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Suppose \(E(X)=5\) and \(E[X(X-1)]=27.5\). What is a. \(E\left(X^{2}\right)\) ? \(\left[\right.\) Hint: \(E[X(X-1)]=E\left[X^{2}-X\right]=\) \(\left.E\left(X^{2}\right)-E(X)\right] ?\) b. \(V(X)\) ? c. The general relationship among the quantities \(E(X)\), \(E[X(X-1)]\), and \(V(X) ?\)

Short Answer

Expert verified
a. \(E(X^2) = 32.5\), b. \(V(X) = 7.5\), c. \(E[X(X-1)] = E(X^2) - E(X)\).

Step by step solution

01

Understanding the Problem

We are given that \(E(X) = 5\) and \(E[X(X-1)] = 27.5\). We need to find \(E(X^2)\), \(V(X)\), and discuss the relationship among these quantities. We also have a hint that \(E[X(X-1)] = E(X^2) - E(X)\).
02

Finding \(E(X^2)\)

Using the hint provided, we know that \(E[X(X-1)] = E(X^2) - E(X)\). Substituting the given values, we have: \[E(X^2) - 5 = 27.5\]Solving for \(E(X^2)\), we add 5 to both sides: \[E(X^2) = 27.5 + 5 = 32.5\]
03

Calculating Variance, \(V(X)\)

The variance \(V(X)\) is given by the formula: \[V(X) = E(X^2) - [E(X)]^2\]We found that \(E(X^2) = 32.5\) and we know \(E(X) = 5\). Calculate \(E(X)^2:\) \[E(X)^2 = 5^2 = 25\]Now, calculating variance:\[V(X) = 32.5 - 25 = 7.5\]
04

Exploring the Relationship

From the steps, we see a notable relationship among the quantities. \(E[X(X-1)]\) represents a different measure of variance-related calculation often used in combinatorics. The given equation \(E[X(X-1)] = E(X^2) - E(X)\) is a derivation to easily compute \(E(X^2)\) when \(E(X)\) and \(E[X(X-1)]\) are known.

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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 is a core concept in probability and statistics. Think of it as the weighted average of all possible values of a random variable. It represents the long-term average if an experiment is repeated many times. Expected value is denoted as \( E(X) \), where \( X \) is our random variable.
  • It's calculated by multiplying each possible outcome by its probability, then summing these products. For discrete variables, this is expressed as \( E(X) = \sum x_i P(x_i) \), where \( x_i \) are the possible values and \( P(x_i) \) their probabilities.
  • In continuous variables, we use integration: \( E(X) = \int x f(x) \, dx \).

In the exercise, we are given \( E(X) = 5 \). This value plays a crucial role in the calculations of \( E(X^2) \) and variance. It's important to understand how it interacts with other measures such as variance, highlighting the balance between average value and distribution spread.
Variance
Variance is a statistic that measures how a set of numbers (or a random variable) is spread out. While the expected value gives us an idea of the central tendency, the variance tells us how much the data can differ from this expected value. It’s denoted by \( V(X) \) or sometimes \( \sigma^2 \).

To calculate variance, we use the formula:
  • \( V(X) = E(X^2) - [E(X)]^2 \)
  • Here, \( E(X^2) \) is the expected value of the square of \( X \).
  • \( [E(X)]^2 \) means we square the expected value of \( X \).

In our exercise, the calculation of \( V(X) \) was simplified through use of previously determined values, \( E(X^2) = 32.5 \) and \( E(X) = 5 \). Substituting in the formula gives us the variance \( V(X) = 7.5 \). This method demonstrates how variance offers insight into potential deviations from the mean, essential for assessing data reliability and consistency.
Combinatorics
Combinatorics is the branch of mathematics dealing with combinations of objects belonging to a finite set in accordance with certain constraints, such as those of graph theory. It's commonly associated with counting principles.

One specific application involves the calculation \( E[X(X-1)] \), which is linked to understanding relationships within sets or sequences.
  • This is a binomial moment, used particularly in statistical distributions arising in combinatorical settings.
  • In our exercise, this moment is valuable for deriving other moments, like \( E(X^2) \), revealing complexities in distributions.

Understanding how \( E[X(X-1)] \) interacts with other moments helps us explore variance in combinatorial contexts. This insight can be valuable for tasks requiring probability distributions, such as calculating permutations or combinations. This formula, \( E[X(X-1)] = E(X^2) - E(X) \), ties neatly into broader probabilistic theories and demonstrates the inherent elegance in mathematical relationships.

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Most popular questions from this chapter

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