A desk that needs a European option value across many spot levels and dates, together with delta and gamma, can solve the pricing PDE once on a grid instead of evaluating an expectation at every point. Whether the grid numbers can be trusted depends on the step sizes. In a constructed case with volatility and a log-price step of , the simplest scheme with a time step of year keeps the smoothing of the PDE, while a step of year makes the values oscillate and grow without bound. This lesson reduces the Black-Scholes equation to the heat equation, derives the explicit, implicit and Crank-Nicolson schemes, and explains the stability condition and the discrete maximum principle that separate them.
Key terms
Heat equation
Definition Assume the Black-Scholes setting with constant , and , and let be the value of a European claim with payoff , solving
for and , with . In the remaining time and the log price , the chain rule gives and , so
Writing with gives , and . The discount and drift terms cancel and leave the heat equation
which runs forward in , starting from the payoff. With it is well posed forward in time and smooths its initial data, while running it backward would be the ill-posed backward heat problem. When grows at most exponentially in , the solution given by Feynman-Kac is the Gaussian average with standard normal, since . With coefficients that depend on the state this reduction is not available, and a scheme has to discretize the full operator.
In plain words Measuring time backward from maturity turns the pricing equation into one that starts from the payoff and moves forward, and the log price together with a moving coordinate absorbs the discounting and the carry. What remains is pure diffusion, which spreads the payoff out more and more as the time to maturity grows.
Example A long forward with delivery price has the value . Since ,
Then , and is the payoff at . With the constructed inputs , and , the coefficients are and . The rates have disappeared from the heat equation itself and enter only through the coordinate and the factor .
Finite difference method
Definition The finite difference method replaces derivatives by difference quotients on a grid and solves the resulting algebraic equations. A uniform spatial grid cuts the line down to an interval with the nodes for , together with the time levels , and denotes the approximation of . For a function with four continuous derivatives, Taylor expansion gives
for some between and , and a one-sided difference in has an error of order . The values at and come from boundary conditions that approximate the contract far from the strike. The computed answer therefore carries a domain truncation error, which shrinks as the interval widens, besides the errors in and , which shrink as the grid is refined. A scheme is consistent when its difference equations approach the PDE as , and stable when, over a fixed horizon, it amplifies perturbations of its data by at most a fixed factor that does not grow as the grid is refined. For a well-posed linear initial value problem, the Lax equivalence theorem, stated here without proof, says that a consistent scheme converges exactly when it is stable.
In plain words The grid turns the PDE into a large set of linear equations, one for each node and time level. Three separate errors arise, from the spacing in space, from the step in time and from cutting off the unbounded range of prices, and each one has to be controlled on its own.
Example For and , the second difference equals
a relative error close to . For a call with , a grid in from to covers prices from about to about at . The boundary values at and at , the forward value of the previous card, differ from the exact call values there by less than for and .
Explicit scheme
Definition The explicit scheme, forward Euler in , evaluates the second difference at the old level,
With the ratio this becomes
for , with boundary values at and . Each new value comes directly from three old ones, with no linear system, and the local truncation error is of order for smooth solutions. The stability condition of this scheme, in this heat scaling, is
equivalently . Von Neumann analysis on an infinite grid inserts the Fourier mode , with the imaginary unit, and finds the amplification factor
which satisfies for every exactly when . Above one half the alternating mode , the pattern , is multiplied by at every step and grows without bound. On a fixed bounded grid with fixed boundary values the largest stable ratio is slightly above one half, and it tends to one half as , so is the condition that works for every grid.
In plain words The explicit step replaces each value by a weighted average of itself and its two neighbors. If the time step is too large for the space step, the center weight turns negative and the scheme amplifies the finest zigzag in the data instead of smoothing it. Halving forces down by a factor of four.
Example With , so , a space step and a time step year, the ratio is . Starting from the values with zeros beyond them, one step gives , and the alternating mode is multiplied by per step. With year the ratio is , the same data give the center value , and the alternating mode is multiplied by per step, so after ten steps it has grown by the factor . For this grid the largest stable step is year, and with it is year.
Discrete maximum principle
Definition A scheme satisfies the discrete maximum principle when every new interior value lies between the smallest and the largest of the values it is computed from, which are the old interior values and the boundary values used in the step. For the explicit scheme with , the weights , and are nonnegative and add to , so each new value is a convex combination of old values and
where and are the boundary values. As consequences, nonnegative data stay nonnegative, data between two bounds stay between them when the boundary values do too, and with zero boundary values the largest absolute value never increases, which proves stability in the maximum norm. For the center weight is negative and the principle fails. The implicit scheme satisfies the principle for every , and for Crank-Nicolson is a sufficient condition. The property is especially useful for comparison arguments and for American obstacle problems, whose algorithms lie beyond this lesson.
In plain words A heat flow cannot create a temperature hotter than the hottest point it started from, and a good grid scheme should not either. Option values computed by such a scheme never turn negative when the payoff and the boundary values are nonnegative, and never leave the range that those values set.
