Derivation of Greeks in European Option Pricing
The European Vanilla call and put options with dividend payments have the values in formulas with:
\[\begin{align*}
c_{t} & = S_{t} e^{-y\tau} N(d_{1}) - K e^{-r\tau}N(d_{2}) \\
p_{t} & = K e^{-r\tau}N(-d_{2}) - S_{t} e^{-y\tau} N(-d_{1})
\end{align*}
\]
where \(c_{t}, ~ p_{t}\) represent the values of European call option and European put option respectively, \(y\) is the continuous rate of dividend payment, \(\tau\) represent the time to the exercise date (i.e. \(\tau = T - t\)), \(N\) is the cumulative distribution function of standard normal distribution \(N(0, 1)\), such that:
\[N^{'}(x) = f_{X}(x) = \frac{1}{\sqrt{2\pi}} \cdot e^{-\frac{x^{2}}{2}}, \qquad \qquad X \sim N(0, 1)
\]
\(S_{t}\) is the price of the underlying asset at time \(t\), \(K\) is the strike, and \(d_{1}, ~ d_{2}\) are calculated by:
\[\begin{align*}
d_{1} & = \frac{\log{ \big( \frac{S_{t}}{K}} \big) + (r - y + \frac{\sigma^{2}}{2})\tau}{\sigma\sqrt{\tau}} \\
d_{2} & = d_{1} - \sigma \sqrt{\tau} = \frac{\log{ \big( \frac{S_{t}}{K}} \big) + (r - y - \frac{\sigma^{2}}{2})\tau}{\sigma\sqrt{\tau}} \\
\end{align*}
\]
where \(\sigma\) is the volatility of the prices of the underlying asset.
Derivation of Delta (\(\Delta\))
Delta (\(\Delta\)) in option pricing usually refers to the partial derivative of option value to the price of underlying asset, which means:
\[\Delta = \frac{\partial V}{\partial S_{t}}
\]
where \(V\) is the option value.
Therefore, the delta for European vanilla call option is derived by:
\[\begin{align*}
\Delta_{call} & = \frac{\partial c_{t}}{\partial S_{t}} = \frac{\partial}{\partial S_{t}} \Big( S_{t} e^{-y\tau} N(d_{1}) - K e^{-r\tau}N(d_{2}) \Big)
\end{align*}
\]
It should be noticed that, it is not correct to simply conclude this partial derivative as \(e^{-y\tau} N(d_{1})\) (although the answer is correct) by reducing the \(S_{t}\) in the first component, because \(N(d_{1})\) and \(N(d_{2})\) are both functions of \(S_{t}\), and therefore the chain rule should be applied here, we continue the derivation as:
\[\begin{align*}
\Delta_{call} & = \frac{\partial}{\partial S_{t}} \Big( S_{t} e^{-y\tau} N(d_{1}) - K e^{-r\tau}N(d_{2}) \Big) \\
& = e^{-y \tau} N(d_{1}) + S_{t} e^{-y \tau} N^{'}(d_{1}) \frac{\partial d_{1}}{\partial S_{t}} - K e^{-r \tau} N^{'}(d_{2}) \frac{\partial d_{2}}{\partial S_{t}}
\end{align*}
\]
where:
\[\begin{align*}
\frac{\partial d_{1}}{\partial S_{t}} & = \frac{\partial}{\partial S_{t}} \Big[ \frac{\log{ \big( \frac{S_{t}}{K}} \big) + (r - y + \frac{\sigma^{2}}{2})\tau}{\sigma\sqrt{\tau}} \Big] \\
& = \frac{1}{\sigma \sqrt{\tau}} \frac{K}{S_{t}} \frac{1}{K} \\
& = \frac{1}{\sigma S_{t} \sqrt{\tau}} \\
\end{align*}
\]
and:
\[\begin{align*}
\frac{\partial d_{2}}{\partial S_{t}} & = \frac{\partial}{\partial S_{t}} \big( d_{1} - \sigma \sqrt{\tau} \big) \\
& = \frac{\partial d_{1}}{\partial S_{t}} \\
& = \frac{1}{\sigma S_{t} \sqrt{\tau}} \\
\end{align*}
\]
and:
\[N^{'}(d_{1}) = \frac{1}{\sqrt{2 \pi}} e^{-\frac{d_{1}^{2}}{2}}
\]
and:
\[\begin{align*}
N^{'}(d_{2}) & = \frac{1}{\sqrt{2 \pi}} e^{-\frac{d_{2}^{2}}{2}} \\
& = \frac{1}{\sqrt{2 \pi}} e^{-\frac{(d_{1} - \sigma \sqrt{\tau})^{2}}{2}} \\
