Problem 4 - Calculus for Biology and Medicine

Question

Determine \(c\) such that $$ f(x)=\left\\{\begin{array}{ll} \frac{c}{x^{2}} & \text { for } x>1 \\ 0 & \text { for } x \leq 1 \end{array}\right. $$ is a density function.

Step-by-Step Solution

Answer

The value of \( c \) should be 1 for \( f(x) \) to be a density function.

Step 1: Understanding the Problem

We need to determine the value of \( c \) so that \( f(x) \) is a density function. For a function \( f(x) \) to be a density function, it must satisfy two criteria: it must be non-negative for all \( x \), and the integral over its entire range must equal 1.

Step 2: Check Non-Negativity

The function \( f(x) \) is given by \( \frac{c}{x^2} \) for \( x > 1 \) and is 0 for \( x \leq 1 \). Both expressions are non-negative given that \( c > 0 \), because division of a positive number by another positive number remains positive.

Step 3: Set Up the Integral of the Density Function

As \( f(x) = 0 \) for \( x \leq 1 \), we only need to integrate over the range \( x > 1 \). The integral of \( \frac{c}{x^2} \) from 1 to infinity must equal 1:\[\int_{1}^{\infty} \frac{c}{x^2} \, dx = 1.\]

Step 4: Compute the Integral

Calculate the integral:\[\int \frac{c}{x^2} \, dx = c \int x^{-2} \, dx = c \left[-x^{-1}\right],\]evaluated from 1 to infinity. This becomes:\[-c \left[ \frac{1}{x} \right]_{1}^{\infty}.\]

Step 5: Evaluate the Definite Integral

Evaluate the expression from step 4:\[-c \left( \lim_{x \to \infty} \frac{1}{x} - \frac{1}{1} \right) = -c(0 - 1) = c.\]Equating this result to 1 gives \( c = 1 \).

Step 6: Conclusion

With \( c = 1 \), the function \( f(x) \) satisfies both the non-negativity and normalization conditions to be a probability density function.

Key Concepts

Understanding Non-Negativity in a Probability Density Function

In the realm of probability density functions (PDFs), non-negativity is a crucial concept. A function

• must be non-negative for it to be considered a valid PDF.
• ensures that probabilities are never negative because probabilities cannot be negative in real-world terms.

For our given function, \(f(x) = \frac{c}{x^2}\) when \(x > 1\), and \(f(x) = 0\) when \(x \leq 1\), non-negativity is achieved by setting \(c > 0\).
Division of a positive constant \(c\) by a positive number such as \(x^2\) results in a positive value, thus satisfying the non-negativity requirement of PDFs.

The Crucial Normalization Condition

The normalization condition for a probability density function means that the total area under the curve of the function must equal one. This reflects the certainty that a probability over all possibilities sums to 1.
To satisfy the normalization condition, we set up the integral of the function over its range. For the given function:

• \(f(x) = \frac{c}{x^2}\) for \(x > 1\)
• we integrate from 1 to infinity

The integral \[\int_{1}^{\infty} \frac{c}{x^2} \, dx\]must be equal to 1.
Computing this integral and setting it equal to 1 helps us solve for the constant \(c\) to maintain the normalization condition.

The Role of Improper Integrals

The term improper integral comes into play when dealing with integrals that have infinite limits or involve infinite discontinuities. In the exercise, we deal with an improper integral
since the integration limit extends to infinity: \[\int_{1}^{\infty} \frac{c}{x^2} \, dx.\]To address an improper integral, we compute it as a limit. Specifically, we compute \[\lim_{t \to \infty} \int_{1}^{t} \frac{c}{x^2} \, dx\]
Doing so, we find it converges, meaning it results in a finite value that helps in confirming the normalization condition.

Calculus in Probability

Calculus serves as a bridge to solving problems in probability, especially when working with continuous data. It allows us to:

• calculate areas under curves for PDFs
• solve and understand improper integrals
• handle the conditions necessary for a function to serve as a probability model

For instance, in this problem, calculus enables the evaluation of the integral of \(\frac{c}{x^2}\), turning the task of confirming \(f(x)\) as a valid PDF into one of computation. The tools of differentiation and integration provide the means to flesh out the required conditions for validity in probability density functions, verifying both non-negativity and normalization.