A Digital Option With Negative Vega, and the Exact Strike Where the Sign Turns

More volatility is worth more is a theorem about convex payoffs, and a step function is not convex. An at-the-money cash-or-nothing digital falls from 46.02 to 42.07 cents when the swing doubles, and the sensitivity is positive only below X exp(-(r + sigma squared over two)(T-t)). The cap on the payoff is only half the explanation; the falling median is the other half.

A share sits at 100. Someone offers you a ticket that pays exactly one dollar if the share finishes the year above 100, and nothing otherwise. Interest rates are zero and the share swings 20 percent a year, so the ticket is worth 46.02 cents. Now double the swing to 40 percent. What happens to the price?

It falls, to 42.07 cents. The extra volatility cost the holder about four cents, and no parameter choice reverses that at this strike. The reflex says otherwise, and the reflex is one of the most reliable in derivatives: more volatility means more value. For a vanilla call that is a theorem. For this contract it is false in and at the money, and true only well out of the money.

The contract, precisely

Everything below concerns one specific instrument, and the specificity matters because the sign of the answer changes if you switch contract.

Cash-or-nothing digital call — the contract under discussion

Pays a fixed notional, taken here to be one dollar, if $S(T) > X$ and zero otherwise. Under lognormal dynamics its value at time $t$ with $\tau = T - t$ is $e^{-r\tau}\,\Phi(d_2)$ where $d_2 = \bigl[\ln(S/X) + (r - \tfrac{1}{2}\sigma^2)\tau\bigr] / (\sigma\sqrt{\tau})$ .

The payoff is bounded above by the notional. That is the structural fact everything else hangs from, and it is worth noticing how different it makes this instrument from a call. Volatility buys a vanilla holder a chance at a bigger payment. Here there is no bigger payment to be had, so extra volatility can only move probability around.

At the money the answer is forced

Set $S = X$ and $r = 0$ . The logarithm vanishes and $d_2$ collapses to a single term:

c \;=\; \Phi\!\left(-\frac{\sigma\sqrt{\tau}}{2}\right).

(1)

The normal cumulative function is increasing, its argument here is decreasing in $\sigma$ , so the price is decreasing in volatility for every volatility and every maturity. At 20 percent over a year it is $\Phi(-0.10)$ , between 0.4600 and 0.4602. At 40 percent it is $\Phi(-0.20)$ , between 0.4200 and 0.4208. The difference is 3.943 cents.

Equation (1) also explains why an at-the-money digital is worth less than fifty cents, which surprises people who expect a coin flip. With a zero rate the risk-neutral expectation of $S(T)$ is pinned at today's price, and pinning a mean of a lognormal pushes its median below that mean. Writing $S(T) = S\exp(-\tfrac{1}{2}\sigma^2\tau + \sigma\sqrt{\tau}\,Z)$ , the share finishes above its start exactly when

Z \;>\; \frac{\sigma\sqrt{\tau}}{2},

(2)

so the probability of finishing up is strictly below a half and shrinks as the swing grows. The median finishing price is $S e^{-\sigma^2\tau/2}$ , which runs 0.9802, 0.9231 and 0.8353 times the starting price at swings of 20, 40 and 60 percent. More volatility drags the typical outcome further below the average, and the digital is paid on the typical outcome rather than the average one.

Fig. 1 — Both densities have mean 100. The wider one has more mass far above the strike, but the payoff is capped there, and its median has slid further left.

Where the sign flips, exactly

Away from the money the answer is a genuine calculation rather than a monotonicity remark. Differentiating the price gives $\partial c/\partial\sigma = e^{-r\tau}\varphi(d_2)\,\partial d_2/\partial\sigma$ , and since a normal density is strictly positive the sign is carried entirely by the second factor. Write $A = \ln(S/X) + r\tau$ so that $d_2 = A/(\sigma\sqrt{\tau}) - \sigma\sqrt{\tau}/2$ . Then

\frac{\partial d_2}{\partial\sigma} \;=\; -\frac{A}{\sigma^2\sqrt{\tau}} \;-\; \frac{\sqrt{\tau}}{2}.

