European call on a share paying no dividends, with current price S, strike K, time to maturity T, continuously compounded risk-free rate r and volatility σ:

c = S · Φ(d1) − K e−rT · Φ(d2)

d1 = [ ln(S/K) + (r + ½σ2)T ] / (σ√T)
d2 = d1 − σ√T

In words: the call is worth what you expect to receive — the share, if you end up owning it — minus what you expect to pay for it, which is the strike, discounted, and only in the cases where you actually pay it.

The corresponding put, which can be written down directly or derived from parity:

p = K e−rT · Φ(−d2) − S · Φ(−d1)

Reading the two terms

The structure is easier to hold on to than the algebra. Both terms are of the form "something you get or give up, multiplied by a probability-like weight".

The second term is the easier one. Φ(d2) is the risk-neutral probability that the option is exercised — that the share finishes above the strike. So K e−rTΦ(d2) is the present value of the strike, paid only in the cases where it is paid at all. That reading is exact rather than a loose analogy.

The first term is the one to be careful about. Φ(d1) is not a probability of anything in the way Φ(d2) is. It is the delta of the option, and SΦ(d1) is the present value of receiving the share conditional on exercise — which is bigger than "the share price times the chance of exercise", because the cases where the option finishes in the money are also the cases where the share is worth more than average. The two effects are bundled together in that one factor, and that is why Φ(d1) > Φ(d2) always.

The gap between them is σ√T, which is the standard deviation of the log of the share price over the life of the option. The whole of the option's value above its intrinsic value lives in that gap, which is why an option on a quiet share with a day to run is worth almost nothing.

What the formula assumes

A question asking you to comment on the realism of the assumptions is asking for a judgement on each, not a list.

  • The share price is lognormal with constant volatility. The structure is right — positive prices, proportional moves — but real returns have fatter tails than a normal distribution, and volatility is plainly not constant. The market's response is to quote a different volatility for different strikes, the volatility smile, which is an admission that the model is being used as a quoting convention rather than believed.
  • No dividends. Easily repaired: for a share with a continuous dividend yield q, replace S by S e−qT throughout. The same substitution prices a currency option, with the foreign interest rate in place of the dividend yield.
  • No transaction costs, and continuous trading. This is the one the derivation actually depends on. The formula is built from a hedge that must be rebalanced continuously, and with real trading costs that is impossible. It is also why gamma matters commercially: it measures how often the hedge has to be adjusted.
  • A constant, known risk-free rate, and unlimited borrowing and lending at it. Reasonable for a short-dated equity option, increasingly strained for anything long-dated.
  • No arbitrage, and the share can be shorted freely. Short-selling restrictions break the replication argument in one direction, which shows up as put prices that will not come down to the no-arbitrage bound.

The honest summary is that every assumption is false and the formula is used everywhere anyway, because it is approximately right in the middle of the distribution and because everyone has agreed on it. Being able to say which assumption matters for which purpose is the examinable skill.

Put-call parity

This is the most useful single relationship in option theory, and it owes nothing to Black-Scholes. It holds for any European options on the same share with the same strike and maturity, whatever the model.

c − p = S − K e−rT

In words: holding a call and writing a put is the same thing as owning the share and owing the strike.

The proof is one table. Buy the call, sell the put, and whatever the share does at expiry you end up holding the share and having paid K:

  • If ST > K, the call is exercised, you pay K and get the share. The put expires worthless.
  • If ST < K, the put is exercised against you, you pay K and get the share. The call expires worthless.
  • If ST = K, the same position for the same money.

So the combination is a commitment to buy the share for K at time T, whose value today is S − K e−rT. No distribution was assumed, no volatility appeared, and nothing was said about how the share behaves — which is exactly why the relationship is so hard to break and so useful as a check.

Three ways it is used, all examinable. It gives the put from the call without a second formula. It gives any one of the five quantities from the other four. And a departure from it in the market is a straightforward arbitrage, which is why real quotes obey it closely.

Delta and gamma

Δcall = Φ(d1)     Δput = Φ(d1) − 1
Γ = φ(d1) / (Sσ√T)    (the same for a call and a put)

In words: delta is how much the option price moves when the share moves by one rupee. Gamma is how much the delta itself moves.