& = \frac{1}{\sqrt{2 \pi}} e^{-\frac{d_{1}^{2} - 2 d_{1} \sigma \sqrt{\tau} + \sigma^{2} \tau}{2}} \\
& = \frac{1}{\sqrt{2 \pi}} e^{-\frac{d_{1}^{2}}{2}} e^{d_{1} \sigma \sqrt{\tau}} e^{-\frac{\sigma^{2} \tau}{2}} \\
\end{align*}
\]
where the second exponential component can be simplified as:
\[\begin{align*}
e^{d_{1} \sigma \sqrt{\tau}} & = e^{\frac{\log{ \big( \frac{S_{t}}{K}} \big) + (r - y + \frac{\sigma^{2}}{2})\tau}{\sigma\sqrt{\tau}} \sigma \sqrt{\tau}} \\
& = e^{\log{ \big( \frac{S_{t}}{K}} \big) + (r - y + \frac{\sigma^{2}}{2})\tau} \\
& = \frac{S_{t}}{K} e^{(r - y + \frac{\sigma^{2}}{2}) \tau}
\end{align*}
\]
Hence \(N^{'}(d_{2})\) becomes:
\[\begin{align*}
N^{'}(d_{2}) & = \frac{1}{\sqrt{2 \pi}} e^{-\frac{d_{1}^{2}}{2}} e^{d_{1} \sigma \sqrt{\tau}} e^{-\frac{\sigma^{2} \tau}{2}} \\
& = \frac{1}{\sqrt{2 \pi}} \frac{S_{t}}{K} e^{(r - y + \frac{\sigma^{2}}{2}) \tau} e^{-\frac{d_{1}^{2}}{2}} e^{-\frac{\sigma^{2} \tau}{2}} \\
& = \frac{S_{t}}{\sqrt{2 \pi} K} e^{(r - y) \tau} e^{-\frac{d_{1}^{2}}{2}} \\
\end{align*}
\]
So that:
\[\begin{align*}
\Delta_{call} & = e^{-y \tau} N(d_{1}) + S_{t} e^{-y \tau} N^{'}(d_{1}) \frac{\partial d_{1}}{\partial S_{t}} - K e^{-r \tau} N^{'}(d_{2}) \frac{\partial d_{2}}{\partial S_{t}} \\
& = e^{-y \tau} N(d_{1}) + \frac{S_{t} e^{-y \tau} e^{-\frac{d_{1}^{2}}{2}}}{\sqrt{2 \pi \tau} \cdot \sigma S_{t}} - \frac{S_{t} K e^{-r \tau} e^{(r - y) \tau} e^{-\frac{d_{1}^{2}}{2}}}{\sqrt{2 \pi \tau} \cdot \sigma S_{t} K} \\
& = e^{-y \tau} N(d_{1}) + \frac{e^{-y \tau} e^{-\frac{d_{1}^{2}}{2}}}{\sqrt{2 \pi \tau} \cdot \sigma } - \frac{ e^{-y \tau} e^{-\frac{d_{1}^{2}}{2}}}{\sqrt{2 \pi \tau} \cdot \sigma} \\
& = e^{-y \tau} N(d_{1})
\end{align*}
\]
Therefore, the delta of European call option with dividend payments is derived as above, that is:
\[\Delta_{call} = e^{-y \tau} N(d_{1})
\]
The delta of European put option with dividend payments can be derived in a similar way.
Derivation of Gamma (\(\Gamma\))
Gamma (\(\Gamma\)) is the second partial derivative of the European option with respect to the price of underlying asset, that is:
\[\Gamma = \frac{\partial^{2} V}{\partial S_{t}^{2}}
\]
Taking European call option with dividend payments as an example, since the delta:
\[\Delta_{call} = e^{-y \tau} N(d_{1})
\]
The gamma would therefore be:
\[\begin{align*}
\Gamma & = \frac{\partial}{\partial S_{t}} \big( e^{-y \tau} N(d_{1}) \big) \\
& = e^{-y \tau} N^{'}(d_{1}) \frac{\partial d_{1}}{\partial S_{t}} \\
& = \frac{e^{-y \tau} N^{'}(d_{1})}{\sigma S_{t} \sqrt{\tau}} \\
\end{align*}
\]
That is:
\[\Gamma = \frac{e^{-y \tau} N^{'}(d_{1})}{\sigma S_{t} \sqrt{\tau}}
\]
Derivation of Theta (\(\Theta\))
Theta (\(\Theta\)) is the partial derivative of the value of European option with respect to the time \(t\) (NOT the time to maturity \(\tau\) !), that is:
\[\Theta = \frac{\partial V}{\partial t}
\]
In the following steps, I will replace the time to maturity \(\tau\) by its original form \(T - t\), in order to show a much more clear version of chain rule. Here, I will write down each component of the chain, and make all of them up together, due to the sophistication of chain equation.
