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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 \(a \leq E(X) \leq b\) because expectation maintains bounds of a random variable.

Step by step solution

01

Understand the Problem

We need to show that if a random variable \(X\) is bounded between \(a\) and \(b\), then its expected value \(E(X)\) is also bounded between \(a\) and \(b\). This relies on understanding and applying properties of expected values of random variables.
02

Apply the Bounded Variable Property

The property of expected value states that, for any random variable \(X\) with bounds \(a\) and \(b\), the expected value \(E(X)\) will also respect those bounds. This is because expectation is a linear operator and the expected value takes into account all possible values of \(X\) weighted by their probabilities.
03

Mathematical Expression of Bounds

Since \(X\) is always between \(a\) and \(b\), for every possible outcome \(x\), we have \(a \leq x \leq b\). The expectation \(E(X)\) is defined as the summation/integration of \(x\) times its probability. This constraint on \(x\) ensures \(E(X)\) also lies between \(a\) and \(b\).
04

Use Linear Expectation Property

The property \(E(a) \leq E(X) \leq E(b)\) simplifies to \(a \leq E(X) \leq b\) because expectations of constants are the constants themselves. This step confirms that when \(a \leq X \leq b\), due to the linearity of expectation, \(E(a) = a\) and \(E(b) = b\).
05

Conclusion

Using the properties and definitions of expectation and bounds, we conclude that the expected value \(E(X)\) indeed lies between \(a\) and \(b\). Thus, \(a \leq E(X) \leq b\) under the given conditions of \(X\).

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Key Concepts

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

Linear Operator
The expected value, often denoted as \( E(X) \), is a fundamental concept in probability theory and statistics. It is considered a linear operator. This means its calculation obeys specific linear properties that greatly simplify its manipulation. In simpler terms, no matter how you shuffle things around, certain mathematical truths hold.

For instance, say we have two random variables, \( X \) and \( Y \), and two constants, \( a \) and \( b \). A key property of the expected value as a linear operator is:
  • \( E(aX + bY) = aE(X) + bE(Y) \)
This tells us that if you multiply a random variable by a constant and calculate the expected value, it’s the same as multiplying the expected value of the variable by the constant.

In practical terms, understanding the linear nature allows for flexible approaches when dealing with complex problems, like linear transformations. This also connects directly with showing that if random variable \( X \) is bound between two values \( a \) and \( b \), so is \( E(X) \). The linear operator property simplifies how we can add, subtract, or multiply elements affecting \( X \) without changing the fundamental bounds of the expectation.
Bounded Variable Property
The bounded variable property concept is much easier than it might sound. Imagine a random variable \( X \) that’s pinned between two values, \( a \) and \( b \). In our scenario, these bounds mean any outcome \( x \) of \( X \) always falls somewhere between \( a \) and \( b \).

What does this imply for the expectation, or \( E(X) \)? Because all possible outcomes \( x \) respect these bounds, and because each outcome is weighted by how likely it is to happen (probability),
  • the expectation will always land between \( a \) and \( b \) too.
In this light, the term "bounded" is quite literal: the random variable's behavior is boxed in by the limits. Mathematically, the expectation respects these boundaries and doesn’t exceed them, regardless of the distribution's shape within \( a \) and \( b \).

This bounded properties concept is incredibly significant when working in real-world scenarios. For instance, with measurements, a bounded variable ensures there's predictability and control over how the expectation behaves, adhering to the set limits.
Probability
Probability is the lifeblood of expectation and affects how we calculate the expected value of a random variable. At the heart of it, probability assigns a likelihood, or weight, to each possible outcome \( x \) of a random variable \( X \). It's essentially the chance that a particular outcome will occur.

When calculating the expected value \( E(X) \), each outcome gets its say depending on its probability:
  • \( E(X) = \sum{x_i P(x_i)} \) for discrete variables
  • \( E(X) = \int{x f(x) dx} \) for continuous variables
Here, \( P(x_i) \) and \( f(x) \) are the probability of outcomes and the probability density function, respectively.

A solid grasp of probability helps in anticipating how altering outcome likelihoods impacts the expectation. For example, if a certain outcome has a high probability, it greatly influences \( E(X) \). Conversely, outcomes with low probabilities exert less effect. Understanding these concepts provides insight into why certain expected values are larger or smaller based on the climbed probability scale.

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Most popular questions from this chapter

In proof testing of circuit boards, the probability that any particular diode will fail is .01. Suppose a circuit board contains 200 diodes. a. How many diodes would you expect to fail, and what is the standard deviation of the number that are expected to fail? b. What is the (approximate) probability that at least four diodes will fail on a randomly selected board? c. If five boards are shipped to a particular customer, how likely is it that at least four of them will work properly? (A board works properly only if all its diodes work.)

A certain type of digital camera comes in either a 3megapixel version or a 4-megapixel version. A camera store has received a shipment of 15 of these cameras, of which 6 have 3 -megapixel resolution. Suppose that 5 of these cameras are randomly selected to be stored behind the counter; the other 10 are placed in a storeroom. Let \(X=\) the number of 3 -megapixel cameras among the 5 selected for behindthe-counter storage. a. What kind of a distribution does \(X\) have (name and values of all parameters)? b. Compute \(P(X=2), P(X \leq 2)\), and \(P(X \geq 2)\). c. Calculate the mean value and standard deviation of \(X\).

Write a general rule for \(E(X-c)\) where \(c\) is a constant. What happens when you let \(c=\mu\), the expected value of \(X\) ?

Suppose the number \(X\) of tornadoes observed in a particular region during a 1-year period has a Poisson distribution with \(\lambda=8\). a. Compute \(P(X \leq 5)\). b. Compute \(P(6 \leq X \leq 9)\). c. Compute \(P(10 \leq X)\). d. What is the probability that the observed number of tornadoes exceeds the expected number by more than 1 standard deviation? 81\. Suppose that the number of drivers who travel between a particular origin and destination during a designated time period has a Poisson distribution with parameter \(\lambda=20\) (suggested in the article "Dynamic Ride Sharing: Theory and Practice," J. of Transp. Engr., 1997: 308-312). What is the probability that the number of drivers will a. Be at most 10 ? b. Exceed 20? c. Be between 10 and 20 , inclusive? Be strictly between 10 and 20 ? d. Be within 2 standard deviations of the mean value?

Customers at a gas station pay with a credit card \((A)\), debit card \((B)\), or cash \((C)\). Assume that successive customers make independent choices, with \(P(A)=.5, P(B)=.2\), and \(P(C)=.3 .\) a. Among the next 100 customers, what are the mean and variance of the number who pay with a debit card? Explain your reasoning. b. Answer part (a) for the number among the 100 who don't pay with cash.

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