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When the regressand is in logarithmic form, the model looks like this:

$$log(y)=X\beta+u$$

When we are ultimately interested in predicting $y$, we cannot simply use $exp(X_0\hat\beta)$ as a predictor. Rather, we have to solve the following expression:

$$\begin{align} E(y|X) &= E(exp(X\beta+u)|X) \\ &=E(exp(X\beta)*exp(u)|X) \\ &=exp(X\hat\beta)*E(exp(u)|X) \end{align}$$

Thus, all we need is an estimator for $E(exp(u)|X)$. There are two ways of going about it:

  • If $u$ is normally distributed, $exp(u)$ will follow a log normal distribution. We know that the mean of a log normal distribution is $exp(\mu+\sigma^2/2)$. In our case, we know that $E(u)=0$ because the model includes an intercept, hence $\mu=0$. Thus, we can use $exp(\hat\sigma^2/2)$ as an estimator for $E(exp(u)|X)$.
  • If $u$ is not normally distributed, the following method-of-moments estimator for $E(exp(u)|X)$ will be consistent: $n^{-1}\sum{exp(\hat u_i)}$.

See Wooldridge's "Introductory Econometrics: A Modern Approach", 5th edition, p. 212-215 for more details on the above.

My question is: can I use the method-of-moments estimator for generic transformations of the regressand?

In my specific case, $y$ represents stock market returns over a multi-year horizon: $y=p_{t+w}/p_t$. As before, the model looks like this:

$$log(y)=X\beta+u$$

However, I am ultimately interested in predicting annualized returns. Thus, I try to solve the following expression:

$$\begin{align} E(y^{1/w}-1|X) &= E(y^{1/w}|X)-1 \\ &=E(exp(X\beta+u)^{1/w}|X)-1 \\ &=E((e^{X\beta+u})^{1/w}|X)-1 \\ &=E(e^{(X\beta+u)*1/w}|X)-1 \\ &=E(exp(1/w*X\beta+1/w*u)|X)-1 \\ &=exp(X\hat\beta/w)*E(exp(u/w)|X)-1 \end{align}$$

Thus, all we need is an estimator for $E(exp(u/w)|X)$. Remember, $w$ is simply the number of years over which the returns ($y$) are measured, i.e. a constant.

Will the following method-of-moments estimator for $E(exp(u/w)|X)$ be consistent?

$$\hat\alpha=n^{-1}\sum{exp(\hat{u_i}/w)}$$

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