The Strike Quietly Absorbed a Square Root

A European call, one year, on a stock at 12. Its payoff is not $\max(S - X, 0)$ but $\max(S^2 - X, 0)$, and the strike written into the contract is 100. The rate is zero and the volatility is 20%. Where does this thing start paying?

The answer is 10, and it is worth 53.8168. Getting the first number wrong loses the whole of the second one.

The reflex, and how much it costs

Read the contract, see the number 100, and conclude that the option pays once the stock reaches 100. A stock at 12 would have to grow more than eightfold in a year. Under 20% volatility that is not going to happen, so the option is worthless and you can hand the sheet back.

Priced that way, with the kink placed at the written strike, the option comes out at $5 \times 10^{-23}$ dollars, and the probability of ever exercising is below one part in $10^{20}$. So the reflex does not produce an approximation. It produces zero where the answer is fifty-four dollars.

The strike is a level of the payoff, not of the stock

The mistake is a category error rather than an arithmetic one. The number 100 is a level of $S^2$, and the stock is $S$. For $\alpha > 0$ the map $s \mapsto s^{\alpha}$ is strictly increasing, so it preserves inequalities, and the two exercise events are the same event written twice:

Nothing about the exercise region has changed shape. Only the threshold has moved, from 100 to 10, and the stock at 12 is not eight times away from paying, it is already in the money by 44.

Why it is still the ordinary formula

Once the effective strike is in place, the rest follows from a single observation: $S^{\alpha}$ is itself lognormal. If $\ln S_T$ is normal then $\alpha \ln S_T$ is normal too, with $\alpha$ times the standard deviation. That gives the transformed asset a volatility of $\alpha\sigma$, which is the only place $\alpha$ shows up in the distance between the two normal arguments: they differ by $\alpha\sigma\sqrt{T-t}$ rather than by $\sigma\sqrt{T-t}$.

The transformed asset also has its own drift, and this is where the second trap lives. The lognormal moment gives $\mathbb{E}\!\left[S_T^{\alpha}\right] = S^{\alpha}\exp\!\left(T\left[\alpha r + \tfrac12\alpha(\alpha-1)\sigma^2\right]\right)$, which is not $S^{\alpha}e^{rT}$. The transformed asset does not grow at the riskless rate, so define

Then $S^{\alpha}e^{m(T-t)}$ is the discounted risk-neutral mean of the terminal transformed asset, which is precisely the role the spot price plays in the ordinary formula. Drop it in:

The strike leg is the second half of the standard error. $X$ is a fixed number of dollars, paid at expiry if the option finishes in the money, and it has no idea that a transform happened. Its present value is $e^{-r(T-t)}X$ and it discounts at $r$, never at $r + m$. Feeding the transformed rate into the strike leg is the mistake that survives a careful reader, because everything around it looks consistent.

The numbers, and where they came from

At $S = 12$, $X = 100$, $\alpha = 2$, $r = 0$, $\sigma = 20\%$ and $T = 1$: the effective strike is 10, the growth term is $m = 0.04$, and $d_1 = 1.21160778$ with $d_2 = 0.81160778$. Equation (3) returns 53.816802. A simulation over 4,000,000 paths returns 53.810650 with a standard error of 0.029045, so the closed form and the simulation agree inside a quarter of one standard error.

The position of the kink was measured rather than assumed, by bisecting the payoff itself. It lands on 10.000000000 for $\max(S^2 - 100, 0)$, on 4.641588834 for $\max(S^3 - 100, 0)$, on 8 for $\max(S^2 - 64, 0)$ and on 3 for $\max(S^4 - 81, 0)$. Each time it is the appropriate root of the written strike.

Above intrinsic, and not steeper than it

There is a tidy remark that usually closes this problem: for $\alpha > 1$ the value curve sits above intrinsic and is steeper than intrinsic. The first half is true everywhere and worth stating. At 9 the value is 8.0502 against nothing; at 11, 33.7193 against 21; at 12, 53.8168 against 44; at 15, 134.5436 against 125; at 20, 316.3279 against 300; at 30, 836.7297 against 800.

The second half is false at the spot the problem asks about. Differentiate equation (3):

So the ratio of the two slopes is $e^{m(T-t)}N(d_1)$, a product of one factor above one and one factor below it. At $S = 12$ that is $1.040811 \times 0.8872 = 0.9234$, so the value curve is shallower than the payoff: a slope of 22.161 against 24. It only becomes steeper once $N(d_1)$ climbs past $e^{-mT} = 0.9608$, which happens near a stock of 13.4, and by 20 the value slope is 41.629 against 40. Far out in the money the ratio settles on $e^{mT} = 1.040811$.

The distinction is not pedantry. A statement about the level of the value and a statement about its delta are different statements, and here they disagree across a range of spots that includes the one being priced. Anyone hedging a power call from the steepness claim would be short too much stock over exactly the region where the contract is interesting.

Where equation (3) stops working

At $\alpha = 1$ everything collapses correctly. The growth term $m = (1-1)(\cdot) = 0$, the effective strike is $X^{1} = X$, and the two arguments differ by $\sigma\sqrt{T-t}$, so equation (3) is the ordinary call. A transformation that adds nothing when there is nothing to transform is the right kind of degenerate.

Negative $\alpha$ breaks equation (1) rather than equation (3). For $\alpha < 0$ the map $s \mapsto s^{\alpha}$ is decreasing, so the exercise event becomes $\left\{S_T < X^{1/\alpha}\right\}$: the inequality flips and the contract is put-like. At $\alpha = 0$ the payoff is constant and the question dissolves. Everything in this article assumes $\alpha > 0$, which is exactly the condition equation (1) needs.

The remaining hazard is model risk in $m$, and it is bigger than it looks. Since $m = (\alpha-1)\left(r + \tfrac12\alpha\sigma^2\right)$ grows like $\tfrac12\alpha^2\sigma^2$, the growth term is quadratic in the exponent. At $\alpha = 2$ and 20% volatility it is a mild 4%, and tripling $\alpha$ makes it an order of magnitude larger. Volatility now enters the price twice, through $\alpha\sigma$ in the arguments and through $m$ in the level, so a power call is far more sensitive to a wrong volatility input than a vanilla is. Equation (4) makes the hedging side of that concrete: the delta grows like $S^{\alpha-1}$, without bound, so the assumption of costless continuous rebalancing is carrying much more weight here than in the ordinary formula.

Sources and further reading

• The property that makes equation (1) an equality of events Monotonic function
• Why $S^{\alpha}$ stays in the same family Log-normal distribution
• The formula equation (3) is a change of variables away from Black-Scholes model
• The quantity equation (4) computes, and the claim it refutes Greeks
• The wider family this contract belongs to Exotic option

The kink was located by bisection at four different pairs of strike and exponent rather than assumed, the growth term was re-derived from the lognormal moment and cross-checked against the simulated transformed mean, and the value was compared with intrinsic at six spots. The steepness claim was then tested by finite-differencing the value at three of them, where it fails at the one the problem actually asks about, which is why this article states it in two halves.