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A family of pdf's that has been used to approximate the distribution of income, city population size, and size of firms is the Pareto family. The family has two parameters, \(k\) and \(\theta\), both \(>0\), and the pdf is $$ f(x ; k, \theta)=\left\\{\begin{array}{cl} \frac{k \cdot \theta^{k}}{x^{k+1}} & x \geq \theta \\ 0 & x<\theta \end{array}\right. $$ a. Sketch the graph of \(f(x, k, \theta)\). b. Verify that the total area under the graph equals 1 . c. If the rv \(X\) has pdf \(f(x ; k, \theta)\), for any fixed \(b>\theta\), obtain an expression for \(P(X \leq b)\). d. For \(\theta

Short Answer

Expert verified
a) Sketch a hyperbola starting at \(x = \theta\). b) Integral from \(\theta\) to \(\infty\) is 1. c) \(1 - \left(\frac{\theta}{b}\right)^{k}\) d) \(\left(\frac{\theta}{a}\right)^{k} - \left(\frac{\theta}{b}\right)^{k}\)

Step by step solution

01

Understand the Pareto PDF

The probability density function (PDF) for the Pareto distribution is given as \(f(x; k, \theta) = \frac{k \cdot \theta^{k}}{x^{k+1}}\) for \(x \geq \theta\) and 0 for \(x < \theta\). Here, \(k\) and \(\theta\) are parameters of the distribution, both greater than zero.
02

Sketch the Graph of the PDF

To sketch \(f(x; k, \theta)\), begin by plotting the function \(f(x; k, \theta) = \frac{k \cdot \theta^{k}}{x^{k+1}}\) starting from \(x = \theta\). The function asymptotically approaches zero as \(x\) increases and takes the shape of a hyperbola starting at \(x = \theta\).
03

Verify Total Area Equals 1

Calculate the integral of \(f(x; k, \theta)\) from \(\theta\) to infinity: \[\int_{\theta}^{\infty} \frac{k \cdot \theta^{k}}{x^{k+1}} \, dx = \left[ -\frac{\theta^{k}}{x^{k}} \right]_{\theta}^{\infty} = 1.\]This confirms that the total area under the curve is 1, proving \(f(x; k, \theta)\) is a valid PDF.
04

Calculate \(P(X \leq b)\)

For \(b > \theta\), find \(P(X \leq b)\):\[P(X \leq b) = \int_{\theta}^{b} \frac{k \cdot \theta^{k}}{x^{k+1}} \, dx = 1 - \left(\frac{\theta}{b}\right)^{k}.\]
05

Calculate \(P(a \leq X \leq b)\)

For \(\theta < a < b\), determine \(P(a \leq X \leq b)\):\[P(a \leq X \leq b) = \int_{a}^{b} \frac{k \cdot \theta^{k}}{x^{k+1}} \, dx = \left(\frac{\theta}{a}\right)^{k} - \left(\frac{\theta}{b}\right)^{k}.\]

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Key Concepts

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

Probability Density Function (PDF)
The Probability Density Function, or PDF, is a fundamental concept in statistics, especially when dealing with continuous probability distributions like the Pareto distribution. A PDF describes the likelihood of a random variable to take on a particular value. For continuous distributions, the PDF technically gives the relative likelihood for the random variable to occur at each point in the allowable range, but not the exact probability. This is due to the nature of continuous variables, which can take infinitely many values within any interval.

The Pareto distribution is defined by its PDF:
  • For values of the random variable, \(x\), greater than or equal to the threshold parameter, \(\theta\), the PDF is given by \(f(x; k, \theta) = \frac{k \cdot \theta^k}{x^{k+1}}\).
  • For values less than \(\theta\), the PDF is zero, as those values are outside the range of the Pareto-distributed variable.
The PDF helps us understand how probabilities are distributed over different values of \(x\). It is important to note that the area under the PDF over all possible values of \(x\) equals 1, which signifies the total probability of the occurrence of all possible outcomes.
Integral Calculus in Statistics
Integral calculus is a powerful tool in statistics used to find probabilities associated with continuous probability distributions. With a continuous random variable, probabilities are found over intervals rather than for specific values, and this is where integration comes into play.

To confirm that a given PDF, like that of the Pareto distribution, is valid, we calculate the integral over its entire range. For the Pareto PDF, this is from \(\theta\) to infinity:
  • We verify the total area under the curve by computing:\[ \int_{\theta}^{\infty} \frac{k \cdot \theta^{k}}{x^{k+1}} \, dx = 1. \]
This calculation ensures that the sum total of probabilities equals 1, confirming the PDF represents a valid distribution.

