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I have been trying to derive equation 14 from Paul Krugman's 1980 AER paper. I keep getting a different result, and further, I find that equation 14 from the paper does not balance payments when I simulate the model numerically. I assume that I am doing something wrong, and would like someone to point out my error. Here is my derivation:

Having balanced payments means that the value of the exports from the home country to the foreign country should be exactly equal to the value of exports from the foreign country to the home country (which are the home country's imports). Krugman divides exports and imports by the wage of the foreign country, $w^*$, to write down the balance of payments in wage units of the foreign country. I will use $h$ and $f$ subscripts to indicate home and foreign and use $c_{residency,origin}$ to denote the demand of an individual in country $residency$ for a good produced in $origin$ that has been shipped to $residency$. The individual demand for a representative home good (local good) by a resident of the home country is $c_{hh}$ and individual demand for a representative foreign good (import) by a home country resident is $c_{hf}$. Individual demand from foreign country residents for local and imported goods is likewise $c_{ff}$ and $c_{fh}$. Dividing imports by the iceberg transportation cost, $g$, gives total demand.

The values of home exports to the foreign country and imports in the home country are:

$Exports_h=p\frac{c_{fh}}{g}nL^*$, $Imports_h=p^*\frac{c_{hf}}{g}n^*L$

$$B=\frac{Exports_h}{w^*}-\frac{Imports_h}{w^*}=\frac{p\frac{c_{fh}}{g}nL^*}{w^*}-\frac{p^*\frac{c_{hf}}{g}n^*L}{w^*}$$

Individuals in each country spend their entire wage on local and imported goods:

$w=p{}c_{hh}n+p^*\frac{c_{hf}}{g}n^*$, $w^*=p^*c_{ff}n^*+p\frac{c_{fh}}{g}n$

Substitute the above for $w^*$ in the denominator of exports, and the equation for $w$ into the denominator for imports, after multiplying by $\frac{w^*}{w}\frac{w}{w^*}=\frac{w^*}{w}\omega$:

$$B=\frac{Exports_h}{w^*}-\frac{Imports_h}{w^*}\frac{w^*}{w}\omega=\frac{p\frac{c_{fh}}{g}nL^*}{p^*c_{ff}n^*+p\frac{c_{fh}}{g}n}-\frac{p^*\frac{c_{hf}}{g}n^*L}{p{}c_{hh}n+p^*\frac{c_{hf}}{g}n^*}\omega$$

By definition (Krugman eq 12), we have:

$\sigma=\frac{c_{hf}/g}{c_{hh}}=\left(\frac{p}{p^*}\right)^{\frac{1}{1-\theta}}g^{\frac{\theta}{1-\theta}}$ and $\sigma^*=\frac{c_{fh}/g}{c_{ff}}=\left(\frac{p}{p^*}\right)^{\frac{-1}{1-\theta}}g^{\frac{\theta}{1-\theta}}$

Multiply exports by $\frac{1/c_{ff}}{1/c_{ff}}$ and imports by $\frac{1/c_{hh}}{1/c_{hh}}$ to rewrite the equations using $\sigma$ and $\sigma^*$:

$$B=\frac{1/c_{ff}}{1/c_{ff}}\frac{Exports_h}{w^*}-\frac{1/c_{hh}}{1/c_{hh}}\frac{Imports_h}{w}\omega=\frac{p\sigma^*nL^*}{p^*n^*+p\sigma^*n}-\frac{p^*\sigma{}n^*L}{pn+p^*\sigma{}n^*}\omega$$

Now using the fact that $\omega=p/p^*$, multiply exports by $\frac{1/p^*}{1/p^*}$ and imports by $\frac{1/p}{1/p}$ to replace the prices with $\omega$:

$$B=\frac{1/(p^*c_{ff})}{1/(p^*c_{ff})}\frac{Exports_h}{w^*}-\frac{1/(p{}c_{hh})}{1/(p{}c_{hh})}\frac{Imports_h}{w}\omega=\frac{\omega\sigma^*nL^*}{n^*+\omega\sigma^*n}-\frac{(1/\omega)\sigma{}n^*L}{n+(1/\omega)\sigma{}n^*}\omega$$

Simplifying the denominator of the imports term, I can rewrite the expression as:

$$B=\frac{\sigma^*n\omega{}}{\omega\sigma^*n+n^*}L^*-\frac{\sigma{}n^*}{\omega{}n+\sigma{}n^*}\omega{}L$$

Finally, using $n=\frac{L(1-\theta)}{\alpha}$ and $n^*=\frac{L^*(1-\theta)}{\alpha}$ (Krugman eq 13), I can further simplify the above as:

$$B=\omega{}L{}L^*\left[\frac{\sigma^*}{\omega\sigma^*L+L^*}-\frac{\sigma}{\omega{}L+\sigma{}L^*}\right]$$

This equation is similar, but not the same as Krugman's equation 14 below:

$$B=\omega{}L{}L^*\left[\frac{\sigma^*}{\sigma^*L+L^*}-\frac{\sigma}{{}L+\sigma{}L^*}\right]$$

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    $\begingroup$ You should probably consider the possibility of a typo in the original article. I followed the slides here, until slide 22, then made some substitutions using the definitions of $\sigma$ and $\sigma^\ast$ of Krugman on page 953, divided everything by $w^\ast$ and also arrived at your final conditions. $\endgroup$
    – tdm
    Commented Jun 3, 2021 at 12:00
  • $\begingroup$ Thanks very much, @tdm. I decided to follow your strategy and just write an answer suggesting that it is likely to be a simple typo (and one that apparently is well known, since no one is using Krugman's original balance of payments equation in their own lecture notes). $\endgroup$ Commented Jun 9, 2021 at 15:04

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Following the strategy of @tdm, I found lecture notes by prominent trade economists Dave Donaldson, Gregory Corcos and Isabelle Mejean, and Alexander Tarasov. I have written their equations below. Since their equations for the wage ratio $\omega$ that balances trade led to the same solution as my equation, I think that the original Krugman paper must have a typo, as @tdm suggested. However, I was unable to find a definitive source documenting this typo, despite the fact that multiple lecturers seems to be using a different equation. I'm posting this answer to help anyone else who might have my same question. However, I would be happy to accept an answer showing that this is a well-known typo documented somewhere.

Equations from Trade lectures slides: Donaldson, and Corcos and Mejean (co-teachers of a Trade course), use the same equation for $\omega$. In their formulation they replace Krugman's CES parameter $\theta$ with $\sigma$, where $\sigma=1/(1-\theta)$. They also use $\tau=1/g$ for iceberg transportation costs:

$\left(\frac{w}{w^*}\right)^{\sigma}=\frac{\tau^{1-\sigma}+(L/L*)(w/w*)^{1-\sigma}}{1+\tau^{1-\sigma}(L/L*)(w/w*)^{1-\sigma}}\tag{Donaldson,Corcos+Mejean}$

Tarasov uses the same CES notation as Krugman, but also replaces the iceberg transportation parameter with $\tau=1/g$:

$\frac{w}{w*}=\frac{L*+(\frac{w}{w*})^{\frac{\theta}{\theta-1}}L\tau^{\frac{\theta}{1-\theta}}}{L+(\frac{w}{w*})^{\frac{\theta}{1-\theta}}L*\tau^{\frac{\theta}{1-\theta}}}\tag{Tarasov}$

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