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If \(a \leq X \leq b\), show that \(a \leq E(X) \leq b\).

Short Answer

Expert verified
If \(a \leq X \leq b\), then \(a \leq E(X) \leq b\) due to linearity of expectation.

Step by step solution

01

Understanding the Expectation Concept

The expectation or expected value of a random variable \(X\), denoted as \(E(X)\), is essentially the weighted average of all possible values that \(X\) can take on, with the weights being the probabilities of these values. This concept is critical in understanding what is being asked.
02

Analyze Given Inequality Property

We are given that \(a \leq X \leq b\). This tells us that the variable \(X\) always takes values between \(a\) and \(b\) inclusively. This is a property that must hold for all random outcomes of \(X\).
03

Utilize Linearity of Expectation

Use the linearity of expectation theorem: if \(a\leq X \leq b\), then the expectation respects this inequality as follows: \(E(a) \leq E(X) \leq E(b)\). Note that since \(a\) and \(b\) are constants, \(E(a) = a\) and \(E(b) = b\).
04

Substitute Expected Values of Constants

Substitute the expected values of the constants in the inequalities. Since \(E(a) = a\) and \(E(b) = b\), substitute these into the inequality, which results in: \(a \leq E(X) \leq b\).
05

Conclusion

We've demonstrated that because \(X\) is always between \(a\) and \(b\), its expected value must also lie between \(a\) and \(b\), inclusive. Hence, \(a \leq E(X) \leq b\) holds true.

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Key Concepts

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

Random Variable
A random variable is a fundamental concept in probability and statistics that represents a numerical outcome of a random process. Essentially, it associates a number with each outcome in a probability experiment. Random variables can be either discrete or continuous.
Discrete random variables take on a finite number of distinct values. For example, rolling a die can result in one of six outcomes, each corresponding to a number from 1 to 6. Continuous random variables, on the other hand, can take on any value within a continuous range, such as measuring the height of individuals.
Random variables are crucial in statistical modeling and predictions, allowing us to quantify and analyze uncertainty. Understanding the behavior of random variables helps in calculating probabilities, expectations, and variances, which are key in making informed decisions based on data.
Linearity of Expectation
Linearity of expectation is a powerful property in probability theory that asserts that the expected value of the sum of random variables is equal to the sum of their expected values, regardless of whether the variables are independent. In mathematical terms, for any two random variables, \(X\) and \(Y\), the linearity of expectation states that:
\[E(X+Y) = E(X) + E(Y)\]
This principle simplifies calculations significantly, especially when dealing with complex problems involving multiple random variables. For example, if we have the inequality \(a \leq X \leq b\), by applying the linearity of expectation, we find that the expected value of \(X\) respects the inequality, leading to \(E(a) \leq E(X) \leq E(b)\).
This property is incredibly helpful when computing expected values, as it allows us to break down complex expressions into simpler components and sum their expectations individually. Since constants have their expected values equal to themselves, \(E(a) = a\) and \(E(b) = b\), thus supporting our original exercise condition.
Inequality in Probability
Inequality in probability deals with the relationship between random variables and their probabilities. These inequalities help in setting bounds and constraints on the potential values and expectations of random variables.
These kinds of inequalities are pivotal when we want to establish safe and reliable expectations about random variables, especially when actual outcomes can vary widely.
In the context of the original exercise, the inequality \(a \leq X \leq b\) specifies that the random variable \(X\) is constrained between \(a\) and \(b\). This means, regardless of the outcomes of the random process, \(X\) cannot exceed these limits. The corresponding conclusion that \(a \leq E(X) \leq b\) borrows heavily from the understanding of probability constraints and how they translate to expected values.
Probability inequalities like these provide essential insight into statistical analyses, allowing us to bound potential variability and better predict real-world phenomena.

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Most popular questions from this chapter

Organisms are present in ballast water discharged from a ship according to a Poisson process with a concentration of 10 organisms/ \(\mathrm{m}^{3}\) [the article "4 Counting at Low Concentrations: The Statistical Challenges of Verifying Ballast Water Discharge Standards" (Ecological Applications, 2013: 339-351) considers using the Poisson process for this purpose]. a. What is the probability that one cubic meter of discharge contains at least 8 organisms? b. What is the probability that the number of organisms in \(1.5 \mathrm{~m}^{3}\) of discharge exceeds its mean value by more than one standard deviation? c. For what amount of discharge would the probability of containing at least 1 organism be \(.999\) ?

A contractor is required by a county planning department to submit one, two, three, four, or five forms (depending on the nature of the project) in applying for a building permit. Let \(Y=\) the number of forms required of the next applicant. The probability that \(y\) forms are required is known to be proportional to \(y\)-that is, \(p(y)=k y\) for \(y=1, \ldots, 5\). a. What is the value of \(k\) ? [Hint: \(\left.\Sigma_{y=1}^{5} p(y)=1\right]\) b. What is the probability that at most three forms are required? c. What is the probability that between two and four forms (inclusive) are required? d. Could \(p(y)=y^{2} / 50\) for \(y=1, \ldots, 5\) be the pmf of \(Y\) ?

Some parts of California are particularly earthquakeprone. Suppose that in one metropolitan area, \(25 \%\) of all homeowners are insured against earthquake damage. Four homeowners are to be selected at random; let \(X\) denote the number among the four who have earthquake insurance. a. Find the probability distribution of \(X\). [Hint: Let \(S\) denote a homeowner who has insurance and \(F\) one who does not. Then one possible outcome is SFSS, with probability \((.25)(.75)(.25)(.25)\) and associated \(X\) value 3. There are 15 other outcomes.] b. Draw the corresponding probability histogram. c. What is the most likely value for \(X\) ? d. What is the probability that at least two of the four selected have earthquake insurance?

Compute the following binomial probabilities directly from the formula for \(b(x ; n, p)\) : a. \(b(3 ; 8,-35)\) b. \(b(5 ; 8, .6)\) c. \(P(3 \leq X \leq 5)\) when \(n=7\) and \(p=.6\) d. \(P(1 \leq X)\) when \(n=9\) and \(p=.1\)

Suppose that trees are distributed in a forest according to a two-dimensional Poisson process with parameter \(\alpha\), the expected number of trees per acre, equal to 80 . a. What is the probability that in a certain quarter-acre plot, there will be at most 16 trees? b. If the forest covers 85,000 acres, what is the expected number of trees in the forest? c. Suppose you select a point in the forest and construct a circle of radius \(.1\) mile. Let \(X=\) the number of trees within that circular region. What is the pmf of \(X\) ? [Hint: 1 sq mile \(=640\) acres.]

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