/*! 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 24 Let \(X\) have the Pareto pdf ... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 24

Let \(X\) have the Pareto pdf $$ f(x ; k, \theta)=\left\\{\begin{array}{cc} \frac{k \cdot \theta^{k}}{x^{k+1}} & x \geq \theta \\ 0 & x<\theta \end{array}\right. $$ introduced in Exercise \(10 .\) a. If \(k>1\), compute \(E(X)\). b. What can you say about \(E(X)\) if \(k=1\) ? c. If \(k>2\), show that \(V(X)=k \theta^{2}(k-1)^{-2}(k-2)^{-1}\). d. If \(k=2\), what can you say about \(V(X)\) ? e. What conditions on \(k\) are necessary to ensure that \(E\left(X^{n}\right)\) is finite?

Short Answer

Expert verified
a. \(E(X) = \frac{k\theta}{k-1}\) if \(k>1\); b. Infinite if \(k=1\); c. \(V(X) = \frac{k\theta^2}{(k-1)^2(k-2)}\) if \(k>2\); d. Infinite if \(k=2\); e. \(n < k-1\) for finiteness.

Step by step solution

01

Understand the Pareto Distribution

The Pareto distribution is characterized by parameters \(k > 0\) and \(\theta > 0\). It is defined for \(x \geq \theta\), and its probability density function is \(f(x; k, \theta) = \frac{k \cdot \theta^{k}}{x^{k+1}}\).
02

Compute E(X) for k > 1

The expected value \(E(X)\) of a Pareto distribution is given by the integral \(E(X) = \int_{\theta}^{\infty} x \cdot f(x; k, \theta) \, dx\). Compute the integral:\[ E(X) = \int_{\theta}^{\infty} x \cdot \frac{k \cdot \theta^{k}}{x^{k+1}} \, dx = k \cdot \theta^{k} \int_{\theta}^{\infty} x^{-k} \, dx \]This simplifies and evaluates to:\[ E(X) = \frac{k \theta^{k}}{k-1} \cdot \theta^{1-k} = \frac{k \cdot \theta}{k-1}, \quad \text{for } k>1 \]
03

Analyze E(X) for k = 1

For \(k = 1\), the expected value\[ E(X) = \frac{k \cdot \theta^{k}}{k-1} \]becomes undefined since division by zero occurs. Therefore, \(E(X)\) is infinite.
04

Compute V(X) for k > 2

The variance \(V(X)\) is calculated knowing that \(V(X) = E(X^2) - (E(X))^2\). Given \(k > 2\), compute \(E(X^2)\):\[ E(X^2) = \int_{\theta}^{\infty} x^2 \cdot \frac{k \cdot \theta^{k}}{x^{k+1}} \, dx = k \cdot \theta^{k} \int_{\theta}^{\infty} x^{1-k} \, dx = \frac{k \cdot \theta^2}{k-2} \]Hence, \(V(X) = \frac{k \cdot \theta^2}{k-2} - \left(\frac{k \cdot \theta}{k-1}\right)^2\).Simplifying gives:\[ V(X) = \frac{k \cdot \theta^2 \cdot (k-1)^2 - k^2 \cdot \theta^2\cdot(k-2)}{(k-1)^2 (k-2)} = \frac{k \cdot \theta^2}{(k-1)^2(k-2)}, \quad k>2 \]
05

Consider V(X) for k = 2

For \(k = 2\), the term \(\frac{k \cdot \theta^2}{k-2}\) is undefined due to division by zero. Thus, \(V(X)\) is infinite.
06

Conditions for E(X^n) to be Finite

To ensure \(E(X^n)\) is finite, the integral\[ E(X^n) = \int_{\theta}^{\infty} x^n \cdot \frac{k \cdot \theta^{k}}{x^{k+1}} \, dx = k \cdot \theta^{k} \int_{\theta}^{\infty} x^{n-k-1} \, dx \]must converge. This requires \(n-k-1 < -1\) or \(n < k-1\).

