When a Gap Option Can Have a Negative Price

Summary

The document examines a gap option that pays the difference between the underlying price and one strike only when the price exceeds a different, lower trigger strike. Because the payoff can be negative whenever the trigger is crossed but the underlying remains below the higher strike, its expected value need not be positive. The answer decomposes the payoff into a vanilla call struck at the trigger and a short position in a digital call at that same strike.

This yields a pricing condition: the magnitude of the negative digital component must exceed the vanilla call value for the total premium to be negative. Under a Black–Scholes setting, the document gives a closed-form expression using the forward price, volatility, time to expiry, and normal distribution terms. It notes that a sufficiently high higher strike can make the premium negative. The result concerns the stated payoff and model assumptions; market skew can affect the condition when using market prices.

Key ideas

• A gap option can have a negative payoff after its trigger is crossed.
• Its payoff decomposes into a vanilla call and a short digital call at the trigger strike.
• The price is negative when the digital component's magnitude exceeds the call value.
• A Black–Scholes setting provides a closed-form premium expression.
• Market skew can affect whether the negative-price condition holds.

Tags

# Can the price of a gap option be negative?


# Can the price of a gap option be negative?












Consider a gap option whose payoff is

$$f(x)=(x-K_1)\mathbb{1}_{\left(x>K_2\right)},$$

where $K_1>K_2$ and $\mathbb{1}_{A}$ is the indicator function of the event A. Can its price, $\mathbb{E}^\mathbb{Q}\left[f(S_T)\right]$, be negative? What about the case when $S_t$ is a geometric Brownian motion (B&S)? In the latter case there should be a closed form expression

## Answer by river_rat (score 7, accepted)

https://quant.stackexchange.com/a/84049

You can decompose this payoff into a call struck at $K_2$ and $K_2-K_1$ binary calls with strike at $K_2$. This shows that the gap option has undiscounted premium $$F\Phi(d_1) - K_1\Phi(d_2)$$ where $d_{1,2} = \frac{\ln\left(\frac{F}{K_2}\right) \pm \frac{1}{2}\sigma^2T}{\sigma\sqrt{T}}$ and then if $K_1$ is sufficiently large then the premium can be negative

## Answer by Frido (score 7)

https://quant.stackexchange.com/a/84050

To add to @river_rat 's answer:

Your payoff can be written as $f(x) = (x-K_2) 1_{x>K_2} + (K_2 - K_1) 1_{x>K_2}$.

Hence $E[f(x)] = C(K_2) + (K_2-K_1) D(K_2)$ where $C(K_2)$ is a vanilla call struck at $K_2$ and $D(K_2)$ is a digital (binary) struck at $K_2$. So the price is negative iff $(K_1-K_2) D(K_2) > C(K_2)$. Given the market skew, or a model if there is no skew, you should be able to see if the condition holds. And in particular, as mentioned by river_rat, if $K_1 \to \infty$ the condition will hold.

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This summary was written by Stratmill's research agent from the original; it is not a copy of the source.