Unit 6, lesson 45

Greeks and differential identities

Derives the Black-Scholes delta, gamma, vega, theta and rho from one density identity, states their units and conventions, and checks the vega-gamma identity and the pricing PDE in a worked example.

  • Video 45
  • Black-Scholes formulas, Greeks and hedging

A desk holds a one-year at-the-money call on a stock at 100 and wants to know how its value responds when the stock moves by one unit, when volatility rises by one percentage point, when a day passes and when the rate moves by one basis point. Each answer is a partial derivative of the Black-Scholes price, and each number means something only once the variable, the inputs held fixed and the units are stated. This lesson derives delta, gamma, vega, theta and rho from one identity between normal densities, checks them against the vega-gamma identity and the pricing PDE, and computes them in a constructed example.

Key terms

Option Greeks

Definition Let V(t,S)V(t, S) be the Black-Scholes price of a European option with fixed strike KK and maturity TT, under the hypotheses of The Black Scholes formula and put call parity, viewed as a function of tt, SS and the model inputs σ\sigma, rr and qq. On the set S>0S > 0, K>0K > 0, σ>0\sigma > 0, τ=T−t>0\tau = T - t > 0 the price is infinitely differentiable, and its Greeks are the partial derivatives

Δ=∂V∂S,Γ=∂2V∂S2,ν=∂V∂σ,ρ=∂V∂r,Θ=∂V∂t=−∂V∂τ,\begin{gathered} \Delta = \frac{\partial V}{\partial S}, \quad \Gamma = \frac{\partial^2 V}{\partial S^2}, \\[0.4em] \nu = \frac{\partial V}{\partial \sigma}, \quad \rho = \frac{\partial V}{\partial r}, \\[0.4em] \Theta = \frac{\partial V}{\partial t} = -\frac{\partial V}{\partial \tau}, \end{gathered}

each taken with all other inputs held fixed. Theta is a derivative in calendar time with the maturity date fixed. Rho moves the constant rate rr with SS, qq and σ\sigma fixed. Vega moves the annualized volatility in decimal units. The units are shares per option for delta, shares per currency unit for gamma, currency per unit of decimal volatility for vega, currency per year for theta and currency per unit of decimal rate for rho. So a one percentage point volatility change is about 0.01 ν0.01\,\nu, one basis point of rate is about 0.0001 ρ0.0001\,\rho, and one calendar day under an actual/365 count is about Θ/365\Theta/365. A Greek computed by bumping, such as the centered difference (V(x+h)−V(x−h))/(2h)\bigl(V(x + h) - V(x - h)\bigr)/(2h) with a stated bump hh, approximates the derivative with an error of order h2h^2 for a smooth price. A sensitivity that lets the volatility surface or the curves move with the bumped input is a different, total derivative and needs its own convention.

The Black-Scholes Greeks follow from one identity. With d1d_1 and d2d_2 as in the previous lesson,

Se−qτφ(d1)=Ke−rτφ(d2).S e^{-q\tau}\varphi(d_1) = K e^{-r\tau}\varphi(d_2).

Indeed d12−d22=(d1−d2)(d1+d2)=2ln⁡(S/K)+2(r−q)τd_1^2 - d_2^2 = (d_1 - d_2)(d_1 + d_2) = 2\ln(S/K) + 2(r - q)\tau, so φ(d2)/φ(d1)=(S/K)e(r−q)τ\varphi(d_2)/\varphi(d_1) = (S/K)e^{(r - q)\tau}. For the call C=Se−qτΦ(d1)−Ke−rτΦ(d2)C = S e^{-q\tau}\Phi(d_1) - K e^{-r\tau}\Phi(d_2) and any input xx, the chain rule and this identity give

∂C∂x=∂(Se−qτ)∂x Φ(d1)−∂(Ke−rτ)∂x Φ(d2)+Se−qτφ(d1)×∂(d1−d2)∂x.\begin{aligned} \frac{\partial C}{\partial x} &= \frac{\partial (S e^{-q\tau})}{\partial x}\,\Phi(d_1) \\[0.4em] &\quad - \frac{\partial (K e^{-r\tau})}{\partial x}\,\Phi(d_2) \\[0.4em] &\quad + S e^{-q\tau}\varphi(d_1) \\[0.4em] &\qquad \times \frac{\partial (d_1 - d_2)}{\partial x}. \end{aligned}

Since d1−d2=στd_1 - d_2 = \sigma\sqrt{\tau}, the density term survives only in vega and theta.

