Deriving Payoff Bounds for Black-Scholes Numerical Integration

Summary

The document asks how to price a derivative in the Black-Scholes model whose payoff is the cosine of the terminal asset price, paid only when that price lies between specified barriers. The question focuses on applying a four-point trapezium rule to a normal-variable integral and on how to find the integration endpoints. The accepted answer clarifies that the payoff conditions determine those endpoints: solve the lognormal mapping from the standard normal variable to the asset price at each barrier, then integrate over the resulting interval.

The trapezium rule itself only approximates an integral once its domain is known; it does not determine the payoff bounds. The document does not provide the numerical quadrature steps or final option value, and it does not discuss discretization error. Its main lesson is the separation of payoff-bound derivation from the numerical integration method used to evaluate the price.

Key ideas

• A payoff restricted to a range of terminal asset prices induces corresponding bounds on the normal variable.
• Find those bounds by inverting the Black-Scholes lognormal mapping at each payoff threshold.
• Apply the trapezium rule only after establishing the correct integration interval.
• The discussion explains the domain setup but does not calculate the final numerical price.

Tags

# 4-point Trapezium rule for numerical integration


# 4-point Trapezium rule for numerical integration












Background:

This is in reference to Mark Joshi's concepts of mathematical finance ch.7 problem 11.

Question:

> We have in the Black-Scholes model: $S_0 = 1, T = 1, \sigma = 0.1, r = 0$. A derivative pays $\cos(S_1)$ at time $1$ if $S_1$ is between $1$ and $2$. Find the price implied by a $4$-point trapezium rule numeric integration.

Now I understand that we need to evaluate

$$\mathbb{E}\left(\cos(S_1)\right) = \int \cos\left(e^{(r - 0.5\sigma^2)T + \sigma\sqrt{T}z})\right)e^{-z^2/2}dz$$

The odd part of his solution is he is evaluating this integral from $z_1$ to $z_2$ and states that this $z_j$ mapping $z$to $j$ and solving for $z_1 = 0.05, z_2 = 6.981$.

The formula in the book states that to solve using the trapezium method. If we wish to integrate a function $g(x)$ over an interval $[a, b]$ then we divide the interval into $N$ pieces of equal length. Thus we set

$$x_j = a + \frac{j}{N}(b-a)$$

for $j = 0,\ldots, N$

I don't see how any reader can take that information and solve for $z_1$ and $z_2$. Any suggestions on this are greatly appreciated.

## Answer by Andrew (score 3, accepted)

https://quant.stackexchange.com/a/38016

This has nothing to do with the trapezium rule. The derivative pays $cos(S_1)$ if $1<=S_1<=2$. Solve $e^{(r-0.5\sigma^2)T+\sigma\sqrt{T}z)}=1$ and $e^{(r-0.5\sigma^2)T+\sigma\sqrt{T}z)}=2$ to obtain the bounds of the integral.

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