0
$\begingroup$

I missed one of the lectures in my stats class and there was one slide that I could not understand.

enter image description here

In this example where $p^* = \sum_{j=1}^n l{\{...}\}/B$, I'm more familiar with the l indicating the function as a likelihood function for MLE. However, what does the l mean in this case for bootstrapping?

$\endgroup$
  • 1
    $\begingroup$ If you look carefully, you'll see that's an "I", not a lower-case "L". $\endgroup$ – jbowman Aug 17 '19 at 14:03
0
$\begingroup$

I don't know why it was notated like this but what is meant is that you count the elements in the set that fulfills the condition in {}. That is, for

$p* = \frac{\Sigma_{j=1}^n I\{\bar{X}^*_j ≥ \bar{X}\}}{B}$

You count, how many $\bar{X}^*_j$ are greater or equal to $\bar{X}$. The typical notation in the statistical literature on bootstraps is:

$p* = \frac{\#\{\bar{X}^*_j ≥ \bar{X}\}}{B}$ where $\#$ denotes the counting operation.

From this knowledge, I infer that what is meant on the slide is that by doing $I\{\bar{X}^*_j ≥ \bar{X}\}$ you create a binary variable, like this:

$I\{\bar{X}^*_j ≥ \bar{X}\} = \begin{cases} 1, & \text{if } \bar{X}^*_j ≥ \bar{X}\\\ 0, & \text{otherwise} \end{cases}$

Summing over this variable is a technical definition of counting.

Does this make sense for you?

$\endgroup$

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.