3
$\begingroup$

This is my first post so I will try to be as clear and concise as possible. I am doing a course in statistics and we define the true mean of a random gaussian variable to be as follows:

$\mu$ = $\int_{R}xdF_{N(\mu, \sigma^2)}(x)$

As the Lesbesgue-Stieljes integral.

In the notes we defined the empirical CDF as follows:

${F_{n}}$ = $\frac{1}{n}\sum_{i=1}^{n}\mathbb{1}_{\{X_{i} \le x \}}$

We then proceed to show that:

$\mu$ = $\int_{R}xd(\frac{1}{n}\sum_{i=1}^{n}\mathbb{1}_{\{X_{i} \le x \}})$ = $\frac{1}{n}\sum_{i=1}^{n}X_{i}$

I am aware that this makes perfect sense as an empirical estimator for the mean, however I am quite confused as to how about deducing this from the integral. We are integrating with respect to a differential of the ECDF, and deduce that the expression is $X_{i}$?

I am struggling to even deduce what the result of $\int_{R}\mathbb{1}_{\{Xi \le x\}}$ evaluates to.

Any help on the above would be much appreciated.

$\endgroup$
1
  • $\begingroup$ I believe the differential of the indicator (step) function would be a dirac-delta? $\endgroup$
    – GeoMatt22
    Commented Oct 29, 2020 at 20:13

1 Answer 1

1
$\begingroup$

This is hinted at in Choosing the number of bootstrap resamples but not written out there. First, look at the definition of Stieltjes integral, from Wikipedia, the integral $$ \int_a^b g(x)\; dF(x) $$ is defined wia limits of approximants of the form, for partitions $a=x_0<x_1\dotsm <x_n=b$, $$ \sum_{i=0}^{n-1} g(x_i)\left[ F(x_{i+1}) - F(x_i)\right] $$ If you integrate with respect to a unit step function, say $\int g(x) \;dI\{x\le z\}$ that differencing can only be non-zero with $x_{i}< z \le x_{i+1}$ so the integral must be $g(z)$, the value of the integrand at the step. Also note that by the above definition we have $$ \int g(x)\;dF(x) + \int g(x) \; dG(x) =\int g(x)\; d\{(F+G)(x)\} $$ and using this with the empirical distribution function $$ \frac{1}{n}\sum_{i=1}^{n}\mathbb{1}_{\{X_{i} \le x \}} $$ and using this $$ \int x \;d\hat{F}_n(x) =\\ \int x \;d \bigl\{ \frac{1}{n}\sum_{i=1}^{n}\mathbb{1}_{\{X_{i} \le x \}} \bigr\}=\\ \frac1n \sum_{i=1}^{n} \int x \; d\mathbb{1}_{\{X_{i} \le x \}}=\frac1n \sum_{i=1}^{n} x_i = \bar{x} $$ And yes, the typical calculus course will not show this kind of integrals ...

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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