<p>Certainly the true variance is not equal to the squared residual.  </p>

<p>In the case or robust standard errors, it has been proven that if we use $\hat{e}^2_i$ we obtain a consistent estimator <strong>of the variance-covariance matrix as a whole</strong>, <em>even though</em> $\hat{e}^2_i$ is not equal to $\sigma^2_i$ and <em>even though</em>  each single $\hat{e}^2_i$ does <em>not</em> converge in probability to $\sigma^2_i$ (it converges to the true squared error, which is a random variable). </p>

<p>The fundamental reason for this result is that </p>

<p>$$E(\hat{e}^2_i) \to_{n \to \infty} E(e^2_i) =\sigma^2_i$$</p>

<p>So the OP's phrase </p>

<blockquote>
  <p><em>When we calculate the robust robust standard error, we simply repalce
  $\sigma^2_i$ with $\hat{e}^2_i$. <strong>Thus, theses two are considered to be equal.</strong></em></p>
</blockquote>

<p>is simply wrong.</p>

CC BY-SA 3.0
2018-04-20T16:15:11.960Z
2018-04-20T18:22:20.243Z
2018-04-20T18:22:20.243Z
Alecos Papadopoulos
<p>What is the difference between $\hat{e}^2_i$ and $\sigma^2_i$?</p>

<p>In regression, we assume that $var(e_i)=E(e_i^2)=var(y_i)=\sigma_i^2$.</p>

<p>Does this imply that $\sigma_i^2$(the sample variance) is equal to $\hat{e}_i^2$?</p>

<p>When we calculate FGLS (feasible generalized least squres), we consider $\ln(\hat{e}_i^2)=\ln(\sigma_i^2)+v_i$, where $v_i$ is just the error term. This seems that we distinguish $\sigma_i^2$ from $\hat{e}_i^2$.</p>

<p>But, when we calculate the robust robust standard error, we simply repalce $\sigma_i^2$ with $\hat{e}_i^2$. Thus, theses two are considered to be equal. </p>

<p>I am confused with these two terms. Can anybody clarify this?  </p>

<p>Thank you in advance. </p>

<econometrics>
the difference between $\hat{e}^2_i$ and $\sigma^2_i$
shk910
# the difference between $\hat{e}^2_i$ and $\sigma^2_i$

(Asked by: username_0 on 2018-04-20)

What is the difference between $\hat{e}^2\_i$ and $\sigma^2\_i$?

In regression, we assume that $var(e\_i)=E(e\_i^2)=var(y\_i)=\sigma\_i^2$.

Does this imply that $\sigma\_i^2$(the sample variance) is equal to $\hat{e}\_i^2$?

When we calculate FGLS (feasible generalized least squres), we consider $\ln(\hat{e}\_i^2)=\ln(\sigma\_i^2)+v\_i$, where $v\_i$ is just the error term. This seems that we distinguish $\sigma\_i^2$ from $\hat{e}\_i^2$.

But, when we calculate the robust robust standard error, we simply repalce $\sigma\_i^2$ with $\hat{e}\_i^2$. Thus, theses two are considered to be equal.

I am confused with these two terms. Can anybody clarify this?

Thank you in advance.

**Tags:** econometrics


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## 1 answers

### Answer by username_1 on 2018-04-20. Score: 2

Certainly the true variance is not equal to the squared residual.

In the case or robust standard errors, it has been proven that if we use $\hat{e}^2\_i$ we obtain a consistent estimator **of the variance-covariance matrix as a whole**, *even though* $\hat{e}^2\_i$ is not equal to $\sigma^2\_i$ and *even though* each single $\hat{e}^2\_i$ does *not* converge in probability to $\sigma^2\_i$ (it converges to the true squared error, which is a random variable).

The fundamental reason for this result is that

$$E(\hat{e}^2\_i) \to\_{n \to \infty} E(e^2\_i) =\sigma^2\_i$$

So the OP's phrase

> *When we calculate the robust robust standard error, we simply repalce
> $\sigma^2\_i$ with $\hat{e}^2\_i$. **Thus, theses two are considered to be equal.***

is simply wrong.


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