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Let \(X, Y,\) and \(Z\) be independent random variables, each with mean \(\mu\) and variance \(\sigma^{2}\) (a) Find the expected value and variance of \(S=X+Y+Z\). (b) Find the expected value and variance of \(A=(1 / 3)(X+Y+Z)\). (c) Find the expected value of \(S^{2}\) and \(A^{2}\).

Short Answer

Expert verified
(a) \(E(S) = 3\mu, \text{Var}(S) = 3\sigma^2\); (b) \(E(A) = \mu, \text{Var}(A) = \frac{\sigma^2}{3}\); (c) \(E(S^2) = 3\sigma^2 + 9\mu^2, E(A^2) = \frac{\sigma^2}{3} + \mu^2\).

Step by step solution

01

Understanding the Problem

We need to find the expected value and variance of the sum and average of three independent random variables, and also find the expected value of their squares.
02

Expected Value of S

The expected value of a sum of random variables is the sum of their expected values. Therefore, for random variables \(X, Y,\) and \(Z\), we have:\[ E(S) = E(X) + E(Y) + E(Z) = 3\mu. \]
03

Variance of S

The variance of a sum of independent random variables is the sum of their variances. Since \(X, Y,\) and \(Z\) are independent, we find:\[ \text{Var}(S) = \text{Var}(X) + \text{Var}(Y) + \text{Var}(Z) = 3\sigma^2. \]
04

Expected Value of A

The expected value of \(A\) can be calculated by using the linearity of expectation:\[ E(A) = E\left(\frac{1}{3}(X + Y + Z)\right) = \frac{1}{3}(E(X) + E(Y) + E(Z)) = \mu. \]
05

Variance of A

To find the variance of the mean of independent random variables, we use:\[ \text{Var}(A) = \text{Var}\left(\frac{1}{3}(X + Y + Z)\right) = \left(\frac{1}{3}\right)^2 \text{Var}(S) = \frac{1}{9} \times 3\sigma^2 = \frac{\sigma^2}{3}. \]
06

Expected Value of S²

By using the formula for the expectation of a square, which is \( E(V^2) = \text{Var}(V) + [E(V)]^2 \), we find:\[ E(S^2) = \text{Var}(S) + [E(S)]^2 = 3\sigma^2 + (3\mu)^2 = 3\sigma^2 + 9\mu^2. \]
07

Expected Value of A²

For variable \(A\), we use the same approach as for \(S^2\), adjusting for the coefficient:\[ E(A^2) = \text{Var}(A) + [E(A)]^2 = \frac{\sigma^2}{3} + \mu^2. \]

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Key Concepts

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

Random Variables
In probability theory, random variables are essential for understanding various scenarios involving uncertainty. A random variable is essentially a variable whose possible values are numerical outcomes of a random phenomenon. They can take on different values according to the specific distribution they belong to.

  • Discrete Random Variables: These are variables that have a countable number of distinct values. Typically, these arise from counting scenarios, like the roll of a die. The probability distribution of a discrete random variable is characterized by a probability mass function (PMF).
  • Continuous Random Variables: These variables take on a continuous range of values, such as the height of a person. Their distribution is described by a probability density function (PDF).
Random variables are crucial because they allow us to model real-world phenomena and analyze outcomes using probability distributions. They help in calculating probabilities, expectations, and variances, which are used to make informed predictions and decisions.
Expected Value
The expected value is a fundamental concept in probability theory, often described as the 'mean' or 'average' of a random variable. It provides a measure of the central tendency of a random variable's distribution.

For a random variable, the expected value is calculated differently based on whether it is discrete or continuous:
  • Discrete Random Variables: The expected value is the sum of the products of all possible values the variable can take and their respective probabilities. For a discrete random variable X with possible values xi and probability P(xi), its expected value is: \[ E(X) = \sum x_i P(x_i) \]
  • Continuous Random Variables: The expected value is the integral of the product of the variable value and its probability density function (PDF) over the entire range of the variable. \[ E(X) = \int x f(x) dx \]
In the context of the original exercise, using the linearity of expectation, we calculated expected values for random variables combined in different ways, demonstrating how this property simplifies such calculations.
Variance
Variance is an important measure that quantifies the spread or dispersion of a set of values in a probability distribution. It tells us how much the values of a random variable differ from the expected value.

