I’m studying the BSM model and having a look at the greeks. I was reading Derivatives, by Paul Wilmott, and he gives the closed form solutions without making the reader see where these solutions come from.

Is there a good book that explains in detail how each greek, from Delta to Rho, is derived and computed?

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    $\begingroup$ The real question about the Greeks these days is not where do they come from but where do they go to... (sorry, couldn't resist :-) $\endgroup$
    – vonjd
    Commented Jun 29, 2015 at 16:20
  • 1
    $\begingroup$ I think some interview prep books like "A Practical Guide to Quantitative Finance Interviews" by Zhou have such a derivation. $\endgroup$
    – bcf
    Commented Jun 29, 2015 at 23:28
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    $\begingroup$ I've gave a read at the book you've suggestested. Even there the greeks are "given", instead I want to know if I can compute them from a PDE with known boundary conditions. $\endgroup$
    – james42
    Commented Jun 30, 2015 at 0:57

2 Answers 2


First, my notation. $K$ is the strike price, $S$ is the stock price, $r$ is the continuously compounded risk-free rate, $T$ is time at expiration, $t$ is time at issue, $\sigma$ is volatility, $\delta$ is continuously compounded dividend rate.

The Black-Scholes formula for a European call is

$C = Se^{-\delta (T-t)} N(d_1) - Ke^{-r(T-t)} N(d_2)$

$d_1 = \dfrac{\ln(S/K) + (r - \delta + 0.5\sigma^2)(T-t)}{\sigma \sqrt{(T-t)}}$ and $d_2 = d_1 - \sigma \sqrt{(T-t)}$.

Some common greeks are

$\Delta$ = $\dfrac{ \partial C}{\partial S}$, $\Gamma = \dfrac{ \partial^2 C}{\partial S^2}$, $\rho = \dfrac{\partial C}{\partial r}$, $v = \dfrac{\partial C}{\partial \sigma}$, $\theta = \dfrac{\partial C}{\partial t}$ and $\psi = \dfrac{\partial C}{\partial \delta}$.

Note that $\theta$ is often equivalenty defined as $- \dfrac{\partial C}{\partial T}$. You can derive the greeks by taking the partial derivatives.

As an example, I will derive $\Delta = e^{-\delta (T-t)} N(d_1)$

\begin{align*} \Delta &= \dfrac{\partial C}{\partial S} \\ &= e^{-\delta (T-t)} N(d_1) + \dfrac{\partial C}{\partial S} Se^{-\delta (T-t)} N(d_1) - \dfrac{\partial C}{\partial S} Ke^{-r(T-t)} N(d_2) \\ &= e^{-\delta (T-t)} N(d_1) \end{align*}

It is not obvious that the last two terms cancel out. I prove this below.

\begin{align} &\dfrac{\partial C}{\partial S} Se^{-\delta (T-t)} N(d_1) - Ke^{-r(T-t)} N(d_2) \\ &= Se^{-\delta (T-t)} \dfrac{\partial}{\partial S} d_1 \dfrac{1}{\sqrt{2\pi}} e^{-0.5d_1^2} - Ke^{-r(T-t)} \dfrac{\partial}{\partial S} d_2 \dfrac{1}{\sqrt{2\pi}}e^{-0.5d_2^2} \\ &\propto \dfrac{\partial}{\partial S}d_1 \left( Se^{-\delta (T-t)} e^{-0.5d_1^2} - Ke^{-r(T-t)} e^{-0.5 d_2^2} \right) \\ \end{align}

Note that

\begin{align*} &\ln\left(Se^{-\delta(T-t)}e^{-0.5d_1^2}\right) - \ln\left(Ke^{-r(T-t)} e^{0.5d_2^2} \right) \\ &= \ln(S) - \delta(T-t) - \ln(K) + rT + 0.5d_2^2 - 0.5d_1^2 \\ &= \ln(S/K) + (r-\delta)(T-t) - 0.5(d_1^2 - (d_1 - \sigma\sqrt{T-t})^2) \\ &= \ln(S/K) + (r-\delta)(T-t) - 0.5(2d_1\sigma\sqrt{T-t} - \sigma^2(T-t)) \\ &= \ln(S/K) + (r-\delta)(T-t) - d_1 \sigma\sqrt{T-t} + 0.5\sigma^2(T-t) \\ &= \ln(S/K) + (r-\delta)(T-t) - \left(\ln(S/K) + (r-\delta + 0.5\sigma^2)(T-t)\right) + 0.5\sigma^2(T-t) \end{align*}

So $\left( Se^{-\delta (T-t)} e^{-0.5d_1^2} - Ke^{-r(T-t)} e^{-0.5 d_2^2} \right) = 0$ and $\Delta$ = $e^{-\delta(T-t)} N(d_1)$ as shown above.

Note that $\Delta > 0$ for a European call. Hedging greeks is a common topic in financial economics. To hedge a European call, short sell $\Delta$ shares of stock. This protects a portfolio against small changes in the stock price.

EDIT 1 By the BSE, $V_t + 0.5V_ss(\sigma S)^2 = rV - VsS(r-\delta)$ can be written as $\theta + 0.5\Gamma(\sigma S)^2 = rV - \Delta S(r-\delta)$. The solution to the BSE depends on the terminal conditions and payoff.

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    $\begingroup$ Easy as a pie. I want to know deeper if we can see the greeks as closed form solutions of PDE written by imposing some conditions... $\endgroup$
    – james42
    Commented Jun 29, 2015 at 15:50

The following paper gives a simple derivation of the BSM (via a simple integration approach instead of the classical PDE approach) and the Greeks plus some intuition for each:

Derivation and Comparative Statics of the Black-Scholes Call and Put Option Pricing Formulas by Garven, J.

You find the derivation of the Greeks in chapter 4 (called "comparative statics") on p. 12ff.

  • $\begingroup$ Thank you. But I've takled the problem in a different way: if we take the original BS PDE, then differentiate it w.r.t. to $r$ for example, we can prove that the solution to this PDE is exactly the greek rho. The problem is that I don't know how the boundary conditions in this case would change. $\endgroup$
    – james42
    Commented Jul 5, 2015 at 15:20
  • $\begingroup$ @ale42: I answered your question about references how to derive the Greeks. Why do you ask something when you mean something else? $\endgroup$
    – vonjd
    Commented Jul 5, 2015 at 15:25
  • $\begingroup$ Maybe in the main post wasn't so clear, but in my comment below I wrote that I'd preferred using a PDE-based approach because I was looking for mathematical insights. Maybe next time I'll explicit better my needs. ;) $\endgroup$
    – james42
    Commented Jul 5, 2015 at 15:37

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