The partial derivative of \(d_{1}\) with respect to time \(t\) is:
\[\begin{align*}
\frac{\partial d_{1}}{\partial t} & = \frac{\partial}{\partial t} \Big[ \frac{\log{ \big( \frac{S_{t}}{K}} \big) + (r - y + \frac{\sigma^{2}}{2}) (T - t)}{\sigma\sqrt{T - t}} \Big] \\
& = - (r - y + \frac{\sigma^{2}}{2}) \sigma \sqrt{T - t} + \frac{\sigma}{2} \sqrt{T - t} (r - y + \frac{\sigma^{2}}{2}) \\
& = - \frac{\sigma}{2} (r - y + \frac{\sigma^{2}}{2}) \sqrt{T - t}
\end{align*}
\]
The partial derivative of \(d_{2}\) with respect to time \(t\) is:
\[\begin{align*}
\frac{\partial d_{2}}{\partial t} & = \frac{\partial}{\partial t} \big( d_{1} - \sigma \sqrt{T - t} \big) \\
& = \frac{\partial d_{1}}{\partial t} - \frac{\partial}{\partial t} \big( \sigma \sqrt{T - t} \big) \\
& = - \frac{\sigma}{2} (r - y + \frac{\sigma^{2}}{2}) \sqrt{T - t} + \frac{\sigma}{2} (T - t)^{-\frac{1}{2}} \\
\end{align*}
\]
Taking European call option with dividend payments as an example, the theta would be:
\[\begin{align*}
\Theta_{call} & = \frac{\partial c_{t}}{\partial t} \\
& = \frac{\partial}{\partial t} \big[ S_{t} e^{-y\tau} N(d_{1}) - K e^{-r\tau}N(d_{2}) \big] \\
~ \\
& = y S_{t} e^{-y (T - t)} N(d_{1}) + S_{t} e^{-y (T - t)} N^{'}(d_{1}) \Big( -\frac{\sigma}{2} (r - y + \frac{\sigma^{2}}{2}) \sqrt{T - t} \Big) \\
- & r K e^{-r (T - t)} N(d_{2}) - K e^{-r (T - t)} N^{'}(d_{2}) \Big( -\frac{\sigma}{2} (r - y + \frac{\sigma^{2}}{2}) \sqrt{T - t} ~ + \\
& \frac{\sigma}{2} (T - t)^{-\frac{1}{2}} \Big) \\
~ \\
& = y S_{t} e^{-y (T - t)} N(d_{1}) - \frac{\sigma}{2} S_{t} (r - y + \frac{\sigma^2}{2}) \sqrt{T - t} \cdot e^{-y (T - t)} N^{'}(d_{1}) \\
- & r K e^{-r (T - t)} N(d_{2}) + \frac{\sigma}{2} K (r - y + \frac{\sigma^{2}}{2}) \sqrt{T - t} \cdot e^{-r (T - t)} N^{'}(d_{2}) \\
- & \frac{\sigma}{2} K (T - t)^{-\frac{1}{2}} e^{-r (T - t)} N^{'}(d_{2}) \\
~ \\
& = y S_{t} e^{-y (T - t)} N(d_{1}) - r K e^{-r (T - t)} N(d_{2}) - \frac{\sigma}{2} S_{t} (r - y + \frac{\sigma^{2}}{2}) \sqrt{T - t} ~ \cdot \\
& e^{-y (T - t)} \frac{1}{\sqrt{2 \pi}} e^{-\frac{d_{1}^{2}}{2}} + \frac{\sigma}{2} K (r - y + \frac{\sigma^{2}}{2}) \sqrt{T - t} ~ \cdot e^{-r (T - t)} \frac{1}{\sqrt{2 \pi}} e^{-\frac{(d_{1} - \sigma \sqrt{T - t})^{2}}{2}} \\
- & \frac{\sigma}{2} K (T - t)^{-\frac{1}{2}} e^{-r (T - t)} \frac{1}{\sqrt {2 \pi}} e^{-\frac{(d_{1} - \sigma \sqrt{T - t})^{2}}{2}} \\
~ \\
& = y S_{t} e^{-y (T - t)} N(d_{1}) - r K e^{-r (T - t)} N(d_{2}) - \frac{\sigma S_{t}}{2 \sqrt{2 \pi}} (r - y + \frac{\sigma^{2}}{2}) \sqrt{T - t} ~ \cdot \\
