Why Digital Option Delta Can Become Unbounded Near Expiry
Summary
The document addresses the apparent paradox of a digital call's delta growing without bound as expiration approaches the strike. A payoff capped at one unit may seem unable to have a price sensitivity greater than that amount, but delta measures a local slope with respect to the underlying price, not the option's total payoff or its full price range.
The answer cautions that a first-order delta approximation to a price change is accurate only when both the underlying move and the time interval are small. Near expiration, a digital option's payoff changes abruptly around the strike, so its local price slope can become very steep even though the option value remains bounded. The answer states that delta is bounded by the slopes of the payoff function; for a digital call that permits an unbounded upper slope. It offers a conceptual resolution, without deriving the formula or discussing hedge instability and discrete rebalancing in detail.
Key ideas
- Delta is a local sensitivity, so it is not capped by the option's maximum payoff.
- The delta approximation works best for small underlying moves over short time intervals.
- A digital option's abrupt payoff transition near the strike can produce a steep price slope near expiry.
- For a digital call, the payoff's slope allows delta to grow without a finite upper bound.
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# Justification for Binary Option's Infinite Delta?
# Justification for Binary Option's Infinite Delta?
First time poster here. Glad to be here. I just graduated with an MSc in computational finance.
I recently read a question by another user about the delta of an at-the-money binary option as it approaches expiration. After writing a quick Matlab script I have confirmed that the delta in this situation explodes to infinity as the option approaches expiration.
How can this be valid? If we understand delta as the change in derivative price for every 1 dollar change in the price of the underlying asset, how could the delta ever exceed the payoff of the binary option? If the payoff on the derivative is 1 dollar when the price of the underlying asset exceeds the strike price, then a rational investor would be willing to pay 1 dollar AT MOST for that derivative, and would be willing to pay 0 dollars AT LEAST for that derivative. Therefore, the maximum range of fair values for the option would be 1 dollar right? How could the value of the derivative change by anything greater than 1 dollar then?
It's almost as if the pricing model broke in this case. Considering I found values for d2 that were negative, when d2 is supposed to be a probability measure that the asset price expires above the strike, I would say that the model somehow broke. Can anyone explain why it breaks in this case?
## Answer by LocalVolatility (score 2, accepted)
https://quant.stackexchange.com/a/30357
Your question is essentially the same as this. The approximation
\begin{equation} V_{t + 1} \approx V_t + \Delta_t \left( S_{t + 1} - S_t \right) \end{equation}
is only accurate when $S_{t + 1} - S_t$ and $\Delta t$ are small.
See also my answer to this question. It provides some references that show that the delta is bounded by the slopes of the payoff function, i.e. in case of a European digital call $[0, \infty)$.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.