Unit 6, lesson 42

Black Scholes replication and the pricing PDE

Derives the Black-Scholes PDE from a self-financing delta hedge under frictionless assumptions, and explains the replication argument, why the drift drops out, and the terminal and boundary conditions.

  • Video 42
  • The Black-Scholes pricing equation

A dealer who sells a European claim on a stock wants to know how much money it takes to reproduce the claim exactly. Is there a portfolio of shares and cash, traded continuously with no money added or withdrawn, whose value equals the payoff at maturity on every path? If there is, its starting cost is the only price that leaves no arbitrage, and that cost does not depend on the stock’s expected return. This lesson builds the portfolio in the frictionless Black-Scholes market, derives the Black-Scholes PDE by matching its random and predictable parts with Itô’s formula, and states the terminal and boundary information that turn the equation into a pricing problem.

Key terms

Frictionless market assumptions

Definition The Black-Scholes market on a horizon [0,T][0, T] has two traded assets. The money-market account is Bt=ertB_t = e^{rt} with a constant continuously compounded rate rr. The stock has an ex-dividend price with

dSt=μSt dt+σSt dWtdS_t = \mu S_t\,dt + \sigma S_t\,dW_t

under the physical measure P\mathbb{P}, with a constant drift μ\mu, a constant volatility σ>0\sigma > 0, a starting price S0>0S_0 > 0 and a constant continuous dividend yield qq, so that one share pays cash at the rate qStq S_t. The filtration is the one generated by the single Brownian motion WW, and stock paths are continuous. The frictionless assumptions are that trading is continuous in time, any real number of shares may be held, short sales are allowed with full use of the proceeds, cash is lent and borrowed at the same rate rr, and there are no transaction costs, bid-ask spreads, taxes, margin limits or price impact. Trading strategies are predictable, satisfy the integrability that their stochastic integrals need, and are admissible. Admissibility is a technical restriction that rules out doubling strategies, for example a lower bound on the value or a square-integrability condition on the discounted gains under a pricing measure. The claim to be priced is European and pays g(ST)g(S_T) at TT.

In plain words The model removes every reason for a hedge to fail except price risk itself. One Brownian motion drives the only risky price, trades cost nothing and can happen at every instant, and cash is available without limit at a single rate. These are hypotheses of the model, not a description of real markets, and removing any one of them changes the mathematics.

Example The constructed claim of the later cards pays ST2/100S_T^2/100, with r=0.05r = 0.05, σ=0.2\sigma = 0.2, q=0q = 0 and one year to maturity. At S=100S = 100 its hedge holds about 2.18832.1883 shares worth 218.8349218.8349 and borrows 109.4174109.4174 in the cash account. If borrowing cost 22 percentage points more than lending, carrying that loan would cost about 109.4174⋅0.02≈2.1883109.4174 \cdot 0.02 \approx 2.1883 more per year at current levels, and the single rate rr in the PDE would no longer describe the hedge. A proportional cost per share traded would make continuous rebalancing unaffordable, because the holding moves with the Brownian path and the total number of shares bought and sold over any time interval is infinite. Discrete rebalancing instead leaves a hedging error, studied in the lesson on discrete delta hedging.

Self-financing delta hedge

Definition A trading strategy holds Δt\Delta_t shares and βt\beta_t units of the money-market account, so its value is Πt=ΔtSt+βtBt\Pi_t = \Delta_t S_t + \beta_t B_t and its cash holding is βtBt=Πt−ΔtSt\beta_t B_t = \Pi_t - \Delta_t S_t. It is self-financing when

dΠt=Δt(dSt+qSt dt)+r(Πt−ΔtSt) dt,\begin{aligned} d\Pi_t &= \Delta_t \bigl(dS_t + q S_t\,dt\bigr) \\[0.4em] &\quad + r\bigl(\Pi_t - \Delta_t S_t\bigr)\,dt , \end{aligned}

understood in integral form, with Δ\Delta predictable and ∫0TΔt2St2 dt<∞\int_0^T \Delta_t^2 S_t^2\,dt < \infty almost surely. The three terms are the gain on the shares, the dividends they receive and the interest on the cash holding, which is negative when cash is borrowed. This extends the discrete accounting of the lesson on self-financing portfolios. For strategies that are themselves Itô processes, the product rule gives

