<p>I have this derivation typed up as part of a PowerPoint slide. Forgive my laziness, but I'm going to just post an image of that slide.
<a href="https://i.sstatic.net/flWlq.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/flWlq.png" alt="enter image description here" /></a></p>

CC BY-SA 4.0
2024-01-21T09:23:14.267Z
2024-01-21T09:23:14.267Z
Michael Gmeiner
<p>Suppose <span class="math-container">$X_{it}^*$</span> is the demeaned version of <span class="math-container">$X_{it}$</span>, i.e.,
<span class="math-container">$$X_{it}^* = X_{it}- \bar{X}_i$$</span></p>
<p>In Hansen's econometrics textbook, he says the demeaned variable has reduced variation relative to the original observations, what is the precise meaning for this?</p>
<p>Indeed, can we prove the following inequality:
<span class="math-container">$$ Var[X_{it}^*] \leq Var[X_{it}].$$</span></p>

CC BY-SA 4.0
2024-01-20T11:36:42.670Z
It’s about panel data. The within-group variability is the same. It’s the between-group variability that is eliminated by the within transformation.
chan1142
CC BY-SA 4.0
2024-01-20T11:38:26.217Z
Average of yit minus y double bar squared versus the average of yit minus y_i bar squared.
chan1142
<econometrics><fixed-effects>
In a one-way fixed effect model, why do we say that the demeaned variable has reduced variation?
Mondayisgood
# In a one-way fixed effect model, why do we say that the demeaned variable has reduced variation?

(Asked by: username_0 on 2024-01-20)

Suppose $X\_{it}^\*$ is the demeaned version of $X\_{it}$, i.e.,
$$X\_{it}^\* = X\_{it}- \bar{X}\_i$$

In Hansen's econometrics textbook, he says the demeaned variable has reduced variation relative to the original observations, what is the precise meaning for this?

Indeed, can we prove the following inequality:
$$ Var[X\_{it}^\*] \leq Var[X\_{it}].$$

**Tags:** econometrics, fixed-effects

> **Comments:**
> **username_1**: It’s about panel data. The within-group variability is the same. It’s the between-group variability that is eliminated by the within transformation.  
> **username_1**: Average of yit minus y double bar squared versus the average of yit minus y_i bar squared.  

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## 1 answers

### Answer by username_2 on 2024-01-21. Score: 3

I have this derivation typed up as part of a PowerPoint slide. Forgive my laziness, but I'm going to just post an image of that slide.
[![enter image description here](https://i.sstatic.net/flWlq.png)](https://i.sstatic.net/flWlq.png)


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