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Key Concepts

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

Expected Value
The expected value, or mean, of a random variable represents a central or average value that the variable might take. In terms of the Pareto distribution, this value helps us understand where we might expect most observations to fall in statistical terms.
The expected value for a Pareto distribution with parameters \( k \) and \( \theta \) is given by:
  • \( E(X) = \int_{\theta}^{\infty} x \cdot f(x; k, \theta) \, dx \)
  • This works out to \( E(X) = \frac{k \cdot \theta}{k-1} \) when \( k > 1 \).
If \( k = 1 \), \( E(X) \) becomes undefined because the integral diverges, as division by zero is not possible. In simpler terms, this means the average value does not exist, as it would theoretically stretch to infinity.
Variance
Variance measures how much the values of a random variable, like our Pareto distribution, are spread out from their mean (expected value). High variance means numbers are more distributed, while low variance implies numbers are closer to the mean.
For the Pareto distribution, variance \( V(X) \) can be calculated by:
  • \( V(X) = E(X^2) - (E(X))^2 \)
For \( k > 2 \), the computations show:
  • \( E(X^2) = \frac{k \cdot \theta^2}{k-2} \)
  • Thus, \( V(X) = \frac{k \cdot \theta^2}{(k-1)^2(k-2)} \)
In cases where \( k = 2 \), the term becomes undefined because it results in division by zero, indicating infinite variance. This means observations can be scattered over an unlimited range.
Probability Density Function
A probability density function (PDF) gives the likelihood of a random variable taking on a particular value. For continuous variables like those in the Pareto distribution, PDFs indicate value density rather than specific probabilities.
In the Pareto distribution:
  • The PDF is expressed as \( f(x; k, \theta) = \frac{k \cdot \theta^{k}}{x^{k+1}} \), valid for \( x \geq \theta \).
  • Values are zero for \( x < \theta \), meaning these values are impossible within its context.
This function highlights that more significant values of \( x \) become less probable, as seen from the \( x^{-(k+1)} \) term. The PDF ensures the total probability over all possible outcomes equals 1.
Integral Convergence
Integral convergence is a vital concept in determining whether an integral (used to calculate expectations, variances, etc.) bears a finite value. Being integral to expectations, variance, and moments calculations, it indicates stability in calculations.
In the Pareto distribution, ensuring that expressions like \( E(X^n) \) are finite requires that the integral converges. Specifically:
  • We compute \( E(X^n) = \int_{\theta}^{\infty} x^n \cdot \frac{k \cdot \theta^k}{x^{k+1}} \, dx \).
  • For this to be finite, we need \( n-k-1 < -1 \), or equivalently, \( n < k-1 \). This ensures the function decreases rapidly enough for the integral to yield a finite result.
If this condition isn't met, the expected value or moment becomes infinite, which might imply unrealistic outcomes like unbounded mean or variance.

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Most popular questions from this chapter

Suppose the reaction temperature \(X\) (in \({ }^{\circ} \mathrm{C}\) ) in a certain chemical process has a uniform distribution with \(A=-5\) and \(B=5\). a. Compute \(P(X<0)\). b. Compute \(P(-2.5

Vehicle speed on a particular bridge in China can be modeled as normally distributed ("Fatigue Reliability Assessment for Long-Span Bridges under Combined Dynamic Loads from Winds and Vehicles" \(J\). of Bridge Engr., 2013: 735-747). a. If \(5 \%\) of all vehicles travel less than \(39.12 \mathrm{~m} / \mathrm{h}\) and \(10 \%\) travel more than \(73.24 \mathrm{~m} / \mathrm{h}\), what are the mean and standard deviation of vehicle speed? [Note: The resulting values should agree with those given in the cited article.] b. What is the probability that a randomly selected vehicle's speed is between 50 and \(65 \mathrm{~m} / \mathrm{h}\) ? c. What is the probability that a randomly selected vehicle's speed exceeds the speed limit of \(70 \mathrm{~m} / \mathrm{h}\) ?

The error involved in making a certain measurement is a continuous rv \(X\) with pdf $$ f(x)=\left\\{\begin{array}{cc} .09375\left(4-x^{2}\right) & -2 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array}\right. $$ a. Sketch the graph of \(f(x)\). b. Compute \(P(X>0)\). c. Compute \(P(-1.5)\).

An ecologist wishes to mark off a circular sampling region having radius \(10 \mathrm{~m}\). However, the radius of the resulting region is actually a random variable \(R\) with pdf $$ f(r)=\left\\{\begin{array}{cl} \frac{3}{4}\left[1-(10-r)^{2}\right] & 9 \leq r \leq 11 \\ 0 & \text { otherwise } \end{array}\right. $$ What is the expected area of the resulting circular region?

Sales delay is the elapsed time between the manufacture of a product and its sale. According to the article "Warranty Claims Data Analysis Considering Sales Delay" (Quality and Reliability Engr. Intl., 2013: 113-123), it is quite common for investigators to model sales delay using a lognormal distribution. For a particular product, the cited article proposes this distribution with parameter values \(\mu=2.05\) and \(\sigma^{2}=.06\) (here the unit for delay is months). a. What are the variance and standard deviation of delay time? b. What is the probability that delay time exceeds 12 months? c. What is the probability that delay time is within one standard deviation of its mean value? d. What is the median of the delay time distribution? e. What is the 99 th percentile of the delay time distribution? f. Among 10 randomly selected such items, how many would you expect to have a delay time exceeding 8 months?
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