Problem 28 - Calculus Single Variable

Question

In each of Exercises \(23-28,\) a function \(g\) and an interval I are specified. The function \(g\) is nonnegative on \(I .\) Find a number \(c\) such that \(f(x)=\operatorname{cg}(x)\) is a probability density function on \(I .\) $$ g(x)=1 /\left(1+x^{2}\right) \quad I=(-\infty, \infty) $$

Step-by-Step Solution

Answer

The constant \( c = \frac{1}{\pi} \).

Step 1: Understand the Probability Density Function

A probability density function (PDF) must satisfy two conditions: for any function \( f(x) \) to be a PDF over an interval \( I \), it must be non-negative for all \( x \) in \( I \), and the integral of \( f(x) \) over \( I \) must equal 1. Therefore, for \( f(x) = c \cdot g(x) \) to be a PDF, we need \( \int_{I} f(x) \, dx = 1 \).

Step 2: Set Up the Integral for Normalization

To find \( c \), we must ensure that \[ \int_{-\infty}^{\infty} c \cdot \frac{1}{1+x^2} \, dx = 1. \] Since \( c \) is a constant, it can be factored outside of the integral: \[ c \cdot \int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx = 1. \] We need to solve for \( c \).

Step 3: Calculate the Integral of g(x) Over I

The function \( g(x) = \frac{1}{1+x^2} \) is known to be the integrable form of the arctangent function. Thus, \[ \int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) + C. \] Therefore, the definite integral over the interval \((-\infty, \infty)\), \[ \int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx = \tan^{-1}(\infty) - \tan^{-1}(-\infty) = \frac{\pi}{2} - \left(-\frac{\pi}{2}\right) = \pi. \]

Step 4: Solve for the Constant c

Using the result from Step 3, substitute back into the equation: \[ c \cdot \pi = 1. \] Solve for \( c \) by dividing both sides by \( \pi \): \[ c = \frac{1}{\pi}. \] Hence, the constant \( c \) makes \( f(x) = c \cdot g(x) \) a valid PDF.

Key Concepts

Normalization

Normalization is a crucial concept when working with probability density functions (PDFs). To satisfy the conditions of a PDF, a function must fulfill two key criteria: it should be non-negative over its domain, and its integral over the entire interval must equal 1. This ensures that the total probability is distributed and sums appropriately to 1.

In the context of our exercise, we're dealing with the function \[ g(x) = \frac{1}{1+x^2} \]which is non-negative over the interval from \(-\infty\) to \(\infty\).

To transform this function into a probability density function, a constant \(c\) must be determined such that the entire area under the curve from \(\int_{-\infty}^{\infty} c \cdot g(x) \, dx = 1\).

This process of scaling the function with a constant is termed as normalization. In simpler terms, normalization adjusts the height of the function to ensure the entire probability adds up to one, adhering to the guidelines of a PDF.

Definite Integral

Definite integrals play a vital role when finding areas under curves, particularly when dealing with probability density functions. They allow us to calculate exactly how much area lies under a function between two specified limits.

In the problem at hand, our goal is to integrate \(\frac{1}{1+x^2}\) over the interval from \(-\infty\) to \(\infty\). The definite integral of \(\frac{1}{1+x^2}\) is pivotal because it determines the cumulative probability that must equal 1 for it to serve as a valid PDF.

Computing the definite integral over an infinite interval results in using limits effectively. Here, the definite integral becomes \(\int_{-\infty}^{\infty} \frac{1}{1+x^2} \, dx = \pi\). This value of a helps us in the normalization process where we divide the whole integral by a to discover the normalization constant \(c\).

• Without the help of definite integrals, translating functions into probability density functions would be considerably arduous.

Arctangent Function

The arctangent function, represented as \(\tan^{-1}(x)\), is an inverse trigonometric function that unravels interesting trigonometric properties into integrable expressions.

In our problem, we find \(\frac{1}{1+x^2}\) to be integrable as \(\tan^{-1}(x)\). Understanding this relationship transforms what may initially appear as a complex integral into a recognizable and calculable form.

When we compute the integral \(\int \frac{1}{1+x^2} \, dx = \tan^{-1}(x) + C\), we consider the indefinite version which features a constant \(C\), representing any constant value due to indefinite integration. However, when computing over an entire interval like \(-\infty\) to \(\infty\), these constants effectively cancel out.

• The arctangent function helps provide a mathematical bridge between algebraic expressions and their geometric or probabilistic interpretations.