dΠt=Δt dSt+βt dBt+St dΔt+d[Δ,S]t+Bt dβt,\begin{aligned} d\Pi_t &= \Delta_t\,dS_t + \beta_t\,dB_t \\[0.4em] &\quad + S_t\,d\Delta_t + d[\Delta, S]_t \\[0.4em] &\quad + B_t\,d\beta_t , \end{aligned}

and the self-financing condition says that the last three terms add up to the dividends qΔtSt dtq \Delta_t S_t\,dt. Shares bought at the market price are paid for out of the cash account, so rebalancing by itself creates no gain. Given a pricing function V(t,S)V(t, S) with a continuous derivative ∂SV\partial_S V, the delta hedge is the self-financing strategy that holds

Δt=∂SV(t,St)\Delta_t = \partial_S V(t, S_t)

shares and keeps the rest of its value in cash. Its Brownian part is σStΔt dWt\sigma S_t \Delta_t\,dW_t, the same as the Brownian part σSt ∂SV(t,St) dWt\sigma S_t\,\partial_S V(t, S_t)\,dW_t of V(t,St)V(t, S_t) in Itô’s formula. The holding is chosen from the current time and price before the next Brownian increment, so it is predictable.

In plain words Self-financing means that nothing is added or taken out after the start. Purchases are paid from cash, sales and dividends go into cash, and cash earns or pays the rate rr. Holding ∂SV\partial_S V shares makes the portfolio and the claim respond to the same small price move by the same amount.

Example For the constructed claim with q=0q = 0, r=0.05r = 0.05, σ=0.2\sigma = 0.2 and one year to maturity, the pricing function verified in the Black-Scholes PDE card is

V(t,S)=S2100 e(r+σ2)(T−t).V(t, S) = \frac{S^2}{100}\, e^{(r + \sigma^2)(T - t)} .

At S=100S = 100 the value is 100e0.09≈109.4174100 e^{0.09} \approx 109.4174 and the hedge holds ∂SV=2V/S≈2.1883\partial_S V = 2V/S \approx 2.1883 shares. Their value is 218.8349218.8349, so the cash holding is 109.4174−218.8349=−109.4174109.4174 - 218.8349 = -109.4174, a loan. The claim and the hedge both have the Brownian coefficient σS ∂SV=0.4 V≈43.7670\sigma S\,\partial_S V = 0.4\,V \approx 43.7670 currency units per square root of a year. A mistaken accounting that differentiates Π=ΔS+cash\Pi = \Delta S + \text{cash} and books the term S dΔS\,d\Delta as a gain would count the purchase of shares as profit, although the cash paid for them leaves the same portfolio.

Replication argument

Definition Work under the frictionless assumptions with σ>0\sigma > 0, and let VV have continuous derivatives ∂tV\partial_t V, ∂SV\partial_S V and ∂SSV\partial_{SS} V on [0,T)×(0,∞)[0, T) \times (0, \infty). Suppose a self-financing strategy has value Πt=V(t,St)\Pi_t = V(t, S_t) for t<Tt < T. Itô’s formula gives

dV(t,St)=(∂tV+μS ∂SV+12σ2S2 ∂SSV) dt+σS ∂SV dWt,\begin{aligned} &dV(t, S_t) \\[0.4em] &= \bigl(\partial_t V + \mu S\,\partial_S V \\[0.4em] &\qquad + \tfrac12 \sigma^2 S^2\,\partial_{SS} V\bigr)\,dt \\[0.4em] &\quad + \sigma S\,\partial_S V\,dW_t , \end{aligned}

with the derivatives at (t,St)(t, S_t). The decomposition of a continuous Itô process into a dtdt part and a dWtdW_t part is unique, so the coefficients can be compared with those of the self-financing equation. The dWtdW_t coefficients give σStΔt=σSt ∂SV\sigma S_t \Delta_t = \sigma S_t\,\partial_S V, and since σSt>0\sigma S_t > 0 the strategy must be the delta hedge. With that holding the dtdt coefficients give

∂tV+μS ∂SV+12σ2S2 ∂SSV=(μ+q)S ∂SV+r(V−S ∂SV).\begin{aligned} &\partial_t V + \mu S\,\partial_S V \\[0.4em] &\quad + \tfrac12\sigma^2 S^2\,\partial_{SS} V \\[0.4em] &= (\mu + q) S\,\partial_S V \\[0.4em] &\quad + r\bigl(V - S\,\partial_S V\bigr). \end{aligned}

