Black Scholes Greeks

We derive the formulae for the Price and Greeks (derivatives with respect to inputs) of the European options under the Black-Scholes assumptions.

Dynamics of the Underlying Asset

We derive the formula for the price of a European option under Black-Scholes' assumptions, which assumes the following dynamics for the underlying under the risk neutral measure (Note we extended this formula under the summary to a dividend paying stock, essentially replacing r with \( r_d-r_f\) ):

$$ d S_t = r S_t dt + \sigma S_t d W_t $$

The solution of which is

$$ S_t = S_0 e^{\sigma W_t + r t- \frac{\sigma^2 t}{2} } $$

As can be easily verified by applying Ito's lemma to \( y= ln{S_t} \) and integrating the resulting equation from 0 to t:

$$ y = ln{S_t} $$ $$ dy= \frac{\partial y}{\partial S_t} d S_t + \frac{1}{2} \frac{\partial^2 y}{\partial S_t^2} d S_t^2 $$ $$ d ln{S_t} = \frac{1}{S_t} d S_t - \frac{1}{2} \frac{1}{S_t^2} d S_t^2 $$ $$ \quad = \frac{1}{S_t} \left( r S_t dt + \sigma S_t d W_t \right)- \frac{1}{2} \frac{1}{S_t^2} \sigma^2 S_t^2 dt $$ $$ \quad = r dt + \sigma d W_t - \frac{\sigma^2}{2}dt $$ $$ \int_0^t{d ln{S_u}}=r \int_0^t {dt}+ \sigma \int_0^t {d W_u} - \frac{\sigma^2}{2}\int_0^t{du} $$ $$ ln{S_t}-ln{S_0}=rt - \frac{\sigma^2}{2}t +\sigma W_t $$ $$ ln{\frac{S_t}{S_0}}=\sigma W_t +r t- \frac{\sigma^2}{2}t $$ $$ S_t=S_0 e^{r t- \frac{\sigma^2}{2}t + \sigma W_t}=e^{ln S_0 + r t- \frac{\sigma^2}{2}t + \sigma W_t } $$

Identifying the Disribution of the Underlying Asset

Now we derive the formula for the probability density of \( S_t \). Notice that \(y = ln{S_t} \) is normally distributed, and its mean and variance can be calculated as:

$$ y = ln{S_t} = ln S_0 + r t- \frac{\sigma^2}{2}t + \sigma W_t $$ $$ E \left[ y \right] = E \left[ ln S_0 + r t- \frac{\sigma^2}{2}t + \sigma W_t \right] $$ $$ \quad = ln S_0 + r t- \frac{\sigma^2}{2}t $$ $$ V \left[ y \right] = V \left[ ln S_0 + r t- \frac{\sigma^2}{2}t + \sigma W_t \right] $$ $$ \quad = \sigma^2 \int_0^t {du} = \sigma^2 t$$

And we can now derive the probability distribution of \( S_t \) by variable transformation:

$$ f_{S_t}(S)=f_{ln S_t}(ln S) \left| \frac{d ln S}{d S} \right| $$ $$ f_{S_t}(S)=f_{ln S_t}(ln S) \frac{1}{S} $$ $$ \quad=\frac{1}{\sqrt{2\pi V[ln S]}} {e^{-\frac{{(ln S-E[ln S])}^{2}}{2 V[ln S]}}} \frac{1}{S} $$ $$ \quad=\frac{1}{S \sqrt{2\pi \sigma^2 t}} {e^{-\frac{{(ln S-ln S_0 - r t+ \frac{\sigma^2 t}{2})}^{2}}{2 \sigma^2 t}}} $$ $$ \quad=\frac{1}{S \sigma \sqrt{2\pi t}} {e^{-\frac{{(ln S-ln S_0 - r t + \frac{\sigma^2 t}{2})}^{2}}{2 \sigma^2 t}}} $$


We are now ready to derive the formula for the price of an option. To simplify the presentation we derive the formula for the price of a call option, and we assume that the price is calculated at t=0. The formula for a put can be easily derived by following the same steps but changing the payoff.

The call option pays, at maturity of the option, the difference between the price of the underlying and the strike if the difference is positive:

$$ Payoff={(S_T-K)}^{+} $$

The present value of the payoff can be calculated using risk neutral valuation approach as follow:

$$ Price_0=e^{-r T} E^Q \left[ {(S_T-K)}^{+} \right] $$ $$ \quad =e^{-r T} \left( E^Q \left[ S_T 1_{S_T > K} \right] - E^Q \left[ K 1_{S_T > K} \right] \right)$$

We now simplify the two sub-components separately, whereby 'simplify' means something we can calculate 'analytically'.The aim is to use variable transformation to translate the above expectations into standard normal probabilities so that one can use the normal probability lookup tables or the standard normal distribution function implemented in one's favourite software.

