Ten Percent Up Is a Smaller Move Than Ten Percent Down

A share is at 100. One contract lets you buy it at 110, another lets you sell it at 90. Both strikes are ten percent from the spot. Under Black-Scholes, which of the two is worth more?

The call, and by a wide margin: 4.2920 against 3.5891 at a zero rate, twenty percent volatility and one year to run. What makes this a good problem is that the most natural argument points the other way, and the second most natural argument gets the right answer for a reason that does not survive inspection.

The argument that points the wrong way

The put reaches its strike more often. That is not a feeling, it is a fact at these inputs. The risk-neutral chance of finishing in the money is 0.3348 for the put and 0.2821 for the call, so the put pays off about five percentage points more frequently.

Frequency is the wrong unit. A contract is not worth how often it pays, it is worth the average of what it pays, and the two come apart whenever the payoffs differ in size across the events being counted. Anyone reasoning from the probability of finishing in the money has quietly replaced the price with a count.

Ten percent down is the bigger move

The reason the call wins starts with the fact that the two strikes are not equally far away, even though the problem describes them as though they were. Ten percent is a percentage of the starting point, and moves in a share price compound rather than add.

From 90 back to 100 takes a rise of $10/90 = 1/9$, about 11.1 percent. From 100 up to 110 takes a rise of ten percent, and the reverse trip from 110 back to 100 is a fall of only $10/110 = 1/11$, about 9.09 percent. The distances are asymmetric. The model measures them in logarithms, which is where the asymmetry becomes a number:

So 90 is the farther strike on the only scale the model cares about. A contract struck farther out is worth less, other things equal, which is the first half of the answer and the half a candidate can produce out loud in an interview.

The prices themselves

A ratio of 1.196. The call is nearly twenty percent dearer than the contract that pays more often, which is a large enough gap that no amount of squinting at the tails will produce it by accident.

The exact reason, which is a symmetry

Here is the statement that actually settles it, rather than merely pointing in the right direction. At a zero rate the Black-Scholes call and put are related by exchanging the spot and the strike, and both are homogeneous of degree one in that pair. Putting those two facts together:

This holds to better than one part in $10^{12}$ at every zero-rate node checked. Now read it. The call struck at 110 is worth 1.1 times a put struck at 90.909. A put is increasing in its strike, and the put we are comparing against is struck at 90, which is below 90.909. So:

The call wins twice over, once because the mirror strike is farther out than 90 and once because of the factor 1.1. Two effects, both in the same direction, which is why the informal distance argument is sufficient rather than merely suggestive.

The popular explanation, and why it is wrong

The usual story is that a share can rise without limit while it can only fall to zero, so the call's payoff has a long tail the put's truncated one cannot match. It sounds compelling and it proves too much.

Take the same argument to a call and a put struck at the same place, at the money. Unbounded upside against bounded downside would predict a gap there too, and at a zero rate there is none: put-call parity makes them equal to the penny. An argument that predicts a gap where there is provably none is not the mechanism, whatever else it is.

What is really going on is that the payoff of a call is unbounded in a way that has already been priced into where the density sits, and the effect that survives is the ratio asymmetry of equation (1), formalised as the mirror identity in (4). Nothing about unlimited payoffs is needed.

Two claims worth striking from the record

Two statements circulate alongside this problem and neither holds for strikes at 110 and 90.

The first says that at $r = \sigma^2/2$ the two contracts have exactly the same chance of finishing in the money. They do not: at that rate the figures are 0.3168 for the call and 0.2992 for the put. Equal chances would require the two strikes to be symmetric in the logarithm, which means placing them at $100 \times 1.1$ and $100 / 1.1 = 90.909$ rather than at 110 and 90. Choosing round numbers on both sides destroys the symmetry that claim relies on.

The second says that at that rate the call is worth about twice the put. It is worth 1.63 times the put, which is a large discrepancy to leave standing next to a factor of two. Both figures were recomputed here from the formulas in the definition above.

Where the ordering can change

The verdict is robust in the parameters but not in every direction, and it is worth knowing which. Over a sweep of eight volatilities, six maturities and three non-negative rates, a hundred and forty-four combinations in all, the call is dearer at every single node. In the limits it stays dearer: as the volatility grows the call converges to the spot itself, and as the volatility vanishes both collapse toward zero with the call larger by many orders of magnitude.

What does flip is the frequency comparison from Fig. 1, and it flips at a threshold. For $\sigma\sqrt{T}$ below about 0.1002 the call becomes the likelier payer, not the put. So a reader who memorised the tail probabilities at these inputs has memorised something that changes sign, while the price ordering does not. That asymmetry between the two facts is the reason the price argument should never be routed through the payoff frequency.

Two conventions also matter and both were pinned above. Ten percent out was read in level terms, giving strikes of 110 and 90; the log-symmetric reading would give 110 and 90.909, which is exactly the mirror case, and there the answer is the clean factor of 1.1. And the rate was set to zero, which is what makes (3) exact. At a positive rate the identity picks up discount factors, the algebra is uglier, and the call's advantage grows.

Sources and further reading

• The pricing formulas used throughout — Black-Scholes model
• The relation that makes the at-the-money pair equal — Put-call parity
• The distribution the shaded tails belong to — Log-normal distribution
• The reason logarithms are the right ruler for a return — Rate of return

Every price above was produced from a Black-Scholes implementation built on the complementary error function alone, so nothing depends on a statistics library. The mirror identity in (4) was verified to a worst relative error of $9 \times 10^{-15}$ across all forty-eight zero-rate nodes of the sweep, and an antithetic Monte Carlo over five hundred thousand paths returned 4.30408 and 3.59687 against the closed forms in (2).