(3)

Both terms are explicit, and setting the sum to zero gives the boundary in closed form:

\frac{\partial c}{\partial\sigma} > 0 \iff S \;<\; X\exp\!\bigl(-(r + \tfrac{1}{2}\sigma^{2})(T-t)\bigr).

(4)

At the money $A = 0$ , so (3) reduces to $-\sqrt{\tau}/2$ , which for one year is exactly $-\tfrac{1}{2}$ . That is negative for every volatility and every maturity, which is what makes the at-the-money case airtight rather than parameter-dependent. And look how tight the boundary in (4) is: with a zero rate, a 20 percent swing and one year, it sits at 0.980199 times the strike. The sensitivity is negative from 2 percent below the strike all the way up, and only in that thin band and below does volatility help.

Fig. 2 — The same instrument, two strikes, opposite slopes. Neither curve is an anomaly; the boundary in equation (4) separates them.

The cap is the wrong explanation

There is a one-line argument in circulation that gets the right answer here for the wrong reason, and it is worth dismantling because it will mislead you on the next contract. The argument runs: the payoff is capped, therefore extra volatility cannot buy a larger payment, therefore the price falls.

That is not a theorem. A digital struck far above the share has a capped payoff and a strictly positive sensitivity to volatility. Take the share at 100 against a strike of 150. At a 20 percent swing the ticket is worth 1.670 cents; at 40 percent it is worth 11.244 cents, nearly seven times as much. Nothing about the cap changed.

The cap does real work, but only half the work. It explains why there is no upside to be gained from a wider distribution. What makes the sensitivity strictly negative at and in the money is the falling median from (2), and that is a separate fact about lognormal dynamics with a pinned mean. Drop either half of the argument and you can construct a counterexample to what remains.

Why the vanilla intuition does not transfer

A vanilla call has volatility sensitivity $S\sqrt{\tau}\,\varphi(d_1)$ , a share price multiplied by a root of time multiplied by a density. Every factor is positive, so the sensitivity is never negative, at any strike and any maturity. There is no thin band and no boundary condition to remember.

The reason runs deeper than the formula. A vanilla payoff $\max(S-X,0)$ is a convex function of the terminal price, and a convex function has larger expectation under a wider distribution with the same mean. That is the general statement the desk intuition is really about. A step function is not convex, so the theorem simply has nothing to say, and neither does the intuition built on it.

A cleaner way to see the same thing: a digital is the limit of a call spread, long a call at $X - h$ and short one at $X + h$ , scaled by $1/2h$ . Its volatility sensitivity is therefore a differenceof two positive quantities, and a difference of positive numbers has no sign. Whichever of the two vanilla sensitivities is larger decides the answer, and near the money that is the one you are short.

The practical residue is small and useful. A trader long an at-the-money digital is short volatility, whatever the position looks like on a risk sheet that only knows about calls. The exposure changes sign a couple of percent below the strike, and it changes sign again if the rate is large enough to move the boundary in (4), which is why the formula and not the story is what belongs on the sheet.

Sources and further reading

Wikipedia: Binary option for the cash-or-nothing payoff, and Greeks (finance) for the vanilla sensitivities quoted above.
Wikipedia: Log-normal distribution, where the gap between the mean and the median is the mechanism behind equation (2).
Fischer Black and Myron Scholes, “The Pricing of Options and Corporate Liabilities”, Journal of Political Economy81 (1973), 637–654, cited for the model itself.

The sign rule in (4) was checked against a grid of 2,340 parameter sets in $r$ , $\tau$ , $\sigma$ , $X$ and $S$ before publication, with no disagreements, and the prices were reproduced on a binomial lattice that contains no normal distribution anywhere in its code path.