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Chapter 4: Problem 38

If \(E[X]=1\) and \(\operatorname{Var}(X)^{\prime}=5\), find (a) \(E\left[(2+X)^{2}\right]\) (b) \(\operatorname{Var}(4+3 X)\)

Short Answer

Expert verified
(a) \(E\left[(2+X)^{2}\right] = 14\) (b) \(\operatorname{Var}(4+3 X) = 45\)

Step by step solution

01

Solving for E[(2+X)^2]

To find E[(2+X)^2], we can apply the linearity of expectation and the property that E[c(X)] = cE[X] for a constant c. First, expand (2+X)^2: \((2+X)^{2} = 4 + 4X + X^{2}\) Now, take the expected value: \(E\left[(2+X)^{2}\right]= E[4 + 4X + X^{2}]\) Apply the linearity of expectation and the property for E[c(X)]: \(E\left[(2+X)^{2}\right]= E[4] + 4E[X] + E[X^{2}]\) We are given E[X] = 1. Now we need to find E[X^2]. E[X^2] can be found using the relationship between variance, standard deviation, and mean: \(\operatorname{Var}(X) = E[X^{2}] - (E[X])^{2}\) Solve for E[X^2]: \(E[X^{2}] = \operatorname{Var}(X) + (E[X])^{2}\) We are given Var(X) = 5. Substitute the given values: \(E[X^{2}] = 5 + (1)^{2} = 6\) Now we can substitute E[X^2] and E[X] into our previously derived expression for E[(2+X)^2]: \(E\left[(2+X)^{2}\right] = E[4] + 4E[X] + E[X^{2}] = 4 + 4(1) + 6 = 14\) So, \(E\left[(2+X)^{2}\right] = 14\).
02

Solving for Var(4+3X)

To find Var(4+3X), we can use the property that Var(cX) = c^2Var(X) for a constant c: Var(4 + 3X) = Var(3X) Since 3 is a constant, we have: Var(4 + 3X) = 3^2 * Var(X) We are given Var(X) = 5: Var(4 + 3X) = 9 * 5 = 45 So, Var(4 + 3X) = 45. To summarize: (a) \(E\left[(2+X)^{2}\right] = 14\) (b) \(\operatorname{Var}(4+3 X) = 45\)

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Key Concepts

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

Expectation of a Random Variable
The expectation, or mean, of a random variable gives us a sense of the 'average' outcome we might expect if we were to observe the random variable's outcomes many times. Think of it as the long-term average value. Mathematically, the expectation of a random variable, denoted as E[X], is the weighted average of all possible values that the random variable can take, each value being weighted by its probability of occurrence.

In practical situations, such as in the exercise where we have E[X] = 1, this informs us that if we repeated the random process that yields X numerous times, the average result would trend toward 1 as the number of trials goes up. This concept is pivotal in understanding how random variables behave and is widely applied in fields such as economics, engineering, and science.
Variance of a Random Variable
While expectation gives us the average outcome, variance, denoted as Var(X), tells us about the spread of a random variable's outcomes. It's essentially a measure of how much the values of the random variable X deviate from the mean (expectation). High variance means the values are spread out widely, low variance means they are clustered close to the mean.

In the solution provided, Var(X) is used to find the expectation of X squared, E[X^2]. We are given that Var(X) = 5, which means there's a moderate spread in the values of X about the mean. Understanding the variance is crucial because it provides insight into the reliability of the expectation and the risk associated with different outcomes.
Linearity of Expectation
One of the most beautiful properties of the expectation operator is its linearity. The linearity of expectation states that for any two random variables X and Y, and any two constants a and b, the expectation of their linear combination is E[aX + bY] = aE[X] + bE[Y]. It's important to note this property holds regardless of whether X and Y are independent or not.

Our exercise exemplifies this with E[(2+X)^2], which after expansion and application of this property simplifies the process of finding the expected value. Grasping the concept of linearity allows for simplification in solving complex probability problems and is especially useful when dealing with sums of random variables.
Transformation of Random Variables
Transformation of random variables is a process that involves changing a random variable into a new random variable, typically by applying a mathematical function. For example, if Y = g(X) where g is a function, then Y is a transformation of X. This concept is vital in probability theory because it allows us to understand how the distribution, expectation, and variance of X affect the distribution, expectation, and variance of Y.

In the solution to the problem Var(4+3X), we see that the variance of a transformed variable, 3X in this case, can be found by squaring the constant multiplier and applying it to the original variance. Such transformations are useful in data analysis and for creating models that better fit the observed data, making them fundamental tools for statisticians and data scientists.

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Most popular questions from this chapter

To determine whether or not they have a certain disease, 100 people are to have their blood tested. However, rather than testing each individual separately, it has been decided first to group the people in groups of \(10 .\) The blood samples of the 10 people in each group will be pooled and anayzed together. If the test is negative, one test will suffice for the 10 people; whereas, if the test is positive each of the 10 people will also be individually tested and, in all, 11 tests will be made on this group. Assume the probability that a person has the disease is .1 for all people, independently of each other, and compute the expected number of tests necessary for each group. (Note that we are assuming that the pooled test will be positive if at least one person in the pool has the disease.)

A communications channel transmits the digits 0 and 1 . However, due to static, the digit transmitted is incorrectly received with probability 2 . Suppose that we want to transmit an important message consisting of one binary digit. To reduce the chance of error, we transmit 00000 instead of 0 and 11111 instead of 1 . If the receiver of the message uses "majority" decoding, what is the probability that the message will be wrong when decoded? What independence assumptions are you making?

A coin that when flipped comes up heads with probability \(p\) is flipped until either heads or tails has occurred twice. Find the expected. number of flips.

\(A\) and \(B\) play the following game: \(A\) writes down either number 1 or number 2 and \(B\) must guess which one. If the number that \(A\) has written down is \(i\) and \(B\) has guessed correctly, \(B\) receives \(i\) units from \(A\). If \(B\) makes a wrong guess, \(B\) pays \(\frac{3}{4}\) unit to \(A\). If \(B\) randomizes his decision by guessing 1 with probability \(p\) and 2 with probability \(1-p\), determine his expected gain if (a) \(A\) has written down number 1 and (b) \(A\) has written down number \(2 .\) What value of \(p\) maximizes the minimum possible value of \(B\) 's expected gain and what is this maximin value? (Note that \(B\) 's expected gain depends not only on \(p\) but also on what \(A\) does.) Consider now player \(A\). Suppose that she also randomizes her decision, writing dowh number 1 with probability \(q\). What is A's expected loss if (c) \(B\) chooses number 1 and (d) \(B\) chooses number \(2 ?\) What value of \(q\) minimizes A's maximum expected loss? Show that the minimum of \(A\) 's maximum expected loss is equal to the maximum of \(B\) 's minimum expected gain. This result, known as the minimax theorem, was first established in generality by the mathematician John von Neumann and is the fundamental result in the mathematical discipline known as the theory of games. The common value is called the value of the game to player \(B\).

Let \(X\) be a Poisson random variable with parameter \(\lambda\). What value of \(\lambda\) maximizes \(P[X=k\\}, k \geq 0 ?\)
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