To determine variance:
  • Formula for Discrete Variables: For a discrete random variable, the variance is the average of the squares of the differences between each possible value and the expected value, weighted by the probability of each value. \[ \text{Var}(X) = \sum (x_i - E(X))^2 P(x_i) \]
  • Formula for Continuous Variables: For continuous random variables, it is the integral of the squared differences from the mean, multiplied by the PDF across the distribution. \[ \text{Var}(X) = \int (x - E(X))^2 f(x) dx \]
In step 3 of the original solution, the variance of the sum of independent random variables was calculated as the sum of their variances, which is a crucial property of variance useful in real-world applications.
Linearity of Expectation
The linearity of expectation is a powerful property in probability theory. It states that the expected value of a sum of random variables is equal to the sum of their expected values, regardless of whether the variables are dependent or independent.

This means that for any random variables X and Y:
  • \[ E(X + Y) = E(X) + E(Y) \]
  • For any constant "a": \[ E(aX) = a \, E(X) \]
This property significantly simplifies the calculation of expected values for sum and linear transformations of random variables, as seen in our exercise solution.

Because the linearity of expectation holds even when variables are correlated, it offers flexibility in theoretical and practical calculations. It is foundational for understanding concepts like the expected value of sums, which simplifies the analysis of complex random processes.

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Most popular questions from this chapter

Let \(X\) be Poisson distributed with parameter \(\lambda\). Show that \(V(X)=\lambda\).

Let \(X\) and \(Y\) be two random variables defined on the finite sample space \(\Omega\). Assume that \(X, Y, X+Y,\) and \(X-Y\) all have the same distribution. Prove that \(P(X=Y=0)=1\).

If the first roll in a game of craps is neither a natural nor craps, the player can make an additional bet, equal to his original one, that he will make his point before a seven turns up. If his point is four or ten he is paid off at 2: 1 odds; if it is a five or nine he is paid off at odds \(3: 2 ;\) and if it is a six or eight he is paid off at odds 6 : 5. Find the player's expected winnings if he makes this additional bet when he has the opportunity.

Let \(X\) be the first time that a failure occurs in an infinite sequence of Bernoulli trials with probability \(p\) for success. Let \(p_{k}=P(X=k)\) for \(k=1,2, \ldots .\) Show that \(p_{k}=p^{k-1} q\) where \(q=1-p .\) Show that \(\sum_{k} p_{k}=1 .\) Show that \(E(X)=1 / q .\) What is the expected number of tosses of a coin required to obtain the first tail?

A deck of ESP cards consists of 20 cards each of two types: say ten stars, ten circles (normally there are five types). The deck is shuffled and the cards turned up one at a time. You, the alleged percipient, are to name the symbol on each card before it is turned up. Suppose that you are really just guessing at the cards. If you do not get to see each card after you have made your guess, then it is easy to calculate the expected number of correct guesses, namely ten. If, on the other hand, you are guessing with information, that is, if you see each card after your guess, then, of course, you might expect to get a higher score. This is indeed the case, but calculating the correct expectation is no longer easy. But it is easy to do a computer simulation of this guessing with information, so we can get a good idea of the expectation by simulation. (This is similar to the way that skilled blackjack players make blackjack into a favorable game by observing the cards that have already been played. See Exercise 29.) (a) First, do a simulation of guessing without information, repeating the experiment at least 1000 times. Estimate the expected number of correct answers and compare your result with the theoretical expectation. (b) What is the best strategy for guessing with information? (c) Do a simulation of guessing with information, using the strategy in (b). Repeat the experiment at least 1000 times, and estimate the expectation in this case. (d) Let \(S\) be the number of stars and \(C\) the number of circles in the deck. Let \(h(S, C)\) be the expected winnings using the optimal guessing strategy in (b). Show that \(h(S, C)\) satisfies the recursion relation $$h(S, C)=\frac{S}{S+C} h(S-1, C)+\frac{C}{S+C} h(S, C-1)+\frac{\max (S, C)}{S+C}$$ and \(h(0,0)=h(-1,0)=h(0,-1)=0 .\) Using this relation, write a program to compute \(h(S, C)\) and find \(h(10,10)\). Compare the computed value of \(h(10,10)\) with the result of your simulation in (c). For more about this exercise and Exercise 26 see Diaconis and Graham. \(^{11}\)

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