A Hedge That Loses on Both Legs at Once

You own one-month calls struck at 110 with the share at 100, and you short 0.1452 shares against each one. If the share rallies to exactly 110 and stops, the calls expire worthless while the short has lost ten dollars a share, so the hedged position is down 2.074208 where the unhedged one would have lost only its 0.622212 premium. The worst case sits at the strike because the profit is piecewise linear with slopes of -0.1452 and +0.8548, and a rebalanced hedge on the same path loses 3.058738.

You own calls on a share. They run one month and they are struck at 110, while the share sits at 100. Rather than carry the directional exposure you short stock against them, sized by the hedge ratio the pricing model gives you. Can that leave you worse off than doing nothing?

Yes, and the move that does it is unremarkable. The share rallies to 110 and stops there. Your calls expire worthless and your short has lost ten dollars a share on the way up. Both legs lose on the same move.

The position, priced

Fix the inputs so the arithmetic can be followed: share 100, strike 110, one month, no dividends, a zero rate and a thirty percent volatility. The call costs 0.622212 and its hedge ratio is 0.1451997, so you short 14.52 of stock against each call.

Static delta hedge

One trade, sized at the outset from the hedge ratio and held to expiry without adjustment. The alternative is to resize it as the share moves, which is the more realistic policy and, on the path in this article, the more expensive one.

Write $c$ for the premium, $\Delta$ for the hedge ratio and $S_0$ for today's share price. The profit at expiry, as a function of where the share finishes, is three terms:

\Pi(S_T) = \underbrace{(S_T - X)^+}_{\text{payoff}} - \underbrace{c}_{\text{premium}} - \underbrace{\Delta\,(S_T - S_0)}_{\text{the short}}

(1)

The one move that hurts both legs

Evaluate (1) at $S_T = X = 110$ . The payoff term is zero, because a call struck at 110 with the share at exactly 110 pays nothing. The premium is gone. And the short was opened at 100 and is being closed at 110, so it has lost ten dollars on each of 0.1451997 shares:

\Pi(110) = 0 - 0.622212 - 0.1451997 \times 10 = -2.074208

(2)

Left unhedged, the same position loses the premium and not a cent more, at every terminal price at or below the strike. So the hedge multiplied the worst case by 3.3336.

Fig. 1 — At one terminal price both legs are underwater at once. Neither number is large; the point is that they add.

Why the worst case is exactly at the strike

Nothing about the strike being the worst place is a coincidence, and finding it needs no search. Below the strike the payoff term in (1) is flat, so the whole expression has slope $-\Delta$ . Above the strike the payoff has slope one, so the whole expression has slope $1 - \Delta$ :

\Pi'(S_T) = \begin{cases} -0.1452 & S_T < 110 \\[2pt] +0.8548 & S_T > 110 \end{cases}

(3)

Negative and then positive, with one kink between them. A piecewise linear function that falls and then rises is convex, and a convex function attains its minimum where its slope changes sign. Since the hedge ratio is strictly between zero and one, that sign change always happens, and it always happens at the strike. The one number in (2) that could be argued about is the hedge ratio; the location of the worst case cannot.

A 150,001-point sweep of terminal prices from 50 to 200 was run anyway, and it put the minimum at 110.0000. Brute force and the argument agree, which is the ideal outcome, because the argument is the part that generalises.

How much worse, and over what range

Setting (1) to zero on each side of the kink gives the two prices where the hedged position breaks even: $100 - c/\Delta = 95.7148$ below, and $110 + 2.074208/0.8548 = 112.4265$ above. Between them the hedge is a net cost, and the window is nearly seventeen dollars wide around a share trading at 100.

Fig. 2 — The hedged profit, with the unhedged floor drawn in as a dashed line. The hedge trades the tails for the middle, and the middle is where the share usually finishes.

The shape is worth reading in both directions. Outside the window the hedge pays: if the share collapses to zero the short returns 14.52 against a premium of 0.62, so the hedged position makes 13.90 where the unhedged one lost everything it could lose. The hedge did not create unbounded risk. It moved money out of the middle of the distribution and into the tails, and then charged 1.45 for the privilege at the one point where the option pays nothing and the share has moved the most.

The static hedge is not the weak version

A reasonable objection: nobody sizes a hedge once and walks away, and the pathology might be an artefact of refusing to adjust. It is not. Walk the share from 100 up to 110 over the month in twenty-one steps, resizing the short at each one, and the loss on the same terminal path is 3.058738 instead of 2.074208.

The mechanism is easy to see once stated. The hedge ratio rises as the share approaches the strike, from 0.145 at 100 to about 0.517 at 110 with a month gone, so a disciplined hedger keeps selling more stock at higher and higher prices, and then the option dies anyway. Every one of those sales is closed out above where it was opened. The version told in this article is the kind one.

What the hedge did do

None of this makes hedging increase risk as a general sentence. That sentence is false, and it is worth saying so plainly. The hedge does exactly what it advertises: around a share price of 100 the hedged position is insensitive to small moves, while the unhedged one is not. First order exposure is what a delta hedge removes, and it removes it.

What it cannot do is cap a loss that was already capped. A long option position has a floor built into the contract, namely the premium, so any additional position can only lower that floor. The slogan comes from the other side of the trade, where a short option position has a genuinely unbounded loss and the hedge takes the unbounded part away. Which side you are on decides whether the slogan is true.

Two edges of the result. The extra loss in (2) is exactly $\Delta\,(X - S_0)$ , which vanishes when the option is struck at the money, since the short then closes where it opened. It also vanishes far out of the money, because the hedge ratio decays like a Gaussian tail while $X - S_0$ only grows linearly. The pathology therefore has a worst strike somewhere in between, and 110 on a hundred-dollar share with a month to run is close to it. Separately, the zero rate is doing a small favour here: with a positive rate the proceeds of the short earn interest, which trims the loss without moving the kink.

Sources and further reading

The practice being examined — Delta neutral
The hedge ratio and its neighbours — Greeks (finance)
The shape that (3) describes — Piecewise linear function
Why a convex function bottoms out at its kink — Convex function
The diagram in Fig. 2 — Option (finance)

Twenty-two checks stand behind the numbers here. The sweep located the minimum rather than assuming it, both slopes in (3) were measured against their closed forms to nine decimals, the unhedged loss was confirmed never to exceed the premium at any terminal price, and the rebalanced control was rebuilt as a sum of position times price change after an earlier version of it liquidated the short with the wrong sign and reported a profit.