The term μS ∂SV\mu S\,\partial_S V appears on both sides and cancels, which leaves the Black-Scholes PDE along the path. For every t>0t > 0 the law of StS_t has a positive density on all of (0,∞)(0, \infty), so by continuity the PDE holds at every point of [0,T)×(0,∞)[0, T) \times (0, \infty). Conversely, suppose VV is also continuous on [0,T]×(0,∞)[0, T] \times (0, \infty), solves the PDE and satisfies V(T,S)=g(S)V(T, S) = g(S). Then Itô’s formula and the PDE show that the delta hedge started with V(0,S0)V(0, S_0) is self-financing with value V(t,St)V(t, S_t) at every t<Tt < T, and this value tends to g(ST)g(S_T) as t→Tt \to T. If the hedge and the opposite position are both admissible, no arbitrage forces the claim’s price at time tt to be V(t,St)V(t, S_t). At any other price, buying the cheaper of the claim and the hedge and selling the dearer locks in the difference with no risk, because the two pay the same at TT. This lesson derives the PDE in this way and verifies a smooth example. A general existence theory for nonsmooth payoffs, and a general representation of every claim as a hedge, have further hypotheses and are stated as boundaries rather than proved.

In plain words Matching the random parts picks the hedge, and matching the predictable parts then gives an equation for the price. The stock’s expected return enters the claim and the hedge in exactly the same way, so it cancels. A forecast of the drift can matter for investing, but it cannot change a price that is fixed by replication.

Example For the constructed claim at S=100S = 100 one year before maturity, S ∂SV=2VS\,\partial_S V = 2V and 12σ2S2 ∂SSV=0.04 V\tfrac12\sigma^2 S^2\,\partial_{SS} V = 0.04\,V. Take two analysts with the drift estimates μ=0.05\mu = 0.05 and μ=0.13\mu = 0.13. The Itô drift of VV is (−0.09+2μ+0.04)V(-0.09 + 2\mu + 0.04)V and the drift of the hedge is 2μV+0.05(V−2V)2\mu V + 0.05(V - 2V), and both equal (2μ−0.05)V(2\mu - 0.05)V for every μ\mu. That is 0.05 V≈5.47090.05\,V \approx 5.4709 per year for the first analyst and 0.21 V≈22.97770.21\,V \approx 22.9777 per year for the second. The two analysts disagree about how fast the claim’s value is expected to grow, but they compute the same price 109.4174109.4174.

Black-Scholes PDE

Definition Under the frictionless assumptions with constants rr, qq and σ>0\sigma > 0, the Black-Scholes PDE for a pricing function V(t,S)V(t, S) is

∂tV+12σ2S2 ∂SSV+(r−q)S ∂SV−rV=0\begin{aligned} &\partial_t V + \tfrac12\sigma^2 S^2\,\partial_{SS} V \\[0.4em] &\quad + (r - q) S\,\partial_S V \\[0.4em] &\quad - rV = 0 \end{aligned}

for 0≤t<T0 \le t < T and S>0S > 0, with q=0q = 0 for a dividend-free stock. It is linear, second order and parabolic, it is solved backward from maturity, and its coefficients do not contain μ\mu. The four terms are the time derivative in calendar time, the gamma term, the carry of the stock position at the rate r−qr - q and the financing of the claim’s value at the rate rr. In the log price x=ln⁡Sx = \ln S, where S ∂S=∂xS\,\partial_S = \partial_x and S2 ∂SS=∂xx−∂xS^2\,\partial_{SS} = \partial_{xx} - \partial_x, the equation has constant coefficients,

∂tV+12σ2 ∂xxV+(r−q−12σ2) ∂xV−rV=0,\begin{aligned} &\partial_t V + \tfrac12\sigma^2\,\partial_{xx} V \\[0.4em] &\quad + \bigl(r - q - \tfrac12\sigma^2\bigr)\,\partial_x V \\[0.4em] &\quad - rV = 0 , \end{aligned}

the form from which numerical schemes and the reduction to the heat equation start.

