Greeks in the Black–Scholes Model¶
Recall (see § Delta, Gamma, Vega, Theta, Rho): Greeks are partial derivatives of the pricing map \(V(t,S;\sigma,r,\dots)\). We now specialize these definitions to the Black–Scholes model and compute the Greeks explicitly from the closed-form European call/put formulas.
Recall (see § Black–Scholes formula): \(C = SN(d_1) - Ke^{-r\tau}N(d_2)\) — this section differentiates that formula.
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 d₁ and d₂¶
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:
- Put (by put-call parity \(P = C - S + Ke^{-r\tau}\)):
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. Recall (see § BS PDE structure): from the Black–Scholes PDE,
Substituting the explicit forms:
- Call:
- Put:
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:
- Put:
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).
Higher-Order Greeks¶
For completeness, we present three commonly used second-order cross-Greeks:
Charm (delta decay):
For a call:
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):
Vanna measures how delta changes with volatility, or equivalently how vega changes with spot.
Volga (vega convexity):
Volga is important for volatility surface dynamics and smile risk management. Long volga positions benefit from volatility-of-volatility.
Forward reference. Deferred — see § Vega and the smile for the role of vanna/volga in smile dynamics, and § Greek asymptotics for limiting behavior.
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.