Alternative Methods to Derive Option Pricing Models

Abstract

The main purposes of this paper are: (1) to review three alternative methods for deriving option pricing models, (2) to discuss the relationship between binomial option pricing model and Black–Scholes model, (3) to compare Cox et al. method and Rendleman and Bartter method for deriving Black–Scholes model, (4) to discuss lognormal distribution method to derive Black–Scholes model, and (5) to show how the Black–Scholes model can be derived by stochastic calculus. This paper shows that the main methodologies used to derive the Black–Scholes model are: binomial distribution, lognormal distribution, and differential and integral calculus. If we assume risk neutrality, then we do not need stochastic calculus to derive the Black–Scholes model. However, the stochastic calculus approach for deriving the Black–Scholes model is still presented in this chapter. In sum, this paper can help statisticians and mathematicians understand how alternative methods can be used to derive the Black–Scholes option pricing model.

This chapter draws upon Lee et al. (2016) which was published in Review of Quantitative Finance and Accounting.

Appendix: The Relationship Between Binomial Distribution and Normal Distribution

In this appendix, we will use the de Moivre–Laplace Theorem to prove that the best fit between the binomial and normal distributions occurs when the binomial probability is $\frac{1}{2}$.

de Moivre–Laplace Theorem

As n grows larger and approaches infinity, for k in the neighborhood of np we can approximate

$$\begin{aligned} \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right)p^{k} q^{n - k} \simeq \frac{1}{{\sqrt {2\pi npq} }}e^{{ - \frac{{(k - np)^{2} }}{2npq}}} , \hfill \\ \quad p + q = 1,\;p,q > 0. \hfill \\ \end{aligned}$$

(21.120)

Proof

According to Stirling’s approximation (or Stirling’s formula) for factorials approximation, we can replace the factorial of large number n with the following:

$$n! \simeq n^{n} e^{ - n} \sqrt {2\pi n} \,{\text{as}}\,n \to \infty$$

(21.121)

Then, $\left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right)p^{k} q^{n - k}$ can be approximated as shown in the following procedures.

$$\begin{aligned} \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right)p^{k} q^{n - k} & = \frac{n!}{{k!\left( {n - k} \right)!}}p^{k} q^{n - k} \\ & \simeq \frac{{n^{n} e^{ - n} \sqrt {2\pi n} }}{{k^{k} e^{ - k} \sqrt {2\pi k} \left( {n - k} \right)^{n - k} e^{ - k} \sqrt {2\pi k} }}p^{k} q^{n - k} \\ & = \sqrt {\frac{n}{2\pi k(n - k)}} \left( {\frac{k}{np}} \right)^{ - k} \left( {\frac{n - k}{nq}} \right)^{{ - \left( {n - k} \right)}} \\ \end{aligned}$$

(21.122)

Let $x = \frac{k - np}{{\sqrt {npq} }}$, we obtain:

$$\begin{aligned} & \sqrt {\frac{n}{{2\pi k\left( {n - k} \right)}}} \left( {\frac{k}{np}} \right)^{ - k} \left( {\frac{n - k}{nq}} \right)^{{ - \left( {n - k} \right)}} \\ & \quad = \sqrt {\frac{n}{{2\pi k\left( {n - k} \right)}}} \left( {1 + x\sqrt {\frac{q}{np}} } \right)^{ - k} \left( {1 - x\sqrt {\frac{p}{nq}} } \right)^{ - (n - k)} \\ & \quad = \sqrt {\frac{{n^{ - 1} }}{{2\pi \tfrac{k}{n}\tfrac{{\left( {n - k} \right)}}{n}}}} \left( {1 + x\sqrt {\frac{q}{np}} } \right)^{ - k} \left( {1 - x\sqrt {\frac{p}{nq}} } \right)^{{ - \left( {n - k} \right)}} \\ & \quad = \sqrt {\frac{{n^{ - 1} }}{{2\pi \tfrac{k}{n}\left( {1 - \tfrac{k}{n}} \right)}}} \left( {1 + x\sqrt {\frac{q}{np}} } \right)^{ - k} \left( {1 - x\sqrt {\frac{p}{nq}} } \right)^{{ - \left( {n - k} \right)}} \\ \end{aligned}$$

(21.123)

