期权定价之希腊值推导（Derivation of Greeks in European Option Pricing）

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.

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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.

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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:

$$\Gamma = \frac{e^{-y \tau} N^{'}(d_{1})}{\sigma S_{t} \sqrt{\tau}}$$

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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:

$$\Theta_{call} = y S_{t} e^{-y \tau} N(d_{1}) - r K e^{-r \tau} N(d_{2}) - \frac{\sigma S_{t} e^{-y \tau} N^{'}(d_{1})}{2 \sqrt{\tau}}$$

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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:

$$\nu_{call} = S_{t} e^{-y \tau} \sqrt{\tau} N^{'}(d_{1})$$

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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:

$$\rho_{call} = \tau K \cdot N(d_{2})$$