# Pricing binary options under volatility smile

I was asked to show that the price of a digital/binary option $$D$$ while a volatility smile $$\sigma(K)$$ is present is given by

$$D= \exp(-rT)( \Phi(d_2) - K \sqrt{T} \phi(d_2) \sigma ' (K))$$

Where $$\Phi$$ is the standard normal CDF and $$\phi$$ is the standard normal pdf. I started with showing that the price of $$D$$ under B-S is given by $$-\frac{\partial C}{\partial K}$$, where $$C$$ is the standard B-S pricing for a call option. Additionally, I have also shown that this can also be expressed as (with no vol smile) $$-\frac{\partial C}{\partial K} = \Phi(d_2)$$

Finally, I was able to show the following $$D = -\frac{\partial C}{\partial K} - \frac{\partial C}{\partial \sigma} \frac{\partial \sigma}{\partial K}$$

Now, the $$\frac{\partial C}{\partial \sigma}$$ term is Vega, which is explicitly given by $$\frac{\partial C}{\partial \sigma} = S\phi(d_1)\sqrt{T}$$

And now I have absolutely no idea what to do, because the question involves a $$K\phi(d_2)$$ term, but I have an $$S\phi(d1)$$ term. Are they related somehow?

• Also $-\frac{\partial C}{\partial K} = e^{-rT}\Phi(d_2)$ which along with my answer below gets the form for $D$ in your first line.
– RRL
Feb 7, 2022 at 5:33

Note that $$d_1 = d_2 +\sigma \sqrt{T}$$, and

$$\tag{1}S\sqrt{T}\phi(d_1) = \frac{S\sqrt{T}}{\sqrt{2\pi}}e^{-(d_2 + \sigma \sqrt{T})^2/2} = \frac{S\sqrt{T}}{\sqrt{2\pi}}e^{-d_2^2/2}e^{-d_2\sigma\sqrt{T}}e^{-\sigma^2T/2}\\ = S\sqrt{T}\phi(d_2)e^{-d_2\sigma\sqrt{T}}e^{-\sigma^2T/2}$$

Since

$$d_2 = \frac{\log \frac{S}{K}+rT -\frac{1}{2}\sigma^2T}{\sigma \sqrt{T}},$$

we have

$$\tag{2}e^{-d_2\sigma\sqrt{T}}= \frac{K}{S}e^{-rT}e^{\sigma^2T/2}$$

Substituting into (1) with (2) we get

$$S\sqrt{T}\phi(d_1) = e^{-rT}K\sqrt{T} \phi(d_2)$$

Note also that for the Black-Scholes price $$C = S \Phi(d_1)-Ke^{-rT}\Phi(d_2)$$ we have $$\tag{3}\frac{\partial C}{\partial K}= - e^{-rT}\Phi(d_2)$$

The derivation is

$$\tag{4}\frac{\partial C}{\partial K}= S\Phi'(d_1)\frac{\partial d_1}{\partial K} - e^{-rT}\Phi(d_2) - Ke^{-rT}\Phi'(d_2)\frac{\partial d_2}{\partial K}\\ = -e^{-rT}\Phi(d_2) + S\phi(d_1)\frac{\partial d_1}{\partial K} - Ke^{-rT}\phi(d_2)\frac{\partial d_2}{\partial K}$$

Since $$\frac{\partial d_1}{\partial K} = \frac{\partial d_2}{\partial K}$$ and using the previous result $$S\phi(d_1) = e^{-rT}K \phi(d_2)$$, the second and third terms on the RHS of (4) cancel, and we get the desired result (3).

• This is very clear. Thank you very much. Feb 7, 2022 at 6:43