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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?

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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.

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  • $\begingroup$ 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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