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Suppose \(E(X)=5\) and \(E[X(X-1)]=27.5\). What is a. \(E\left(X^{2}\right)\) ? [Hint: \(E[X(X-1)]=E\left[X^{2}-X\right]=\) \(E\left(X^{2}\right)-E(X)\).] 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. The relationship involves three expressions: \(E[X^2] = E[X(X-1)] + E(X)\) and \(V(X) = E[X^2] - (E[X])^2\).

Step by step solution

01

Set Up Initial Equations

Given the problem, we know two things: \(E(X) = 5\) and \(E[X(X-1)] = 27.5\). From the hint, we have the equation \(E[X(X-1)] = E[X^2] - E(X)\). We can use this information to find \(E(X^2)\).
02

Solve for \(E(X^2)\)

We know that \(E[X(X-1)] = E[X^2] - E(X)\). Substitute the given values: \(27.5 = E[X^2] - 5\). Solve for \(E[X^2]\): \[ E[X^2] = 27.5 + 5 = 32.5 \] Thus, \(E[X^2] = 32.5\).
03

Use Variance Formula

The formula for variance is \(V(X) = E[X^2] - (E[X])^2\). We already have \(E[X^2] = 32.5\) and \(E[X] = 5\). Substitute these into the variance formula: \[ V(X) = 32.5 - (5)^2 = 32.5 - 25 = 7.5 \] Thus, \(V(X) = 7.5\).
04

General Relation Among Quantities

From the given information and calculations, we have the relationship between the quantities: - \(E[X(X-1)] = E[X^2] - E(X)\), which was used to find \(E[X^2]\).- The variance formula \(V(X) = E[X^2] - (E[X])^2\) connects all three quantities.

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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, often denoted as \( E(X) \), is a core statistical concept that represents the average or mean value of a random variable \( X \) over many instances or trials. It is essentially the center of the probability distribution of \( X \), offering a sense of the "central tendency" of the outcomes. For a discrete random variable, this is calculated as the sum of all possible values each multiplied by their respective probabilities. Imagine you have a dice, and you want to know the expected value of the numbers when you roll it. You would add up the results of each die and divide by the number of times you rolled. For a fair six-sided die, this would be 3.5 because each number from 1 to 6 has an equal probability of appearing, and the mean of these numbers is 3.5.
  • The formula for expected value is \( E(X) = \sum [x_i \cdot P(x_i)] \).
  • In the context of the problem, \( E(X) \) is given as 5.
Variance
Variance, denoted as \( V(X) \), measures how much values differ from the expected value. In simpler terms, it's a way to quantify the spread or dispersion of a set of values. If the numbers are far from the expected value, you will have a high variance; if they are close, the variance is low. The concept is crucial in assessing the reliability of the data being studied. For example, data with high variance might suggest inconsistency or data volatility.
  • The formula for variance is \( V(X) = E[X^2] - (E[X])^2 \).
  • From the problem, you calculated \( V(X) = 7.5 \), indicating a moderate spread of values around the mean.
Variance tells us how much the data set is expected to spread out from the mean. Here, the data set value of 7.5 indicates that the values are fairly spread around the mean of 5. This is a key metric for statisticians when they are trying to understand data distributions.
Polynomial Expectation
Polynomial expectation deals with scenarios where we work with expectations of polynomial functions of random variables. Given the expressions like \( E[X(X-1)] \) or \( E[X^2] \), polynomial expectation provides insights into higher moments of a statistical distribution, beyond the first moment, which is just the simple expected value of \( X \). The key is recognizing how these expectations relate to known values.
  • We use the equation \( E[X(X-1)] = E[X^2] - E(X) \) to derive needed polynomial terms like \( E[X^2] \).
  • In the exercise, solving \( E[X^2] = 32.5 \) was crucial for deriving both the polynomial expectation results and subsequently the variance.
This illustrates the importance of polynomial expectation in statistical problems, providing deeper insights that are vital for further analysis, such as computing variance or higher-order statistics.

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