Derivations of trade balance

I am facing some problems in deriving the two equations for trade balance (from which we derive that the trade balance is zero within each country) in the article "A SIMPLE FRAMEWORK FOR INTERNATIONAL MONETARY POLICY ANALYSIS" (2002) that you can find here. My objective is to derive equations 28 and 29 of that paper. In order to do that, I used the following equations:

$equation\; 7: P_{h}C_{h}=(1-\gamma)P_{t}C_{t}$

$equation\; 7a: P_{h}^{*}C_{h}^{*}=(1-\gamma)P_{t}^{*}C_{t}^{*}$

$equation\; 8: P_{f}C_{f}=(\gamma)P_{t}C_{t}$

$equation\; 8a: P_{f}^{*}C_{f}^{*}=(\gamma)P_{t}^{*}C_{t}^{*}$

$LOOP: P_{t}=\epsilon P_{t}^{*}$

$C_{t}=C_{t}^{*}$

$equation \; 25: (1-\gamma)Y_{t}=(1-\gamma)C_{h}+\gamma C_{h}^{*}$

$equation \; 26: \gamma Y_{t}^{*}=(1-\gamma)C_{f}+\gamma C_{f}^{*}$

Firstly, I tried to substitute the value of $C_{h}^{*}$ given by equation 7a in equation 25, but what I got was not right:

$P_{h}^{*}Y_{t}=P_{h}^{*}C_{h}+ \gamma P_{t}^{*}C_{t}^{*}$

Then I tried to get the value of $C_{h}$ from equation 25 and substitute in equation 7, but this was not right either:

$P_{t}C_{t}=(1- \gamma)P_{h}Y_{t}- \gamma P_{h}C_{h}^{*}$

I also tried to substitute for gamma, but the results were not correct. I cannot think of anything else. How would you do that?

The equations that I am looking for:

Equation 28: $P_{h}Y_{t}=P_{t}C_{t}$

Equation 29: $P_{f}^{*}Y_{t}^{*}=P_{t}^{*}C_{t}^{*}$

• To make the question more self contained, perhaps also include equations 28 and 29? May 7 at 12:02
• @Giskard You are right, I will write them immediately. May 7 at 12:03

Dividing equation 25 by $$\gamma$$ gives: $$Y_t = C_h + \frac{\gamma}{1-\gamma} C_h^\ast.$$ Then using 7 and 7a to substitute for $$C_h$$ and $$C_h^\ast$$, we get: $$Y_t = (1-\gamma) \frac{P_t}{P_h} C_t + \frac{\gamma}{1-\gamma}(1-\gamma) \frac{P_t^\ast}{P_h^\ast} C_t^\ast.$$ Then multiply both sides by $$P_h$$ and use $$C_t^\ast = C_t$$ to get: \begin{align*} P_h Y_t &= (1-\gamma)P_t C_t + \gamma \left(\frac{P_t^\ast}{P_t}\frac{P_h}{P_h^\ast}\right) P_t C_t,\\ &= P_t C_t \left(1- \gamma + \gamma \left(\frac{P_t^\ast}{P_t}\frac{P_h}{P_h^\ast}\right)\right) \end{align*} This is equal to equation 28 if: $$\frac{P_t^\ast}{P_t} \frac{P_h}{P_h^\ast} = 1.$$ Now $$\dfrac{P_t^\ast}{P_t} = \dfrac{1}{\varepsilon}$$, which would require that: $$\frac{P_h}{P_h^\ast} = \varepsilon.$$
Dividing 7 by 7a gives: $$\frac{P_h}{P_h^\ast} \frac{C_h}{C_h^\ast} = \frac{P_t}{P_t^\ast} = \varepsilon.$$ So this means that $$C_h$$ has to equal $$C_h^\ast$$. I don't know if this makes sense (as I don't know the meaning of all these variables).
Similarly, you can show that 29 is satisfied if $$C_f = C_f^\ast$$.