Unlimited Upside, Identical Price

A share trades at 100. Two options expire in a year, both struck at 100. The call lets you buy the share at 100, the put lets you sell it at 100, and the interest rate is exactly zero. Which one costs more?

The call, surely. If the share goes to 400 the call pays 300, and there is no ceiling on how good it can get. The put cannot pay more than 100, because the share cannot fall below zero. One payoff is unbounded and the other is capped, so the prices should differ. They do not. The two options are worth exactly the same amount, and the reason has nothing to do with volatility, skew, or which model you like.

The step the reflex skips

The unbounded-upside argument compares the two payoff functions and then quietly concludes something about two prices. Prices are expectations, and an expectation weights every payoff by the probability of getting it. Saying that the call’s payoff has no ceiling tells you about the support of the distribution. It says nothing about the integral.

A lottery ticket that pays a billion with probability $10^{-12}$ also has a payoff far larger than anything a put can produce, and it is worth a thousandth of a penny. Unbounded and large on average are separate properties, and confusing them is the whole content of this question.

Subtract one payoff from the other

Rather than compare the two payoffs, subtract them. Write $S_T$ for the share price at expiry and $X$ for the common strike. Above the strike the call pays $S_T - X$ and the put pays nothing. Below it the call pays nothing and the put pays $X - S_T$. So on the upper branch the difference is $(S_T - X) - 0$, and on the lower branch it is $0 - (X - S_T)$. Both equal the same thing:

Two kinked functions, and their difference has no kink at all. It is one straight line through the strike, valid on both sides and at the strike itself, where both payoffs are zero. Nothing about the identity is probabilistic; it is an algebraic fact about the two formulas, and it can be checked at whatever share price you like.

That straight line is the payoff of a forward contract to buy the share at 100. You can build it without any options at all: buy the share today, and owe 100 at expiry. At expiry you hold something worth $S_T$ and you owe $X$, which is the line. The cost of building it today is the share price minus the present value of the debt, which at a zero rate is $100 - 100 = 0$. The forward is free.

One expectation, and no model

The pricing step is a single line. Take risk-neutral expectations of equation (1) and discount:

The second equality uses only that the discounted share price is a martingale under $\mathbb{Q}$, which is what makes $\mathbb{Q}$ risk neutral in the first place. Set $r = 0$ and $S_0 = X = 100$, and the right side is zero:

Notice which words never appeared. No volatility. No lognormal. No skew. No distribution of any kind. Equation (2) holds for every terminal law with the right mean, and the only fact used about the law is that mean. This is worth saying loudly, because there is a tempting wrong explanation in circulation: that the lognormal’s right skew happens to offset the payoff asymmetry. It sounds sophisticated and it is not the reason. If lognormality were load bearing, the equality would fail for other distributions, and it does not.

Five laws, and a control that breaks it

Claims like this one deserve to be tested where they could fail, so the equality was evaluated on five different terminal distributions, each with mean 100. A lognormal, a uniform law on the interval from 0 to 200, a two-point law that lands on 200 or 0 with equal chance, a right-tailed law that pays 1000 five per cent of the time and about 52.63 otherwise, and a left-tailed law that pays 0 thirty per cent of the time and about 142.86 otherwise. In all five the call and the put have equal value, and the common value moves wildly between them: 11.92 for the lognormal at thirty per cent volatility, 25 for the uniform, 50 for the two-point law.

Then the control. Take the two-point law again, but tilt it so the share lands on 200 sixty per cent of the time, which puts the mean at 120 rather than 100. Now the call is worth 60 and the put is worth 40, and the difference is exactly 20, which is the mean minus the strike, exactly as equation (2) says it must be. The equality is not a coincidence that survives some laws and not others. It tracks one number, and when you move that number the gap opens by precisely the amount predicted.

What lognormality adds, and what it does not

Assume the standard model and you get a number rather than a relation. At a zero rate with the strike at the money, $d_1 = \sigma\sqrt{T}/2$ and $d_2 = -\sigma\sqrt{T}/2$, so the two formulas collapse to one:

At $\sigma = 30\%$ and one year that is 11.9235 per cent of the share price for both options, which quadrature and a two-million-path simulation both confirm. The symmetry in equation (4) is visible in the formula: the two normal probabilities are $\Phi(a)$ and $\Phi(-a)$, and they sum to one.

Lognormality does buy one genuine extra statement. Under it the call finishes in the money with probability $\Phi(-\sigma\sqrt{T}/2)$, which is 44.0 per cent at those parameters and is strictly below one half for every positive volatility, because the median of a lognormal sits below its mean. So in that model the put pays off more often and pays less when it does, and the two effects cancel to the last decimal.

Do not promote that sentence into the general explanation. The frequency claim is model dependent and easy to break: in the right-tailed law above the call pays off five per cent of the time, and in the left-tailed one it pays off seventy per cent of the time. The prices are equal in both. Frequency of payoff is not what the parity argument is about.

Where the equality stops

Equation (2) is the general statement, so it also tells you exactly which assumptions were load bearing. Raise the rate above zero and $C - P = S_0 - Xe^{-rT} > 0$, so the call is genuinely dearer, by the interest you save on the strike you have not yet paid. Add a continuous dividend yield $q$ and the relation becomes $C - P = S_0 e^{-qT} - X e^{-rT}$, which pushes the other way and can make the put the more expensive one. Move the strike off the money and the gap opens immediately.

American exercise is the one that surprises people, and in this particular configuration it changes nothing. At a zero rate with no dividend there is no interest to be earned on an early-exercised strike, so exercising the put early gains nothing, and an American call on a share with no dividend is never worth exercising early anyway. The American prices coincide with the European ones here. Restore a positive rate or a dividend and early exercise acquires value, at which point parity weakens into a pair of inequalities rather than an equation.

The pattern generalises past this problem. When a quantity looks hard to compute, look for a combination of it with something else that is easy. Here the difference of two awkward kinked expectations is the expectation of a straight line, and a straight line needs only a mean. The individual prices still depend on the whole distribution, and you cannot get either one without a model. Their difference never did.

Sources and further reading

• The relation in equation (2), and its arbitrage proof — Put-call parity
• The closed form used in equation (4) — Black-Scholes model
• Why the discounted share price is a martingale under the pricing measure — Risk-neutral measure
• Where the gap between median and mean in the last section comes from — Log-normal distribution

The payoff identity was checked at 400,001 share prices between 0 and 400 with a worst error of zero, the price equality on the five laws described above, and the lognormal figures by closed form, by quadrature and by simulation.