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When computing changes in $\lambda_{ij}$ defined as $\hat \lambda_{ij} = {\lambda_{ij}^{'} \over \lambda_{ij}}$, where does the $\lambda_{lj}$ in the sum in the denominator come from?

$$\lambda_{ij} = {\chi_i (\tau_{ij} w_i)^{-\epsilon} \over \sum_{l=1}^N{ \chi_l (\tau_{lj} w_l)^{-\epsilon}}}$$

$$\lambda_{ij}^{'} = {\chi_i^{'} (\tau_{ij}^{'} w_i^{'})^{-\epsilon} \over \sum_{l=1}^N{ \chi_l^{'} (\tau_{lj}^{'} w_l^{'})^{-\epsilon}}}$$

$$\hat \lambda_{ij} = {\hat\chi_i (\hat \tau_{ij} \hat w_i)^{-\epsilon} \over \sum_{l=1}^N{\lambda_{lj} \hat \chi_l (\hat \tau_{lj} \hat w_l)^{-\epsilon}}}$$

Isn't $ \sum^N_{l=1} { \lambda_{lj} } = \sum^N_{l=1} { \chi_l (\tau_{lj} w_l)^{-\epsilon} \over \sum_{l=1}^N {\chi_l (\tau_{lj} w_l)^{-\epsilon}}}$ just equal to one?

mfauth
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1 Answers1

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So simplifying the notation of the expressions, you have $b_i=\frac{a_i}{\sum_{l=1}^na_l}$ and $b_i'=\frac{a_i'}{\sum_{l=1}^na_l'}$. Now in computing the quotient you get \begin{align} \hat b_i&=\frac{a_i'}{a_i}\frac{\sum_{l=1}^na_l}{\sum_{l=1}^na_l'} =\hat a_i\frac{\sum_{m=1}^na_m}{\sum_{l=1}^na_l\hat a_l} =\frac{\hat a_i}{\sum_{l=1}^n\frac{a_l}{\sum_{m=1}^na_m}\hat a_l} =\frac{\hat a_i}{\sum_{l=1}^nb_l\hat a_l} \end{align} There is no deep transformation in there, just the elimination of all primed for the hatted variables.

Lutz Lehmann
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