Problem 3 - Calculus for Biology and Medicine

Question

Determine \(c\) such that \(f(x)=\frac{c}{1+x^{2}}, \quad x \in \mathbf{R}\) is a density function.

Step-by-Step Solution

Answer

The value of \(c\) is \(\frac{1}{\pi}\).

Step 1: Understand the problem

For a function to be a probability density function (pdf), it must satisfy two conditions: it must be non-negative, and the integral over its entire domain must equal 1. For this function, we need to integrate \(f(x)\) from negative infinity to positive infinity and set it equal to 1.

Step 2: Set up the integral

We need to compute the integral of \(f(x) = \frac{c}{1+x^2}\) over \(x\) from \(-\infty\) to \(+\infty\). This integral is expressed as \(\int_{-\infty}^{\infty} \frac{c}{1+x^2} \,dx\).

Step 3: Evaluate the integral

Recognize that \(\int_{-\infty}^{\infty} \frac{1}{1+x^2} \,dx\) is a standard result, which equals \(\pi\) because it is the integral of the standard Cauchy distribution's density function. Therefore, \(\int_{-\infty}^{\infty} \frac{c}{1+x^2} \,dx = c\pi\).

Step 4: Solve for \(c\)

Set the integral equal to 1, according to the requirement for a density function: \(c\pi = 1\). Solve this equation for \(c\) to find \(c = \frac{1}{\pi}\).

Key Concepts

Cauchy distribution

The Cauchy distribution is a fascinating concept in probability theory. Unlike the normal distribution, the Cauchy distribution features thicker tails and a peak that's not as sharply defined. This type of distribution is notable because it lacks a mean and variance, distinguishing it from many common probability distributions.

One of the defining features of the Cauchy distribution is its probability density function (pdf), which is represented as:
\[f(x) = \frac{1}{\pi(1 + x^2)}\]This pdf shows that the Cauchy distribution focuses more probability in its tails, which means it's more likely to generate extremely large or small values compared to the normal distribution. In practical terms, the Cauchy distribution can sometimes better model phenomena with a higher frequency of extreme deviations.

When dealing with the Cauchy distribution, particularly if you're asked to set up a pdf, it’s important to recognize its characteristic formula. This will guide you in identifying whether the function you're working with fits this unique distribution.

Integration

Integration is a fundamental concept in calculus and is especially crucial when dealing with probability density functions. Essentially, integration helps us find the area under a curve, which in probability, corresponds to the likelihood of outcomes within a range.

To integrate the function \( f(x) = \frac{c}{1+x^2} \), you're assessing the area under this curve from \(-\infty\) to \(+\infty\). The result of the integral gives the total probability across the entire real line. For the standard Cauchy distribution, this integral without the constant \(c\) simplifies neatly to \(\pi\).

• A common task in probability is to adjust the constant \(c\) such that the integral over all possibilities equals 1, fulfilling the criteria of a probability density function.
• Recognizing when to use integration involves understanding its role as a tool for summarizing data dispersed across ranges.

Understanding integration in this context enables you to approach complex problems with clarity and determine the validity of probability distributions.

Density function condition

To determine whether a function is a valid probability density function (pdf), it needs to fulfill two key conditions:

• The function must be non-negative for all inputs. This means \(f(x) \geq 0\) for all \(x\); if any part of the function dips below zero, it cannot represent a probability.
• The integral of the function over its entire range must equal 1. This represents the total probability of all possible outcomes and confirms that probabilities sum up to the whole, ensuring the function is a valid pdf.

In this exercise, after recognizing and integrating the function \( f(x) = \frac{c}{1+x^2} \), the step to ensure it's a proper pdf is setting \\(c\pi = 1\), which leads to solving for \(c\) as \( c=\frac{1}{\pi} \).

Achieving this reflects that the function meets the total probability requirement, a crucial step in confirming its status as a pdf. This adjustment ensures that while working with such functions, you are reflecting sensible real-world scenarios in terms of probability distribution.