Greeks in the Black–Scholes Model¶
In the previous sections, Greeks were defined abstractly as partial derivatives of a pricing operator \(V(t,S;\sigma,r,\dots)\). We now specialize these definitions to the Black–Scholes model and compute the Greeks explicitly from the closed-form European call/put formulas.
Cross-reference. The Black–Scholes formula \(C = SN(d_1) - Ke^{-r\tau}N(d_2)\) was derived in Section 5.6 using risk-neutral pricing. This section differentiates that formula to obtain closed-form Greeks.
Notation and Black–Scholes parameters¶
Let
- \(S\): current underlying price (at time \(t\))
- \(K\): strike
- \(r\): constant risk-free rate
- \(\sigma\): volatility
- \(T\): maturity
- \(t\): current time
- \(\tau := T-t\): time to maturity
- \(N(\cdot)\): standard normal CDF
- \(N'(x) = \dfrac{1}{\sqrt{2\pi}}e^{-x^2/2}\): standard normal PDF
Define
The Black–Scholes prices are
Preliminary: Derivatives of \(d_1\) and \(d_2\)¶
Before computing Greeks, we establish the key derivatives:
A crucial identity linking \(N'(d_1)\) and \(N'(d_2)\):
Proof. Since \(d_1 - d_2 = \sigma\sqrt{\tau}\), we have
Now \(d_1 + d_2 = 2d_1 - \sigma\sqrt{\tau}\), and using \(d_1 = \frac{\ln(S/K) + (r + \frac12\sigma^2)\tau}{\sigma\sqrt{\tau}}\):
Thus \(\frac{N'(d_1)}{N'(d_2)} = \frac{K e^{-r\tau}}{S}\), giving \(S N'(d_1) = K e^{-r\tau} N'(d_2)\). \(\square\)
Delta¶
Derivation for a call. Differentiate \(C = SN(d_1) - Ke^{-r\tau}N(d_2)\) with respect to \(S\):
Since \(\frac{\partial d_1}{\partial S} = \frac{\partial d_2}{\partial S} = \frac{1}{S\sigma\sqrt{\tau}}\):
Using the identity \(SN'(d_1) = Ke^{-r\tau}N'(d_2)\), the last two terms cancel:
-
Call: [ \Delta_{\text{call}} = N(d_1) ]
-
Put (by put-call parity \(P = C - S + Ke^{-r\tau}\)): [ \Delta_{\text{put}} = N(d_1)-1 = -N(-d_1) ]
Delta lies in \((0,1)\) for calls and \((-1,0)\) for puts.
Gamma¶
Derivation. Differentiate \(\Delta_{\text{call}} = N(d_1)\) with respect to \(S\):
For both calls and puts,
Gamma is always positive (convexity in \(S\)) and becomes large near maturity (small \(\tau\)). Note that \(\Gamma_{\text{call}} = \Gamma_{\text{put}}\) since they differ by a linear function of \(S\).
Vega¶
Derivation for a call. Differentiate \(C = SN(d_1) - Ke^{-r\tau}N(d_2)\) with respect to \(\sigma\):
Using \(\frac{\partial d_1}{\partial \sigma} = -\frac{d_2}{\sigma}\) and \(\frac{\partial d_2}{\partial \sigma} = -\frac{d_1}{\sigma}\):
Using \(SN'(d_1) = Ke^{-r\tau}N'(d_2)\) and \(d_1 - d_2 = \sigma\sqrt{\tau}\):
For both calls and puts,
Vega is typically largest near-the-money and for moderate maturities.
Theta¶
Convention. Some texts define theta as \(-\partial_t V\) (time-to-maturity derivative).
Here we use calendar-time theta \(\Theta=\partial_t V\). Since \(\tau = T - t\), we have \(\frac{\partial}{\partial t} = -\frac{\partial}{\partial \tau}\).
Derivation for a call. Using the Black–Scholes PDE:
Substituting the explicit forms:
-
Call: [ \Theta_{\text{call}} = -\frac{S\sigma N'(d_1)}{2\sqrt{\tau}} - rK e^{-r\tau}N(d_2) ]
-
Put: [ \Theta_{\text{put}} = -\frac{S\sigma N'(d_1)}{2\sqrt{\tau}} + rK e^{-r\tau}N(-d_2) ]
Theta is typically negative (time decay), with exceptions in certain regimes (e.g., deep ITM puts where the interest earned on the strike exceeds time decay).
Rho¶
Derivation for a call. Differentiate \(C = SN(d_1) - Ke^{-r\tau}N(d_2)\) with respect to \(r\):
Since \(\frac{\partial d_1}{\partial r} = \frac{\partial d_2}{\partial r} = \frac{\sqrt{\tau}}{\sigma}\), the terms involving \(N'\) cancel:
-
Call: [ \rho_{\text{call}} = K\tau e^{-r\tau}N(d_2) ]
-
Put: [ \rho_{\text{put}} = -K\tau e^{-r\tau}N(-d_2) ]
Rho reflects the present-value effect of the strike and is usually small for short maturities.
