<p>I'm not at home in the literature, so I can not tell you if the two decompositions are the same. I can only help you with the derivations.</p>
<p>For the BBDF expression, you can simply obtain the first from the second by substituting out <span class="math-container">$\rho$</span> and <span class="math-container">$\rho_\varphi$</span>.</p>
<p>For the SPGBV expression things are a bit more tricky. Let <span class="math-container">$i$</span> represent the unit and <span class="math-container">$g$</span> the group variable (which is <span class="math-container">$j$</span> in your equation).</p>
<p>I'm guessing that
<span class="math-container">$$
\bar \theta^g = \dfrac{1}{n_g} \sum_{i \in g} \theta^i
$$</span>
where <span class="math-container">$n_g$</span> is the number of elements in group <span class="math-container">$g$</span>. Let there be <span class="math-container">$N ( = \sum_g n_g)$</span> observations in total. Let <span class="math-container">$p_g = \dfrac{n_g}{n}$</span> be the probability of <span class="math-container">$\bar \theta^g$</span> (I take that <span class="math-container">$\dfrac{1}{N}$</span> is the probability of the unit <span class="math-container">$\theta^i$</span>).</p>
<p>First look at the expression <span class="math-container">$var(\theta^i - \bar \theta^g)$</span>​​. We can decompose it in the following way:
<span class="math-container">$$
var(\theta^i- \bar \theta^g)  = var(\theta^i) + var(\bar \theta^g) - 2 cov(\theta^i, \bar \theta^g).
$$</span>
Now, we can expand the covariance term:
<span class="math-container">$$
\begin{align*}
cov(\theta^i, \bar \theta^g) &amp;= \frac{1}{N}\sum_{g}\sum_{i \in g} (\theta^i - \bar \theta)(\bar \theta^g - \bar \theta),\\
&amp;= \frac{1}{N} \sum_g (\bar \theta^g - \bar \theta) \sum_{i \in g} (\theta^i - \bar \theta),\\
&amp;= \frac{1}{N} \sum_g n_g (\bar \theta^g - \bar \theta)(\bar \theta^g - \bar \theta),\\
&amp;= \sum_g p_g(\bar \theta^g - \bar \theta)^2,\\
&amp;= var(\bar \theta^g).
\end{align*}
$$</span>
This gives:
<span class="math-container">$$
var(\theta^i - \bar \theta^g) = var(\theta^i) - var(\bar \theta^g)
$$</span>
Rewriting gives:
<span class="math-container">$$
var(\theta^i) = var(\theta^i - \bar \theta^g) + var(\bar \theta^g). \tag{1}
$$</span>
Next, let's have a look at the covariance term <span class="math-container">$cov(\theta^i, \psi^g)$</span>.
<span class="math-container">$$
\begin{align*}
cov(\theta^i, \psi^g) &amp;= \frac{1}{N} \sum_g \sum_{i \in g} (\theta^i - \bar \theta)(\psi^g - \bar \psi),\\
&amp;= \frac{1}{N} \sum_g (\psi^g - \bar \psi) \left( \sum_{i \in g} (\theta^i - \bar \theta)\right),\\
&amp;= \frac{1}{N} \sum_g (\psi^g - \bar \psi) n_g(\bar \theta^g - \bar \theta),\\
&amp;= \sum_g \frac{n_g}{N} (\bar \theta^g - \bar \theta) (\psi^g - \bar \psi),\\
&amp;= \sum_g p_g (\bar \theta^g - \bar \theta)(\psi^g - \bar \psi),\\
&amp;= cov(\bar \theta^g, \psi^g)
\end{align*}
$$</span>
So:
<span class="math-container">$$
cov(\theta^i, \psi^g) = cov(\bar \theta^g, \psi^g). \tag{2}
$$</span>
Subsituting <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> into the first SPGBV condition should give you the second one.</p>

