# Black-Scholes and solving for both $r$ and $\sigma$ ; Do I have a unique solution?

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.

• Why do you assume that your non-intuitive model outputs aren't the optimal solution to the problem posed? Are you sure that these aren't indeed the implicit assumptions baked into real prices? Aug 29, 2020 at 23:24
• Unsure why you assume the solution is not right. Alternatively, you can try to implement a quadtree decomposition algorithm yourself to double-check. Let's say you assume some bounds for your parameters: $r\in[r_l,r_h]$, $\sigma\in[\sigma_l,\sigma_h]$. Then your solution region is a rectangle. You then divide successively the area in 4 equal-area subregions. You carry on this decomposition for every subregion for which at least one vertex is of different sign than the others, until you've reached your tolerance level. Aug 29, 2020 at 23:55

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.