As $k \to np$, we get $\tfrac{k}{n} \to p$. Then, Eq. (21.123) can be approximated as:

$$\begin{aligned} & \sqrt {\frac{{n^{ - 1} }}{{2\pi \tfrac{k}{n}\left( {1 - \tfrac{k}{n}} \right)}}} \left( {1 + x\sqrt {\frac{q}{np}} } \right)^{ - k} \left( {1 - x\sqrt {\frac{p}{nq}} } \right)^{{ - \left( {n - k} \right)}} \\ & \simeq \sqrt {\frac{{n^{ - 1} }}{2\pi pq}} \left( {1 + x\sqrt {\frac{q}{np}} } \right)^{ - k} \left( {1 - x\sqrt {\frac{p}{nq}} } \right)^{{ - \left( {n - k} \right)}} \\ & = \sqrt {\frac{1}{2\pi npq}} \exp \left\{ {\ln \left[ {\left( {1 + x\sqrt {\frac{q}{np}} } \right)^{ - k} } \right]} \right. \\ & \left. {\quad + \ln \left[ {\left( {1 - x\sqrt {\frac{p}{nq}} } \right)^{{ - \left( {n - k} \right)}} } \right]} \right\} \\ & = \sqrt {\frac{1}{2\pi npq}} \exp \left\{ { - k\ln \left( {1 + x\sqrt {\frac{q}{np}} } \right)} \right. \\ & \left. {\quad - \left( {n - k} \right)\ln \left( {1 - x\sqrt {\frac{p}{nq}} } \right)} \right\} \\ & = \sqrt {\frac{1}{2\pi npq}} \exp \left\{ { - \left( {np + x\sqrt {npq} } \right)\ln \left( {1 + x\sqrt {\frac{q}{np}} } \right)} \right. \\ & \left. {\quad - \left( {nq - x\sqrt {npq} } \right)\ln \left( {1 - x\sqrt {\frac{p}{nq}} } \right)} \right\} \\ \end{aligned}$$

(21.124)

We are considering the term in exponential function, i.e.,

$$- \left( {np + x\sqrt {npq} } \right)\ln \left( {1 + x\sqrt {\frac{q}{np}} } \right) - \left( {nq - x\sqrt {npq} } \right)\ln \left( {1 - x\sqrt {\frac{p}{nq}} } \right)$$

(21.125)

Here, we are using the Taylor series expansions of functions $\ln \left( {1 \pm x} \right)$:

$$\begin{aligned} \ln (1 + x) & = x - \frac{{x^{2} }}{2} + \frac{{x^{3} }}{3} + o\left( {x^{3} } \right) \\ \ln (1 - x) & = - x - \frac{{x^{2} }}{2} - \frac{{x^{3} }}{3} + o\left( {x^{3} } \right) \\ \end{aligned}$$

Then, we expand Eq. (21.125) with respect to x and obtain:

$$\begin{aligned} & - \left( {np + x\sqrt {npq} } \right)\left( {x\sqrt {\frac{q}{np}} - \frac{{x^{2} q}}{2np} + \frac{{x^{3} q^{{\tfrac{3}{2}}} }}{{3n^{{^{{\tfrac{3}{2}}} }} p^{{^{{\tfrac{3}{2}}} }} }} + o\left( {x^{3} } \right)} \right) \\ & - \left( {nq - x\sqrt {npq} } \right)\left( { - x\sqrt {\frac{p}{nq}} - \frac{{x^{2} p}}{2nq} - \frac{{x^{3} p^{{\tfrac{3}{2}}} }}{{3n^{{^{{\tfrac{3}{2}}} }} q^{{^{{\tfrac{3}{2}}} }} }} + o\left( {x^{3} } \right)} \right) \\ & \quad = - \left( {x\sqrt {npq} + x^{2} q - \frac{1}{2}x^{2} q - \frac{{x^{3} p^{{ - \tfrac{1}{2}}} q^{{\tfrac{3}{2}}} }}{2\sqrt n } + \frac{{x^{3} p^{{ - \tfrac{1}{2}}} q^{{\tfrac{3}{2}}} }}{3\sqrt n }} \right) \\ & \quad - \left( { - x\sqrt {npq} + x^{2} p - \frac{1}{2}x^{2} q + \frac{{x^{3} p^{{\tfrac{3}{2}}} q^{{ - \tfrac{1}{2}}} }}{2\sqrt n } - \frac{{x^{3} p^{{\tfrac{3}{2}}} q^{{ - \tfrac{1}{2}}} }}{3\sqrt n }} \right) \\ & \quad + o\left( {x^{3} } \right) \\ & \quad = - \frac{1}{2}x^{2} \left( {p + q} \right) - \frac{1}{{6\sqrt {npq} }}x^{3} \left( {p^{2} - q^{2} } \right) + o\left( {x^{3} } \right) \\ \end{aligned}$$