Delta is between 0 and 1 for a call and between −1 and 0 for a put, and the difference between them is exactly 1 — which is put-call parity differentiated, since the right-hand side S − Ke−rT has a derivative of 1 with respect to S. Every relationship between a call and a put in this topic is parity in another costume.

A delta-neutral portfolio holds Δ shares for every option sold, so that a small move in the share leaves the total value unchanged. It stops being neutral as soon as the share moves, because delta has changed, and gamma says how fast. A low gamma means a hedge that holds its shape and can be rebalanced rarely; a high gamma — which is where an at-the-money option close to expiry lives — means a hedge that must be adjusted constantly, and in a market with transaction costs that is expensive. Gamma is the number that decides what hedging a book actually costs.

Where marks are lost

Discounting the share price. The first term is SΦ(d1), with S undiscounted. Only the strike is discounted, because only the strike is paid in the future.

Using an annual effective rate for r. Everything here is continuously compounded. If a question gives an effective rate i, convert with r = ln(1 + i) before starting.

Getting the sign in d1 wrong. It is ln(S/K), not ln(K/S), and (r + ½σ2), not (r − ½σ2). The minus version is the drift of the log price under the risk-neutral measure and appears in the derivation, which is precisely what makes it a tempting wrong answer.

Reading Φ for the wrong sign of d. Normal tables start at zero, so a negative d needs Φ(−x) = 1 − Φ(x). Write that line down rather than doing it in your head.

Writing parity with the strike undiscounted. c − p = S − K is wrong and is wrong by the interest on the strike. The discount factor is the whole content of the relationship.

Applying parity to American options. It holds for European options. For American options only an inequality survives, because early exercise breaks the exact replication.

Questions students ask

Why does the expected return on the share not appear?

For the same reason it did not appear in the binomial model: the option is priced relative to the share by a hedging argument, and the share's price already contains everything the market believes about its expected return. The only parameter of the share's behaviour that survives is its volatility, because that is what determines how much the hedge has to be adjusted. This is the single deepest point in the topic and a question will ask it in some form.

Is volatility the only input that has to be estimated?

Yes, and that is both the formula's strength and where all the argument is. S, K, T and r are observable. σ is not, so the practice is to work backwards: take the market price of a traded option and find the volatility that reproduces it. That is the implied volatility, and it is what option markets actually quote. The fact that it differs across strikes and maturities on the same share is the clearest evidence that the model's assumptions are not satisfied.

What is the formula worth if the assumptions all fail?

A great deal, used properly. It gives the right structural answers — which way a price moves when an input changes, how a hedge should be set, roughly what an option is worth — and it provides the common language of implied volatility in which option markets are quoted. It is a model to think with, not a measurement, and saying so is the right answer to a question about its limitations.

Can the formula handle an American option?

Not directly. For a call on a non-dividend-paying share it does not need to, because early exercise is never optimal and the American and European prices are equal. For an American put, or for anything paying dividends, a numerical method is required — a binomial tree, or simulation — which is one of the main reasons trees remain in use alongside a closed-form formula.

A worked example

A share that pays no dividend trades at ₹500. Its volatility is 30% a year and the continuously compounded risk-free rate is 5% a year.

Price a six-month European call and a six-month European put, both with a strike of ₹520. Then find the call's delta and gamma, and set up a hedge.

The share price, volatility and strike are invented for this example. They are not a real security.

Step 1: d1 and d2

ln(500/520) = ln(0.9615385) = −0.0392207
(r + ½σ2)T = (0.05 + 0.045) × 0.5 = 0.095 × 0.5 = 0.0475000
σ√T = 0.30 × 0.7071068 = 0.2121320
d1 = (−0.0392207 + 0.0475000) / 0.2121320 = 0.0082793 / 0.2121320 = 0.0390290
d2 = 0.0390290 − 0.2121320 = −0.1731030

Note the sign of d2, and note that the gap between them is exactly σ√T. A quick sanity check before going on: the share is below the strike, so the option is out of the money, and d1 has come out only just above zero. A d1 of 1.5 on these numbers would be a signal that something had gone wrong.

Step 2: the normal distribution function

Φ(0.0390290) = 0.5155664     Φ(−0.1731030) = 0.4312852
Φ(−0.0390290) = 1 − 0.5155664 = 0.4844336
Φ(0.1731030) = 1 − 0.4312852 = 0.5687148

Step 3: the call

e−rT = e−0.025 = 0.9753099
K e−rT = 520 × e−0.025 = 507.161154
c = 500 × 0.5155664 − 507.161154 × 0.4312852
   = 257.783200 − 218.731100 = ₹39.052100

Call that ₹39.05.

