# A supposedly straight forward proof

The authors of this paper claim that equation (1) is equivalent to equation (2). I really don't see how this is possible. There even go further to say, "it is easy to check". Can anyone help?. First note that: $$e^{\ln(x)} = x$$

and

$$e^{-\ln(x)} = \frac{1}{x}$$

$$\prod_{i=1}^G (X_g/\nu)^{1/G} = Z_{pool}$$

\begin{align*} \overline{\ln(X_g)} &= \sum_g \ln(X_g)/G \\ &= \ln(\prod_g X_g)/G\\ &= \ln(\prod_g X_g^{1/G})\\ &= \ln(\prod (Z_g\nu)^{1/G})\\ \end{align*}

\begin{align*} e^{-\overline{\ln(X_g)}} &= \frac{1}{\prod_g (Z_g\nu)^{1/G}}\\ &=\frac{1}{Z_{pool}\nu^{G/G}}\\ &=\frac{1}{Z_{pool}\nu} \end{align*}

\begin{align*} \exp((1-\alpha) \times (\ln(X_g) - \overline{\ln(X_g)}) &= \big(e^{(\ln(X_g) - \overline{\ln(X_g))}} \big)^{(1-\alpha)}\\ &= \exp(\ln(X_g))^{(1-\alpha)} \times \exp(-\overline{\ln(X_g)})^{(1-\alpha)}\\ &= (X_g)^{(1-\alpha)} \times \big(\frac{1}{Z_{pool}\nu} \big)^{(1-\alpha)}\\ &= (X_g/\nu)^{(1-\alpha)}\times \big(\frac{1}{Z_{pool}} \big)^{(1-\alpha)}\\ &= Z_g^{(1-\alpha)}\big(\frac{1}{Z_{pool}} \big)^{(1-\alpha)} \end{align*}

Thus,

\begin{align*} \prod_{i=1}^G (X_g/\nu)^{1/G} \times \exp((1-\alpha) \times (\ln(X_g) - \overline{\ln(X_g)}) &= Z_{pool} \times Z_g^{(1-\alpha)}\big(\frac{1}{Z_{pool}} \big)^{(1-\alpha)}\\ &= Z_{pool}^{\alpha} (Z_g)^{(1-\alpha)} \end{align*}

• Thanks very much @righskewed. This is exactly what I wanted, have a great day. Oct 13, 2015 at 9:06