Options (I): General Description, Parity Relations, Basic Concepts, and Valuation Using the Binomial Model

Abstract

Option theory, which was developed at the start of the 1970s by Black, Scholes, and Merton, constituted a major advance in economic and financial theory. The applications of this theory extend well beyond its use for options. Not only do numerous financial products have option components (convertible bonds, caps and floors, hybrid products, …, and even the bonds and shares issued by limited companies where there is a risk of bankruptcy) but many decisions have an aspect that can only be understood in terms of options (investments, analysis of credit risk, etc.). Option theory provides tools that not only allow to price optional components but also to manage portfolios of assets and liabilities that may include them. By greatly improving our understanding of financial mechanisms and risk management, option theory has significantly contributed to the increase in activity on financial markets.

Appendices

Appendix 1

1.1 Calibration of the Binomial Model

We start by computing the first three moments of a Bernoulli variable X that can take on two values u and d with respective probabilities q and 1 – q.

$$ {\displaystyle \begin{array}{l}E(X)= qu+\left(1\hbox{--} q\right)d\\ {}\mathit{\operatorname{var}}(X)=E\left({X}^2\right)\hbox{--} {\left[E(X)\right]}^2\\ {}={qu}^2+\left(1\hbox{--} q\right){d}^2\hbox{--} {q}^2{u}^2\hbox{--} {\left(1\hbox{--} q\right)}^2{d}^2+2q\left(1\hbox{--} q\right) ud\\ {}={u}^2q\left(1\hbox{--} q\right)+{d}^2\left(1\hbox{--} q\right)q+2q\left(1\hbox{--} q\right) ud,\end{array}} $$

(10.33)

so

$$ \operatorname{var}(X)=q\left(1-q\right){\left(u-d\right)}^2E{\left[X-E(X)\right]}^3=q{\left[u- qu-\left(1-q\right)d\right]}^3+\left(1-q\right)\ {\left[d- qu-\left(1-q\right)d\right]}^3=q{\left(1-q\right)}^3{\left(u-d\right)}^3+\left(1-q\right){q}^3{\left(d-u\right)}^3={\left(u-d\right)}^3\left[q\left(1-q\right)\left({\left(1-q\right)}^2-{q}^2\right)\right], $$

(10.34)

and it follows that

$$ E{\left[X\hbox{--} E(X)\right]}^3={\left(u\hbox{--} d\right)}^3q\left(1\hbox{--} q\left)\right(1\hbox{--} 2q\right). $$

(10.35)

We justify Calibration 2 (Jarrow and Rudd) given in Eqs. (10.32).

– On the one hand, in the Black-Scholes model, the logarithm of the stock price moves according to a Brownian process with independent increments of variance = σ² and mean = $\rho -\frac{\sigma^2}{2}$ per unit of time (i.e., σ² T/N and ($\rho -\frac{\sigma^2}{2}$)T/N for a period of length T/N). Furthermore, the third moment of each increment of lnS vanishes since it is a normal random variable (and hence has a symmetric distribution). Letting Δ_ilnS = ln(S_i + 1) – ln(S_i) (for the increment in an interval of duration T/N), we can therefore find binomial model parameters such that Δ_ilnS has the mean $\left(\rho -\frac{1}{2}{\sigma}^2\right)\frac{T}{N}$ and the variance ${\sigma}^2\frac{T}{N}$ (and, possibly, a vanishing third moment).

– The discretized binomial model means that Δ_ilnS = ln(S_i + 1) – ln(S_i) = ln(S_i + 1/S_i) is a binomial variable X taking on the two values lnu_N and lnd_N with respective probabilities q_N and 1 – q_N. The first two moments of Δ_ilnS are thus given by Eqs. (10.33) and (10.34) (when we replace u and d by lnu_N and lnd_N, and q by q_N). Equate the first two moments of the binomial process with those of the Black-Scholes process:

$$ {q}_N\ln {u}_N+\left(1\hbox{--} {q}_N\right)\ln {d}_N=\left(\rho -\frac{1}{2}{\sigma}^2\right)\frac{T}{N} $$

(10.36)

$$ {q}_N\left(1-{q}_N\right){\left(\ln {u}_N-\ln {d}_N\right)}^2={\sigma}^2\frac{T}{N}. $$

(10.37)

An infinity of triples q_N, lnu_N, lnd_N can satisfy these two equations.

If we impose the additional condition that the third moment is zero, the three variables sought will have to obey a third equation:

$$ {\left(\ln {u}_N\hbox{--} \ln {d}_N\right)}^3\ {q}_N\ \left(1\hbox{--} {q}_N\right)\ \left(1\hbox{--} 2{q}_N\right)=0. $$

This last equation requires the choice q_N =$\frac{1}{2}$.