& e^{-y (T - t)} e^{- \frac{d_{1}^{2}}{2}} + \frac{\sigma K}{2 \sqrt{2 \pi}} (r - y + \frac{\sigma^{2}}{2}) \sqrt{T - t} \cdot e^{-r (T - t)} e^{- \frac{d_{1}^{2} - 2 \sigma d_{1} \sqrt{T - t} + \sigma^{2} (T - t)}{2}} \\
- & \frac{\sigma K}{2 \sqrt{2 \pi}} (T - t)^{-\frac{1}{2}} e^{-r (T - t)} e^{- \frac{d_{1}^{2} - 2 \sigma d_{1} \sqrt{T - t} + \sigma^{2}(T - t)}{2}} \\
~ \\
& = y S_{t} e^{-y (T - t)} N(d_{1}) - r K e^{-r (T - t)} N(d_{2}) - \frac{\sigma S_{t}}{2 \sqrt{2 \pi}} (r - y + \frac{\sigma^{2}}{2}) \sqrt{T - t} ~ \cdot \\
& e^{-y (T - t)} e^{- \frac{d_{1}^{2}}{2}} + \frac{\sigma K}{2 \sqrt{2 \pi}} (r - y + \frac{\sigma^{2}}{2}) \sqrt{T - t} \cdot e^{-r (T - t)} \frac{S_{t}}{K} e^{-\frac{d_{1}^{2}}{2}} e^{(r - y)(T - t)} \\
- & \frac{\sigma K}{2 \sqrt{2 \pi}} (T - t)^{-\frac{1}{2}} e^{-r (T - t)} \frac{S_{t}}{K} e^{-\frac{d_{1}^{2}}{2}} e^{(r - y)(T - t)} \\
~ \\
& = y S_{t} e^{-y (T - t)} N(d_{1}) - r K e^{-r (T - t)} N(d_{2}) - \frac{\sigma S_{t}}{2 \sqrt{2 \pi}} (T - t)^{-\frac{1}{2}} e^{-\frac{d_{1}^{2}}{2}} e^{-y(T - t)} \\
& = y S_{t} e^{-y (T - t)} N(d_{1}) - r K e^{-r (T - t)} N(d_{2}) - \frac{\sigma S_{t} e^{-y (T - t)} N^{'}(d_{1})}{2 \sqrt{T - t}} \\
& = y S_{t} e^{-y \tau} N(d_{1}) - r K e^{-r \tau} N(d_{2}) - \frac{\sigma S_{t} e^{-y \tau} N^{'}(d_{1})}{2 \sqrt{\tau}}
\end{align*}
\]
That is:
\[\Theta_{call} = y S_{t} e^{-y \tau} N(d_{1}) - r K e^{-r \tau} N(d_{2}) - \frac{\sigma S_{t} e^{-y \tau} N^{'}(d_{1})}{2 \sqrt{\tau}}
\]
Derivation of Vega (\(\nu\))
Vega (\(\nu\)) is the partial derivative of European option value with respect to the volatility \(\sigma\), that is:
\[\nu = \frac{\partial V}{\partial \sigma}
\]
Taking European call option with dividend payments as an example, due to the complication of applying the chain rule, I will firstly derive the partial derivative of \(d_{1}\) and \(d_{2}\) with respect to \(\sigma\) respectively, thus:
\[\begin{align*}
\frac{\partial d_{1}}{\partial \sigma} & = \frac{\sigma^{2} \tau \sqrt{\tau} - \big( \log(\frac{S_{t}}{K}) + (r - y + \frac{\sigma^{2}}{2})\tau \big) \tau}{\sigma^{2}{\tau}} \\
& = \frac{\sigma^{2}\sqrt{\tau} - \big( \log(\frac{S_{t}}{K}) + (r - y + \frac{\sigma^{2}}{2}) \tau \big)}{\sigma^{2}} \\
\end{align*}
\]
and:
\[\begin{align*}
\frac{\partial d_{2}}{\partial \sigma} & = \frac{\partial d_{1}}{\partial \sigma} - \sqrt{\tau} \\
& = - \frac{\log(\frac{S_{t}}{K}) + (r - y + \frac{\sigma^{2}}{2}) \tau}{\sigma^{2}}
