Problem 13 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

Find a number \(c\) so that the following function \(f(x)\) is a probability density function: $$ f(x)= \begin{cases}c x e^{-x / 4} & \text { if } x \geq 0 \\ 0 & \text { otherwise }\end{cases} $$

Step-by-Step Solution

Answer

The value of \(c\) is \( \frac{1}{16} \).

Step 1 - Understand the problem

To find a number \(c\) that makes \(f(x)\) a probability density function (PDF), the integral of \(f(x)\) over its domain must equal 1.

Step 2 - Set up the integral

Set up the integral of \(f(x)\) from 0 to infinity: \[ \ \int_{0}^{\infty} c x e^{-x/4} \, dx = 1 \]

Step 3 - Solve the integral

Solve the integral. Let \( u = x/4 \), then \( du = 1/4 \, dx \), therefore, \( dx = 4 \, du \). The integral becomes: \[ \ \int_{0}^{\infty} c x e^{-x/4} \, dx = c \int_{0}^{\infty} 4u e^{-u} \, 4 \, du = 16c \int_{0}^{\infty} u e^{-u} \, du \]

Step 4 - Use the Gamma function

The integral \( \int_{0}^{\infty} u e^{-u} \, du \) is a known Gamma function \( \Gamma(2) \) and equals 1. Thus: \[ \ 16c \cdot 1 = 1 \]

Step 5 - Solve for c

Set the equation from Step 4 equal to 1 and solve for \(c\): \[ \ 16c = 1 \ \rightarrow c = \frac{1}{16} \]

Key Concepts

Integral Calculus

Integral calculus helps us find the area under curves. In probability, it is commonly used to ensure that a probability density function (PDF) sums to 1 over its entire range. This is vital to satisfy the properties of a PDF.

In the given exercise, we have \(f(x) = c x e^{-x / 4}\) for \(x \geq 0\). To determine if this is a valid PDF, we must integrate it from 0 to infinity and set this integral equal to 1. This process is known as 'normalization'. We start by setting up the integral:

\[- \int_{0}^{\infty} c x e^{-x/4} \, dx = 1 \-\]

Substitutions can simplify these integrals. In this case, we use \(u = x/4\), then \(du = 1/4 dx\) or \(dx = 4 du\). The integral becomes:

\[- \int_{0}^{\infty} c x e^{-x/4} \, dx = c \int_{0}^{\infty} 4u e^{-u} \, 4 \, du \Rightarrow 16c \int_{0}^{\infty} u e^{-u} \, du\-\]

The above process converts your problem into a simpler integral involving the variable u. This substitution helps in evaluating the integral more efficiently.

Gamma Function

The Gamma function is a special function that extends the factorial function to real and complex numbers. It is denoted \(\Gamma(n)\) and is defined by the integral:

\[- \Gamma(n) = \int_{0}^{\infty} t^{n-1} e^{-t} \, dt\-\]

This function is particularly useful in probability and statistics, especially when dealing with integrals.

In our exercise, we encounter an integral that looks like the Gamma function when n is 2:

\[- \Gamma(2) = \int_{0}^{\infty} u e^{-u} \, du\-\]

We know \(\Gamma(2) = 1\) from the properties of the Gamma function. Hence:

\[- 16c \cdot 1 = 1\-\]

Recognizing the Gamma function in integrals can greatly simplify our work. By equating it to its known value, we streamline the process of finding our constant 'c'.

Normalization Condition

The normalization condition is essential for a function to be recognized as a probability density function (PDF). This condition requires that the total area under the curve defined by the PDF equals 1.

For the problem at hand, the normalization condition takes the form of ensuring:

\[- \int_{0}^{\infty} f(x) \, dx = 1\-\]

Given our function \(f(x) = c x e^{-x / 4}\), we solve:

\[- 16c \int_{0}^{\infty} u e^{-u} \, du = 1\-\]

Substituting the value \(\Gamma(2) = 1\), we have:

\[- 16c = 1 \Rightarrow c = \frac{1}{16}\-\]

Therefore, \(c = \frac{1}{16}\) satisfies the normalization condition, transforming our function into a valid PDF. This ensures the area under the curve equals 1, fulfilling the requirement for all PDFs.