CC BY-SA 4.0
2021-09-15T00:13:33.513Z
Thanks so much for your answer. Now I can confirm that yes these two decompositions are exactly the same one.
Alalalalaki
CC BY-SA 4.0
2021-09-14T13:47:23.227Z
2021-09-14T13:47:23.227Z
tdm
<p>I am reading two recent papers studying between-firm and within-firm wage inequality, <a href="https://www.journals.uchicago.edu/doi/full/10.1086/684045" rel="nofollow noreferrer">Barth et al 2016</a> (hereafter BBDF) and <a href="https://academic.oup.com/qje/article/134/1/1/5144785" rel="nofollow noreferrer">Song et al 2019</a> (hereafter SPGBV). I am confused by the different variance decomposition methods used in these two papers.</p>
<p>Both of these two paper first yield a simple log wage variance decomposition.</p>
<p>BBDF: <span class="math-container">$$V(\ln w)=V(s)+V(\varphi)+2 \operatorname{Cov}(s, \varphi)+V(u)$$</span></p>
<p>SPGBV: <span class="math-container">$$\operatorname{var}\left(y_{t}^{i, j}\right)=\operatorname{var}\left(\theta^{i}\right)+\operatorname{var}\left(\psi^{j}\right)+2 \operatorname{cov}\left(\theta^{i}, \psi^{j}\right)+\operatorname{var}\left(\epsilon_{t}^{i, j}\right)$$</span></p>
<p>Where the <span class="math-container">$s$</span> or <span class="math-container">$\theta$</span> is person effects, the <span class="math-container">$\varphi$</span> or <span class="math-container">$\psi$</span> is firm effects, and <span class="math-container">$u$</span> or <span class="math-container">$\epsilon$</span> is the match error.</p>
<p>However they then both rewrite this simple decomposition to a more complicated decomposition that distinguish the between-firm component and within-firm component, in somehow different ways.</p>
<p>BBDF: <span class="math-container">$$V(\ln w) = \underbrace{V(s)(1-\rho)+V(u)}_{\text {Within-firm component }} + \underbrace{V(s)\left(\rho+2 \rho_{\varphi}\right)+V(\varphi)}_{\text {Between-firm component }}$$</span> , where <span class="math-container">$\rho=\operatorname{Cov}(s, S) / V(s)$</span>, <span class="math-container">$\rho_{\varphi}=\operatorname{Cov}(s, \varphi) / V(s)$</span>, and <span class="math-container">$S$</span> is defined as the establishment's average level of the predicted wage from <span class="math-container">$(s)$</span>.</p>
<p>SPGBV: <span class="math-container">$$\begin{aligned} \operatorname{var}\left(y_{t}^{i, j}\right)= \underbrace{\operatorname{var}\left(\theta^{i}-\bar{\theta}^{j}\right)+\operatorname{var}\left(\epsilon_{t}^{i, j}\right)}_{\text {Within-firm component }} +\underbrace{\operatorname{var}\left(\psi^{j}\right)+2 \operatorname{cov}\left(\bar{\theta}^{j}, \psi^{j}\right)+\operatorname{var}\left(\bar{\theta}^{j}\right)}_{\text {Between-firm component }}, \end{aligned}$$</span></p>
<p>Are these two decompositions the same thing but written in different ways? I try some calculations but fail to show that they are the same. Moreover while it is very clear how BBDF get their second decomposition (add and subtract one <span class="math-container">$\operatorname{Cov}(s, S)$</span> from the first decomposition), it is unclear to me where does the second formula in SPGBV come from? However in terms of interpretation, SPGBV seems to be a more intuitive way to explain the within- and between- components than the one in BBDF. I also wonder what is the principle behind a decomposition that separate the between- and within-firm effects?</p>

<labor-economics><applied-econometrics>
Variance Decomposition of Wage Equation
Alalalalaki
# Variance Decomposition of Wage Equation

(Asked by: username_0 on 2021-09-11)

I am reading two recent papers studying between-firm and within-firm wage inequality, [Barth et al 2016](https://www.journals.uchicago.edu/doi/full/10.1086/684045) (hereafter BBDF) and [Song et al 2019](https://academic.oup.com/qje/article/134/1/1/5144785) (hereafter SPGBV). I am confused by the different variance decomposition methods used in these two papers.

Both of these two paper first yield a simple log wage variance decomposition.

BBDF: $$V(\ln w)=V(s)+V(\varphi)+2 \operatorname{Cov}(s, \varphi)+V(u)$$

SPGBV: $$\operatorname{var}\left(y\_{t}^{i, j}\right)=\operatorname{var}\left(\theta^{i}\right)+\operatorname{var}\left(\psi^{j}\right)+2 \operatorname{cov}\left(\theta^{i}, \psi^{j}\right)+\operatorname{var}\left(\epsilon\_{t}^{i, j}\right)$$

Where the $s$ or $\theta$ is person effects, the $\varphi$ or $\psi$ is firm effects, and $u$ or $\epsilon$ is the match error.

However they then both rewrite this simple decomposition to a more complicated decomposition that distinguish the between-firm component and within-firm component, in somehow different ways.

BBDF: $$V(\ln w) = \underbrace{V(s)(1-\rho)+V(u)}\_{\text {Within-firm component }} + \underbrace{V(s)\left(\rho+2 \rho\_{\varphi}\right)+V(\varphi)}\_{\text {Between-firm component }}$$ , where $\rho=\operatorname{Cov}(s, S) / V(s)$, $\rho\_{\varphi}=\operatorname{Cov}(s, \varphi) / V(s)$, and $S$ is defined as the establishment's average level of the predicted wage from $(s)$.

