What is the Delta of an at-the-money binary option with a payout $0$ at $S(T)<100$ dollars, and payout of $1$ at $S(T)>100$ dollars, as it approaches expiry?

This is from a sample interview exam. I understand that Delta essentially measures the change in the derivative price relative to the change in the asset price, as trading on the open market.

How do I actually go about computing Delta for a particular situation like the one above? I've been unable to find a formula for it on Google which is a bit weird? My naive guess is that the answer should be 0.5 but I'm not sure why?


4 Answers 4


If it wasn't clear from the previous answers, the answer they want is that the delta becomes infinite. That's because a tiny move in the stock will change the payout by $100 so your delta hedge must be enormous.

  • $\begingroup$ That being said, you can hedge with something other than stock, like a call spread that is ITM at \$100. An ideal hedge would be 100 call spreads from \$99.99/\$100. Since that doesn't exist and fees will kill you, something like the \$99/\$100 spread would work hedge fully, though it will cost more than your binary option. $\endgroup$ Oct 19, 2020 at 16:00

The value of European binary call, paying \$1 if $S_T > K$ or nothing otherwise, is $$c_t=e^{-r(T-t)}N(d_2)$$ where, $d_2=\frac{ln(S_t/K)+(r-\sigma^2/2)(T-t)}{\sigma \sqrt{T-t}}$

Delta of your binary call option is $$\Delta_t=\frac{\partial c_t}{\partial S_t}=\frac{e^{-r(T-t)}N'(d_2)}{\sigma S_t \sqrt{T-t}}$$


We need to compute $$\Delta_t=\frac{\partial c_t}{\partial S_t}$$

$$\frac{\partial c_t}{\partial S_t}=\frac{\partial}{\partial S_t}\bigg(e^{-r(T-t)}N(d_2)\bigg)=e^{-r(T-t)}\frac{\partial}{\partial S_t}N(d_2)$$

$$\frac{\partial}{\partial S_t}N(d_2)=\frac{\partial}{\partial S_t} \int_{-\infty}^{d_2} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}dx$$

where $d_2=f(S_t)$. Using Leibniz integral rule

$$ \frac{\mathrm{d}}{\mathrm{d}x} \left (\int_{a(x)}^{b(x)}f(x,t)\,\mathrm{d}t \right) = f(x,b(x))\cdot b'(x) - f(x,a(x))\cdot a'(x) + \int_{a(x)}^{b(x)} \frac{\partial}{\partial x}f(x,t)\; \mathrm{d}t. $$

So, $$\frac{\partial}{\partial S_t} \int_{-\infty}^{d_2} \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}dx=\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}d_2^2} \frac{\partial}{\partial S_t} (d_2)$$

You may check it yourself that $$\frac{\partial d_2}{\partial S_t}=\frac{1}{S_t \sigma \sqrt{T-t}}$$

Putting all the results together

$$\frac{\partial c_t}{\partial S_t}=\frac{e^{-r(T-t)}N'(d_2)}{\sigma S_t \sqrt{T-t}}$$ where $N'(d_2)$ denote the standard normal probability density function,

Relationship between Binary option delta and Time to expiry @dm63 already provided a brief answer to your question how delta will respond as option will approach its expiry, below I have shown more accurate relationship enter image description here
Ref: http://www.binaryoptions.com/binary-option-greeks/binary-call-option-delta

You can see as the time to expiry decrease the delta of an at-the-money option approaches to infinity. Because a small change in stock price ($\epsilon$), assume $S_t=K$ and option is near maturity, will cause the option payoff to change its value by \$1 (as information provided in OP). So, option delta $\Delta_t= \frac{1}{\epsilon} \to \infty$. You may also check this result from formula derived above.

  • $\begingroup$ Do you have a reference where I could read up on this? I'm not familiar with all the notation (applying for finance jobs from a maths degree). Also, does this mean that the answer they want is a formula rather than a number? $\endgroup$
    – user11128
    Feb 13, 2016 at 10:58
  • $\begingroup$ @user11128 You can find computation of delta in any standard text book and there(Quant.SE) is sufficient information on price of binary call option. I just followed the two and provided you entire formula for delta of Binary option. $\endgroup$
    – Neeraj
    Feb 13, 2016 at 18:12
  • $\begingroup$ @user11128 I just used most basics and standard notations. Since, you have been asked such question in interview so I was expecting such basic knowledge from you. $\endgroup$
    – Neeraj
    Feb 13, 2016 at 18:14
  • $\begingroup$ Hi there! The deduction here is brilliant, however I do have one question. Is it not in fact the case that $\Delta_t\approx \frac{1}{\sqrt{T-t}}\to\infty$ as $t\to T$? It seems to me that $N'(d_2)$ actually close to $1$ when $S_t$ is close to $K$, so the blow up is really in $t$ and not $\varepsilon$? $\endgroup$
    – Ellya
    Jan 28, 2021 at 11:08

Delta of a digital (or binary) option is like the normal distribution probability function , approaching 0 at far OTM / ITM conditions and representing a very high peak at ATM.

The peak at ATM approaches infinity as we approach the maturity. This is never 0.5 like a vanilla option since the payoff never simulates the payoff of the underlying.

If you want to have an approximation for delta at ATM , I'd suggest you to either use longer dated options , or to use a spread to smoothen out the delta at ATM. That's how the traders smoothen out the deltas of digital products while hedging. That structure may be slightly costly though !

  • $\begingroup$ @Jonathon Thanks for the reply. I wrote the question exactly as it appears on the sample test. No other information was provided. How can I answer this? Is a number answer not possible? $\endgroup$
    – user11128
    Feb 13, 2016 at 10:57
  • 1
    $\begingroup$ Straight Answer to this would be :- Infinite / or Not defined, if that helps $\endgroup$
    – HyperVol
    Feb 14, 2016 at 12:57

A fun thing about binary options is that ATM close to expiration the delta turns into a Dirac Delta which is a function originally created in theoretical physics.

Nassim Taleb explains it in pg 286 of this link: http://docs.finance.free.fr/Options/Dynamic_Hedging-Taleb.pdf


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