<p><span class="math-container">$π = φ^{(σ−1)}*\frac{f_E+f_D}{φ^{(σ−1)}*(1+τ^{(1-σ)})}+φ^{(σ−1)}\frac{f_E+f_D}{φ^{(σ−1)}*(1+τ^{(1-σ)})}-f_E-2f_D$</span></p>
<p><span class="math-container">$π = 2\frac{f_E+f_D}{(1+τ^{(1-σ)})}-f_E-2f_D$</span></p>
<p><span class="math-container">$(1+τ^{(1-σ)})π = 2(f_E+f_D)-(1+τ^{(1-σ)})f_E-2(1+τ^{(1-σ)})f_D$</span></p>
<p><span class="math-container">$(1+τ^{(1-σ)})π = 2f_E+2f_D-f_E-τ^{(1-σ)}f_E-2f_D-2τ^{(1-σ)}f_D$</span></p>
<p><span class="math-container">$(1+τ^{(1-σ)})π = f_E-τ^{(1-σ)}f_E-2τ^{(1-σ)}f_D$</span></p>
<p><span class="math-container">$(1+τ^{(1-σ)})π = f_E(1-τ^{(1-σ)})-2τ^{(1-σ)}f_D$</span></p>
<p>use <span class="math-container">$\pi = 0$</span></p>
<p><span class="math-container">$0 = f_E(1-τ^{(1-σ)})-2τ^{(1-σ)}f_D $</span></p>
<p><span class="math-container">$2τ^{(1-σ)}f_D = f_E(1-τ^{(1-σ)}) $</span></p>
<p><span class="math-container">$\frac{2f_D}{f_E} = \frac{(1-τ^{(1-σ)}) }{τ^{(1-σ)}} = \frac{1}{τ^{(1-σ)}} - 1 = τ^{(σ-1)} -1$</span></p>
<p>in last identity I use  <span class="math-container">$\frac{1}{τ^{(1-σ)}} = τ^{(σ-1)}$</span> but you write <span class="math-container">$τ^{(1-σ)}$</span> in expression (3) so this is not quite what you are looking for but maybe you have sign error in this exponent in (3)?</p>

CC BY-SA 4.0
2021-05-04T07:58:31.030Z
thx for accepting, hope the answer provided some help.
Jesper Hybel
CC BY-SA 4.0
2021-05-03T23:35:48.590Z
2021-05-03T23:35:48.590Z
Jesper Hybel
<p>Hi everyone I am running into a stone wall and don't seem to be able to solve this, can someone show me the steps how I can insert 1 into 2 and get 3.</p>
<ol>
<li><p><span class="math-container">$B=\frac{f_E+f_D}{φ^{(σ−1)}*(1+τ^{(1-σ)})}$</span></p>
</li>
<li><p><span class="math-container">$π = φ^{(σ−1)}*B+φ^{(σ−1)}B-f_E-2f_D$</span></p>
</li>
<li><p><span class="math-container">$\frac{2f_D}{f_E} = τ^{(1-σ)} -1$</span></p>
</li>
</ol>
<p>I am trying to understand how the symmetric model of the proximity-concentration is derived. Would love to show you guys my steps but to be honest I haven't gotten much further than</p>
<p><span class="math-container">$π = φ^{(σ−1)}*\frac{f_E+f_D}{φ^{(σ−1)}*(1+τ^{(1-σ)})}+φ^{(σ−1)}\frac{f_E+f_D}{φ^{(σ−1)}*(1+τ^{(1-σ)})}-f_E-2f_D$</span>
and then setting <span class="math-container">$π = 0$</span>. The paper sets <span class="math-container">$π&lt;0$</span> but for the purpose of question and my issues setting <span class="math-container">$π = 0$</span> difficult enough.</p>
<p>Appreciate any and all help.</p>

<microeconomics><international-trade><investment><firm><multinational-companies>
Proximity-concentration hypothesis: homogeneous firms and symmetric countries
James
# Proximity-concentration hypothesis: homogeneous firms and symmetric countries

(Asked by: username_0 on 2021-05-03)

Hi everyone I am running into a stone wall and don't seem to be able to solve this, can someone show me the steps how I can insert 1 into 2 and get 3.

1. $B=\frac{f\_E+f\_D}{φ^{(σ−1)}\*(1+τ^{(1-σ)})}$
2. $π = φ^{(σ−1)}\*B+φ^{(σ−1)}B-f\_E-2f\_D$
3. $\frac{2f\_D}{f\_E} = τ^{(1-σ)} -1$

I am trying to understand how the symmetric model of the proximity-concentration is derived. Would love to show you guys my steps but to be honest I haven't gotten much further than

$π = φ^{(σ−1)}\*\frac{f\_E+f\_D}{φ^{(σ−1)}\*(1+τ^{(1-σ)})}+φ^{(σ−1)}\frac{f\_E+f\_D}{φ^{(σ−1)}\*(1+τ^{(1-σ)})}-f\_E-2f\_D$
and then setting $π = 0$. The paper sets $π<0$ but for the purpose of question and my issues setting $π = 0$ difficult enough.

Appreciate any and all help.

**Tags:** microeconomics, international-trade, investment, firm, multinational-companies

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## 1 answers

### Answer by username_1 on 2021-05-03. Score: 3 | Accepted answer

$π = φ^{(σ−1)}\*\frac{f\_E+f\_D}{φ^{(σ−1)}\*(1+τ^{(1-σ)})}+φ^{(σ−1)}\frac{f\_E+f\_D}{φ^{(σ−1)}\*(1+τ^{(1-σ)})}-f\_E-2f\_D$

$π = 2\frac{f\_E+f\_D}{(1+τ^{(1-σ)})}-f\_E-2f\_D$

$(1+τ^{(1-σ)})π = 2(f\_E+f\_D)-(1+τ^{(1-σ)})f\_E-2(1+τ^{(1-σ)})f\_D$

$(1+τ^{(1-σ)})π = 2f\_E+2f\_D-f\_E-τ^{(1-σ)}f\_E-2f\_D-2τ^{(1-σ)}f\_D$

$(1+τ^{(1-σ)})π = f\_E-τ^{(1-σ)}f\_E-2τ^{(1-σ)}f\_D$

$(1+τ^{(1-σ)})π = f\_E(1-τ^{(1-σ)})-2τ^{(1-σ)}f\_D$

use $\pi = 0$

$0 = f\_E(1-τ^{(1-σ)})-2τ^{(1-σ)}f\_D $

$2τ^{(1-σ)}f\_D = f\_E(1-τ^{(1-σ)}) $

$\frac{2f\_D}{f\_E} = \frac{(1-τ^{(1-σ)}) }{τ^{(1-σ)}} = \frac{1}{τ^{(1-σ)}} - 1 = τ^{(σ-1)} -1$

in last identity I use $\frac{1}{τ^{(1-σ)}} = τ^{(σ-1)}$ but you write $τ^{(1-σ)}$ in expression (3) so this is not quite what you are looking for but maybe you have sign error in this exponent in (3)?

> **Comments:**
> **username_1**: thx for accepting, hope the answer provided some help.  

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