From this, lnu_N and lnd_N are fixed by (10.36) and (10.37) with q_N = 0.5, which gives

$$ \ln {u}_N=\left(\rho -\frac{1}{2}{\sigma}^2\right)\frac{T}{N}+\sigma \sqrt{\frac{T}{N}}\kern1em \mathrm{and}\kern1em \ln {d}_N=\left(\rho -\frac{1}{2}{\sigma}^2\right)\frac{T}{N}-\sigma \sqrt{\frac{T}{N}}. $$

(Without necessarily solving these equations, the reader could just check they are satisfied for these two values). The triple q_N, lnu_N, lnd_N is indeed that of Calibration 2 (Eqs. (10.32)).

Let us note that ln(S_N) – ln(S₀) = $\sum \limits_{i=1}^N$ [ln(S_i) – ln(S_i–1)] is a sum of independent random variables that, by the central limit theorem, converges in law as N goes to infinity to a normal random variable with variance σ² T and mean $\left(\rho -\frac{1}{2}{\sigma}^2\right)T$. These are precisely the properties characteristic of the continuous Black-Scholes model (see the next chapter).

To finish, let us remark that for the values of u_N and d_N in Calibration 1 (Eqs. (10.31)), the variance of Δ_ilnS only equals ${\sigma}^2\frac{T}{N}$ for q_N = 0.5, which is not the exact value one has from Eqs. (10.31). However, the reader can check that the desired variance is attained asymptotically (since $\underset{N\to \infty }{\lim }{q}_N$ = 0.5).

*Appendix 2

Proof of Proposition 12

(convergence of the Cox-Ross-Rubinstein formula to the Black-Scholes formula)

We assume that the binomial model is calibrated using Eqs. (10.32) (Calibration 2 justified in Appendix 1), start from the results of Proposition 12, and show that ${\mathrm{Proba}}^{q_N}\left({S}_N\ge K\right)\to N\left({d}_2\right)$ and $\Pr {\mathrm{oba}}^{q{\hbox{'}}_N}\left({S}_N\ge K\right)\to N\left({d}_1\right)$ if N → ∞, with ${d}_1=\frac{\mathrm{In}\left({S}_0/K\right)+\left(\rho +{\sigma}^2/2\right)T}{\sigma \sqrt{T}}$ and ${d}_2=\frac{\mathrm{In}\left({S}_0/K\right)+\left(\rho -{\sigma}^2/2\right)T}{\sigma \sqrt{T}}.$

First, consider the value of S_N as a function of the discrete random variable j, the number of increases during the period [0, T]; since S_N = S₀ u_N^j d_N^N–j and u_N and d_N are given by Eq. (10.32), we have

$$ {S}_N={S}_0{e}^{\left(\rho -\frac{1}{2}{\sigma}^2\right)T}{e}^{2 j\sigma \sqrt{\frac{T}{N}}- N\sigma \sqrt{\frac{T}{N}}}. $$

We deduce from this that the event S_N ≥ K translates for j as

$$ j\ge \frac{\ln \left(K/{S}_0\right)-\left(\rho -\frac{1}{2}{\sigma}^2\right)T}{\sigma \sqrt{T}}\frac{\sqrt{N}}{2}+\frac{N}{2}. $$

(10.38)

Now let us examine the properties of the random variable j which is a binomial variable since it is the sum of N independent and identically distributed Bernoulli random variables. Indeed, define X_i for i = 1, …, N by: X_i = 1 if there is an increase between i and i + 1, and X_i = 0 otherwise.

Then ${}_{i=1}^N{X}_i$ counts the number of increases occurring during the N periods between dates 0 and T, and we have

$$ j={\sum \limits}_{i=1}^N{X}_i. $$

The Central Limit Theorem applied to $\frac{1}{N}{\sum \limits}_{i=1}^N{X}_i$ gives:

$$ \frac{\frac{1}{N}{\sum \limits}_{i=1}^N{X}_i-\frac{1}{N}E\left({}_{i=1}^N{X}_i\right)}{\frac{1}{N}\sigma \left({\sum \limits}_{i=1}^N{X}_i\right)}=\frac{{\sum \limits}_{i=1}^N{X}_i- NE\left({X}_i\right)}{\sqrt{N}\sigma \left({X}_i\right)}\to N\left(0,1\right). $$

In addition, E(X_i) = q_N and σ²(X_i) = q_N (1 – q_N). We then have

$$ \frac{j-{Nq}_N}{\sqrt{Nq_N\left(1-{q}_N\right)}}\to N\left(0,1\right). $$

We will make $\frac{j-{Nq}_N}{\sqrt{Nq_N\left(1-{q}_N\right)}}$ appear in Eq. (10.38):

$$ \frac{j-{Nq}_N}{\sqrt{Nq_N\left(1-{q}_N\right)}}\ge \left(\frac{\ln \frac{K}{S_0}-\left(\rho -\frac{1}{2}{\sigma}^2\right)T}{2\sigma \sqrt{T}}\sqrt{N}+\frac{N}{2}-{Nq}_N\right)\;\frac{1}{\sqrt{Nq_N\left(1-{q}_N\right)}}. $$

