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Chapter 3: Problem 33

Suppose that the number of plants of a particular type found in a rectangular region (called a quadrat by ecologists) in a certain geographic area is an rv \(X\) with pmf $$ p(x)=\left\\{\begin{array}{cc} c / x^{3} & x=1,2,3, \ldots \\ 0 & \text { otherwise } \end{array}\right. $$ Is \(E(X)\) finite? Justify your answer (this is another distribution that statisticians would call heavy-tailed).

Short Answer

Expert verified
Yes, \(E(X)\) is finite because \(\sum_{x=1}^{\infty} \frac{1}{x^2}\) converges.

Step by step solution

01

Understand the Problem

We need to determine if the expected value \(E(X)\) of the random variable \(X\), which represents the number of plants in a quadrat, is finite. Given that \(X\) is a discrete random variable with probability mass function (pmf) \(p(x) = \frac{c}{x^3}\) for \(x = 1, 2, 3, \ldots\), our task is to determine if the series used to calculate \(E(X)\) converges.
02

Determine the Constant c

To find the expected value, we first need to determine the constant \(c\) in the pmf by using the condition that the sum of all probabilities must equal 1. This gives us \(\sum_{x=1}^{\infty} \frac{c}{x^3} = 1\). Therefore, \(c = \left(\sum_{x=1}^{\infty} \frac{1}{x^3}\right)^{-1}\). The series \(\sum_{x=1}^{\infty} \frac{1}{x^3}\) is known to be convergent.
03

Write the Definition of E(X)

The expected value \(E(X)\) is given by the sum \(E(X) = \sum_{x=1}^{\infty} x \cdot p(x) = \sum_{x=1}^{\infty} x \cdot \frac{c}{x^3} = c \cdot \sum_{x=1}^{\infty} \frac{1}{x^2}\).
04

Evaluate the Series for E(X)

Now we evaluate the series \(\sum_{x=1}^{\infty} \frac{1}{x^2}\). This series converges to \(\frac{\pi^2}{6}\), which is a known result. So \(E(X) = c \cdot \sum_{x=1}^{\infty} \frac{1}{x^2} = c \cdot \frac{\pi^2}{6}\).
05

Confirm E(X) is Finite

Since \(c = \left(\sum_{x=1}^{\infty} \frac{1}{x^3}\right)^{-1}\) and \(\sum_{x=1}^{\infty} \frac{1}{x^3}\) converges, the constant \(c\) is finite. Therefore, \(E(X) = c \cdot \frac{\pi^2}{6}\) is also finite.

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Key Concepts

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

Probability Mass Function
The Probability Mass Function (pmf) is a key concept in probability theory, especially when dealing with discrete random variables like the random variable \(X\) in our exercise. The pmf provides the probability that a discrete random variable is exactly equal to a specific value. Simply put, it describes how the probabilities are distributed across the possible values.
In this example, the pmf is given by \( p(x) = \frac{c}{x^3} \) for \( x = 1, 2, 3, \ldots \). This means that the probability of finding \(x\) plants in the quadrat decreases with larger values of \(x\), following a cubic decay.
  • The pmf must satisfy two conditions: \( p(x) \geq 0 \) for all \( x \), and the sum over all possible \( x \) must equal 1, i.e., \( \sum_{x} p(x) = 1 \).
  • In our scenario, \(c\) acts as a constant to ensure the total probability sums to one.
  • The pmf is used to find important measures like the expected value, which is crucial for understanding the average outcome in probabilistic terms.
Convergence of Series
Convergence of series is a mathematical concept that plays a critical role in determining whether certain summations result in a finite number. In the context of expected value, we need to sum the probabilities of all outcomes, weighted by their values, to find the expected value \(E(X)\).
In our exercise, to determine \(E(X)\), we use the series \( \sum_{x=1}^{\infty} x \cdot \frac{c}{x^3} \). This transforms into \( c \cdot \sum_{x=1}^{\infty} \frac{1}{x^2} \), where \(c\) ensures the series converges to a value because \( \sum_{x=1}^{\infty} \frac{1}{x^3} \) determines \(c\).
  • Convergence indicates that adding more terms in the series doesn't change the final sum significantly - it stabilizes as more terms are added.
  • For \( \sum_{x=1}^{\infty} \frac{1}{x^3} \), the convergence means the series approaches a particular value, allowing us to find \(c\) that normalizes the pmf.
  • The series \( \sum_{x=1}^{\infty} \frac{1}{x^2} \) converges to \( \frac{\pi^2}{6} \), a well-known mathematical result which helps us calculate a finite \(E(X)\).
Heavy-Tailed Distribution
A heavy-tailed distribution refers to probability distributions with tails that are not exponentially bounded, meaning they have a higher probability of extreme values compared to distributions with light tails like the normal distribution.
The random variable \(X\) in our problem is an example of such a distribution, because the pmf \( p(x) = \frac{c}{x^3} \) results in probabilities that decrease slowly as \(x\) increases.
  • Heavy-tailed distributions are common in risk management or reliability analysis, where extreme values have significant impacts.
  • The 'heaviness' in the tail suggests greater probability of encountering large values than in typical distributions.
  • For our given pmf, while individual probabilities of large \(x\) values are small, their contribution to the expected value still matters.
  • The series convergence for \( \sum_{x=1}^{\infty} \frac{1}{x^3} \) and \( \sum_{x=1}^{\infty} \frac{1}{x^2} \) ensures we can still calculate meaningful measures like \(E(X)\) despite the heavy tails.

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