# 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?

• It depends exactly how you want to define delta - if you want it to be the change in value for a \$1 change in underlying, then use$h=1$in your finite difference. If you do this, you'll get the answer you want. On the other hand, If you're looking at eurusd, a \$1 change is not going to happen, so delta defined this way is meaningless - it makes more sense to use a value of $h$ in the finite difference that is applicable to the underlying you're interested in. Mathematically, the gradient of a step function is a dirac delta function, so infinite delta is actually correct (but useless). – will Sep 29 '16 at 7:21

$$V_{t + 1} \approx V_t + \Delta_t \left( S_{t + 1} - S_t \right)$$
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 $[0, \infty)$.
• +1 for this answer. @Alex Ockenden, claiming that $\delta V = \Delta \delta S$ (whence for $\delta S = 1(€)$, $\delta V = \Delta(€)$ and the interpretation you gave) is essentially a first order approximation. – Quantuple Sep 29 '16 at 7:17