Step 4: the put, two ways

First from the formula:

p = 507.161154 × 0.5687148 − 500 × 0.4844336
   = 288.430054 − 242.216800 = ₹46.213254

Then from put-call parity, which uses the call and no normal distribution function at all:

p = c + K e−rT − S = 39.052100 + 507.161154 − 500 = ₹46.213254

The two agree to every decimal place, and that is not luck. Subtract the put formula from the call formula and the Φ terms pair up as Φ(d) + Φ(−d) = 1, leaving S − Ke−rT exactly — which is parity. So the agreement is an arithmetic check rather than a theoretical one — but it is a thorough arithmetic check, because it would fail if any one of the four normal values had been read off wrongly.

A second check on the same figures: c − p = 39.052100 − 46.213254 = −7.161154, and S − Ke−rT = 500 − 507.161154 = −7.161154. The put is dearer than the call by ₹7.16, which is what parity says it must be when the discounted strike exceeds the share price.

Do the answers pass the bounds?

  • The call is out of the money, so its intrinsic value is zero and all of its ₹39.05 is time value. It must sit between 0 and S, and it does.
  • The put has an intrinsic value of 520 − 500 = ₹20, so 46.213254 − 20 = ₹26.21 of its ₹46.21 is time value.
  • A European put must be worth at least Ke−rT − S = ₹7.16 and at most Ke−rT = ₹507.16. ₹46.21 is comfortably inside.

Step 5: delta and gamma

Δcall = Φ(d1) = 0.5156
Δput = 0.5155664 − 1 = −0.4844
φ(d1) = 0.3989423 × e−0.03902902/2 = 0.3989423 × 0.9992387 = 0.3986386
Γ = 0.3986386 / (500 × 0.2121320) = 0.3986386 / 106.0660 = 0.003758

The deltas differ by exactly 1, as parity requires. The call's delta just above a half is what an almost-at-the-money option should give.

Step 6: hedging a book of 10,000 written calls

An investor who has written 10,000 of these calls is short 10,000 × 0.5155664 = 5,156 shares of delta, so the hedge is to buy 5,156 shares.

The hedge is only good for small moves, and gamma says how quickly it decays. Take a ₹10 rise in the share:

delta predicted by gamma: 0.5155664 + 10 × 0.003758 = 0.553146
exact delta at S = 510: 0.552658

So after a ₹10 move the hedge is short about 0.0371 of delta per option — 0.552658 − 0.515566 = 0.037092, or roughly 371 shares across the book — and it has to be topped up. On a book of 10,000 options that is a real trade, and it would have to be made again on the next move, and in the opposite direction if the share came back — buying high and selling low each time, which is the cost gamma measures.

The same arithmetic applied to the option price shows why the second-order term is not optional:

exact call price at S = 510: ₹44.394106
estimate from delta alone: 39.052100 + 10 × 0.5155664 = ₹44.207764
estimate from delta and gamma: 44.207764 + ½ × 0.003758 × 102 = ₹44.395664

Delta alone is 19 paise out; delta and gamma together are under 2 paise out. For a single option that hardly matters. For a book of 10,000, 19 paise an option is ₹1,863 of unhedged exposure from a single ₹10 move, which is exactly the sort of number a risk report is built to show.

A note on using four-figure tables

Every normal value above carries seven decimal places. A student working from a four-figure normal table would read Φ(d1) = 0.5156 and Φ(d2) = 0.4313 and get a call of ₹39.061394 rather than ₹39.052100 — a difference of 0.93 paise, which is rounding and not an error. The habit worth forming is the one that makes the difference visible: carry the intermediate figures unrounded, round once at the end, and say what precision you have used.

Where this fits in CM2

This page explains one idea from CM2 Financial Engineering & Loss Reserving. The full CM2 syllabus, with what each part is examined on, sits on the course page, and past-paper style mock exams with full solutions are where you find out whether the idea has stuck under time pressure.

Other CM2 topics explained

Stuck on a term rather than the idea? The actuarial glossary defines the notation used above. If something here is wrong or unclear, tell us — these pages get corrected.