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

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

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

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

CC BY-SA 4.0
2019-08-03T13:14:56.370Z
A good supplement to support the Answer.
Subhash C. Davar
CC BY-SA 4.0
2019-05-29T16:18:11.383Z
2019-05-29T16:18:11.383Z
chan1142
<p>Take <span class="math-container">$\sigma^2 = 1$</span>. If <span class="math-container">$E[u^2|x] = f(x)$</span> for some <span class="math-container">$f(x) &gt; 0$</span> , then, trivially,
<span class="math-container">$$
u^2 = f(x) \frac{u^2}{f(x)},
$$</span>
where <span class="math-container">$v = \frac{u^2}{f(x)}$</span> has mean equal to 1. In this particular case, <span class="math-container">$f(x) = exp(\cdots)$</span>.</p>

CC BY-SA 4.0
2020-09-11T21:12:29.663Z
2020-09-11T21:12:29.663Z
Michael
<p>I'm using the 4th edition, but the content should be the same.  It sounds like you are asking how he got from</p>

<p><span class="math-container">$$
Var(u|\boldsymbol{x}) = \sigma^2exp(\delta_0 +\delta_1x_1+\delta_2x_2+...+\delta_kx_k)
$$</span></p>

<p>to</p>

<p><span class="math-container">$$
u^2=\sigma^2exp(\delta_0 +\delta_1x_1+\delta_2x_2+...+\delta_kx_k)v
$$</span></p>

<p>As you correctly noted, <span class="math-container">$Var(u|\boldsymbol{x}) = E(u^2|\boldsymbol{x})$</span> under the zero conditional mean expectation assumption (<span class="math-container">$E(u|\boldsymbol{x}) = 0$</span>).  So that allows us to replace the left-hand side of the equation, as Wooldridge did.  <span class="math-container">$E(u^2|\boldsymbol{x})$</span> turned into <span class="math-container">$u^2$</span> because <span class="math-container">$u^2 = E(u^2|\boldsymbol{x}) + \epsilon$</span>, where <span class="math-container">$\epsilon$</span> is the component of <span class="math-container">$u^2$</span> that's uncorrelated with any function of <span class="math-container">$\boldsymbol{x}$</span>.  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 <span class="math-container">$log(u^2)$</span> on <span class="math-container">$\boldsymbol{x}$</span> (i.e., part of <span class="math-container">$e$</span> in the next equation in the book).</p>

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

<p><span class="math-container">$$
log(u^2) = \alpha_0 + \delta_1x_1+\delta_2x_2+...+\delta_kx_k+e
$$</span></p>

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

<p>P.S. The <span class="math-container">$v$</span> in this book is not the same <span class="math-container">$v$</span> that you found in Assumption 6 in the linked paper.  There it is a function, not a variable.</p>

CC BY-SA 4.0
2019-05-29T06:41:45.567Z
2019-05-29T06:41:45.567Z
AlexK
<p>I am following Chapter 8 ("Heteroskedasticity" p. 259) in the 6th edition of Woolridge <em>Introductory Econometrics: A Modern Approach</em> and I don't understand one piece of the transformation of our model.</p>

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

<p>Then, we use the definition of conditional variance to say <span class="math-container">$Var(u|\boldsymbol{x}) = E(u^2|\boldsymbol{x}$</span>), since our zero conditional mean assumption tells us that <span class="math-container">$(E(u))^2$</span> is zero.</p>

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

<p>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 <span class="math-container">$E(u^2|\boldsymbol{x})=\frac{u^2}{v}$</span> and I don't understand why or the properties that allow us to do this.</p>

<p>I've found this paper <a href="https://www.econ.uzh.ch/dam/jcr:e3cddc1b-f89d-4fb4-9474-c2c380355d69/joe_2017.pdf" rel="nofollow noreferrer">https://www.econ.uzh.ch/dam/jcr:e3cddc1b-f89d-4fb4-9474-c2c380355d69/joe_2017.pdf</a> 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.</p>

<p>Thanks-<br>
Maurus</p>

<econometrics><heteroskedasticity>
Heteroskedasticity assumption in fGLS into linear form?
M.M.
# Heteroskedasticity assumption in fGLS into linear form?

(Asked by: username_0 on 2019-05-28)

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

**Tags:** econometrics, heteroskedasticity

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## 3 answers

### Answer by username_4 on 2019-05-29. Score: 2 | Accepted answer

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.


---

### Answer by username_1 on 2019-05-29. Score: 1

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.

> **Comments:**
> **username_2**: A good supplement to support the Answer.  

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### Answer by username_3 on 2020-09-11. Score: 0

Take $\sigma^2 = 1$. If $E[u^2|x] = f(x)$ for some $f(x) > 0$ , then, trivially,
$$
u^2 = f(x) \frac{u^2}{f(x)},
$$
where $v = \frac{u^2}{f(x)}$ has mean equal to 1. In this particular case, $f(x) = exp(\cdots)$.


---