Symmetries and notes¶
- Gamma and Vega are identical for calls and puts: they depend on the diffusion of \(S\), not the payoff direction.
- Delta, Theta, and Rho differ between calls and puts due to payoff asymmetry and the discounting term.
- All expressions above follow by differentiating the Black–Scholes closed forms for \(C\) and \(P\).
Conceptual interpretation (typical signs)¶
| Greek | Meaning | Call sign | Put sign |
|---|---|---|---|
| Delta | price change per unit \(S\) | \(>0\) | \(<0\) |
| Gamma | convexity / rate of change of Delta | \(>0\) | \(>0\) |
| Theta | calendar-time decay | usually \(<0\) | usually \(<0\) |
| Vega | sensitivity to volatility | \(>0\) | \(>0\) |
| Rho | sensitivity to interest rate | \(>0\) | \(<0\) |
Summary¶
This section shows how the abstract Greeks reduce, in the Black–Scholes model, to explicit closed-form expressions. These formulas are the baseline for practical risk management, hedging, and calibration, and they also highlight where Black–Scholes is too restrictive (e.g., constant \(\sigma\) across strikes/maturities).
Detailed Derivations from First Principles¶
Delta Derivation¶
The delta of a European call option measures the rate of change of the call price with respect to the underlying stock price.
Starting from the Black–Scholes call formula: [ C = S N(d_1) - Ke^{-r\tau} N(d_2) ]
Differentiate with respect to \(S\): [ \frac{\partial C}{\partial S} = N(d_1) + S N'(d_1) \frac{\partial d_1}{\partial S} - Ke^{-r\tau} N'(d_2) \frac{\partial d_2}{\partial S} ]
Since both \(\frac{\partial d_1}{\partial S}\) and \(\frac{\partial d_2}{\partial S}\) equal \(\frac{1}{S\sigma\sqrt{\tau}}\), and using the key identity \(S N'(d_1) = Ke^{-r\tau} N'(d_2)\), the derivative terms simplify:
Thus, the delta of a call simplifies to the probability that the option finishes in-the-money under the risk-neutral measure.
Gamma Derivation¶
Gamma measures the second-order sensitivity to price changes, representing the convexity (curvature) of the option value.
Differentiate the call delta \(\Delta = N(d_1)\) with respect to \(S\):
Substituting \(\frac{\partial d_1}{\partial S} = \frac{1}{S\sigma\sqrt{\tau}}\):
Peak of Gamma. Setting \(\frac{\partial^2 \Gamma}{\partial S^2} = 0\), we find that gamma is maximized when: [ d_1 = -\sigma\sqrt{\tau} \quad \Rightarrow \quad S = Ke^{-rT} \exp\left(-\frac{3}{2}\sigma^2\tau\right) ]
This shows that gamma is largest for at-the-money options and decreases away from the strike.
Vega Derivation¶
Vega measures sensitivity to changes in volatility, the sole parameter in Black–Scholes subject to estimation error.
Differentiate the call price with respect to \(\sigma\):
Using \(\frac{\partial d_1}{\partial \sigma} = -\frac{d_2}{\sigma}\) and \(\frac{\partial d_2}{\partial \sigma} = -\frac{d_1}{\sigma}\):
With the identity \(SN'(d_1) = Ke^{-r\tau}N'(d_2)\) and \(d_1 - d_2 = \sigma\sqrt{\tau}\):
Vega is always positive, indicating that option values increase with volatility, and is typically largest near-the-money.
Higher-Order Greeks¶
For completeness, we present three commonly used second-order cross-Greeks:
Charm (delta decay): [ \text{Charm} := \frac{\partial \Delta}{\partial \tau} = -\frac{\partial \Delta}{\partial t} ]
For a call: [ \text{Charm}_{\text{call}} = -N'(d_1)\left(\frac{2(r-q)\tau - d_2\sigma\sqrt{\tau}}{2\tau\sigma\sqrt{\tau}}\right) ]
where \(q\) is the dividend yield (set \(q=0\) for non-dividend-paying stocks). Charm measures how delta changes as time passes.
Vanna (delta-vol sensitivity): [ \text{Vanna} := \frac{\partial \Delta}{\partial \sigma} = \frac{\partial \nu}{\partial S} ]
Vanna measures how delta changes with volatility, or equivalently how vega changes with spot.
Volga (vega convexity): [ \text{Volga} := \frac{\partial^2 V}{\partial \sigma^2} = \frac{\partial \nu}{\partial \sigma} ]
Volga is important for volatility surface dynamics and smile risk management. Long volga positions benefit from volatility-of-volatility.
