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In Obstfeld and Rogoff (2000), formula (12) states the following:

$$ W = (\frac{\phi}{\phi-1}) \frac{E\{K(L^\nu)\}}{E\{\frac{L}{P}C^{-\rho}\}} $$

where $\phi$, $\rho$ and $\nu$ are parameters, $E$ is the expectation operator, and $K$, $L$, $P$,$C$ are endogenous variables jointly log-normally distributed.

They state that given the log-normality it is equivalent to write equation (12) as:

$$ W = (\frac{\phi}{\phi-1}) \frac{E\{K\}E\{L\}^{\nu-1})}{E\{C\}^{-\rho} E\{ \frac{1}{P} \} } \exp{\psi} $$

where:

$$\psi = \frac{\nu(\nu-1)}{2} \sigma_l^2 - \frac{\rho(\rho+1)}{2} \sigma_c^2 + \nu \sigma_{kl} + \rho \sigma_{cl} - \rho \sigma_{cp} + \sigma_{lp} $$

I tried to derive the expression for psi, I get something slightly different:

$$\psi = \frac{\nu}{2} \sigma_l^2 - \frac{\rho}{2} \sigma_c^2 + \nu \sigma_{kl} + \rho \sigma_{cl} - \rho \sigma_{cp} + \sigma_{lp} $$

I don't understand where those extra terms come from.

Any help?

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    $\begingroup$ Is $\sigma_{kl}$ the co-variance between $K$ and $L$? $\endgroup$ – Gordon Mar 30 '16 at 16:28
  • $\begingroup$ Yes, indeed. And $\sigma_{lp}$ is the covariance between $L$ and $P$ and so on and so forth. $\endgroup$ – phdstudent Mar 30 '16 at 16:44
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Note that \begin{align*} E(K) &= E\big(\exp(\ln K) \big)\\ &=\exp\Big(E(\ln K) + \frac{1}{2}\sigma_k^2 \Big),\\ E(L) &= E\big(\exp(\ln L) \big)\\ &=\exp\Big(E(\ln L) + \frac{1}{2}\sigma_l^2 \Big),\\ E\Big(\frac{1}{P}\Big) &= E\big(\exp(-\ln P) \big)\\ &=\exp\Big(-E(\ln P) + \frac{1}{2}\sigma_p^2 \Big), \end{align*} and \begin{align*} E(C) &= E\big(\exp(\ln C) \big)\\ &=\exp\Big(E(\ln C) + \frac{1}{2}\sigma_c^2 \Big). \end{align*} Then, \begin{align*} E(K L^{\nu}) &= E\Big(\exp\big(\ln K + \nu \ln L\big) \Big)\\ &=\exp\Big(E(\ln K) + v E(\ln L)+ \frac{1}{2}\sigma_k^2 + \frac{1}{2} \nu^2\sigma_l^2 + \nu \sigma_{kl}\Big)\\ &= \exp\Big(E(\ln K) + \frac{1}{2}\sigma_k^2 + \nu \Big(E(\ln L)+ \frac{1}{2} \sigma_l^2\Big) + \frac{1}{2}(\nu^2-\nu)\sigma_l^2 + \nu \sigma_{kl}\Big)\\ &= E(K)(E(L))^{\nu}\exp\Big( \frac{1}{2}\nu(\nu-1)\sigma_l^2 + \nu \sigma_{kl} \Big). \end{align*} Moreover, \begin{align*} E\left( \frac{L}{P}C^{-\rho} \right) &= E\Big(\exp\big(\ln L - \ln P - \rho \ln C \big) \Big)\\ &= \exp\Big(E(\ln L) - E(\ln P) - \rho E(\ln C) \\ &\qquad\qquad\qquad\qquad +\frac{1}{2}\sigma_l^2 + \frac{1}{2}\sigma_p^2 + \frac{1}{2}\rho^2\sigma_c^2 - \sigma_{lp} - \rho \sigma_{cl} +\rho\sigma_{cp}\Big)\\ &=\exp\Big(E(\ln L) +\frac{1}{2}\sigma_l^2 - E(\ln P) +\frac{1}{2}\sigma_p^2 - \rho E(\ln C) - \rho \frac{1}{2}\sigma_c^2 \\ &\qquad\qquad\qquad\qquad + \frac{1}{2}\big(\rho^2+\rho\big)\sigma_c^2 - \sigma_{lp} - \rho \sigma_{cl} +\rho\sigma_{cp}\Big)\\ &= E(L)E\Big(\frac{1}{P}\Big)(E(C))^{-\rho} \exp\Big(\frac{1}{2}\big(\rho^2+\rho\big)\sigma_c^2 - \sigma_{lp} - \rho \sigma_{cl} +\rho\sigma_{cp}\Big). \end{align*} Consequently, \begin{align*} \frac{E(K L^{\nu})}{E\Big( \frac{L}{P}C^{-\rho} \Big)}&=\frac{E(K)(E(L))^{\nu-1}}{E\Big(\frac{1}{P}\Big)(E(C))^{-\rho}}\\ &\qquad\qquad \exp\left(\frac{1}{2}\nu(\nu-1)\sigma_l^2 - \frac{1}{2}\big(\rho^2+\rho\big)\sigma_c^2 + \nu \sigma_{kl} + \rho \sigma_{cl} -\rho\sigma_{cp} +\sigma_{lp}\right). \end{align*} That is, \begin{align*} \psi = \frac{1}{2}\nu(\nu-1)\sigma_l^2 - \frac{1}{2}\rho\big(\rho+1\big)\sigma_c^2 + \nu \sigma_{kl} + \rho \sigma_{cl} -\rho\sigma_{cp} +\sigma_{lp}. \end{align*}

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