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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.

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$π = φ^{(σ−1)}*\frac{f_E+f_D}{φ^{(σ−1)}*(1+τ^{(1-σ)})}+φ^{(σ−1)}\frac{f_E+f_D}{φ^{(σ−1)}*(1+τ^{(1-σ)})}-f_E-2f_D$

$π = 2\frac{f_E+f_D}{(1+τ^{(1-σ)})}-f_E-2f_D$

$(1+τ^{(1-σ)})π = 2(f_E+f_D)-(1+τ^{(1-σ)})f_E-2(1+τ^{(1-σ)})f_D$

$(1+τ^{(1-σ)})π = 2f_E+2f_D-f_E-τ^{(1-σ)}f_E-2f_D-2τ^{(1-σ)}f_D$

$(1+τ^{(1-σ)})π = f_E-τ^{(1-σ)}f_E-2τ^{(1-σ)}f_D$

$(1+τ^{(1-σ)})π = f_E(1-τ^{(1-σ)})-2τ^{(1-σ)}f_D$

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)?

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  • $\begingroup$ thx for accepting, hope the answer provided some help. $\endgroup$ – Jesper Hybel May 4 at 7:58

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