0
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ If $\sigma^2$ isn't defined as $\operatorname{Var}(\varepsilon),$ then what is it? $\endgroup$ – whuber Jan 2 at 23:42
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
1
$\begingroup$

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 $Var(\hat{\beta}_1)$.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.