Question

Average of a linear function. What is the average value of \(x\), given a distribution function \(q(x)=c x\), where \(x\) ranges from zero to one, and \(q(x)\) is normalized?

Short answer

The average value of x is \( \frac{2}{3} \).

Step-by-step solution

- Normalization Condition

To find the constant of normalization, use the condition that the integral of the probability density function over its range must equal 1. So, set up the integral: \[ \int_{0}^{1} q(x) \, dx = 1 \] Substitute the given distribution function: \[ \int_{0}^{1} c x \, dx = 1 \]

- Integrate the Function

Integrate to find the constant c: \[ c \int_{0}^{1} x \, dx = 1 \] The integral of x from 0 to 1 is: \[ \int_{0}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{0}^{1} = \frac{1}{2} - 0 = \frac{1}{2} \] So: \[ c \times \frac{1}{2} = 1 \rightarrow c = 2 \]

- Find Average Value

The average value of a function over an interval is given by: \[ E(x) = \int_{a}^{b} x q(x) \, dx \] Substitute the given distribution function and limits: \[ E(x) = \int_{0}^{1} x (2x) \, dx \] Simplify and integrate: \[ E(x) = 2 \int_{0}^{1} x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_{0}^{1} = 2 \times \frac{1}{3} = \frac{2}{3} \]

Key concepts

Normalization Condition

Understanding the concept of normalization is a great starting point when dealing with probability density functions (PDFs). When a function describes a probability distribution, the total probability represented by the function must be equal to 1. This concept ensures that the function accurately represents a distribution over its range.
In mathematical terms, this is expressed as:
\[\begin{equation} \ \int_{a}^{b} q(x) \, dx = 1 \ \end{equation}\]For the given problem, the distribution function is given by \ \ q(x) = c x \ \, and the range of \ \ x is from 0 to 1. To find the constant c that normalizes the function, we need to integrate \ \ q(x) \ \ over its range and set it equal to 1:\[\begin{equation} \ \int_{0}^{1} c x \, dx = 1 \ \end{equation}\]Solving this integral will provide the necessary normalization constant.

Probability Density Function (PDF)

A Probability Density Function (PDF) describes the likelihood of a random variable taking on a particular value. Unlike a simple probability, a PDF is used for continuous random variables and requires integration to find specific probabilities.
In essence, the PDF is a function that is non-negative and integrates to 1 over its specified range. For our linear function \ \ q(x) = c x \ \, we confirmed the normalization by solving:\[\begin{equation} \ c \, \int_{0}^{1} x \, dx = 1 \ \end{equation}\]After calculating the integral and solving for c, we find that our normalization constant c = 2. This normalized PDF can then be used to find probabilities and other statistical properties like the average value.

Average Value in Probability

Calculating the average value or expected value of a function in probability is a common task. This involves integrating the product of the variable and its PDF over the range of the variable. For any function \ \ q(x) \ \, the average value can be found from:
\[\begin{equation} E(x) = \int_{a}^{b} x q(x) \, dx \ \end{equation}\]In our problem, with the range 0 to 1 and \ \ q(x) = 2x \:\[\begin{equation} E(x) = \ \int_{0}^{1} x (2x) \, dx = 2 \int_{0}^{1} x^2 \, dx = 2 \left[ \frac{x^3}{3} \right]_{0}^{1} = 2 \times \frac{x}{3} = \frac{2}{3} \ \end{equation}\]Thus, the average value of x in this linearly distributed function is \frac{2}{3}\