# Why the larger the sample, the lower standard deviation?

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.

• You seem to be linking to a question without any answers or comments. Commented Jun 15, 2021 at 6:53

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.

• Thank @tdm. So, in short, do you mean firm-level should have a higher standard deviation compared to comparable country-level variables? Commented Jun 15, 2021 at 8:28
• @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
Commented Jun 15, 2021 at 8:53
• 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 Commented Jun 17, 2021 at 23:26
• @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
Commented Jun 18, 2021 at 5:23
• I just came back and let you know that I prove it successfully, thank you so much Commented Jun 18, 2021 at 22:05