Suppose that we have the following equation:

$g = c\left( {\frac{{a + {\sigma ^2}}}{{b + {\sigma ^2}}}} \right)$

Is there anything that can be done to remove the $\sigma ^2$ term from the RHS of the equation so that the RHS $\rightarrow ca/b$, and the LHS of the equation has $\sigma ^2$ instead? Why or why not?

By dividing both sides by $\sigma ^2$, we get:

$\frac{g}{{{\sigma ^2}}} = c\left( {\frac{{\frac{a}{{{\sigma ^2}}} + 1}}{{b + {\sigma ^2}}}} \right)$

But this doesn't seem to get me any closer to my goal of removing the $\sigma ^2$ from the RHS. Maybe someone could point me in the right direction?


Since it's not homework, here's what you can do. You want to put the sigmas on the left, so $$ \begin{align*} g &= \frac{ca+c\sigma^2}{b+\sigma^2}&\text{so}\\ g(b+\sigma^2) &= ca+c\sigma^2&\text{an hence}\\ gb+g\sigma^2 &= ca+c\sigma^2&\text{collect the terms to get}\\ gb+g\sigma^2-c\sigma^2&=ca&\text{and divide both sides by }b\text{ to obtain}\\ g+\frac{g-c}{b}\sigma^2 &= \frac{ca}{b} \end{align*} $$ Presto! We've isolated the sigmas on one side (with a bit of extra stuff) and we've managed to get $ca/b$ on the right. Of course this works in this particular example; in general things might not be arranged so that you can always get the form you want.

  • $\begingroup$ That's wonderful, Rick - thank you for helping me see further! This is actually a simple "toy" problem that will help me put a more complicated equation in another form. $\endgroup$ – Nicholas Kinar Aug 23 '12 at 16:23

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