could anyone please help to explain how equation (4) in this paper is derived? I understand that it's an application of first order conditions using the chain rule from equations (1)-(3), but the notation is very confusing and I can't follow the math at all. Thankyou in advance to anyone that can help

For reference, equation (1) is:

$ Y_t = A_t[\theta_t L_{1t}^\rho + L_{2t}^\rho]^{1/\rho} $

Equation (2) is:

$ L_{et} = (\sum_{a}\alpha_{ea}L_{ea}^\eta)^{1/\eta}, e = (1,2) $

Equation (3) is:

$ L_{eat} = (N_{eat}^\delta + \beta_{eat}M_{eat}^\delta)^{1/\delta} $

And from this, equation 4 is derived as:

$ \ln W_{eat}^S = \ln A_t + 1/\sigma_E \ln Y_t + \ln\theta_{et} + \ln\alpha_{e\alpha} + \ln\beta_{eat}^S + (1/\sigma_A - 1/\sigma_E)\ln L_{et} + (1/\sigma_I - 1/\sigma_A)\ln L_{eat} - (1/\sigma_I) \ln S_{eat}, S = {M,N} $


$ \sigma_I = (1/{1-\delta}), \sigma_A = (1/{1-\eta}), \sigma_E = (1/{1-\rho}) $

  • $\begingroup$ please consider making the question self contained by including the equation in the question - you can do it here the same way as you would do in Latex (if you are not familiar you do it by typing out the equation and enclosing it in dollar signs e.g. $ x+5=y$will be rendered as $x+5=y$, for Greek letters you can use \letter e.g. $y = \beta x + \gamma$ renders as $y = \beta x + \gamma$ $\endgroup$
    – 1muflon1
    Feb 8 at 12:23
  • 1
    $\begingroup$ I have done so now, thankyou $\endgroup$
    – wui
    Feb 8 at 13:00

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