Furthermore, to find the probability that a random variable \(X\) is less than or equal to a point \(b\), or within an interval \([a, b]\), integral calculations allow us to evaluate these probabilities effectively:
  • \(P(X \leq b) = \int_{\theta}^{b} \frac{k \cdot \theta^{k}}{x^{k+1}} \, dx = 1 - \left(\frac{\theta}{b}\right)^{k}.\)
  • \(P(a \leq X \leq b) = \int_{a}^{b} \frac{k \cdot \theta^{k}}{x^{k+1}} \, dx = \left(\frac{\theta}{a}\right)^{k} - \left(\frac{\theta}{b}\right)^{k}.\)
Continuous Probability Distributions
Continuous probability distributions describe the probabilities of outcomes over a continuum of values rather than discrete events. Key examples include the normal distribution and the Pareto distribution.

The Pareto, a type of continuous distribution, is used in various fields such as economics to model phenomena like income or wealth distributions. Key characteristics of continuous distributions include:
  • Defined over an interval with infinite possible values, allowing no way to assign positive probability to individual points.
  • Probabilities are derived from integrals over intervals, which are represented by areas under the curve of their PDFs.
For the Pareto distribution:
  • It is characterized by the parameters \(k\) and \(\theta\), which shape the distribution's form and define the threshold above which all outcomes are considered.
  • This distribution is notable for its 'heavy-tailed' nature, where large values become less probable but never impossible as \(x\) increases.
Understanding continuous distributions like the Pareto is crucial for interpreting real-world data where exact values are indefinite, and outcomes span a continuum.

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Most popular questions from this chapter

Let \(X\) be the total medical expenses (in 1000 s of dollars) incurred by a particular individual during a given year. Although \(X\) is a discrete random variable, suppose its distribution is quite well approximated by a continuous distribution with pdf \(f(x)=k(1+x / 2.5)^{-7}\) for \(x \geq 0\). a. What is the value of \(k\) ? b. Graph the pdf of \(X\). c. What are the expected value and standard deviation of total medical expenses? d. This individual is covered by an insurance plan that entails a \(\$ 500\) deductible provision (so the first \(\$ 500\) worth of expenses are paid by the individual). Then the plan will pay \(80 \%\) of any additional expenses exceeding \(\$ 500\), and the maximum payment by the individual (including the deductible amount) is \(\$ 2500\). Let \(Y\) denote the amount of this individual's medical expenses paid by the insurance company. What is the expected value of \(Y\) ? [Hint: First figure out what value of \(X\) corresponds to the maximum out-of- pocket expense of \(\$ 2500\). Then write an expression for \(Y\) as a function of \(X\) (which involves several different pieces) and calculate the expected value of this function.]

When circuit boards used in the manufacture of compact dise players are tested, the long-run percentage of defectives is \(5 \%\). Suppose that a batch of 250 boards has been received and that the condition of any particular board is independent of that of any other board. a. What is the approximate probability that at least \(10 \%\) of the boards in the batch are defective? b. What is the approximate probability that there are exactly 10 defectives in the batch?

Let \(X=\) the time (in \(10^{-1}\) weeks) from shipment of a defective product until the customer returns the product. Suppose that the minimum return time is \(\gamma=3.5\) and that the excess \(X-3.5\) over the minimum has a Weibull distribution with parameters \(\alpha=2\) and \(\beta=1.5\) (see "Practical Applications of the Weibull Distribution" Industrial Quality Control, Aug. 1964:71-78). a. What is the cdf of \(X\) ? b. What are the expected return time and variance of return time? [Hint: First obtain \(E(X-3.5)\) and \(V(X-3.5)\).] c. Compute \(P(X>5)\). d. Compute \(P(5 \leq X \leq 8)\).

The authors of the article "A Probabilistic Insulation Life Model for Combined Thermal-Electrical Stresses" (IEEE Trans. on Elect. Insulation, 1985: 519-522) state that "the Weibull distribution is widely used in statistical problems relating to aging of solid insulating materials subjected to aging and stress." They propose the use of the distribution as a model for time (in hours) to failure of solid insulating specimens subjected to \(\mathrm{AC}\) voltage. The values of the parameters depend on the voltage and temperature; suppose \(\alpha=2.5\) and \(\beta=200\) (values suggested by data in the article). a. What is the probability that a specimen's lifetime is at most 250 ? Less than 250 ? More than 300 ? b. What is the probability that a specimen's lifetime is between 100 and 250 ? c. What value is such that exactly \(50 \%\) of all specimens have lifetimes exceeding that value?

The accompanying observations are precipitation values during March over a 30 -year period in Minneapolis-St. Paul. \(\begin{array}{rrrrrr}.77 & 1.20 & 3.00 & 1.62 & 2.81 & 2.48 \\ 1.74 & .47 & 3.09 & 1.31 & 1.87 & .96 \\ .81 & 1.43 & 1.51 & .32 & 1.18 & 1.89 \\ 1.20 & 3.37 & 2.10 & .59 & 1.35 & .90 \\ 1.95 & 2.20 & .52 & .81 & 4.75 & 2.05\end{array}\) a. Construct and interpret a normal probability plot for this data set. b. Calculate the square root of each value and then construct a normal probability plot based on this transformed data. Does it seem plausible that the square root of precipitation is normally distributed? c. Repeat part (b) after transforming by cube roots.

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