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Below is a problem that I am working on. I believe that my incomplete solution is correct as far as it goes. I would like to know if my solution is incorrect. I plan to solve the system of two equations by using a computer program such as SciLab. What I am wondering is, does this system of equations have a unique solution? I believe it does but I cannot offer any real proof.
Problem:
Recall the following equations govern the price of a call option according to the Black-Scholes model. \begin{align*} c &= S_0 N(d_1) - Ke^{-rT}N(d_2) \\ d_1 &= \frac{ \ln{ \frac{S_0}{K} } + ( r + \frac{ {\sigma}^2}{2}) T } { \sigma \sqrt{T } } \\ d_2 &= d_1 - \sigma \sqrt{T} \\ \end{align*} Here is an explanation of the $6$ variables of the model: \begin{align*} c &-\text{The price of the call option} \\ S_0 &- \text{ The initial price of the stock } \\ K &- \text{ The strike price of the option } \\ T &- \text{ The time to expiration of the option.} \\ r &- \text{ The interest rate } \\ \sigma &- \text{ A measure of how volatily the price of the under lying stock is.} \end{align*} Suppose we have a stock with an initial price of $100$. There are two call options with strike prices $100$ and $105$ They both expire in exactly one year. The price of these options are $10$ and $8$ respectively. Find $r$ and $\sigma$. \newline Answer: \newline We have: \begin{align*} S_0 &= 100 \\ T &= 1 \\ c_1 &= 10 \\ K_1 &= 100 \\ c_2 &= 8 \\ K_2 &= 105 \\ \end{align*} Now we find $d_{11}$ By $d_{11}$, I mean $d_1$ for the first call option. Similar, by $d_{12}$, I mean $d_2$ for the first call option. \begin{align*} d_{11} &= \frac{ \ln{\left( \frac{100}{100} \right) } + ( r + \frac{ {\sigma}^2}{2}) 1 } { \sigma \sqrt{1 } } \\ d_{11} &= \frac{ \ln{\left( 1 \right) } + ( r + \frac{ {\sigma}^2}{2}) } { \sigma } \\ d_{11} &= \frac{ ( r + \frac{ {\sigma}^2}{2}) } { \sigma } \\ d_{11} &= \frac{ 2r + \sigma^2 } {2 \sigma } \\ d_{12} &= d_{11} - \sigma \sqrt{T} = \frac{ 2r + \sigma^2 } {2 \sigma } - \sigma \sqrt{1} \\ d_{12} &= \frac{ 2r + \sigma^2 } {2 \sigma } - \sigma = \frac{ 2r + \sigma^2 } {2 \sigma } - \frac{ 2\sigma^2}{2\sigma } \\ d_{12} &= \frac{ 2r - \sigma^2 } { 2 \sigma } \end{align*} Let $c_1$ be the price of first call option and $c_2$ be the price of the second call option. \begin{align*} c_1 &= 100 N\left(\frac{ 2r + \sigma^2 } {2 \sigma } \right) - 100e^{-rT} N\left(\frac{ 2r - \sigma^2 } { 2 \sigma }\right) \\ 10 &= 100 N\left(\frac{ 2r + \sigma^2 } {2 \sigma } \right) - 100e^{-rT} N\left(\frac{ 2r - \sigma^2 } { 2 \sigma }\right) \\ c_1 &= 100 N\left(\frac{ 2r + \sigma^2 } {2 \sigma } \right) - 100e^{-rT} N\left(\frac{ 2r - \sigma^2 } { 2 \sigma }\right) \\ 1 &= 10 N\left(\frac{ 2r + \sigma^2 } {2 \sigma } \right) - 10e^{-r} N\left(\frac{ 2r - \sigma^2 } { 2 \sigma }\right) \end{align*} Now we have one equation with two unknowns. We want two equations. \begin{align*} d_{21} &= \frac{ \ln{ \left( \frac{100}{105} \right) } + ( r + \frac{ {\sigma}^2}{2}) 1 } { \sigma \sqrt{1 } } \\ d_{21} &= \frac{ \ln{ \left( \frac{21}{20} \right) } + ( r + \frac{ {\sigma}^2}{2}) } { \sigma } \\ d_{21} &= \frac{ 0.0487902 + ( r + \frac{ {\sigma}^2}{2}) } { \sigma } \\ d_{21} &= \frac{ 2(0.0487902) + 2r + {\sigma}^2 } { 2 \sigma } \\ d_{21} &= \frac{ 2r + {\sigma}^2 + 0.0975804 } { 2 \sigma } \\ d_{22} &= d_{21} - \sigma \sqrt{T} = \frac{ 2r + {\sigma}^2 + 0.0975804 } { 2 \sigma } - \sigma \sqrt{1} \\ d_{22} &= \frac{ 2r - {\sigma}^2 + 0.0975804 } { 2 \sigma } \\ c_2 &= 100 N \left( \frac{ 2r + {\sigma}^2 + 0.0975804 } { 2 \sigma } \right) - 105 e^{-r\left( 1 \right) }N \left( \frac{ 2r - {\sigma}^2 + 0.0975804 } { 2 \sigma } \right) \\ 8 &= 100 N \left( \frac{ 2r + {\sigma}^2 + 0.0975804 } { 2 \sigma } \right) - 105 e^{-r}N \left( \frac{ 2r - {\sigma}^2 + 0.0975804 } { 2 \sigma } \right) \\ \end{align*} Here is the system of two equations that we need to solve: \begin{align*} 10 N\left(\frac{ 2r + \sigma^2 } {2 \sigma } \right) - 10e^{-r} N\left(\frac{ 2r - \sigma^2 } { 2 \sigma }\right) - 1 &= 0 \\ 100 N \left( \frac{ 2r + {\sigma}^2 + 0.0975804 } { 2 \sigma } \right) - 105 e^{-r}N \left( \frac{ 2r - {\sigma}^2 + 0.0975804 } { 2 \sigma } \right) - 8 &= 0 \\ \end{align*}

Using SciLab, I find \begin{align*} r &= -0.152815 \\ \sigma &= 0.396862 \end{align*} This solution cannot be right.

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At an informal level, this is a system of two nonlinear equations in two unknowns, hence you can plot it in the $(r,\sigma)$ plane and see how many times they cross each other.

At a more formal level, you can check if the Jacobian matrix is nonsingular everywhere. Nonsingularity of the Jacobian matrix (i.e., the determinant is not null) is a local argument for the uniqueness of solutions. So, you can take the first derivative of each equation with respect to each variable and compute the determinant of that matrix: \begin{align} \begin{bmatrix} 10 \phi(.)/\sigma - 10 \left( -\exp(-r) N(.) + \exp(-r) \phi(.)/\sigma \right) & 10 \phi(.) \left( \frac{1}{2} - \frac{r}{\sigma^2} \right) \left( 1 - \exp(-r) \right) \\ 100 \phi(.)/\sigma - 105 \left( -\exp(-r)N(.) + \exp(-r)\phi(.)/\sigma \right) & \phi(.) \left( \frac{1}{2} - \frac{r + \alpha}{\sigma^2} \right) \left( 100 - 105 \exp(-r) \phi(.) \right) \end{bmatrix} \end{align} where $\alpha = 0.0975804$, $\phi(.)$ is the standard normal density, $N(.)$ is the standard normal cumulative. Equations are on the lines and I go from the derivatives wrt. $r$ to those wrt. $\sigma$. Now, just check the determinant.

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  • $\begingroup$ Are you saying that if I have a system of non-linear equations and the Jacobian matrix is non-singular everywhere then that system has a unique solution? $\endgroup$ – Bob Apr 1 at 1:15

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