We start with the first component (get ready for some practice with the variables transformation!):

$$ E^Q \left[ S_T 1_{S_T > K} \right] =\int_K^{\infty} {S f_{S_t}(S) dS}$$ $$ E^Q \left[ S_T 1_{S_T > K} \right] =\int_K^{\infty} {S \frac{1}{S \sigma \sqrt{2\pi T}} {e^{-\frac{{(ln S-ln S_0 - r T+ \frac{\sigma^2 T}{2})}^{2}}{2 \sigma^2 T}}} dS}$$

Let \( x= \frac{ln S-ln S_0 - r T+ \frac{\sigma^2 T}{2}}{\sigma \sqrt{T}} \), then \( \frac{dx}{dS}= \frac{1}{S \sigma \sqrt{T}} \Rightarrow dS= S \sigma \sqrt{T} dx= e^{x \sigma \sqrt{T} + ln S_0 + r T- \frac{\sigma^2 T}{2}} \sigma \sqrt{T} dx \), \(S=K \Rightarrow x= \frac{ln K-ln S_0 -r T+ \frac{\sigma^2 T}{2}}{\sigma \sqrt{T}} \), and \(S= \infty \Rightarrow x= \infty \). Making the substitutions,

$$ E^Q \left[ S_T 1_{S_T > K} \right] =\int_K^{\infty} {\frac{1}{\sigma \sqrt{2\pi T}} {e^{-\frac{{(ln S-ln S_0 - r T+ \frac{\sigma^2 T}{2})}^{2}}{2 \sigma^2 T}}} dS}$$ $$ \quad =\int_{\frac{ln K-ln S_0 - r T+ \frac{\sigma^2 T}{2}}{\sigma \sqrt{T}}}^{\infty} {\frac{1}{\sigma \sqrt{2\pi T}} e^{-\frac{x^2}{2}} e^{x \sigma \sqrt{T} + ln S_0 + rT- \frac{\sigma^2 T}{2}} \sigma \sqrt{T} dx}$$ $$ \quad =e^{ln S_0+rT} \int_{\frac{ln K-ln S_0 - rT + \frac{\sigma^2 T}{2}}{\sigma \sqrt{T}}}^{\infty} {\frac{1}{\sqrt{2\pi}} e^{-\frac{x^2 -2 x \sigma \sqrt{T} +\sigma ^2 T}{2}} dx}$$ $$ \quad = S_0 e^{rT}\int_{\frac{ln K-ln S_0 - rT+ \frac{\sigma^2 T}{2}}{\sigma \sqrt{T}}}^{\infty} {\frac{1}{\sqrt{2\pi}} e^{-\frac{{(x - \sigma \sqrt{T})}^2}{2}} dx}$$

Now to simplify further, let \( z= x -\sigma \sqrt{T} \Rightarrow dz=dx\) and the lower limit becomes \( x = \frac{ln K-ln S_0 -rT+ \frac{\sigma^2 T}{2}}{\sigma \sqrt{T}} \Rightarrow z= \frac{ln K-ln S_0 -rT + \frac{\sigma^2 T}{2}}{\sigma \sqrt{T}} -\sigma \sqrt{T}=\frac{ln K-ln S_0 -rT- \frac{\sigma^2 T}{2}}{\sigma \sqrt{T}} \). Hence

$$ E^Q \left[ S_T 1_{S_T > K} \right]=S_0 e^{rT} \int_{\frac{ln K-ln S_0 -rT-\frac{\sigma^2 T}{2}}{\sigma \sqrt{T}}}^{\infty} {\frac{1}{\sqrt{2\pi}} e^{-\frac{{z}^2}{2}} dz}$$ $$ \quad \quad= S_0 e^{rT} \left( 1 -N \left[ \frac{ln K-ln S_0 -rT - \frac{\sigma^2 T}{2}}{\sigma \sqrt{T}} \right] \right) $$ $$ \quad \quad = S_0 e^{rT} N \left[ \frac{ln S_0 -ln K +rT+ \frac{\sigma^2 T}{2}}{\sigma \sqrt{T}} \right] $$

Now lets calculate the second component

$$ E^Q \left[ K 1_{S_T > K} \right]= K E^Q \left[ 1_{S_T > K} \right] $$ $$ \quad \quad= K P \left[ S_T > K \right] $$ $$ \quad \quad= K P \left[ e^{ln S_0 +rT- \frac{\sigma^2}{2}T + \sigma W_T } > K \right] $$ $$ \quad \quad= K P \left[ \sigma W_T > ln K -ln S_0 -rT+ \frac{\sigma^2}{2}T \right] $$ $$ \quad \quad= K P \left[ \frac{W_T}{\sqrt{T}} > \frac{ln K -ln S_0 -rT+ \frac{\sigma^2}{2}T}{\sigma \sqrt{T}} \right] $$ $$ \quad \quad= K \left( 1- N \left[ \frac{ln K -ln S_0 -rT+ \frac{\sigma^2}{2}T}{\sigma \sqrt{T}} \right] \right) $$ $$ \quad \quad= K N \left[ \frac{ln S_0 -ln K +rT- \frac{\sigma^2}{2}T}{\sigma \sqrt{T}} \right] $$

Putting it all together , we get the famous Black-Scholes' formula:

$$ Price = e^{-r T} \left( E^Q \left[ S_T 1_{S_T > K} \right] - E^Q \left[ K 1_{S_T > K} \right] \right)$$ $$ \quad \quad = e^{-r T} \left( S_0 e^{rT} N \left[ \frac{ln S_0 -ln K +rT+ \frac{\sigma^2 T}{2}}{\sigma \sqrt{T}} \right] - K N \left[ \frac{ln S_0 -ln K +rT- \frac{\sigma^2}{2}T}{\sigma \sqrt{T}} \right] \right)$$ $$ \quad \quad = S_0 N \left[ d_1 \right] - K e^{-rT}N \left[ d_2 \right]$$