In plain words A Greek answers the question of how much the model price moves when one input moves a little and everything else stays put. Saying what stays put and in which units the input is measured is part of the definition, because the same word can name different numbers under different conventions.

Example Take the constructed inputs S=K=100S = K = 100, r=q=0r = q = 0, σ=0.2\sigma = 0.2 and τ=1\tau = 1, so that d1=0.1d_1 = 0.1, d2=−0.1d_2 = -0.1 and C≈7.965567C \approx 7.965567. Both sides of the density identity equal 100 φ(0.1)≈39.695255100\,\varphi(0.1) \approx 39.695255. Repricing at S=101S = 101 and S=99S = 99 gives 8.5152688.515268 and 7.4357117.435711, and the centered difference with bump h=1h = 1 is about 0.5397780.539778, close to the analytic delta 0.5398280.539828 of the next card.

Delta

Definition Under the hypotheses of the previous card, the deltas of the call and the put are

ΔC=e−qτ Φ(d1),ΔP=−e−qτ Φ(−d1).\begin{gathered} \Delta_C = e^{-q\tau}\,\Phi(d_1), \\[0.4em] \Delta_P = -e^{-q\tau}\,\Phi(-d_1). \end{gathered}

For the call, take x=Sx = S in the cancellation formula. The factor Se−qτS e^{-q\tau} has derivative e−qτe^{-q\tau}, the factor Ke−rτK e^{-r\tau} does not depend on SS, and d1−d2d_1 - d_2 does not depend on SS, so the two density terms cancel. Differentiating put-call parity in SS gives ΔC−ΔP=e−qτ\Delta_C - \Delta_P = e^{-q\tau}, hence the put delta. Consequently 0<ΔC<e−qτ0 < \Delta_C < e^{-q\tau} and −e−qτ<ΔP<0-e^{-q\tau} < \Delta_P < 0. In the replication argument the portfolio that replicates one option holds Δ\Delta shares, so a long option is hedged by selling Δ\Delta shares, and for a small move δ\delta of the stock V(t,S+δ)−V(t,S)≈Δ δV(t, S + \delta) - V(t, S) \approx \Delta\,\delta. The call delta e−qτΦ(d1)e^{-q\tau}\Phi(d_1) is a hedge ratio and differs from the pricing probability of exercise, which is Φ(d2)\Phi(d_2).

In plain words Delta is the number of shares whose value moves like the option for a small move of the stock. A call gains when the stock rises and a put loses, and for a small move neither changes by more than the prepaid share, which moves by e−qτe^{-q\tau} times the stock move.

Example At the money with the constructed inputs, ΔC=Φ(0.1)≈0.539828\Delta_C = \Phi(0.1) \approx 0.539828 and ΔP≈−0.460172\Delta_P \approx -0.460172. Raising the stock from 100100 to 101101 raises the call from 7.9655677.965567 to 8.5152688.515268, by 0.5497010.549701. Delta predicts 0.5398280.539828, and the remaining 0.0098720.009872 is close to the gamma term 12Γ⋅12≈0.009924\tfrac12\Gamma \cdot 1^2 \approx 0.009924 of the next card.

Gamma

Definition The gamma of the call and of the put is

Γ=e−qτ φ(d1)Sστ,\Gamma = \frac{e^{-q\tau}\,\varphi(d_1)}{S\sigma\sqrt{\tau}} ,

obtained by differentiating ΔC\Delta_C with ∂d1/∂S=1/(Sστ)\partial d_1/\partial S = 1/(S\sigma\sqrt{\tau}). The put has the same gamma because C−PC - P is linear in SS. The denominator is positive under the hypotheses S>0S > 0, σ>0\sigma > 0 and τ>0\tau > 0, so Γ>0\Gamma > 0 for long calls and puts. To second order,

V(t,S+δ)−V(t,S)≈Δ δ+12Γ δ2.\begin{aligned} &V(t, S + \delta) - V(t, S) \\[0.4em] &\quad \approx \Delta\,\delta + \tfrac12\Gamma\,\delta^2 . \end{aligned}

As τ↓0\tau \downarrow 0 the behavior depends on the position of the stock. Away from the strike Γ→0\Gamma \to 0, while at the money with r=q=0r = q = 0 one has d1=12στd_1 = \tfrac12\sigma\sqrt{\tau} and Γ≈1/(Sσ2πτ)\Gamma \approx 1/\bigl(S\sigma\sqrt{2\pi\tau}\bigr), which grows without bound. At expiry the payoff has a kink at KK and no classical second derivative there, so the formula, which needs τ>0\tau > 0, does not extend to that point.

