Chapter 7, Problem 5

[dstefan]

Chapter 7, Problem 5

Solve for a and b the following system of equations:
\left\{ \begin{array}{rcl} 2a + 1 - \frac{2(r-q)}{\sigma^2} &=& 0; \\ b + a^2 + a \left( 1 - \frac{2(r-q)}{\sigma^2} \right) - \frac{2r}{\sigma^2} &=& 0. \end{array} \right.


Solution:

From the first equation, it is easy to see that a ~=~ \frac{r-q}{\sigma^2} - \frac{1}{2}.

Note that the the second equation can be written as
0 = b + a^2 - 2a \left( \frac{r-q}{\sigma^2} - \frac{1}{2} \right) ~-~ \frac{2r}{\sigma^2} = b + \left( a - \left( \frac{r-q}{\sigma^2} - \frac{1}{2} \right) \right)^2 ~-~ \left( \frac{r-q}{\sigma^2} - \frac{1}{2} \right)^2 ~-~ \frac{2r}{\sigma^2} = b + \left( a - \left( \frac{r-q}{\sigma^2} - \frac{1}{2} \right) \right)^2 - \left( \frac{r-q}{\sigma^2} + \frac{1}{2} \right)^2 + \frac{2(r-q)}{\sigma^2} - \frac{2r}{\sigma^2}
= b + \left( a - \left( \frac{r-q}{\sigma^2} - \frac{1}{2} \right) \right)^2 - \left( \frac{r-q}{\sigma^2} + \frac{1}{2} \right)^2 - \frac{2q}{\sigma^2}
Since a = \frac{r-q}{\sigma^2} - \frac{1}{2}, we find that b ~=~ \left( \frac{r-q}{\sigma^2} + \frac{1}{2} \right)^2 + \frac{2q}{\sigma^2}.

[Pravit]

I was a bit confused by this problem; with
a ~=~ \frac{r-q}{\sigma^2} - \frac{1}{2}.
b ~=~ \left( \frac{r-q}{\sigma^2} + \frac{1}{2} \right)^2
I get b + a^2 + a \left( 1 - \frac{2(r-q)}{\sigma^2} \right) - \frac{2r}{\sigma^2} = \frac{-2q}{\sigma^2}, which is only 0 when q=0. You mentioned that the second equation can be written as 0 = b + a^2 - 2a \left( \frac{r-q}{\sigma^2} - \frac{1}{2} \right) ~-~ \frac{2(r-q)}{\sigma^2}, but I didn't get how \frac{2r}{\sigma^2} can be replaced with \frac{2(r-q)}{\sigma^2}. Did I go wrong somewhere?

[doug reich]

I was a bit confused by this problem; with
a ~=~ \frac{r-q}{\sigma^2} - \frac{1}{2}.
b ~=~ \left( \frac{r-q}{\sigma^2} + \frac{1}{2} \right)^2
I get b + a^2 + a \left( 1 - \frac{2(r-q)}{\sigma^2} \right) - \frac{2r}{\sigma^2} = \frac{-2q}{\sigma^2}, which is only 0 when q=0. You mentioned that the second equation can be written as 0 = b + a^2 - 2a \left( \frac{r-q}{\sigma^2} - \frac{1}{2} \right) ~-~ \frac{2(r-q)}{\sigma^2}, but I didn't get how \frac{2r}{\sigma^2} can be replaced with \frac{2(r-q)}{\sigma^2}. Did I go wrong somewhere?

I'm guessing the problem description had a typo...

[dstefan]

I was a bit confused by this problem; with
a ~=~ \frac{r-q}{\sigma^2} - \frac{1}{2}.
b ~=~ \left( \frac{r-q}{\sigma^2} + \frac{1}{2} \right)^2
I get b + a^2 + a \left( 1 - \frac{2(r-q)}{\sigma^2} \right) - \frac{2r}{\sigma^2} = \frac{-2q}{\sigma^2}, which is only 0 when q=0. You mentioned that the second equation can be written as 0 = b + a^2 - 2a \left( \frac{r-q}{\sigma^2} - \frac{1}{2} \right) ~-~ \frac{2(r-q)}{\sigma^2}, but I didn't get how \frac{2r}{\sigma^2} can be replaced with \frac{2(r-q)}{\sigma^2}. Did I go wrong somewhere?

Indeed, the correct solution is
b ~=~ \left( \frac{r-q}{\sigma^2} + \frac{1}{2} \right)^2 ~+~ \frac{2q}{\sigma^2}

The solution above was corrected accordingly.

[dstefan]

I'm guessing the problem description had a typo...

The solution was actually not right - the system of equations was typo-less :)

[Pravit]

Thanks for the explanation, Dr. Stefanica. I also noticed the same typo in equation (7.36) on page 222.