Example With the spike becomes , inside . With the explicit step gives at the center, a negative value from nonnegative data. The implicit scheme with the same , on a long grid with zero values far away, gives about at the center and at the neighbors, all inside , with the total still . Crank-Nicolson with gives about at the center, a negative value from a scheme that is nevertheless stable.
Implicit scheme
Definition The implicit scheme, backward Euler in , evaluates the second difference at the new level. With the discrete operator at the interior nodes it reads
where the boundary values and move to the right-hand side of the first and last rows. Each interior row is
The matrix is tridiagonal with diagonal and off-diagonal entries , so it is strictly diagonally dominant with nonpositive off-diagonal entries. It is therefore an M-matrix, whose inverse has only nonnegative entries, and the system is solved in a number of operations proportional to by the tridiagonal (Thomas) algorithm. The amplification factor
lies in for every , so the scheme is stable with no restriction on , and it satisfies the discrete maximum principle for every . Its error is of order for smooth solutions, only first order in time, so a large stable step can still be inaccurate. Unconditional stability is not unconditional accuracy.
In plain words The implicit step asks which new values, smoothed by one step of diffusion, give back the old values. That needs a linear system at every step, but a cheap one, and in return the time step is limited only by accuracy, never by stability.
Example With , where the explicit scheme failed, one implicit step on a long grid turns the spike into about around the center, positive values below that add to . The alternating mode is multiplied by per step, while the exact heat flow multiplies the same grid mode by
so the implicit step damps the finest zigzag correctly in sign but far too weakly in size.
Crank-Nicolson scheme
Definition The Crank-Nicolson scheme averages the explicit and implicit spatial differences,
with the boundary values moved to the right-hand side. Each interior row is
Its error is of order for sufficiently smooth solutions. The amplification factor
satisfies for every , so the scheme is stable in the discrete norm with no restriction on . Stability does not give the maximum principle. The left matrix is an M-matrix for every , and the right-hand weights , and are nonnegative when , so is a sufficient condition for the discrete maximum principle. For large the alternating mode has , close to , so it flips sign at every step and decays slowly, while the exact heat flow removes it almost at once. Near the kink of a call payoff this appears as oscillations in the value and especially in delta and gamma. A common remedy, Rannacher smoothing, takes a few implicit steps, often with half the step size, before switching to Crank-Nicolson.
In plain words Crank-Nicolson centers the time step, which makes it more accurate than either Euler step for smooth data. Its weakness is rough data with large steps, where it keeps the sign-flipping zigzag alive for many steps instead of smoothing it away.
Example With the alternating mode is multiplied by per step, so after ten steps its amplitude is still about of the original. The implicit factor is per step, and the exact factor is . On a long grid with zero values far away, one Crank-Nicolson step with turns the spike into a profile with center about and neighbors about . With the center stays positive at about , and with it is already negative at about .
Check yourself
- With and , find the heat coefficient , the largest stable time step of the explicit scheme, and the ratio and the three weights for year. How many explicit steps does one year need at least?
Show the answer
The coefficient is . Stability needs , that is
year. For the ratio is , and the weights are , and , nonnegative and adding to , so the step is stable and satisfies the discrete maximum principle. One year needs at least steps, taken with and exactly, where the center weight is . With it takes steps.
- With , a colleague uses and year. Find and decide whether the explicit scheme would be stable. The colleague then switches to Crank-Nicolson and argues that, since it is unconditionally stable, its values near the strike of a call cannot oscillate. Check the claim with the amplification factor of the alternating mode and the condition for the discrete maximum principle.
Show the answer
The ratio is . The explicit scheme would need year, and with its alternating mode would be multiplied by per step, so it would be unstable. For Crank-Nicolson
which is bounded, so the scheme is stable, but negative and close to . A zigzag flips sign at every step and after ten steps still keeps about of its size. The sufficient condition fails, since the center weight on the right-hand side is negative, and indeed one step from the spike gives about at the center. The kink of the call payoff contains such high-frequency components, so oscillations near the strike, most visible in gamma, are to be expected. Stability only bounds their size. A few implicit steps at the start, or smaller steps with there, are the usual ways to damp them.
- Take , a call with in heat coordinates, , and a grid with a node at , and year. Compute one explicit step at the strike node and compare it with the exact heat solution there. What happens when is halved repeatedly with fixed?
Show the answer
Here . At the strike node the payoff and its left neighbor are , and the right neighbor is , so one step gives . The exact solution is the Gaussian average of the payoff, which at the strike, with and , is
The error of one coarse step is about , roughly percent, because the grid is coarse compared with the spread and the payoff has a kink at the node, where the Taylor argument behind the local error estimate does not apply. Keeping , so that falls by four whenever halves, gives with and steps, and with and steps. The errors , and fall by about four per halving in this computation. That is an observation for this grid with the strike on a node, not a convergence theorem. Only enters , so and are not needed.