In plain words Write theta for ∂tV\partial_t V and gamma for ∂SSV\partial_{SS} V. The equation says that theta plus 12σ2S2\tfrac12\sigma^2 S^2 times gamma equals r(V−S ∂SV)+qS ∂SVr(V - S\,\partial_S V) + q S\,\partial_S V, the interest on the hedge’s cash plus the dividends on its shares. Along a delta-hedged position, time decay and convexity together earn exactly what the hedge’s cash and dividends earn. With r=q=0r = q = 0, a claim with positive gamma loses value as time passes at the rate 12σ2S2 ∂SSV\tfrac12\sigma^2 S^2\,\partial_{SS} V.

Example The constructed function V(t,S)=S2100e(r+σ2)(T−t)V(t, S) = \frac{S^2}{100} e^{(r + \sigma^2)(T - t)} with q=0q = 0 has ∂tV=−(r+σ2)V\partial_t V = -(r + \sigma^2)V, ∂SV=2V/S\partial_S V = 2V/S and ∂SSV=2V/S2\partial_{SS} V = 2V/S^2. With r=0.05r = 0.05, σ=0.2\sigma = 0.2, T−t=1T - t = 1 and S=100S = 100, the four terms of the PDE are −0.09 V-0.09\,V, 0.04 V0.04\,V, 0.10 V0.10\,V and −0.05 V-0.05\,V, which are

∂tV≈−9.8476,12σ2S2 ∂SSV≈4.3767,rS ∂SV≈10.9417,−rV≈−5.4709,\begin{aligned} \partial_t V &\approx -9.8476, \\[0.4em] \tfrac12\sigma^2 S^2\,\partial_{SS} V &\approx 4.3767, \\[0.4em] r S\,\partial_S V &\approx 10.9417, \\[0.4em] -rV &\approx -5.4709, \end{aligned}

and they add to zero. The same cancellation holds at every SS and every tt, since each term is a constant multiple of VV. Theta is −9.8476-9.8476 per year, about −0.0270-0.0270 per calendar day in a 365365-day year.

Terminal condition

Definition The terminal condition of a European claim with payoff gg is V(T,S)=g(S)V(T, S) = g(S) for every S>0S > 0, and the pricing function must approach it, V(t,S)→g(S)V(t, S) \to g(S) as t↑Tt \uparrow T. For a call with strike KK the payoff is g(S)=(S−K)+g(S) = (S - K)^{+}, and for a put it is g(S)=(K−S)+g(S) = (K - S)^{+}. The PDE alone has many solutions, and the terminal condition selects the one that belongs to the contract. The data are given at the final time and the equation is solved toward earlier times. In the remaining time τ=T−t\tau = T - t it becomes a forward parabolic equation with initial data gg, while prescribing the values at t=0t = 0 and solving forward in calendar time would be an ill-posed backward heat problem. A payoff need not be smooth. The call payoff has a kink at KK, so the smooth derivation is applied for t<Tt < T, where the model’s call price is smooth when σ>0\sigma > 0, and the payoff is reached as the limit at maturity. Weak and viscosity formulations handle nonsmooth problems without claiming classical derivatives everywhere, and they are not developed here.

In plain words The equation describes how a value changes, and the payoff says where it must end. Without the payoff the equation cannot tell a call from a forward contract or from a bond.

Example With q=0q = 0, the functions SS, Ke−r(T−t)K e^{-r(T - t)} and S2100e(r+σ2)(T−t)\frac{S^2}{100} e^{(r + \sigma^2)(T - t)} all solve the PDE. For the first, rS⋅1−rS=0rS \cdot 1 - rS = 0. For the second, the time derivative rKe−r(T−t)rK e^{-r(T - t)} cancels the financing term −rKe−r(T−t)-rK e^{-r(T - t)}. Their terminal values are SS, KK and S2/100S^2/100, the payoffs of a share, of a bond paying KK at TT and of the constructed claim. Because the PDE is linear, the difference S−Ke−r(T−t)S - K e^{-r(T - t)} is the solution with terminal value S−KS - K, a forward contract with delivery price KK. The constructed claim’s exponential factor equals 11 at t=Tt = T, so that function meets its payoff S2/100S^2/100 exactly at maturity.