Forward reference. These higher-order Greeks become central in Chapter 7 (Implied Volatility) for understanding smile dynamics, and in Chapter 9 (Stochastic Volatility) for analyzing model risk.
Exercises¶
Exercise 1. Using the identity \(SN'(d_1) = Ke^{-r\tau}N'(d_2)\), verify that vega is the same for a European call and a European put by directly computing \(\partial P / \partial \sigma\) from the put formula \(P = Ke^{-r\tau}N(-d_2) - SN(-d_1)\).
Solution to Exercise 1
The put formula is \(P = Ke^{-r\tau}N(-d_2) - SN(-d_1)\). Differentiate with respect to \(\sigma\):
Since \(N'(-x) = N'(x)\) (the standard normal PDF is symmetric), this becomes:
Using \(\frac{\partial d_1}{\partial \sigma} = -\frac{d_2}{\sigma}\) and \(\frac{\partial d_2}{\partial \sigma} = -\frac{d_1}{\sigma}\):
Applying the identity \(SN'(d_1) = Ke^{-r\tau}N'(d_2)\):
This equals the call vega \(\nu = S\sqrt{\tau}\,N'(d_1)\), confirming that \(\nu_{\text{call}} = \nu_{\text{put}}\).
Exercise 2. For a European call with \(S = 100\), \(K = 105\), \(\tau = 0.5\), \(r = 0.03\), \(\sigma = 0.25\), compute \(d_1\), \(d_2\), and all five Greeks: \(\Delta\), \(\Gamma\), \(\nu\), \(\Theta\), \(\rho\). Verify the theta-gamma identity \(\Theta + \frac{1}{2}\sigma^2 S^2 \Gamma + rS\Delta - rC = 0\).
Solution to Exercise 2
Given: \(S = 100\), \(K = 105\), \(\tau = 0.5\), \(r = 0.03\), \(\sigma = 0.25\).
Step 1: Compute \(d_1\) and \(d_2\).
Step 2: Evaluate the normal CDF and PDF.
\(N(d_1) = N(-0.10278) \approx 0.4591\), \(N(d_2) = N(-0.27956) \approx 0.3899\), \(N'(d_1) = \frac{1}{\sqrt{2\pi}}e^{-d_1^2/2} \approx 0.3970\).
Step 3: Compute Greeks.
- \(\Delta = N(d_1) \approx 0.4591\)
- \(\Gamma = \frac{N'(d_1)}{S\sigma\sqrt{\tau}} = \frac{0.3970}{100 \times 0.17678} \approx 0.02246\)
- \(\nu = S\sqrt{\tau}\,N'(d_1) = 100 \times 0.7071 \times 0.3970 \approx 28.07\)
-
\(\Theta = -\frac{S\sigma N'(d_1)}{2\sqrt{\tau}} - rKe^{-r\tau}N(d_2) = -\frac{100 \times 0.25 \times 0.3970}{2 \times 0.7071} - 0.03 \times 105 \times e^{-0.015} \times 0.3899\)
Computing each term: first term \(= -\frac{9.925}{1.4142} = -7.019\); second term \(= -0.03 \times 103.43 \times 0.3899 = -1.210\).
\(\Theta \approx -7.019 - 1.210 = -8.229\)
-
\(\rho = K\tau e^{-r\tau}N(d_2) = 105 \times 0.5 \times e^{-0.015} \times 0.3899 \approx 52.5 \times 0.9851 \times 0.3899 \approx 20.16\)
Step 4: Verify the theta-gamma identity. The call price is:
Check \(\Theta + \frac{1}{2}\sigma^2 S^2 \Gamma + rS\Delta - rC\):
The identity is verified (up to rounding).
Exercise 3. Prove that the peak of gamma occurs near \(S = K\) for short maturities by setting \(d_1 = 0\) and solving for \(S\) in terms of \(K\), \(r\), \(\sigma\), and \(\tau\). Show that as \(\tau \to 0\), the peak location converges to \(S = K\).
Solution to Exercise 3
The gamma is:
Since \(N'(d_1) = \frac{1}{\sqrt{2\pi}}e^{-d_1^2/2}\) is maximized when \(d_1 = 0\), the peak of gamma (as a function of \(S\)) occurs approximately when \(d_1 = 0\).
Setting \(d_1 = 0\):
Solving for \(S\):
As \(\tau \to 0\):
Therefore the peak of gamma converges to \(S = K\) as maturity approaches. For small \(\tau\), the peak is slightly below \(K\) (since the exponent is negative), but the deviation is of order \(\tau\) and vanishes at expiration.
