# Delta of binary option

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?

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. • 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. 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 $$\Delta_t=\frac{\partial c_t}{\partial S_t}=\frac{e^{-r(T-t)}N'(d_2)}{\sigma S_t \sqrt{T-t}}$$ Derivation 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 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.

• 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? Feb 13, 2016 at 10:58
• @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. Feb 13, 2016 at 18:12
• @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. Feb 13, 2016 at 18:14
• 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 $\varepsilon$? 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 !

• @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? Feb 13, 2016 at 10:57
• Straight Answer to this would be :- Infinite / or Not defined, if that helps 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