Boundary conditions

Definition The Black-Scholes PDE is posed on the unbounded domain S>0S > 0, and a GBM price started above zero never reaches zero or infinity in finite time. No boundary value is imposed in the classical sense. Instead, the solution is required to lie in a growth class, for example ∣V(t,S)∣≤A(1+Sm+S−m)\lvert V(t, S) \rvert \le A(1 + S^m + S^{-m}) for constants AA and mm, which excludes solutions that grow too fast toward the edges of the domain. How such a growth class gives uniqueness is shown in the lesson on Feynman-Kac. The behavior near the edges then follows from the payoff. Setting S=0S = 0 in the equation removes the terms with derivatives in SS and leaves ∂tV−rV=0\partial_t V - rV = 0, which is consistent with the limits V(t,0+)=e−rτg(0)V(t, 0+) = e^{-r\tau} g(0) of a call and a put, where τ=T−t\tau = T - t. For a call, V(t,S)→0V(t, S) \to 0 as S→0S \to 0, and as S→∞S \to \infty

V(t,S)−Se−qτ+Ke−rτ→0.\begin{aligned} &V(t, S) - S e^{-q\tau} \\[0.4em] &\quad + K e^{-r\tau} \to 0 . \end{aligned}

For a put, V(t,S)→Ke−rτV(t, S) \to K e^{-r\tau} as S→0S \to 0 and V(t,S)→0V(t, S) \to 0 as S→∞S \to \infty. A numerical method on a truncated interval [Smin⁡,Smax⁡][S_{\min}, S_{\max}] needs conditions at the artificial edges that agree with these limits, such as V(t,Smax⁡)=Smax⁡e−qτ−Ke−rτV(t, S_{\max}) = S_{\max} e^{-q\tau} - K e^{-r\tau} for a call, and the error from the truncation shrinks as the interval widens.

In plain words On the whole half-line there is no edge where the stock can be stopped, so the payoff and a growth condition carry all the information. Far below the strike a call is nearly worthless, and far above it a call is worth almost exactly the discounted stock less the discounted strike. A computer grid has to stop somewhere, and the values it imposes at its edges must match this behavior.

Example Take K=100K = 100, r=0.05r = 0.05, q=0.02q = 0.02, σ=0.2\sigma = 0.2 and τ=1\tau = 1. A grid for the call that stops at Smax⁡=400S_{\max} = 400 sets the edge value

400 e−0.02−100 e−0.05≈392.0795−95.1229=296.9565,\begin{aligned} &400\, e^{-0.02} - 100\, e^{-0.05} \\[0.4em] &\quad \approx 392.0795 - 95.1229 \\[0.4em] &\quad = 296.9565 , \end{aligned}

and a grid that stops at Smax⁡=300S_{\max} = 300 sets about 198.9367198.9367. The exact call value from the formula of the lesson on the Black-Scholes formula differs from these edge values by about 5⋅10−85 \cdot 10^{-8} at 300300 and by less than 10−1110^{-11} at 400400. For the put, the edge value at S=0S = 0 is 100e−0.05≈95.1229100 e^{-0.05} \approx 95.1229, the discounted strike.

Check yourself

  1. With S=100S = 100, K=105K = 105, r=0.05r = 0.05, q=0.02q = 0.02 and τ=0.5\tau = 0.5, show that V(t,S)=Se−qτ−Ke−rτV(t, S) = S e^{-q\tau} - K e^{-r\tau} solves the Black-Scholes PDE with dividends and has terminal value S−KS - K. Find its value, its delta hedge and its cash holding, and compare the value with the forward price.
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Since ∂tτ=−1\partial_t \tau = -1, the derivatives are ∂tV=qSe−qτ−rKe−rτ\partial_t V = q S e^{-q\tau} - r K e^{-r\tau}, ∂SV=e−qτ\partial_S V = e^{-q\tau} and ∂SSV=0\partial_{SS} V = 0. Substituting,

qSe−qτ−rKe−rτ+(r−q)Se−qτ−rSe−qτ+rKe−rτ=0,\begin{aligned} &q S e^{-q\tau} - r K e^{-r\tau} \\[0.4em] &\quad + (r - q) S e^{-q\tau} \\[0.4em] &\quad - r S e^{-q\tau} + r K e^{-r\tau} = 0 , \end{aligned}