SPGBV: $$\begin{aligned} \operatorname{var}\left(y\_{t}^{i, j}\right)= \underbrace{\operatorname{var}\left(\theta^{i}-\bar{\theta}^{j}\right)+\operatorname{var}\left(\epsilon\_{t}^{i, j}\right)}\_{\text {Within-firm component }} +\underbrace{\operatorname{var}\left(\psi^{j}\right)+2 \operatorname{cov}\left(\bar{\theta}^{j}, \psi^{j}\right)+\operatorname{var}\left(\bar{\theta}^{j}\right)}\_{\text {Between-firm component }}, \end{aligned}$$

Are these two decompositions the same thing but written in different ways? I try some calculations but fail to show that they are the same. Moreover while it is very clear how BBDF get their second decomposition (add and subtract one $\operatorname{Cov}(s, S)$ from the first decomposition), it is unclear to me where does the second formula in SPGBV come from? However in terms of interpretation, SPGBV seems to be a more intuitive way to explain the within- and between- components than the one in BBDF. I also wonder what is the principle behind a decomposition that separate the between- and within-firm effects?

**Tags:** labor-economics, applied-econometrics


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## 1 answers

### Answer by username_1 on 2021-09-14. Score: 2 | Accepted answer

I'm not at home in the literature, so I can not tell you if the two decompositions are the same. I can only help you with the derivations.

For the BBDF expression, you can simply obtain the first from the second by substituting out $\rho$ and $\rho\_\varphi$.

For the SPGBV expression things are a bit more tricky. Let $i$ represent the unit and $g$ the group variable (which is $j$ in your equation).

I'm guessing that
$$
\bar \theta^g = \dfrac{1}{n\_g} \sum\_{i \in g} \theta^i
$$
where $n\_g$ is the number of elements in group $g$. Let there be $N ( = \sum\_g n\_g)$ observations in total. Let $p\_g = \dfrac{n\_g}{n}$ be the probability of $\bar \theta^g$ (I take that $\dfrac{1}{N}$ is the probability of the unit $\theta^i$).

First look at the expression $var(\theta^i - \bar \theta^g)$​​. We can decompose it in the following way:
$$
var(\theta^i- \bar \theta^g) = var(\theta^i) + var(\bar \theta^g) - 2 cov(\theta^i, \bar \theta^g).
$$
Now, we can expand the covariance term:
$$
\begin{align\*}
cov(\theta^i, \bar \theta^g) &= \frac{1}{N}\sum\_{g}\sum\_{i \in g} (\theta^i - \bar \theta)(\bar \theta^g - \bar \theta),\\
&= \frac{1}{N} \sum\_g (\bar \theta^g - \bar \theta) \sum\_{i \in g} (\theta^i - \bar \theta),\\
&= \frac{1}{N} \sum\_g n\_g (\bar \theta^g - \bar \theta)(\bar \theta^g - \bar \theta),\\
&= \sum\_g p\_g(\bar \theta^g - \bar \theta)^2,\\
&= var(\bar \theta^g).
\end{align\*}
$$
This gives:
$$
var(\theta^i - \bar \theta^g) = var(\theta^i) - var(\bar \theta^g)
$$
Rewriting gives:
$$
var(\theta^i) = var(\theta^i - \bar \theta^g) + var(\bar \theta^g). \tag{1}
$$
Next, let's have a look at the covariance term $cov(\theta^i, \psi^g)$.
$$
\begin{align\*}
cov(\theta^i, \psi^g) &= \frac{1}{N} \sum\_g \sum\_{i \in g} (\theta^i - \bar \theta)(\psi^g - \bar \psi),\\
&= \frac{1}{N} \sum\_g (\psi^g - \bar \psi) \left( \sum\_{i \in g} (\theta^i - \bar \theta)\right),\\
&= \frac{1}{N} \sum\_g (\psi^g - \bar \psi) n\_g(\bar \theta^g - \bar \theta),\\
&= \sum\_g \frac{n\_g}{N} (\bar \theta^g - \bar \theta) (\psi^g - \bar \psi),\\
&= \sum\_g p\_g (\bar \theta^g - \bar \theta)(\psi^g - \bar \psi),\\
&= cov(\bar \theta^g, \psi^g)
\end{align\*}
$$
So:
$$
cov(\theta^i, \psi^g) = cov(\bar \theta^g, \psi^g). \tag{2}
$$
Subsituting $(1)$ and $(2)$ into the first SPGBV condition should give you the second one.

> **Comments:**
> **username_0**: Thanks so much for your answer. Now I can confirm that yes these two decompositions are exactly the same one.  

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