<p>Intuitively, variables that only vary at the group level should have a lower variance (and therefore lower standard deviation) than <em>comparable</em> variables that vary at the individual level.</p>
<p>As these variables do not vary within each group, their within-group variance is zero, so their variance is solely determined by the between-group variation.</p>
<p>To see this, let <span class="math-container">$i$</span> be individual and <span class="math-container">$g$</span> be the group level (e.g. country). Let us denote the means and conditional means as:
<span class="math-container">$$
\begin{align*}
&amp;x_{..} = \mathbb{E}(x_{i,g}),\\
&amp;x_{.g} = \mathbb{E}(x_{i,g}|g)
\end{align*}
$$</span>
Then, we can write:
<span class="math-container">$$
\begin{align*}
var(x_{ig}) &amp;= \mathbb{E}(x_{ig}^2) - x_{..}^2,\\
&amp;= \sum_g \left(\mathbb{E}(x_{ig}^2|g) - x_{.g}^2 \right)\Pr(g) + \sum_g \left( x_{.,g}^2 - x_{..}^2 \right) \Pr(g),\\
&amp;= \sum_g var(x_{i,g}|g) \Pr(g) + var(x_{.,g})
\end{align*}
$$</span>
So we see that the variance of <span class="math-container">$x_{i,g}$</span> can be written as a weighted sum of the within group variances plus the between-group variance.</p>
<p>Now, consider two variables <span class="math-container">$x_{i,g}$</span> and <span class="math-container">$y_{i,g}$</span> such that <span class="math-container">$x_{.g} = y_{.g}$</span> but <span class="math-container">$y$</span> only varies on the group level, (so <span class="math-container">$y_{i,g} = y_{.g}$</span> for all <span class="math-container">$i$</span>).</p>
<p>As they have the same group mean, we have that:
<span class="math-container">$$
var(x_{.g}) = var(y_{.g}).
$$</span>
also
<span class="math-container">$$
var(y_{i,g}|g) = 0, \text{ while } var(x_{i,g}|g) \ge 0.
$$</span>
As such:
<span class="math-container">$$
var(x_{i,g}) \ge var(y_{i,g}).
$$</span>
This captures the simple fact that averaging across groups can only decrease the variance. In the extreme case where there is only one group, we get that <span class="math-container">$y_{.g} = y_{..}$</span>, so the variance drops to zero.</p>

CC BY-SA 4.0
2021-06-15T08:28:40.603Z
Thank @tdm. So, in short, do you mean firm-level should have a higher standard deviation compared to *comparable* country-level variables?
Phil Nguyen
CC BY-SA 4.0
2021-06-15T08:53:47.913Z
@BeautifulMindset That's indeed what happens if the variables are comparable (in terms of within group mean). Notice that this is different from the st-dev of the estimated coefficients in a regression (which, I guess, was the question in the linked post).
tdm
CC BY-SA 4.0
2021-06-17T23:26:16.143Z
Hi @tdm, I get lost at this place, can you please tell me where it comes from then?$$\begin{align*}
var(x_{ig}) &= \mathbb{E}(x_{ig}^2) - x_{..}^2$$
Phil Nguyen
CC BY-SA 4.0
2021-06-18T05:23:36.333Z
@BeautifulMindset We have that $var(x) = \mathbb{E}((x - \mathbb{E}(x))^2)$. If you work this out, you should get that $var(x) =  \mathbb{E}(x^2) - \mathbb{E}(x)^2$. Here $x_{..} = \mathbb{E}(x)$, which gives the formulat you are referring to.
tdm
CC BY-SA 4.0
2021-06-18T22:05:01.070Z
I just came back and let you know that I prove it successfully, thank you so much
Phil Nguyen
CC BY-SA 4.0
2021-06-15T08:22:02.417Z
2021-06-15T08:22:02.417Z
tdm
<p>I concern about why country-level variables normally have higher standard deviation compared to that in firm-level variables.</p>
<p>Today, my senior friend told me that it seems to be because the <strong>firm-level variables have more observations compared to country-level variables</strong>, leading to such an asymmetric standard deviation.</p>
<p>I am wondering is there any mathematical, reference document, or intuitive way to explain such a justification.</p>

CC BY-SA 4.0
2021-06-15T06:53:07.037Z
You seem to be linking to a question without any answers or comments.
Giskard
<applied-econometrics>
Why the larger the sample, the lower standard deviation?
Phil Nguyen
# Why the larger the sample, the lower standard deviation?

