Sampling Mean and Variance & Lebesgue–Stieltjes Integration

1. General Concept of Sampling Mean and Variance

In probability and statistics, the sampling mean and variance are two essential concepts used to describe data derived from a population sample.

Key Features of Their Distributions:

2. General Idea of Lebesgue–Stieltjes Integration

The Lebesgue–Stieltjes integral generalizes the Riemann integral and is particularly useful in probability theory and measure theory.

The integral is defined with respect to a function \( F \), which is non-decreasing and bounded. For a function \( f \), the Lebesgue–Stieltjes integral is: \[ \int_a^b f(x) \, dF(x) \] Here, \( dF(x) \) can represent a probability measure or a cumulative distribution function (CDF).

Applications to Probability Theory

Applications to Measure Theory

3. Numerical Comparison of Lebesgue and Riemann Integrals

To understand the difference between the Lebesgue and Riemann integrals, consider the task of finding the expected value \( \mathbb{E}[X] \) of a random variable \( X \) with the following probability density function (PDF):

\[ f(x) = \begin{cases} 3x^2 & \text{if } 0 \leq x \leq 1 \\ 0 & \text{otherwise.} \end{cases} \]

Problem:

Compute \( \mathbb{E}[X] \) using both Riemann and Lebesgue approaches.

Solution:

Using Riemann Integral:

Divide the domain of \( f(x) \), \([0, 1]\), into small intervals \( [x_i, x_{i+1}] \) and approximate \( \mathbb{E}[X] \) as: \[ \mathbb{E}[X] = \int_0^1 x f(x) \, dx = \int_0^1 x \cdot 3x^2 \, dx = \int_0^1 3x^3 \, dx. \] Computing this integral: \[ \int_0^1 3x^3 \, dx = 3 \cdot \frac{x^4}{4} \Big|_0^1 = 3 \cdot \frac{1}{4} = \frac{3}{4}. \]

Using Lebesgue Integral:

The Lebesgue integral partitions the range of \( f(x) \) instead of the domain. The expected value is given by: \[ \mathbb{E}[X] = \int_0^1 x \, dF(x), \] where \( F(x) \) is the cumulative distribution function (CDF) of \( f(x) \). First, compute \( F(x) \): \[ F(x) = \int_0^x 3t^2 \, dt = \left[ t^3 \right]_0^x = x^3. \] Now compute \( \mathbb{E}[X] \): \[ \mathbb{E}[X] = \int_0^1 x \, d(x^3). \] Using integration by parts: \[ \int_0^1 x \, d(x^3) = \Big[x \cdot x^3 \Big]_0^1 - \int_0^1 x^3 \, dx = \Big[1^4 - 0^4\Big] - \int_0^1 x^3 \, dx. \] The second term is: \[ \int_0^1 x^3 \, dx = \frac{x^4}{4} \Big|_0^1 = \frac{1}{4}. \] Thus: \[ \mathbb{E}[X] = 1 - \frac{1}{4} = \frac{3}{4}. \]

Conclusion:

Both methods yield the same result, \( \mathbb{E}[X] = \frac{3}{4} \). The Riemann integral divides the domain into intervals, while the Lebesgue integral considers the contributions of ranges of \( f(x) \) directly, which is especially advantageous for discontinuous or more complex functions.