(10.39)

Let us use the calibration q_N = 1/2 (Calibration 2, Eqs. (10.32)). We then obtain E(X_i) = q_N = 1/2 and σ²(X_i) = q_N (1 – q_N) = 1/4. From this we have

$$ \frac{{\sum \limits}_{i=1}^N{X}_i- NE\left({X}_i\right)}{\sqrt{N}\sigma \left({X}_i\right)}=\frac{j-N/2}{\sqrt{N}/2}\to N\left(0,1\right). $$

Equation (10.39) becomes

$$ \frac{j-N/2}{\sqrt{N}/2}\ge \left(\left(\frac{\ln \frac{K}{S_0}-\left(\rho -\frac{1}{2}{\sigma}^2\right)T}{\sigma \sqrt{T}}\frac{\sqrt{N}}{2}+\frac{N}{2}\right)-\frac{N}{2}\right)\;\frac{1}{\sqrt{N}/2}=\left(\frac{\ln \frac{K}{S_0}-\left(\rho -\frac{1}{2}{\sigma}^2\right)T}{\sigma \sqrt{T}}\right)\equiv -{d}_2. $$

(10.40)

Thus we see that

$$ \Pr {\mathrm{oba}}^{q_N}\left({S}_N\ge K\right)=\Pr {\mathrm{oba}}^{q_N}\left(\frac{j-N/2}{\sqrt{N}/2}\ge -{d}_2\right)=\Pr \mathrm{oba}\left(U\ge -{d}_2\right) $$

where U is a reduced centered normal law under the probability being considered. N(u) is the standard notation for the distribution function of such a law:

$$ N(u)= Prob\left(U\le u\right). $$

By the law’s symmetry we have

$$ Prob\left(U\le {d}_2\right)= Prob\ \left(U\ge \hbox{--} {d}_2\right)=N\left({d}_2\right). $$

That finishes the proof of convergence for the second term of the call ${C}_0^{(N)}$. In fact, we have shown

$$ \Pr {\mathrm{oba}}^{q_N}\left({S}_N\ge K\right)\to N\left({d}_2\right). $$

To prove that $\Pr {\mathrm{oba}}^{q_N^{\prime }}\left({S}_N\ge K\right)\to N\left({d}_1\right)$, it is sufficient to repeat the reasoning while assigning q’_N to the event X_i = 1 with ${q}_N^{\prime }=\frac{q_N{u}_N}{e^{\rho T/N}}.$ Using Eq. (10.32) and a Taylor’s series expansion we obtain

$$ {q}_N^{\prime }=\frac{1}{2}\left(1+\sigma \sqrt{\frac{T}{N}}+o\left(\sqrt{\frac{T}{N}}\right)\right);q{'}_N\left(1\hbox{--} q{'}_N\right)=\frac{1}{4}\left(1+o\left(\sqrt{\frac{T}{N}}\right)\right). $$

So we have again inequality Eq. (10.39) with q’_N instead of q_N, by rewriting the second member as:

$$ \left(\left(\frac{\ln \frac{K}{S_0}-\left(\rho -\frac{1}{2}{\sigma}^2\right)T}{\sigma \sqrt{T}}\frac{\sqrt{N}}{2}+\frac{N}{2}\right)-\frac{N}{2}\left(1+\sigma \sqrt{\frac{T}{N}}+o\left(\sqrt{\frac{T}{N}}\right)\right)\right)\;\frac{2}{\sqrt{N}\sqrt{\left(1+o\left(\sqrt{\frac{T}{N}}\right)\right)}}=\left(\left(-{d}_2\frac{\sqrt{N}}{2}\right)-\frac{N}{2}\left(\sigma \sqrt{\frac{T}{N}}+o\left(\sqrt{\frac{T}{N}}\right)\right)\right)\;\frac{2}{\sqrt{N}\sqrt{\left(1+o\left(\sqrt{\frac{T}{N}}\right)\right)}}\kern6.1em =-{d}_2-\sigma \sqrt{T}+\sqrt{N}o\left(\sqrt{\frac{T}{N}}\right) $$

This term goes to $-{d}_2-\sigma \sqrt{T}$ = − d₁ as N tends to infinity, which proves the result.

The proof of convergence for a put can be inferred from this by using the call-put parity, or by replacing the event S_N ≥ K by the event S_N ≤ K.