Once I have derived the variance of $\hat{\beta_1}$ as:

$\text{Var}(\hat{\beta_1})= \frac{\sigma^2}{\sum(x_i-\overline{x})^2}$

I would like to know how are affecting to this formula:

  • the size of the dataset $n$ and
  • the variance of residuals $\text{Var}(\varepsilon)$.

Any idea about how to check it? I mean i.e.: A higher $n$ will make $\text{Var}(\hat{\beta_1})$ increase/decrease, so on. I am really stuck on finding any kind of relationship between them.

  • 1
    $\begingroup$ If $\sigma^2$ isn't defined as $\operatorname{Var}(\varepsilon),$ then what is it? $\endgroup$ – whuber Jan 2 '20 at 23:42

Nick is technically right. But note that if you assume that you have a series of random samples from the underlying population, larger sample sizes (larger $n$) cause $Var(\hat\beta)$ to decrease asymptotically.

To see this, note that $\frac{1}{n} \sum (x_i - \bar{x})^2 \overset{p}{\to} Var(x)$ so $\sum (x_i - \bar{x})^2 \overset{p}{\to} n \,Var(x) = \infty$ (where $\overset{p}{\to}$ denotes convergence in probability). So the denominator goes to infinity as the sample size increases. You can similarly show that the numerator converges to a constant (namely the variance of the errors). So the variance of $\hat\beta$ goes to $0$ as the sample size gets large.


Just look at the formula. A larger n won't definitely make $Var(\hat{\beta}_1)$ increase or decrease. The variance only depends how spread out the data is from the mean. You could have a very large n, with data points that are close to $\bar{x}$ (and therefore, a very large variance) or you could have a very large n with data points that are spread out, and have a small variance. It's not directly influenced by the number of data points.

Then as whuber mentioned, $\sigma^2$ is the variance of the errors (which we usually estimate with the variance of the residuals). According to the formula, a larger error variance will result in a larger variance in $Var(\hat{\beta}_1)$.


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