Question

Let \(X\) be a random variable of the continuous type that has pdf \(f(x)\). If \(m\) is the unique median of the distribution of \(X\) and \(b\) is a real constant, show that $$ E(|X-b|)=E(|X-m|)+2 \int_{m}^{b}(b-x) f(x) d x $$ provided that the expectations exist. For what value of \(b\) is \(E(|X-b|)\) a minimum?

Short answer

\[E(|X-b|) = E(|X-m|) + 2 \int_{m}^{b}(b-x) f(x) dx\]. The expected absolute deviation \(E(|X-b|)\) is minimum for \(b = m\), the median of the distribution of \(X\)

Step-by-step solution

Express \(E(|X-b|)\)

We can express \(E(|X-b|)\) as the integral of \(|x-b|\) multiplied by the pdf \(f(x)\) over the entire real line. Let's break this integral into two parts from negative infinity to \(b\) and from \(b\) to positive infinity: \[E(|X-b|)= \int_{-\infty}^{b}(b-x) f(x) dx + \int_{b}^{\infty}(x-b) f(x) dx\]

Show that \(E(|X-b|)=E(|X-m|)+2 \int_{m}^{b}(b-x) f(x) dx\)

Similarly express \(E(|X-m|\) as two integrals over negative infinity to \(m\) and from \(m\) to positive infinity. We note that the integrand will cancel out to zero for the section \(-\infty\) to \(m\), and \(b\) to \(\infty\) as \(m\) is a unique median. Hence, \[E(|X-m|) = \int_{-\infty}^{m}(m-x) f(x) dx + \int_{m}^{\infty}(x-m) f(x) dx = 0\] The only remaining part of \(E(|X-b|)\) that does not cancel is the \(b-m\) integral, so: \[E(|X-b|)=E(|X-m|)+2 \int_{m}^{b}(b-x) f(x) dx\]

Find value of \(b\) minimising \(E(|X-b|)\)

To minimise \(E(|X-b|)\), we take its derivative and equate it to zero. \[\frac{d}{db} E(|X-b|)=\frac{d}{db}(E(|X-m|)+2 \int_{m}^{b}(b-x) f(x) dx)= 0\] As \(E(|X-m|)\) doesn't depend on \(b\) it contributes 0 to the derivative, and the derivative of the integral part is \(2(b-m)f(b)\). Setting it equal to zero, \[b=m\] So, \(E(|X-b|)\) is minimum when \(b\) is equal to the median \(m\).

Key concepts

Probability Density Function

A probability density function (PDF), denoted here as f(x), is a fundamental concept in theoretical statistics and probability theory. It describes the likelihood of a continuous random variable taking on a particular value. The PDF characterizes the distribution of the variable such that the area under the curve between two points is the probability that the variable falls between those points. Importantly, for a valid PDF, the total area under the curve must be 1, reflecting the fact that a random variable is certain to take on a value within the range of possible outcomes.

When calculating expectations involving a continuous random variable, like E(|X-b|), the integral of the function times the PDF over all possible values gives us the expected value. This process is akin to summing up the products of variable values and their probabilities in discrete distributions, but it's done through integration due to the continuous nature of the variable involved.

Defining Properties of a PDF

• The PDF is non-negative for all possible values of x.
• The integral of the PDF over the entire space is equal to 1.
• The probability of the variable falling between two points is given by the integral of the PDF between these points.

Median of a Distribution

The median of a distribution represents the middle value in the dataset, where one half of the data points are below it and the other half above it. In the context of a continuous random variable, the median, denoted as m in our exercise, is the value that divides the area under the PDF into two equal parts. Therefore, it's also the value for which the probability of the random variable being less than m is the same as it being greater than m, each having a probability of 0.5.

In the provided exercise, the unique median ensures that the integrals of (m-x)f(x) from -fty to m and (x-m)f(x) from m to fty+ are equal, implying that the expectation E(|X-m|) is effectively zero, simplifying the evaluation of the expected value of the absolute deviation.

Importance of the Median

• It is a robust measure of central tendency that is less affected by outliers compared to the mean.
• In skewed distributions, the median provides a better center location than the mean.
• It is the point at which 50% of the data lies on either side.

Theoretical Statistics

Theoretical statistics provides the foundation for making inferences about real-world phenomena through mathematical models. It involves deriving properties of statistical methods, such as estimators and tests, as well as formulating the underlying principles that govern random behavior. Concepts like the expected value of a random variable are part of this theoretical framework, which also encompass the methods employed to analyze the absolute deviation from a specific point like the median.

Understanding the behavior of the expected absolute deviation is essential in identifying measures of variability that are not squared, like variance, and it can give us insights into the distribution's spread around a point of interest, most commonly the mean or median.

Applications in Problem Solving

• It helps in formulating problems like finding the minimum expected absolute deviation, as seen in our exercise.
• It draws connections between different areas of probability, like PDFs and medians, and their use in statistical calculations.
• Theoretical statistics involves proving generalization and laws such as the one depicted in the exercise which shows the relationship between the absolute deviation from a point and the median of the distribution.