0
$\begingroup$

I am following Chapter 8 ("Heteroskedasticity" p. 259) in the 6th edition of Woolridge Introductory Econometrics: A Modern Approach and I don't understand one piece of the transformation of our model.

For fGLS, we assume [1] $Var(u|\boldsymbol{x}) = \sigma^2exp(\delta_0 +\delta_1x_1+\delta_2x_2+...+\delta_kx_k)$, where $x_i, x_2,...,x_k$ are the independent variables appearing in the regression model and the $\delta_j$ are unknown parameters.

Then, we use the definition of conditional variance to say $Var(u|\boldsymbol{x}) = E(u^2|\boldsymbol{x}$), since our zero conditional mean assumption tells us that $(E(u))^2$ is zero.

Now, here is where I'm stuck: Wooldridge says that our assumption [1] above allows us to write $u^2=\sigma^2exp(\delta_0 +\delta_1x_1+\delta_2x_2+...+\delta_kx_k)v$, where $v$ has a mean equal to unity, conditional on $\boldsymbol{x}=x_1,x_2, ...,x_k$

Can someone please help me to develop an intuition for this last step that Wooldridge has taken? Essentially, it seems like we've assumed that $E(u^2|\boldsymbol{x})=\frac{u^2}{v}$ and I don't understand why or the properties that allow us to do this.

I've found this paper https://www.econ.uzh.ch/dam/jcr:e3cddc1b-f89d-4fb4-9474-c2c380355d69/joe_2017.pdf to be useful (specifically, assumption #6 on p. 2), but it doesn't leave me with much intuition for what we're doing and why, especially because I'm fairly new to econometrics.

Thanks-
Maurus

$\endgroup$
0
$\begingroup$

I'm using the 4th edition, but the content should be the same. It sounds like you are asking how he got from

$$ Var(u|\boldsymbol{x}) = \sigma^2exp(\delta_0 +\delta_1x_1+\delta_2x_2+...+\delta_kx_k) $$

to

$$ u^2=\sigma^2exp(\delta_0 +\delta_1x_1+\delta_2x_2+...+\delta_kx_k)v $$

As you correctly noted, $Var(u|\boldsymbol{x}) = E(u^2|\boldsymbol{x})$ under the zero conditional mean expectation assumption ($E(u|\boldsymbol{x}) = 0$). So that allows us to replace the left-hand side of the equation, as Wooldridge did. $E(u^2|\boldsymbol{x})$ turned into $u^2$ because $u^2 = E(u^2|\boldsymbol{x}) + \epsilon$, where $\epsilon$ is the component of $u^2$ that's uncorrelated with any function of $\boldsymbol{x}$. That component could have been left in at this stage, but it can be removed because later on it would turn into a part of the error term anyway that does not affect the fitted values from the regression of $log(u^2)$ on $\boldsymbol{x}$ (i.e., part of $e$ in the next equation in the book).

On the right-hand side of the equation, Wooldridge adds $v$, which has a mean equal to unity, conditional on $\boldsymbol{x} = (x_1, x_2, ..., x_k)$. That means that given some set of values for $\boldsymbol{x}$, $v$ is expected to have a multiplicative factor effect of $1$ on $u^2$ (in other words, no effect). By introducing this term to the equation (and assuming that it is independent of $\boldsymbol{x}$), we can then convert it into the error term $e$ in the subsequent equation:

$$ log(u^2) = \alpha_0 + \delta_1x_1+\delta_2x_2+...+\delta_kx_k+e $$

because taking a log of a variable that has a mean of $1$ will give us a variable that has a mean of $0$ plus some constant (equal to the negative of the mean of the log values of the original variable $v$). That constant is moved to the constant term (that's why this last equation has $\alpha_0$ instead of $\delta_0$) and we are left with an equation that can be estimated using OLS.

P.S. The $v$ in this book is not the same $v$ that you found in Assumption 6 in the linked paper. There it is a function, not a variable.

$\endgroup$
1
$\begingroup$

Let us write $\mathbf{x}\delta = \delta_0 + \delta_1 x_1 + \cdots + \delta_k x_k$ for notational brevity. If $u^2 = \sigma^2 \exp(\mathbf{x}\delta) v$, where $E(v|\mathbf{x})=1$, then it is indeed true that $$E(u^2|\mathbf{x}) = \sigma^2 \exp(\mathbf{x}\delta) E(v|\mathbf{x}) = \sigma^2 \exp(\mathbf{x}\delta).$$ The LHS is $Var(u|\mathbf{x})$ by definition when $E(u|\mathbf{x})=0$ as you noted.

Note that $u^2 = \sigma^2 \exp(\mathbf{x}\delta) v$ is not the only possible path to $E(u^2|\mathbf{x}) = \sigma^2 \exp(\mathbf{x}\delta)$. Another possibility is $u^2 = \sigma^2 \exp(\mathbf{x}\delta) + v$ with $E(v|\mathbf{x})=0$, which also implies $E(u^2|\mathbf{x}) = \sigma^2 \exp(\mathbf{x}\delta)$. Wooldridge chooses the multiplicative model $u^2 = \sigma^2 \exp(\mathbf{x}\delta) v$ with $E(v|\mathbf{x})=1$ because he wants to take log (note $v>0$ is also needed) and have a linear model. AlexK gives a nice explanation on the derivation of the linear model.

Wait. Why is taking log so important? It's because we want the fitted values to be strictly positive (by applying its inverse function, exp) for all $i$. It is also so nice that a linear model is derived, which is extremely easy to estimate. A model like $u^2 = \exp(\mathbf{x}\delta) + \varepsilon$ or $u^2 = (\mathbf{x}\delta)^2 +\varepsilon$ with $E(\varepsilon|\mathbf{x})=0$ also give strictly positive fitted values but it's not a linear model so estimation is hard.

To add a bit more technical details. We need that $v$ and $\mathbf{x}$ are mutually independent so that $\log(v)$ and $\mathbf{x}$ are mutually independent. Then after taking log, we have $$\log (u^2) = \log(\sigma^2) + \mathbf{x} \delta + \log(v),$$ where $E[\log(v)|\mathbf{x}] = E[\log(v)]$ is a (nonzero) constant independent of $\mathbf{x}$, and we have a standard linear model after adjusting the intercept. Actually, we don't even need $E(v|\mathbf{x})=1$. All we need is that $v$ and $\mathbf{x}$ are mutually independent.

$\endgroup$
  • $\begingroup$ A good supplement to support the Answer. $\endgroup$ – Subhash C. Davar Aug 3 at 13:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.