(21.126)

Since we have $p + q = 1$ when we ignore the higher order of x, Eq. (21.126) can be simply approximated to:

$$- \frac{1}{2}x^{2} - \frac{1}{{6\sqrt {npq} }}x^{3} \left( {p - q} \right)$$

(21.127)

Then, we replace Eq. (21.127) in the exponential function in Eq. (21.124), and we obtain:

$$\begin{aligned} \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right)p^{k} q^{n - k} & \simeq \frac{1}{{\sqrt {2\pi pq} }}\exp \left[ { - \frac{1}{2}x^{2} } \right. \\ & \quad \left. { - \frac{1}{{6\sqrt {npq} }}x^{3} \left( {p - q} \right)} \right] \\ \end{aligned}$$

(21.128)

Although $- \frac{1}{{6\sqrt {npq} }}x^{3} \left( {p - q} \right) \to 0$ as $n \to \infty$, the term $- \frac{1}{{6\sqrt {npq} }}x^{3} \left( {p - q} \right)$ will be exactly zero if and only if $p = q$. Under this condition,

$$\begin{aligned} \left( {\begin{array}{*{20}c} n \\ k \\ \end{array} } \right)p^{k} q^{n - k} & \simeq \frac{1}{{\sqrt {2\pi pq} }}\exp \left( { - \frac{1}{2}x^{2} } \right) \\ & = \frac{1}{{\sqrt {2\pi pq} }}\exp \left[ { - \frac{{\left( {k - np} \right)^{2} }}{2npq}} \right]. \\ \end{aligned}$$

Thus, it is shown that the best fit between the binomial and normal distribution occurs when $p = q = \frac{1}{2}$.

If $p \ne q$, then there exists an additional term $- \frac{1}{{6\sqrt {npq} }}x^{3} \left( {p - q} \right)$. It is obvious that $\sqrt {pq} = \sqrt {p\left( {p - q} \right)}$ will reach maximum if and only if $p = q = \frac{1}{2}$. Therefore, when n is fixed, if the difference between p and q becomes larger, the absolute value of an additional term $- \frac{1}{{6\sqrt {npq} }}x^{3} \left( {p - q} \right)$ will be larger. This implies that the magnitude of absolute value of the difference between p and q is an important factor to make the approximation to normal distribution less precise. We use the following figures to demonstrate how the absolute number of differences between p and q affect the precision of using binomial distribution to approximate normal distribution.

From Figs. 21.5 and 21.6, we find that when $p \ne q$, the absolute magnitude does affect the estimated continuous distribution as indicated by red solid curves. For example, when $n = 30$, the red solid curve when $p = 0.9$ is very much different from $p = 0.5$. In other words, when $p = 0.9$, the red solid curve is not as similar to the normal curve as $p = 0.5$. If we increase n from 30 to 100, the solid red curve from $p = 0.9$ is less different from the solid red curve when $p = 0.5$. In sum, both the magnitude of n and p will affect the shape of using normal distribution to approximate binomial distribution.

Fig. 21.5

Binomial distributions to approximate normal distributions $\left( {n = 30} \right)$

Fig. 21.6

Binomial distributions to approximate normal distributions $\left( {n = 100} \right)$

From Eqs. (21.15) and (21.16) in the text, we can define the binomial OPM and the Black–Scholes OPM as follows:

$$C = SB_{1} (a;n,p^{\prime } ) - \frac{X}{{r^{n} }}B_{2} (a;n,p)$$

(21.15)

$$C = SN(d_{1} ) - Xe^{ - rT} N(d_{2} )$$

(21.16)

Both Cox et al. and Rendleman and Bartter tried to show the binomial cumulative functions of Eq. (21.15) will converge to the normal cumulative function of Eq. (21.16) when n approaches infinity. In this appendix, we have mathematically and graphically showed that the relative magnitude between p and q is the important factor to determine this approximation when n is constant. In addition, we also demonstrate the size of n which also affects the precision of this approximation process.