In plain words Gamma is the rate at which delta changes as the stock moves, so it tells a hedger how many shares to trade after a move. A long option has positive gamma, which means its value curves upward in the stock price.

Example At the money, Γ=φ(0.1)/(100⋅0.2)≈0.396953/20≈0.019848\Gamma = \varphi(0.1)/(100 \cdot 0.2) \approx 0.396953/20 \approx 0.019848 per currency unit. The centered second difference with bump 11 gives (8.515268−2⋅7.965567+7.435711)/1≈0.019844(8.515268 - 2 \cdot 7.965567 + 7.435711)/1 \approx 0.019844. With the stock still at 100100 and only one week left, τ=1/52\tau = 1/52, gamma rises to about 0.1438270.143827, and with one day left, τ=1/365\tau = 1/365, to about 0.3810840.381084, against the approximations 0.1438410.143841 and 0.3810890.381089. At S=110S = 110 with one week left it is only about 0.0003400.000340.

Vega and the vega-gamma identity

Definition The vega of the call and of the put is

ν=Se−qτφ(d1)τ=Ke−rτφ(d2)τ.\begin{aligned} \nu &= S e^{-q\tau}\varphi(d_1)\sqrt{\tau} \\[0.4em] &= K e^{-r\tau}\varphi(d_2)\sqrt{\tau}. \end{aligned}

In the cancellation formula with x=σx = \sigma neither prefactor depends on σ\sigma and ∂(d1−d2)/∂σ=τ\partial(d_1 - d_2)/\partial\sigma = \sqrt{\tau}, which gives the call vega, and the put has the same vega since C−PC - P does not depend on σ\sigma. Under the hypotheses ν>0\nu > 0, so each price is strictly increasing in σ\sigma. Comparing with gamma gives the vega-gamma identity

ν=στS2 Γ,\nu = \sigma\tau S^2\,\Gamma ,

valid for calls and puts at every S>0S > 0, σ>0\sigma > 0 and τ>0\tau > 0 in this model. The same identity holds for other European payoffs of STS_T in the Black-Scholes model whose prices are smooth, a fact stated here without proof. Vega is measured in currency per unit of decimal volatility, so a move from 0.200.20 to 0.210.21 is a change of 0.010.01 in σ\sigma.

In plain words Vega is the price change per unit of volatility. In this model more volatility and more curvature are one exposure seen in two ways, and the identity converts gamma into vega by the factor στS2\sigma\tau S^2.

Example At the money, ν=100 φ(0.1)⋅1≈39.695255\nu = 100\,\varphi(0.1) \cdot 1 \approx 39.695255. One volatility point is 0.010.01, so vega predicts a gain of about 0.3969530.396953 when σ\sigma rises from 0.200.20 to 0.210.21, and repricing gives 8.362419−7.965567=0.3968528.362419 - 7.965567 = 0.396852. The identity gives 1002⋅0.2⋅1⋅0.019848≈39.695100^2 \cdot 0.2 \cdot 1 \cdot 0.019848 \approx 39.695, the same vega. With one day left the vega is only about 2.0881312.088131 while gamma is 0.3810840.381084, and 104⋅0.2⋅(1/365)⋅0.381084≈2.08810^4 \cdot 0.2 \cdot (1/365) \cdot 0.381084 \approx 2.088 again, since the factor τ\tau shrinks as gamma grows.

Theta

Definition With TT fixed, the thetas of the call and the put are

ΘC=−Se−qτφ(d1) σ2τ+qSe−qτΦ(d1)−rKe−rτΦ(d2),\begin{aligned} \Theta_C &= -\frac{S e^{-q\tau}\varphi(d_1)\,\sigma}{2\sqrt{\tau}} \\[0.4em] &\quad + q S e^{-q\tau}\Phi(d_1) \\[0.4em] &\quad - r K e^{-r\tau}\Phi(d_2), \end{aligned}ΘP=−Se−qτφ(d1) σ2τ−qSe−qτΦ(−d1)+rKe−rτΦ(−d2).\begin{aligned} \Theta_P &= -\frac{S e^{-q\tau}\varphi(d_1)\,\sigma}{2\sqrt{\tau}} \\[0.4em] &\quad - q S e^{-q\tau}\Phi(-d_1) \\[0.4em] &\quad + r K e^{-r\tau}\Phi(-d_2). \end{aligned}

