Deriving Covariance Corrections for Products of Log-Normal Variables
Summary
The document derives the exponential correction needed when expectations of products and powers of jointly log-normal variables are rewritten using their individual means. The motivating expression involves a numerator with a variable multiplied by a powered labor term and a denominator containing labor, inverse price, and a powered consumption term. The question is why the stated correction includes additional variance contributions beyond the asker’s proposed formula.
The accepted derivation applies the moment-generating formula for a normal variable to the logarithms of the variables. It expands the variance of each relevant linear combination of log variables, which introduces both individual variance terms and covariance terms. Rewriting those expressions in terms of the original variables’ expectations accounts for the extra factors involving the exponent parameters and log variances. The result is an explicit correction combining labor and consumption variance with covariances among the variables. The explanation assumes joint log-normality and addresses the algebra of this formula; it does not discuss estimation or empirical use of the parameters.
Key ideas
- Expectations of products of jointly log-normal variables can be evaluated through linear combinations of their logarithms.
- The variance of each log combination generates both variance and covariance contributions.
- Expressing the result using individual variable means adds exponent-dependent variance corrections.
- The correction term includes covariances between the relevant log variables.
- The derivation relies on joint log-normality and does not address parameter estimation.
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# Covariance of Log-Normal Variables
# Covariance of Log-Normal Variables
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?
## Answer by Gordon (score 3, accepted)
https://quant.stackexchange.com/a/25155
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*}Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.