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.