\end{align*}
\]
Therefore, by the chain rule:
\[\begin{align*}
\nu_{call} & = \frac{\partial c_{t}}{\partial \sigma} \\
& = S_{t} e^{-y \tau} N^{'}(d_{1}) \frac{\partial d_{1}}{\partial \sigma} - K e^{-r \tau} N^{'}(d_{2}) \frac{\partial d_{2}}{\partial \sigma} \\
~ \\
& = S_{t} e^{-y \tau} \frac{1}{\sqrt{2 \pi}} e^{-\frac{d_{1}^{2}}{2}} \frac{\sigma^{2} \tau - \log(\frac{S_{t}}{K}) - (r - y + \frac{\sigma^{2}}{2}) \tau}{\sigma^{2}} \\
+ & K e^{-r \tau} \frac{1}{\sqrt{2 \pi}} e^{-\frac{d_{2}^{2}}{2}} \frac{\log(\frac{S_{t}}{K}) + (r - y + \frac{\sigma^{2}}{2}) \tau}{\sigma^{2}} \\
~ \\
& = S_{t} e^{-y \tau} e^{-\frac{d_{1}^{2}}{2}} \frac{\sigma^{2} \sqrt{\tau} - \log(\frac{S_{t}}{K}) - (r - y + \frac{\sigma^{2}}{2}) \tau}{\sqrt{2 \pi} \sigma^{2}} \\
+ & K e^{-r \tau} e^{-\frac{d_{1}^{2}}{2}} e^{-\frac{\sigma^{2} \tau}{2}} \frac{S_{t}}{K} e^{(r - y + \frac{\sigma^{2}}{2}) \tau} \frac{\log(\frac{S_{t}}{K}) + (r - y + \frac{\sigma^{2}}{2}) \tau}{\sqrt{2 \pi} \sigma^{2}} \\
~ \\
& = S_{t} e^{-y \tau} e^{-\frac{d_{1}^{2}}{2}} \frac{\sigma^{2} \sqrt{\tau}}{\sqrt{2 \pi} \sigma^{2}} \\
& = S_{t} e^{-y \tau} \sqrt{\tau} N^{'}(d_{1})
\end{align*}
\]
That is:
\[\nu_{call} = S_{t} e^{-y \tau} \sqrt{\tau} N^{'}(d_{1})
\]
Derivation of Rho (\(\rho\))
Rho (\(\rho\)) is defined as the partial derivative of the value of the European option with respect to the risk free interest rate (\(r\)).
That is:
\[\rho = \frac{\partial V}{\partial r}
\]
Taking European call option with dividend payments as an example, we firstly derive the partial derivatives of \(d_{1}\) and \(d_{2}\) with respect to \(r\), that is:
\[\frac{\partial d_{1}}{\partial r} = \frac{\sqrt{\tau}}{\sigma}
\]
and:
\[\frac{\partial d_{2}}{\partial r} = \frac{\partial}{\partial r} (d_{1} - \sigma \sqrt{\tau}) = \frac{\sqrt{\tau}}{\sigma}
\]
Therefore, we have:
\[\begin{align*}
\frac{\partial c_{t}}{\partial r} & = S_{t} e^{-y \tau} N^{'}(d_{1}) \frac{\partial d_{1}}{\partial r} + \tau K \cdot N(d_{2}) - K e^{-r \tau} N^{'}(d_{2}) \frac{\partial d_{2}}{\partial r} \\
& = \frac{\sqrt{\tau}}{\sigma} \Big( S_{t} e^{-y \tau} N^{'}(d_{1}) - K e^{-r \tau} N^{'}(d_{2}) \Big) + \tau K \cdot N(d_{2}) \\
& = \sqrt{\frac{\tau}{2 \pi}} \cdot \sigma \Big( S_{t} e^{-y \tau} e^{-\frac{d_{1}^{2}}{2}} - K e^{-r \tau} e^{-\frac{d_{2}^{2}}{2}} \Big) + \tau K \cdot N(d_{2}) \\
& = \tau K \cdot N(d_{2})
\end{align*}
\]
That is:
\[\rho_{call} = \tau K \cdot N(d_{2})
\]
