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.


1 Answer 1


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

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

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

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

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

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

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)?

  • $\begingroup$ thx for accepting, hope the answer provided some help. $\endgroup$ May 4, 2021 at 7:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.