The Greeks

An option's price depends on the stock price, the time left, the volatility and the interest rate. How much does it move when one of them moves?

Each answer is a partial derivative of the Black–Scholes price, called a Greek: delta and gamma for the stock price, theta for time, vega for volatility, rho for the interest rate. They are not independent of each other: the Black–Scholes equation ties theta, gamma and delta together at every point of the price surface.

Before you start · 3 quick checks on partial derivatives, the normal density and Taylor expansion, about 3 min

Setup: a European call in the Black–Scholes model

Model. The stock pays no dividends and follows geometric Brownian motion with constant volatility $\sigma \gt 0$. The interest rate $r$ is a constant, continuously compounded per year, so that $1 grows to $e^{r\tau}$ over a time $\tau$ in years; the theorem in stage 3 allows any real $r$, and the examples use $r \ge 0$. Trading is continuous and frictionless, the call and the stock each trade at one price, short selling and borrowing at $r$ are allowed, and there is no arbitrage. These are the assumptions under which the lesson on delta hedging and the Black–Scholes PDE derives its equation. The price uses only $r$, not the stock's expected return $\mu$ under real-world probabilities.
The call. A European call with strike $K \gt 0$ expires at time $T$ and has no cash flows before then. At a time $t \lt T$, with stock price $S \gt 0$, write $\tau = T - t \gt 0$ for the time left. Its price, from the lesson on the Black–Scholes formula, is \[ \begin{aligned} V(t,S) &= S\,N(d_1) - K e^{-r\tau} N(d_2), \\ d_1 &= \frac{\ln(S/K) + (r + \tfrac12\sigma^2)\tau}{\sigma\sqrt{\tau}}, \\ d_2 &= d_1 - \sigma\sqrt{\tau}, \end{aligned} \] where $N$ is the standard normal CDF and $\varphi = N'$ is its density.

The Greeks are the partial derivatives of $V$, each taken with the other inputs held fixed:

delta $\Delta = \partial V/\partial S$, the slope of the price in the stock price;
gamma $\Gamma = \partial^2 V/\partial S^2$, the curvature in the stock price, which is also how fast delta changes;
theta $\Theta = \partial V/\partial t$, the change as calendar time moves forward, so that $\tau$ falls;
vega $\nu = \partial V/\partial\sigma$ and rho $\rho = \partial V/\partial r$.

Fine print: units, and what vega and rho mean in a model with constant σ and r

$\Theta$ is measured per year. The page also shows it per calendar day, $\Theta/365$. $\nu$ is per unit of $\sigma$, so a change from $\sigma = 0.20$ to $\sigma = 1.20$; divided by 100 it is per volatility point, a change from 20% to 21%. In the same way $\rho/100$ is per percentage point of the rate.

In the model $\sigma$ and $r$ never change, so $\partial V/\partial\sigma$ and $\partial V/\partial r$ compare the prices that two models with different constants give. Traders use them as the sensitivity of the price to a change in those inputs.

1 See

Where is gamma largest?

2 Try

A delta-hedged call over one day

Explorer: the Greeks of one call

Move the inputs until each goal is reached. The tiles show the price and its Greeks at the current inputs, and the figure shows one Greek against the stock price.

More inputs

The three terms of the Black–Scholes equation at the current inputs, per year, rounded to 4 decimals.
Term	Value
$\Theta$
$\tfrac12\sigma^2S^2\Gamma$
$rS\Delta - rV$
Sum

3 Prove

The Greeks of a call, and the equation that links them

Theorem (delta, gamma and theta of a Black–Scholes call)

Assumptions. $S \gt 0$, $K \gt 0$, $\sigma \gt 0$ and $\tau = T - t \gt 0$, and $r$ is any real constant. $V$, $d_1$ and $d_2$ are as in the setup.

Facts the proof uses. $N' = \varphi$ with $\varphi(z) = e^{-z^2/2}/\sqrt{2\pi}$; the product and chain rules; and $\partial/\partial t = -\partial/\partial\tau$, since $\tau = T - t$.

Conclusion.

\[ \begin{aligned} \Delta &= N(d_1), \qquad \Gamma = \frac{\varphi(d_1)}{S\sigma\sqrt{\tau}}, \\ \Theta &= -\frac{S\sigma\varphi(d_1)}{2\sqrt{\tau}} - rKe^{-r\tau}N(d_2), \end{aligned} \]

and at every such $(t, S)$

\[ \Theta + \tfrac12\sigma^2S^2\Gamma + rS\Delta - rV = 0. \]

The last equation is the Black–Scholes PDE $V_t + \tfrac12\sigma^2S^2V_{SS} + rSV_S - rV = 0$, written with the Greeks. The lesson on delta hedging derived it from a replicating portfolio. Here it is checked directly on the formula, by computing the Greeks.

Each option is one sentence about the $S$, $K$, $\tau$, $\sigma$ and $r$ of the theorem. Pick the one that is true for all of them, $r = 0$ and $r \lt 0$ included.

Scratch work, not part of the proof.

Vega, rho, and the Greeks of a put

Key point Two facts carry the computation: $d_1 - d_2 = \sigma\sqrt{\tau}$, and $S\varphi(d_1) = Ke^{-r\tau}\varphi(d_2)$. They make the extra terms of the chain rule cancel in $\Delta$, and they turn $\Theta + \tfrac12\sigma^2S^2\Gamma + rS\Delta - rV$ into 0.

4 Apply

Problems

All problems use a European call on a stock with no dividends, under the Black–Scholes assumptions of the setup. Prices are hypothetical.