/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 45 As mentioned at the beginning of... [FREE SOLUTION] | 91Ó°ÊÓ

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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 $$ \int_{-\infty}^{\infty} p(x) d x=1 $$

Short Answer

Expert verified
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 by step solution

01

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\[\[\int_{-\infty}^{\infty} p(x) d x=\int_{-\infty}^{0}0 d x+\int_{0}^{\infty}\lambda e^{-\lambda x} d x\]\]
02

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\[\[-\int_{0}^{\infty} e^{u} d u\]\]
03

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\[\[- e^{u} \Biggr|_0^\infty = e^0 - e^{-\infty} = 1 - 0 = 1\]\]
04

Combine the Results

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

Evaluate the integral. In many cases it will be advantageous to begin by doing a substitution. For example, in Problem 19, let \(w=\sqrt{x}, w^{2}=x ;\) then \(2 w d w=d x .\) This eliminates \(\sqrt{x}\) by replacing \(x\) with a perfect square. $$ \int_{0}^{\frac{\pi}{2}} x \sin x \cos x d x $$

Evaluate the integral. In many cases it will be advantageous to begin by doing a substitution. For example, in Problem 19, let \(w=\sqrt{x}, w^{2}=x ;\) then \(2 w d w=d x .\) This eliminates \(\sqrt{x}\) by replacing \(x\) with a perfect square. $$ \int_{1}^{e} \ln \sqrt{w} d w $$

Evaluate the integrals. $$ \int \frac{2 x^{3}-2 x^{2}+4 x+8}{(x-2)^{2}\left(x^{2}+3\right)} d x $$

Essay Question. Two of your classmates are having some trouble with improper integrals. Todd believes that improper integrals ought to diverge. He reasons that if \(f\) is positive, then the accumulated area keeps increasing, even if only by a little bit, so how can we get anything other than in nity? Dylan, on the other hand, is convinced that if \(\lim _{x \rightarrow \infty} f(x)=0\), then \(\int_{0}^{\infty} f(x) d x\) ought to converge. After all, he reasons, the rate at which area is accumulating is going to zero. Why isn \(\mathrm{t}\) that enough to assure convergence? Write an essay responding to Todd and Dylan s misconceptions. Your essay should be designed to help your classmates see the errors in their reasoning.

Determine whether the integral is convergent or divergent. Evaluate all convergent integrals. Be efficient. If \(\lim _{x \rightarrow \infty} \neq 0\), then \(\int_{a}^{\infty} f(x) d x\) is divergent. \(\int_{0}^{\infty} \cos x d x\)

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