Problem 45 - Calculus: An Integrated Approach to Functions and Their Rates of Change

Question

As mentioned at the beginning of this section, statisticians use probability density functions to determine the probability of a random variable falling in a certain interval. If \(p(x)\) is a probability density function, then \(p(x) \geq 0\) for all \(x\) and \(\int_{-\infty}^{\infty} p(x) d x=1 .\) A probability density function of the form \(p(x)=\left\\{\begin{array}{ll}\lambda e^{-\lambda x} & \text { for } x \geq 0, \\ 0 & \text { for } x<0\end{array}\right.\) where \(\lambda\) is a positive constant describes what is known as an exponential distribution. Verify that $$ \int_{-\infty}^{\infty} p(x) d x=1 $$

Step-by-Step Solution

Answer

The integral of the given function from negative infinity to positive infinity evaluates to 1, verifying that it is indeed a probability density function

Step 1: Set up the integral

The function \(p(x)\) is defined as \(p(x)=\lambda e^{-\lambda x}\) for \(x \geq 0\), and \(p(x)=0\) for \(x<0\). Thus, we can write the integral over all real numbers as two separate integrals:\n\n\[\[\int_{-\infty}^{\infty} p(x) d x=\int_{-\infty}^{0}0 d x+\int_{0}^{\infty}\lambda e^{-\lambda x} d x\]\]

Step 2: Compute the integrals

The integral of 0 over any range is 0, so the first term of the sum disappears. For the second term:\n\n\[\[\int_{0}^{\infty}\lambda e^{-\lambda x} d x\]\]\n\nLet \(u = -\lambda x\), then \(du = -\lambda dx\), and \(dx = -du/\lambda\). After making the substitution, the integral becomes \n\n\[\[-\int_{0}^{\infty} e^{u} d u\]\]

Step 3: Evaluate the integral

The integral of \( e^{u} \) is \( e^{u} \). So \n\[\[-\int_{0}^{\infty} e^{u} d u =- e^{u} \Biggr|_0^\infty \]\]\nConsidering the bounds, \( u = 0 \) (when \( x = 0 \)) and \( u = -\infty \) (when \( x = \infty \)). Substituting these values gives\n\n\[\[- e^{u} \Biggr|_0^\infty = e^0 - e^{-\infty} = 1 - 0 = 1\]\]

Step 4: Combine the Results

Now, adding the results obtained from the two integrals, we get \(0 + 1 = 1\)

Key Concepts

Probability Density Function

Understanding probability density functions (PDFs) is essential when studying continuous random variables. A PDF, such as the one denoted as \( p(x) \), represents the likelihood of a random variable taking on a specific value within a given range. Mathematically, the function is non-negative over its entire domain, meaning \( p(x) \geq 0 \) for all possible values of \( x \). Furthermore, the integral of a PDF across its entire range is always equal to 1, symbolically shown as \( \int_{-\infty}^{\infty} p(x) dx = 1 \).

This integral criterion is crucial as it ensures that the total probability across all possible outcomes is 1, consistent with the definition of probability. In the case of the exponential distribution, the PDF is characterized by \( \lambda e^{-\lambda x} \) for positive values of \( x \) and 0 otherwise, where \( \lambda \) is a positive constant signifying the rate parameter. By checking the integral of this function over the entire range of \( x \), we validate the defining property of the PDF.

Integration in Probability

Integration plays a pivotal role in probability theory when dealing with continuous distributions. It is the tool used to calculate the probability that a continuous random variable falls within a certain interval. Instead of summing individual probabilities as in discrete distributions, integration allows us to aggregate probabilities over an interval for continuous variables.

To calculate the total probability, we integrate the PDF over the desired range. For the exponential distribution, the PDF is only positive for \( x \geq 0 \), thus in practice, we only integrate from 0 to infinity. Integration in this context allows us to formalize concepts like expected value and variance, which are foundational in statistics.

Calculus in Statistics

Calculus, and in particular the field's principles of integration and differentiation, are instrumental in several statistical concepts. In statistics, calculus helps in creating models for probability distributions, finding the expected values of random variables, and analyzing the changing rates at which events occur.

For example, the exponential distribution is heavily rooted in calculus concepts. Differentiating the distribution function gives the PDF, and integrating this PDF over a given range provides the probabilities for the distribution. The exponential distribution is unique as it models the time between events in a Poisson process, an instance where calculus sheds light on statistical phenomena.

Evaluating Integrals

Evaluating integrals is a fundamental operation in both calculus and statistics, especially for determining probabilities in continuous probability distributions. In the step-by-step solution provided above, the integral was separated into two parts, recognizing the piecewise nature of the exponential PDF. Since the PDF is zero for negative values of \( x \), the integral from \( -\infty \) to 0 contributes nothing to the total probability.

The integral from 0 to \( \infty \) is evaluated using a substitution, which simplifies the exponential part of the function. Ultimately, evaluating the integral yields 1, which confirms that the area under the curve of the exponential PDF over its entire range corresponds to a total probability of 100%, as expected for any probability density function.