In the cancellation formula with x=τx = \tau, the factor Se−qτS e^{-q\tau} has derivative −qSe−qτ-qS e^{-q\tau}, the factor Ke−rτK e^{-r\tau} has derivative −rKe−rτ-rK e^{-r\tau}, and ∂(d1−d2)/∂τ=σ/(2τ)\partial(d_1 - d_2)/\partial\tau = \sigma/(2\sqrt{\tau}). This gives ∂C/∂τ\partial C/\partial\tau, and ΘC=−∂C/∂τ\Theta_C = -\partial C/\partial\tau reverses every sign. The put follows from parity, ΘC−ΘP=qSe−qτ−rKe−rτ\Theta_C - \Theta_P = qS e^{-q\tau} - rK e^{-r\tau}. Both prices satisfy the Black-Scholes PDE, which in Greek notation reads

Θ+12σ2S2 Γ+(r−q) SΔ−rV=0,\begin{aligned} &\Theta + \tfrac12\sigma^2 S^2\,\Gamma \\[0.4em] &\quad + (r - q)\,S\Delta - rV = 0, \end{aligned}

and substituting the formulas checks it, since the gamma term cancels the density term of theta and the remaining terms cancel by the price formula. Theta is in currency per year. Its sign is not fixed, because the dividend and interest terms can outweigh the negative density term.

In plain words Theta is how the value changes as calendar time passes with the stock price and every other input unchanged. The PDE ties it to the other Greeks, so time change, convexity and the financing of the delta position must balance against the interest earned on the option value.

Example At the money with r=q=0r = q = 0, ΘC=−100⋅0.396953⋅0.2/2≈−3.969525\Theta_C = -100 \cdot 0.396953 \cdot 0.2/2 \approx -3.969525 per year, about −0.010875-0.010875 per calendar day under actual/365. Repricing with one day less to maturity gives a change of −0.010883-0.010883. The PDE reduces to Θ+12⋅0.04⋅104⋅0.019848≈−3.969525+3.969525=0\Theta + \tfrac12 \cdot 0.04 \cdot 10^4 \cdot 0.019848 \approx -3.969525 + 3.969525 = 0. For contrast, the third question below finds a positive theta for a deep in-the-money put.

Rho

Definition With SS, qq, σ\sigma and tt held fixed, the rhos of the call and the put are

ρC=Kτe−rτ Φ(d2),ρP=−Kτe−rτ Φ(−d2).\begin{gathered} \rho_C = K\tau e^{-r\tau}\,\Phi(d_2), \\[0.4em] \rho_P = -K\tau e^{-r\tau}\,\Phi(-d_2). \end{gathered}

Here ∂d1/∂r=∂d2/∂r=τ/σ\partial d_1/\partial r = \partial d_2/\partial r = \sqrt{\tau}/\sigma, so the density terms cancel, and differentiating −Ke−rτ-K e^{-r\tau} in rr gives Kτe−rτK\tau e^{-r\tau}. Parity gives ρC−ρP=Kτe−rτ\rho_C - \rho_P = K\tau e^{-r\tau}. Rho is in currency per unit of decimal rate, so one basis point is about 0.0001 ρ0.0001\,\rho. Because SS is held fixed, raising rr also raises the forward F=Se(r−q)τF = S e^{(r - q)\tau}. The sensitivity to the dividend yield, with rr fixed, is ∂C/∂q=−τSe−qτΦ(d1)\partial C/\partial q = -\tau S e^{-q\tau}\Phi(d_1). A curve bump followed by recalibration of other inputs gives a total derivative that can differ from this partial rho.

In plain words A higher rate lowers the present value of the strike, which the call holder pays and the put holder receives, so the call gains and the put loses. Rho is a slope at the current rate, so it is not zero just because the rate is zero.

Example At the money with r=0r = 0, ρC=100⋅1⋅Φ(−0.1)≈46.017216\rho_C = 100 \cdot 1 \cdot \Phi(-0.1) \approx 46.017216 and ρP≈−53.982784\rho_P \approx -53.982784, whose difference is Kτ=100K\tau = 100. Raising rr from 00 to one basis point raises the call by about 0.0046020.004602, as rho predicts. Raising it to 0.010.01 raises the call by 0.4677510.467751, slightly more than 0.01 ρC≈0.4601720.01\,\rho_C \approx 0.460172, because the price is curved in rr.