Note: the precise maximum of \(\Gamma(S)\) for fixed \(\tau\) requires differentiating \(\Gamma\) with respect to \(S\) and solving \(\frac{\partial\Gamma}{\partial S} = 0\), which gives a slightly different condition (\(d_1 = \sigma\sqrt{\tau}\) rather than \(d_1 = 0\)). However, this distinction also vanishes as \(\tau \to 0\), and the conclusion \(S^* \to K\) is the same.
Exercise 4. Using the closed-form expression for charm,
with \(q = 0\), compute the rate at which delta changes over one day for an ATM call with \(S = K = 100\), \(\sigma = 0.20\), \(r = 0.05\), \(\tau = 30/365\). Interpret the sign of charm in terms of the hedger's rebalancing needs.
Solution to Exercise 4
Given: \(S = K = 100\), \(\sigma = 0.20\), \(r = 0.05\), \(\tau = 30/365 \approx 0.08219\), \(q = 0\).
Step 1: Compute \(d_1\) and \(d_2\).
For ATM (\(S = K\)):
Step 2: Compute \(N'(d_1)\).
Step 3: Compute charm.
Numerator of the fraction: \(2(0.05)(0.08219) - (0.04302)(0.20)(0.28669) = 0.008219 - 0.002467 = 0.005752\).
Denominator: \(2(0.08219)(0.20)(0.28669) = 0.009413\).
Step 4: Rate of delta change over one day. One day corresponds to \(\delta\tau = -1/365\) (time to maturity decreases). Since charm is defined as \(-\partial\Delta/\partial t = \partial\Delta/\partial\tau\):
Wait -- we must be careful with signs. Charm \(= \partial\Delta/\partial\tau = -\partial\Delta/\partial t\). As one day passes, \(t\) increases by \(1/365\), so:
Interpretation. The positive sign means delta increases slightly as time passes (for this near-ATM call). This is because an ATM call with \(d_1 > 0\) sees its delta pushed toward \(0.5\) more decisively as the diffusion narrows with less time remaining. The hedger must buy approximately \(0.067\) additional delta-units per 100 shares per day to stay hedged.
Exercise 5. Show that volga \(= \nu \cdot d_1 d_2 / \sigma\) is negative for options that are near-the-money (where \(d_1 > 0\) and \(d_2 < 0\)) and positive for options that are sufficiently far from the money. At what moneyness does volga change sign?
Solution to Exercise 5
Volga is given by:
Since \(\nu = S\sqrt{\tau}\,N'(d_1) > 0\) and \(\sigma > 0\), the sign of volga is determined by the sign of \(d_1 d_2\).
Recall \(d_2 = d_1 - \sigma\sqrt{\tau}\). For near-the-money options, \(d_1 \approx \frac{1}{2}\sigma\sqrt{\tau} > 0\) (slightly positive for ATM), so:
Therefore \(d_1 > 0\) and \(d_2 < 0\), giving \(d_1 d_2 < 0\), and volga is negative for near-the-money options.
Volga changes sign when either \(d_1 = 0\) or \(d_2 = 0\):
- \(d_2 = 0\) when \(d_1 = \sigma\sqrt{\tau}\), i.e., \(\ln(S/K) + (r + \frac{1}{2}\sigma^2)\tau = \sigma^2\tau\), giving \(S = K\exp\!\left((\frac{1}{2}\sigma^2 - r)\tau\right)\).
- \(d_1 = 0\) when \(S = K\exp\!\left(-(r + \frac{1}{2}\sigma^2)\tau\right)\).
For sufficiently far OTM or ITM options, both \(d_1\) and \(d_2\) have the same sign (both positive deep ITM, both negative deep OTM), making \(d_1 d_2 > 0\) and volga positive.
In summary, volga is negative in a band around the money where \(d_1\) and \(d_2\) have opposite signs, and positive outside this band.
Exercise 6. Using put-call parity \(P = C - S + Ke^{-r\tau}\), derive the put formulas for \(\Delta_{\text{put}}\), \(\Theta_{\text{put}}\), and \(\rho_{\text{put}}\) from the corresponding call formulas without differentiating the put pricing formula directly.
Solution to Exercise 6
Starting from put-call parity: \(P = C - S + Ke^{-r\tau}\).
Put delta. Differentiate with respect to \(S\):
Put theta. Differentiate with respect to \(t\) (using \(\frac{\partial\tau}{\partial t} = -1\)):
Since \(\frac{\partial}{\partial t}(Ke^{-r(T-t)}) = rKe^{-r\tau}\):
Substituting \(\Theta_{\text{call}} = -\frac{S\sigma N'(d_1)}{2\sqrt{\tau}} - rKe^{-r\tau}N(d_2)\):
Put rho. Differentiate with respect to \(r\):
Substituting \(\rho_{\text{call}} = K\tau e^{-r\tau}N(d_2)\):
All three results match the standard Black-Scholes put Greek formulas.