# How to decompose the following expression?

1. Jan 18, 2015

### Alec V

This comes from one of the research papers that I'm reading. The authors decompose Eq. 1 and obtain Eq. 2. After several attempts, I have been fairly unsuccessful at obtaining Eq. 2 from Eq. 1.

I have tried using partial fractions decomposition without much success. Any help would be greatly appreciated.

Eq. 1

Eq. 2

Thanks!

2. Jan 18, 2015

### Staff: Mentor

Try to find the other direction, that is easier (and all steps work both ways). You can simplify equation 2 and get equation 1.

3. Jan 18, 2015

### Stephen Tashi

You have to explain the relations that exist among the symbols to get help on this question.

4. Jan 18, 2015

### Staff: Mentor

Yes, especially $\Pi_{s1}$ and $\Pi_{sS2}$. Are they just numbers or is something else going on? The symbol $\Pi$ is usually used to denote a product.

5. Jan 18, 2015

### Alec V

Sorry, I should have clarified. ∏s1 is profit in country 1, ∏s2 is profit in country 2, ∈s1 is risk in country 1, ∈s2 is risk in country 2, and δ is the discount rate.

Eq. 2 tries to separate risk elements (∈s1 and ∈s2) from the profit gap between countries. Hope this helps.

6. Jan 18, 2015

### Staff: Mentor

∏ as variable is a bit unusual, but as long as all symbols are just variables, their meaning does not matter. This just needs the basic rules to work with fractions.

7. Jan 19, 2015

### Stephen Tashi

The results of a simple numerical test did produce identical answers for EPV. (Of course someone should check my program.)

The output

8. Jan 19, 2015

### Stephen Tashi

Getting rid of terms in a denominator that don't cancel with anything in the numerator is a frequent mathematical daydream. Suppose we approach the problem as the desire to express:

$( \frac{A}{W + F} - \frac{B}{W + G})$ as $( \frac{A}{W} - \frac{B}{W}) \ + \ L$

where $L$ is the leftover stuff.

Then
$L = \frac{A}{W+F} - \frac{A}{W} - \frac{B}{W + G} + \frac{B}{W }$

$= \frac{ AW - A(W+F)} {W(W+F)} + \frac{-BW+B(W+G)}{W(W+G)}$

$= \frac{-AF}{W(W+F)} + \frac{BG}{W(W+G)}$

$= \frac{1}{W} ( \frac{BG}{W+G} - \frac{AF}{W+F})$

In the problem at hand, there are relations between $W, F, G$

$W = 1 -\delta$
$F = \delta \epsilon_1 = (1-W)\epsilon_1$
$G = \delta \epsilon_2 = (1-W)\epsilon_2$

Substituting in selected places for $F$ and $G$

$L = \frac{1}{W}( \frac{B(1-W)\epsilon_2}{W+G} - \frac{A(1-W)\epsilon_1}{W+F} )$

$= \frac{1-W}{W} ( \frac{ \epsilon_2 B}{W+G} - \frac{\epsilon_1 A}{W+F})$

$= \frac{1-W}{W} ( \frac{ (W+F)\epsilon_2 B - (W+G)\epsilon_1A}{(W+G)(W+F)})$

In the problem at hand:
$A = \prod_{s1}$
$B = \prod_{s2}$
$W = (1 -\delta)$
$F = \delta \epsilon_{s1}$
$G = \delta \epsilon_{s2}$
$\epsilon_1 = \epsilon_{s1}$
$\epsilon_2 = \epsilon_{s2}$.

9. Jan 31, 2015

### Alec V

Thanks so much! The simplification of L makes sense. Could you please dwell on how to separate $( \frac{A}{W} - \frac{B}{W})$ from $( \frac{A}{W + F} - \frac{B}{W + G})$?

10. Jan 31, 2015

### Stephen Tashi

I'm not sure what you are asking.

11. Feb 1, 2015

### Alec V

So, we expressed $( \frac{A}{W + F} - \frac{B}{W + G})$ as $( \frac{A}{W} - \frac{B}{W}) \ + \ L$. You then proceed to further simplify the leftover stuff. However, how can we get from $( \frac{A}{W + F} - \frac{B}{W + G})$ to $( \frac{A}{W} - \frac{B}{W}) \ + \ L$?

Thanks.

12. Feb 1, 2015

### Stephen Tashi

I just defined $L$ to be $( \frac{A}{W + F} - \frac{B}{W + G}) - ( \frac{A}{W} - \frac{B}{W})$