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This example explores basic arbitrage concepts in a single-period, two-state asset portfolio. The portfolio consists of a bond, a long stock, and a long call option on the stock.

It uses these Symbolic Math Toolbox™ functions:

`equationsToMatrix`

to convert a linear system of equations to a matrix.`linsolve`

to solve the system.Symbolic equivalents of standard MATLAB® functions, such as

`diag`

.

This example symbolically derives the risk-neutral probabilities and call price for a single-period, two-state scenario.

Create the symbolic variable `r`

representing the risk-free rate over the period. Set the assumption that `r`

is a positive value.

syms r positive

Define the parameters for the beginning of a single period, `time = 0`

. Here `S0`

is the stock price, and `C0`

is the call option price with strike, `K`

.

syms S0 C0 K positive

Now, define the parameters for the end of a period, `time = 1`

. Label the two possible states at the end of the period as U (the stock price over this period goes up) and D (the stock price over this period goes down). Thus, `SU`

and `SD`

are the stock prices at states U and D, and `CU`

is the value of the call at state U. Note that $$SD<=K<=SU$$.

syms SU SD CU positive

The bond price at `time = 0`

is 1. Note that this example ignores friction costs.

Collect the prices at `time = 0`

into a column vector.

prices = [1 S0 C0]'

prices =$$\left(\begin{array}{c}1\\ {S}_{0}\\ {C}_{0}\end{array}\right)$$

Collect the payoffs of the portfolio at `time = 1`

into the `payoff`

matrix. The columns of `payoff`

correspond to payoffs for states D and U. The rows correspond to payoffs for bond, stock, and call. The payoff for the bond is `1 + r`

. The payoff for the call in state D is zero since it is not exercised (because $$SD<=K$$).

payoff = [(1 + r), (1 + r); SD, SU; 0, CU]

payoff =$$\left(\begin{array}{cc}r+1& r+1\\ \mathrm{SD}& \mathrm{SU}\\ 0& \mathrm{CU}\end{array}\right)$$

`CU`

is worth `SU - K`

in state U. Substitute this value in `payoff`

.

payoff = subs(payoff, CU, SU - K)

payoff =$$\left(\begin{array}{cc}r+1& r+1\\ \mathrm{SD}& \mathrm{SU}\\ 0& \mathrm{SU}-K\end{array}\right)$$

Define the probabilities of reaching states U and D.

syms pU pD real

Under no-arbitrage, `eqns == 0`

must always hold true with positive `pU`

and `pD`

.

eqns = payoff*[pD; pU] - prices

eqns =$$\left(\begin{array}{c}\mathrm{pD}\hspace{0.17em}\left(r+1\right)+\mathrm{pU}\hspace{0.17em}\left(r+1\right)-1\\ \mathrm{SD}\hspace{0.17em}\mathrm{pD}-{S}_{0}+\mathrm{SU}\hspace{0.17em}\mathrm{pU}\\ -{C}_{0}-\mathrm{pU}\hspace{0.17em}\left(K-\mathrm{SU}\right)\end{array}\right)$$

Transform equations to use risk-neutral probabilities.

syms pDrn pUrn real; eqns = subs(eqns, [pD; pU], [pDrn; pUrn]/(1 + r))

eqns =$$\left(\begin{array}{c}\mathrm{pDrn}+\mathrm{pUrn}-1\\ \frac{\mathrm{SD}\hspace{0.17em}\mathrm{pDrn}}{r+1}-{S}_{0}+\frac{\mathrm{SU}\hspace{0.17em}\mathrm{pUrn}}{r+1}\\ -{C}_{0}-\frac{\mathrm{pUrn}\hspace{0.17em}\left(K-\mathrm{SU}\right)}{r+1}\end{array}\right)$$

The unknown variables are `pDrn`

, `pUrn`

, and `C0`

. Transform the linear system to a matrix form using these unknown variables.

[A, b] = equationsToMatrix(eqns, [pDrn, pUrn, C0]')

A =$$\left(\begin{array}{ccc}1& 1& 0\\ \frac{\mathrm{SD}}{r+1}& \frac{\mathrm{SU}}{r+1}& 0\\ 0& -\frac{K-\mathrm{SU}}{r+1}& -1\end{array}\right)$$

b =$$\left(\begin{array}{c}1\\ {S}_{0}\\ 0\end{array}\right)$$

Using `linsolve`

, find the solution for the risk-neutral probabilities and call price.

x = linsolve(A, b)

x =$$\left(\begin{array}{c}\frac{{S}_{0}-\mathrm{SU}+{S}_{0}\hspace{0.17em}r}{\mathrm{SD}-\mathrm{SU}}\\ -\frac{{S}_{0}-\mathrm{SD}+{S}_{0}\hspace{0.17em}r}{\mathrm{SD}-\mathrm{SU}}\\ \frac{\left(K-\mathrm{SU}\right)\hspace{0.17em}\left({S}_{0}-\mathrm{SD}+{S}_{0}\hspace{0.17em}r\right)}{\left(\mathrm{SD}-\mathrm{SU}\right)\hspace{0.17em}\left(r+1\right)}\end{array}\right)$$

Verify that under risk-neutral probabilities, `x(1:2)`

, the expected rate of return for the portfolio, `E_return`

equals the risk-free rate, `r`

.

E_return = diag(prices)\(payoff - [prices,prices])*x(1:2); E_return = simplify(subs(E_return, C0, x(3)))

E_return =$$\left(\begin{array}{c}r\\ r\\ r\end{array}\right)$$

As an example of testing no-arbitrage violations, use the following values: `r = 5%`

, `S0 = 100`

, and `K = 100`

. For `SU < 105`

, the no-arbitrage condition is violated because `pDrn = xSol(1)`

is negative (`SU >= SD`

). Further, for any call price other than `xSol(3)`

, there is arbitrage.

xSol = simplify(subs(x, [r,S0,K], [0.05,100,100]))

xSol =$$\left(\begin{array}{c}-\frac{\mathrm{SU}-105}{\mathrm{SD}-\mathrm{SU}}\\ \frac{\mathrm{SD}-105}{\mathrm{SD}-\mathrm{SU}}\\ \frac{20\hspace{0.17em}\left(\mathrm{SD}-105\right)\hspace{0.17em}\left(\mathrm{SU}-100\right)}{21\hspace{0.17em}\left(\mathrm{SD}-\mathrm{SU}\right)}\end{array}\right)$$

Plot the call price, `C0 = xSol(3)`

, for `50 <= SD <= 100`

and `105 <= SU <= 150`

. Note that the call is worth more when the "variance" of the underlying stock price is higher for example, `SD = 50, SU = 150`

.

fsurf(xSol(3), [50,100,105,150]) xlabel SD ylabel SU title 'Call Price'

`Advanced Derivatives, Pricing and Risk Management: Theory, Tools and Programming Applications`

by Albanese, C., Campolieti, G.