5
$\begingroup$

This is an excerpt from BW Silverman's Density Estimation for Statistics and Data Analysis:

The oldest and most widely used density estimator is the histogram. Given an origin $x_0$ and a bin width of $h$, we define the bins of the histogram to be the intervals $[x_0+mh,x_0+(m+1)h]$ for integers $m$. The histogram is defined by

$$\hat{f} = \dfrac{1}{nh}(\text{no. of $X_i$ in the same bin as > $x$})$$

Since I am seeing $\hat{f}$ , does this mean that we are talking about an estimator? I also did not understand the formula. I am not sure about the switch from $m$ to $n$, I just assume that they are the same thing.

$\endgroup$
  • $\begingroup$ m takes values 0, 1, ..., M-1 (where M is chosen such that the full range is covered) while n is the number of data points, and yes this would be considered an estimator of the probability density function. $\endgroup$ – tristan Mar 25 '15 at 6:45
5
$\begingroup$

As @tristan comments, $m$ is a counter integer, while $n$ is the total number of data points in the sample, and $h$ is the histogram bin width. The formula is correct.

It may be easier to understand if you consider the case where you have $M$ bins and the same number of data points $\frac{n}{M}$ in each bin. Then your histogram height will be the same for each bin, $\hat{f}=\frac{1}{Mh}$. So you have $M$ bins, each of width $h$ and height $\hat{f}=\frac{1}{Mh}$, for a total area of 1. As a density should be.

In fact, if you count the data points in each bin, you will find that your histogram always has a total area of 1. Again: this is just what a density should be.

And yes, $\hat{f}$ is an estimator. It is an estimate of the density, in the space of step functions. You can approximate most "normal" functions using step functions (in the sense that the integral over the absolute difference between the step function and the function to be approximated goes to zero as $h\to 0$), so step functions are a logical simple approximation.


In fact, histograms can be seen as related to kernel density estimators, with "kernels" that don't only depend on $\frac{x-x_i}{h}$, but additionally on $x$: i.e., "counting kernels" that count how many $x_i$ fall into the interval (bin) containing $x$. This is a somewhat contrived way of looking at histograms, but I actually find it a bit enlightening.

| 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.