I can't seem to solve/simplify step by step to get from equation 3) to 4) as they do in this paper. As per the paper:
3) $p = \frac{p' + (r-R)p}{1+r}$
Because both sides of equation 3) involve the current price, $p$, the equation can be solved as follows:
4) $p = \frac{p'}{1+R}$
Could somebody please step by step show how I'd solve this step please?
Best Answer
We start with the original equation: \begin{align*} p = \frac{p' + (r-R)p}{1+r} \end{align*} We want to obtain $p$ as a function of the other values. So, we first want to collect all terms with $p$ on the same side. We do this by splitting the fraction and bringing one term over. \begin{align*} p &= \frac{p'}{1+r} + \frac{(r-R)p}{1+r} \\ p - \frac{(r-R)p}{1+r} &= \frac{p'}{1+r} \end{align*} Now, we can factor out the $p$ on the left side, and put everything on the same denominator. \begin{align*} p \left( 1 - \frac{(r-R)}{1+r} \right) &= \frac{p'}{1+r} \\ p \left( \frac{1+r}{1+r} - \frac{r-R}{1+r} \right) &= \frac{p'}{1+r} \\ p \left( \frac{1+R}{1+r} \right) &= \frac{p'}{1+r} \end{align*} (Note of course that $(1+r)-(r-R) = (1+r) + (-r +R) = 1+R$.). Finally, we multiply both sides of the equation by $(1+r)/(1+R)$ to get: \begin{align*} p \left( \frac{1+R}{1+r} \right) \cdot \frac{1+r}{1+R} &= \frac{p'}{1+r} \cdot \frac{1+r}{1+R} \\ p = \frac{p'}{1+R} \end{align*} And this is the desired result.