and at τ=0\tau = 0 the function equals S−KS - K. The value is 100e−0.01−105e−0.025≈99.0050−102.4075=−3.4026100 e^{-0.01} - 105 e^{-0.025} \approx 99.0050 - 102.4075 = -3.4026. The hedge holds e−0.01≈0.9900e^{-0.01} \approx 0.9900 shares and has the cash holding −105e−0.025≈−102.4075-105 e^{-0.025} \approx -102.4075. The forward price is Se(r−q)τ=100e0.015≈101.5113S e^{(r - q)\tau} = 100 e^{0.015} \approx 101.5113, the delivery price that would make this value zero. Here the delivery price 105105 is above it, so the long position has negative value. The value of an existing forward and the forward price are different quantities. As tt increases the share holding e−qτe^{-q\tau} rises toward 11, and the dividends received pay for exactly those extra shares, so the loan only accrues interest.

  1. For the constructed claim paying ST2/100S_T^2/100 with r=0.05r = 0.05, σ=0.2\sigma = 0.2 and τ=1\tau = 1, the stock now pays a continuous dividend yield q=0.02q = 0.02. Find the constant cc for which V=S2100ecτV = \frac{S^2}{100} e^{c\tau} solves the PDE, and the value at S=100S = 100. A classmate keeps the dividend-free value 109.4174109.4174. Which part of the hedge’s accounting was left out, and what residual does the dividend-free function leave in the new PDE?
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For this function ∂tV=−cV\partial_t V = -cV, 12σ2S2 ∂SSV=σ2V\tfrac12\sigma^2 S^2\,\partial_{SS} V = \sigma^2 V and (r−q)S ∂SV=2(r−q)V(r - q) S\,\partial_S V = 2(r - q)V, so the PDE requires −c+σ2+2(r−q)−r=0-c + \sigma^2 + 2(r - q) - r = 0, that is

c=r−2q+σ2=0.05.c = r - 2q + \sigma^2 = 0.05 .

The value at S=100S = 100 is 100e0.05≈105.1271100 e^{0.05} \approx 105.1271. The classmate left out the dividend cash qΔtSt dtq\Delta_t S_t\,dt that the hedge’s shares receive. Including it changes the coefficient of S ∂SVS\,\partial_S V from rr to r−qr - q. In the dividend PDE, the dividend-free function gives (−0.09+0.04+0.06−0.05)V=−0.04 V≈−4.3767(-0.09 + 0.04 + 0.06 - 0.05)V = -0.04\,V \approx -4.3767 per year instead of zero.

  1. Suppose σ=0\sigma = 0 and q=0q = 0, so the stock is riskless with dSt=μSt dtdS_t = \mu S_t\,dt. Why does the matching of Brownian coefficients no longer determine the hedge? Show that μ≠r\mu \ne r allows an arbitrage, and for μ=r\mu = r find the value of a call with S=K=100S = K = 100, r=0.05r = 0.05 and τ=1\tau = 1.
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With σ=0\sigma = 0 both the claim and every portfolio have zero Brownian coefficient, so the first matching step holds for every Δ\Delta and fixes nothing. The step that divides by σSt>0\sigma S_t > 0 fails, and the stock is a second riskless asset. If μ>r\mu > r, borrow S0S_0 at the rate rr and buy one share. At TT the share is worth S0eμTS_0 e^{\mu T} and the loan S0erTS_0 e^{rT}, a sure gain at zero cost, for example 100(e0.08−e0.05)≈3.2016100(e^{0.08} - e^{0.05}) \approx 3.2016 with μ=0.08\mu = 0.08 and T=1T = 1. If μ<r\mu < r, the reverse trade gains. With μ=r\mu = r the terminal price is SerτS e^{r\tau} for certain, so the call pays (Serτ−K)+(S e^{r\tau} - K)^{+} for certain and is worth

e−rτ(Serτ−K)+=(S−Ke−rτ)+≈4.8771.\begin{aligned} &e^{-r\tau}\bigl(S e^{r\tau} - K\bigr)^{+} \\[0.4em] &\quad = \bigl(S - K e^{-r\tau}\bigr)^{+} \\[0.4em] &\quad \approx 4.8771 . \end{aligned}

Wherever it is differentiable, this function solves ∂tV+rS ∂SV−rV=0\partial_t V + rS\,\partial_S V - rV = 0, the PDE with σ=0\sigma = 0.

How this lesson connects