Lognormal

Covariance of Log-Normal Variables

  • March 30, 2016

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?

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, $$ \begin{align*} \psi = \frac{1}{2}\nu(\nu-1)\sigma_l^2 - \frac{1}{2}\rho\big(\rho+1\big)\sigma_c^2 + \nu \sigma_{kl} + \rho \sigma_{cl} -\rho\sigma_{cp} +\sigma_{lp}. \end{align*} $$

引用自:https://quant.stackexchange.com/questions/25151