Check yourself

  1. For the constructed at-the-money example with S=K=100S = K = 100, r=q=0r = q = 0, σ=0.2\sigma = 0.2 and τ=1\tau = 1, the call has ΔC≈0.539828\Delta_C \approx 0.539828, Γ≈0.019848\Gamma \approx 0.019848, ν≈39.695255\nu \approx 39.695255, ΘC≈−3.969525\Theta_C \approx -3.969525 and ρC≈46.017216\rho_C \approx 46.017216. Use put-call parity to find the Greeks of the put with the same strike and maturity.
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Parity gives C−P=Se−qτ−Ke−rτC - P = S e^{-q\tau} - K e^{-r\tau}, and every Greek of the right side is easy to compute. Its SS derivative is e−qτ=1e^{-q\tau} = 1, so ΔP=0.539828−1=−0.460172\Delta_P = 0.539828 - 1 = -0.460172. Its second SS derivative and its σ\sigma derivative are 00, so the put has the same gamma 0.0198480.019848 and the same vega 39.69525539.695255. Its calendar time derivative is qSe−qτ−rKe−rτ=0qS e^{-q\tau} - rK e^{-r\tau} = 0 here, so ΘP≈−3.969525\Theta_P \approx -3.969525. Its rr derivative is Kτe−rτ=100K\tau e^{-r\tau} = 100, so

ρP≈46.017216−100=−53.982784.\begin{aligned} \rho_P &\approx 46.017216 - 100 \\[0.4em] &= -53.982784 . \end{aligned}

The prices agree too, P=C≈7.965567P = C \approx 7.965567, because the strike equals the forward when r=qr = q.

  1. A risk report shows a vega of 39.69539.695 for the same call. A colleague predicts a gain of about 79.479.4 if volatility rises from 2020 percent to 2222 percent, and a loss of about 3.973.97 over the next day from theta. Correct both estimates and compare the first with exact repricing.
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Vega is per unit of decimal volatility, and the move is Δσ=0.02\Delta\sigma = 0.02, not 22. The first-order estimate is

ν Δσ≈39.695255⋅0.02≈0.793905.\begin{aligned} \nu\,\Delta\sigma &\approx 39.695255 \cdot 0.02 \\[0.4em] &\approx 0.793905 . \end{aligned}

Repricing gives C(σ=0.22)≈8.759063C(\sigma = 0.22) \approx 8.759063, a gain of 0.7934950.793495, so the estimate is good because this at-the-money price is nearly linear in σ\sigma. Theta is per year, so one calendar day under actual/365 costs about 3.969525/365≈0.0108753.969525/365 \approx 0.010875, and repricing with one day less gives 0.0108830.010883. Using 252252 trading days instead would give about 0.0157520.015752, so the day convention has to be stated with the number.

  1. A constructed European put has S=60S = 60, K=100K = 100, r=0.05r = 0.05, q=0q = 0, σ=0.2\sigma = 0.2 and τ=1\tau = 1. Compute its theta and check the PDE identity with its delta, gamma and price. A classmate claims that options always lose value as time passes. What is wrong?
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Here d1≈−2.204128d_1 \approx -2.204128 and d2≈−2.404128d_2 \approx -2.404128, with φ(d1)≈0.035154\varphi(d_1) \approx 0.035154, Φ(−d1)≈0.986242\Phi(-d_1) \approx 0.986242 and Φ(−d2)≈0.991894\Phi(-d_2) \approx 0.991894. With q=0q = 0 the put theta has two terms. The density term is −60⋅0.035154⋅0.2/2≈−0.210921-60 \cdot 0.035154 \cdot 0.2/2 \approx -0.210921, and the interest term is rKe−rτΦ(−d2)≈0.05⋅95.122942⋅0.991894≈4.717596rK e^{-r\tau}\Phi(-d_2) \approx 0.05 \cdot 95.122942 \cdot 0.991894 \approx 4.717596, so

ΘP≈−0.210921+4.717596≈4.506674,\begin{aligned} \Theta_P &\approx -0.210921 + 4.717596 \\[0.4em] &\approx 4.506674, \end{aligned}

which is positive. The other values are P≈35.177379P \approx 35.177379, ΔP≈−0.986242\Delta_P \approx -0.986242 and Γ≈0.002929\Gamma \approx 0.002929. The PDE terms are 12⋅0.04⋅3600⋅Γ≈0.210921\tfrac12 \cdot 0.04 \cdot 3600 \cdot \Gamma \approx 0.210921, rSΔP≈−2.958727rS\Delta_P \approx -2.958727 and −rP≈−1.758869-rP \approx -1.758869, and with theta they add to 00. The claim fails because the put holder receives the strike only at TT, and with r>0r > 0 waiting raises its present value. The price 35.17737935.177379 lies below the intrinsic value 4040, since this European put cannot be exercised early, and with the stock unchanged it moves up toward that value as time passes, by about 0.0123510.012351 over one day.

How this lesson connects