Problem 27 - 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 \(I .\) $$ g(x)=e^{-2 x} \quad I=[0, \infty] $$

Step-by-Step Solution

Answer

The constant \( c \) such that \( f(x) = c g(x) \) is a probability density function is \( c = 2 \).

Step 1: Understand the Definition of a Probability Density Function (PDF)

A function \( f(x) = c \cdot g(x) \) is a probability density function on an interval \( I \) if the integral over \( I \) equals 1. That is, \( \int_I f(x) \, dx = 1 \). For this problem, \( I = [0, \infty) \).

Step 2: Set Up the Integral for \( f(x) \)

We need to solve the integral \( \int_0^{\infty} f(x) \, dx = 1 \). Given \( f(x) = c \cdot e^{-2x} \), the formula becomes \( \int_0^{\infty} c \cdot e^{-2x} \, dx = 1 \). We can factor out the constant \( c \): \( c \cdot \int_0^{\infty} e^{-2x} \, dx = 1 \).

Step 3: Evaluate the Integral

We now need to evaluate the integral \( \int_0^{\infty} e^{-2x} \, dx \). This is an exponential integral that can be computed as follows: \( \int e^{-2x} \, dx = -\frac{1}{2} e^{-2x} + C \). Evaluating from 0 to \( \infty \), we have:\[ \left[ -\frac{1}{2} e^{-2x} \right]_0^{\infty} = 0 - \left( -\frac{1}{2} e^0 \right) = \frac{1}{2} \].

Step 4: Solve for \( c \)

With the result of the integral being \( \frac{1}{2} \), our equation becomes \( c \cdot \frac{1}{2} = 1 \). Solving for \( c \), we get: \( c = 2 \).

Step 5: Verify the Solution

To confirm our solution, substitute \( c = 2 \) back into the integral equation: \[ \int_0^{\infty} 2 \cdot e^{-2x} \, dx = 2 \cdot \frac{1}{2} = 1 \].This confirms that with \( c = 2 \), the equation holds true, verifying that \( f(x) = 2e^{-2x} \) is a probability density function.

Key Concepts

Exponential Functions

An exponential function is a mathematical expression in which a constant base is raised to a variable exponent, typically seen with the natural base \( e \), where \( e \approx 2.718 \). In our case, we specifically look at functions like \( g(x) = e^{-2x} \). The negative sign in the exponent, \(-2x\), indicates that this specific exponential function decreases as the value of \( x \) increases.
An exponential function is commonly used in various mathematical and real-world contexts, such as population growth, radioactive decay, and continuously compounded interest. It's important because it models situations where rates of change are proportional to current values.

• The base of the natural exponential function, \( e \), is a mathematical constant.
• Exponential decay occurs when the exponent is negative.
• Key properties include rapid growth or decay and the fact that any power of \( e \) is always positive.

Understanding exponential functions is crucial when dealing with probability density functions (PDFs), as they often appear when modeling time-related phenomena like the time between events in a Poisson process.

Integration

Integration is a fundamental concept in calculus that involves finding the area under a curve. When dealing with probability density functions, integration helps us determine the total probability, which should equal \( 1 \). In this exercise, we focused on integrating the function \( f(x) = c \cdot e^{-2x} \) over the interval \([0, \infty)\).
When integrating exponential functions, a useful property is their simple antiderivatives. For example, the integral \( \int e^{-2x} \, dx \) is given by \(-\frac{1}{2} e^{-2x} + C\).

• Integration over an interval adds up the infinitesimal areas, yielding the total area under the curve.
• For a probability density function, this total area corresponds to the total probability.
• Improper integrals like \( \int_0^{\infty} e^{-2x} \, dx \) extend the integration bounds, used here to model continuous random variables over unlimited domains.

In our exercise, evaluating the integral from \( 0 \) to \( \infty \) gave us a value of \( \frac{1}{2} \), which helped us find the constant \( c \) necessary for \( f(x) \) to qualify as a PDF.

Nonnegative Functions

A nonnegative function is one that does not drop below zero for any input within its given interval. This is crucial for probability density functions, as they represent probabilities, which are inherently nonnegative. The exponential function \( g(x) = e^{-2x} \) is nonnegative over its domain.
Nonnegative functions are integral to probability because they ensure that each piece of the domain measured reflects a non-zero contribution to the total probability, helping ensure the full integral equates to \( 1 \).

• The function \( g(x) = e^{-2x} \) ensures nonnegative output since \( e \) raised to any power is always positive.
• These functions are simple yet powerful tools in statistical modeling.
• Ensuring a function is nonnegative in a probability context prevents misinterpretations or mathematical inconsistencies.

In conclusion, the nonnegative nature of \( g(x) = e^{-2x} \) over \([0, \infty)\) confirms it is suitable for transformation into a probability density function upon finding the correct constant \( c \).