(Asked by: username_0 on 2021-06-15)

I concern about why country-level variables normally have higher standard deviation compared to that in firm-level variables.

Today, my senior friend told me that it seems to be because the **firm-level variables have more observations compared to country-level variables**, leading to such an asymmetric standard deviation.

I am wondering is there any mathematical, reference document, or intuitive way to explain such a justification.

**Tags:** applied-econometrics
> **Comments:**
> **username_1**: You seem to be linking to a question without any answers or comments.  

---

## 1 answers

### Answer by username_2 on 2021-06-15. Score: 6 | Accepted answer

Intuitively, variables that only vary at the group level should have a lower variance (and therefore lower standard deviation) than *comparable* variables that vary at the individual level.

As these variables do not vary within each group, their within-group variance is zero, so their variance is solely determined by the between-group variation.

To see this, let $i$ be individual and $g$ be the group level (e.g. country). Let us denote the means and conditional means as:
$$
\begin{align\*}
&x\_{..} = \mathbb{E}(x\_{i,g}),\\
&x\_{.g} = \mathbb{E}(x\_{i,g}|g)
\end{align\*}
$$
Then, we can write:
$$
\begin{align\*}
var(x\_{ig}) &= \mathbb{E}(x\_{ig}^2) - x\_{..}^2,\\
&= \sum\_g \left(\mathbb{E}(x\_{ig}^2|g) - x\_{.g}^2 \right)\Pr(g) + \sum\_g \left( x\_{.,g}^2 - x\_{..}^2 \right) \Pr(g),\\
&= \sum\_g var(x\_{i,g}|g) \Pr(g) + var(x\_{.,g})
\end{align\*}
$$
So we see that the variance of $x\_{i,g}$ can be written as a weighted sum of the within group variances plus the between-group variance.

Now, consider two variables $x\_{i,g}$ and $y\_{i,g}$ such that $x\_{.g} = y\_{.g}$ but $y$ only varies on the group level, (so $y\_{i,g} = y\_{.g}$ for all $i$).

As they have the same group mean, we have that:
$$
var(x\_{.g}) = var(y\_{.g}).
$$
also
$$
var(y\_{i,g}|g) = 0, \text{ while } var(x\_{i,g}|g) \ge 0.
$$
As such:
$$
var(x\_{i,g}) \ge var(y\_{i,g}).
$$
This captures the simple fact that averaging across groups can only decrease the variance. In the extreme case where there is only one group, we get that $y\_{.g} = y\_{..}$, so the variance drops to zero.

> **Comments:**
> **username_0**: Thank @tdm. So, in short, do you mean firm-level should have a higher standard deviation compared to *comparable* country-level variables?  
> **username_2**: @BeautifulMindset That's indeed what happens if the variables are comparable (in terms of within group mean). Notice that this is different from the st-dev of the estimated coefficients in a regression (which, I guess, was the question in the linked post).  
> **username_0**: Hi @tdm, I get lost at this place, can you please tell me where it comes from then?$$\begin{align*} var(x_{ig}) &= \mathbb{E}(x_{ig}^2) - x_{..}^2$$  
> **username_2**: @BeautifulMindset We have that $var(x) = \mathbb{E}((x - \mathbb{E}(x))^2)$. If you work this out, you should get that $var(x) =  \mathbb{E}(x^2) - \mathbb{E}(x)^2$. Here $x_{..} = \mathbb{E}(x)$, which gives the formulat you are referring to.  
> **username_0**: I just came back and let you know that I prove it successfully, thank you so much  

---

