1 The Delta
I assume most readers are familiar with expression for the delta of a call option within the Black-Scholes model:
\[ \Delta = \frac{\partial C(K,T)}{\partial S} = N(d_1) \tag{1}\]
where \(C(K,T)\) is the price of a call option with strike \(K\) and maturity \(T\), on an asset with price \(S\), \(N(x)\) is the cumulative distribution function of the standard normal distribution, and \(d_1\) is given by
\[ d_1 = \frac{\ln(S/K) + (r + \sigma^2/2)T}{\sigma\sqrt{T}}. \]
The derivation of this formula, starting from the Black-Scholes call option price, is accessible to any STEM undergraduate, with a few pieces of paper (go on, try it!).
Beginning with the Black-Scholes call option price,
\[ C(K,T) = S_{0} N(d_{1}) - Ke^{-rT} N(d_{2}), \]
differentiate once with respect to \(S_{0}\) to get the delta:
\[ \begin{align} \Delta C(K,T) &= \frac{ \partial }{ \partial S_{0} } C(K,T) \\ &= \frac{ \partial }{ \partial S_{0} } S_{0} N(d_{1}) - \frac{ \partial }{ \partial S_{0} } Ke^{-rT} N(d_{2}) \\ &= N(d_{1}) + S_{0}\frac{ \partial }{ \partial S_{0} } N(d_{1}) - Ke^{-rT} \frac{ \partial }{ \partial S_{0} } N(d_{2}). \end{align} \]
I will leave it to you to show that \[ \begin{align} \frac{ \partial }{ \partial S_{0} } N(d_{1}) = \frac{\phi(d_{1})}{S_{0} \sigma \sqrt{T}} && \text{and} && \frac{ \partial }{ \partial S_{0} } N(d_{2}) = \frac{\phi(d_{2})}{S_{0} \sigma \sqrt{T}} \end{align} \]
for \(\phi(\cdot)\) the probability density function of the standard normal distribution.
But, working backwards from Equation 1, we still need to show \[ \frac{\phi(d_{1})}{\sigma \sqrt{T}} = Ke^{-rT} \frac{\phi(d_{2})}{S_{0} \sigma \sqrt{T}}. \]
With enough elbow grease, you can do the algebra and make it work out, but this actually has a much more elegant statement as the fact that \(\frac{Ke^{-rT}}{S_{0}}\) is Radon-Nikodym derivative of the risk-neutral measure with respect to the stock-numeraire measure.
Let’s break this down.
2 Risk-Neutral Measure
Define two assets \(S_t\) and \(B_t\), with dynamics \[ dS_t = \mu S_t \, dt + \sigma S_t \, dW_t \] and \[ dB_t = r B_t \, dt \quad \implies \quad B_t = e^{rt}, \] where \(W_t\) is a standard Brownian motion under the physical measure \(\mathbb{P}\).
Defien the discounted stock asset \(\bar{S}_t\) as \[ \bar{S}_t = \frac{S_t}{B_t} = e^{-rt} S_t. \]
Recall Itô’s lemma:
Theorem 1 Itô’s Lemma
Let \(X_t\) be an Itô process given by \[ dX_t = \mu(X_t, t) \, dt + \sigma(X_t, t) \, dW_t, \]
then for any function \(f(X_t, t)\) that is twice differentiable with respect to \(X_t\) and once differentiable with respect to \(t\), we have \[ df(X_t, t) = \left( \frac{\partial f}{\partial t} + \mu \frac{\partial f}{\partial X_t} + \frac{1}{2} \sigma^2 \frac{\partial^2 f}{\partial X_t^2} \right) dt + \sigma \frac{\partial f}{\partial X_t} dW_t. \]
Apply Theorem 1 to \(\bar{S}_t\) to get
\[ \begin{align} d\bar{S}_t &= \left( \frac{\partial \bar{S}_t}{\partial t} + \mu S_t \frac{\partial \bar{S}_t}{\partial S_t} + \frac{1}{2} \left( \sigma S_t \right)^2 \frac{\partial^2 \bar{S}_t}{\partial S_t^2} \right) dt + \sigma S_t \frac{\partial \bar{S}_t}{\partial S_t} dW_t \\ &= \left( \mu - r \right) \bar{S}_t \, dt + \sigma \bar{S}_t \, dW_t. \end{align} \]
If we define a new Brownian motion \(W_t^\mathbb{Q}\) as \[ dW_t^\mathbb{Q} = dW_t + \frac{\mu - r}{\sigma} \, dt, \] then we can rewrite the dynamics of \(\bar{S}_t\) as \[ d\bar{S}_t = \left( \mu - r \right) \bar{S}_t \, dt + \sigma \bar{S}_t \, \left( dW_t^\mathbb{Q} - \frac{\mu - r}{\sigma} \, dt \right). \]