Show that the expectation of the sum of two random variables defined over the same sample space is the sum of the expectations. Hint: Let \(p_{1}, p_{2}, p_{3}, \cdots, P_{*}\) be the probabilitics associated with the \(n\) sample points; let \(x_{1}, x_{2}, \cdots, x_{n}\), and \(y_{1}, y_{2}, \cdots, y_{n}\), be the values of the random variables \(x\) and \(y\) for the \(n\) sample points. Write out \(E(x), E(y)\), and \(E(x+y)\).

Short Answer

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The expectation of the sum of two random variables is the sum of their expectations: \( E(x + y) = E(x) + E(y) \).

Step by step solution

01

Define the Expectation of a Random Variable

The expectation (or expected value) of a random variable is the weighted average of all possible values that the random variable can take, with the probabilities as weights. For a random variable \(x\) with values \(x_1, x_2, \, ... , x_n\) and corresponding probabilities \(p_1, p_2, \, ... , p_n\), the expectation is given by: \[ E(x) = \sum_{i=1}^{n} p_i x_i \]
02

Write Out the Expectation of \(x\)

Using the definition from Step 1, the expectation of the random variable \(x\) is: \[ E(x) = \sum_{i=1}^{n} p_i x_i \]
03

Write Out the Expectation of \(y\)

Similarly, for another random variable \(y\) with values \(y_1, y_2, \, ... , y_n\) and corresponding probabilities \(p_1, p_2, \, ... , p_n\), the expectation is given by: \[ E(y) = \sum_{i=1}^{n} p_i y_i \]
04

Define the Sum of the Two Random Variables

The sum of the two random variables \(x\) and \(y\) at each sample point can be written as \(x_i + y_i\). We need to find the expectation of the sum \( E(x + y) \).
05

Write Out the Expectation of \(x + y\)

Using the linearity of expectation, we have: \[ E(x + y) = \sum_{i=1}^{n} p_i (x_i + y_i) \]
06

Apply the Distributive Property

Apply the distributive property to the sum in the expression for \( E(x + y) \): \[ E(x + y) = \sum_{i=1}^{n} p_i x_i + \sum_{i=1}^{n} p_i y_i \]
07

Use Definition of Expectation

Recognize that the two sums can be rewritten as the expectations of \(x\) and \(y\) respectively: \[ E(x + y) = E(x) + E(y) \]
08

Conclusion: State the Result

We have shown that the expectation of the sum of two random variables is equal to the sum of the expectations of those two random variables, i.e., \[ E(x + y) = E(x) + E(y) \]

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Key Concepts

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

Probability Theory
Probability theory is a branch of mathematics that deals with the analysis of random phenomena. The outcome of a random event can be one of several possible results, and the likelihood of each result is quantified by a probability. Probability values range from 0 (indicating an impossible event) to 1 (indicating a certain event). For instance, flipping a fair coin has two possible outcomes, heads or tails, each with a probability of 0.5.
A key part of probability theory involves understanding how different probabilities relate and interact, such as through the concepts of independent and dependent events. It's foundational for comprehending random variables and their behaviors, which leads us into the next important concept.
Linearity of Expectation
Linearity of expectation is a fundamental and very useful principle in probability theory. It states that for any two random variables, the expected value (also known as the expectation) of their sum is equal to the sum of their individual expected values. Mathematically, this can be expressed as: Note that this property does not require the random variables to be independent. This normalization makes expectation calculations much easier and is crucial in fields like statistics, economics, and data science.
Random Variables
A random variable is a numerical description of the outcome of a statistical experiment. There are two main types of random variables: discrete and continuous.

  • Discrete random variables can only take on a finite or countably infinite number of distinct values. Examples include the number of heads in a series of coin flips, or the outcome of rolling a six-sided die.
  • Continuous random variables can take an infinite number of possible values within a given range. Examples are the height of students in a class, or the time it takes for a computer algorithm to run.
For both types, the expectation (or expected value) is a weighted average of the possible values, with the weights being the probabilities of those values. Mathematically, for a discrete random variable, the expectation is given by:

For a continuous random variable, it is given by: Understanding these concepts is essential for analyzing how random variables behave and for calculating important metrics such as variances, covariances, and standard deviations.

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Most popular questions from this chapter

(a) One box contains one die and another box contains two dice. You select a box at random and take out and toss whatever is in it (that is, toss both dice if you have picked box 2 ). Let \(x=\) number of 3 's showing. Set up the sample space and associated probabilities for \(x\). (b) What is the probability of at least one 3 ? (c) If at least one 3 turns up, what is the probability that you picked the first box? (d) Find \(\bar{x}\) and \(\sigma\).

Suppose 520 people each have a shuffled deck of cards and draw one card from the deck. What is the probability that exactly 13 of the 520 cards will be aces of spades? Write the binomial formula and approximate it. Which is best, the normal or the Poisson approximation?

Suppose you have 3 nickels and 4 dimes in your right pocket and 2 nickels and a quarter in your left pocket. You pick a pocket at random and from it select a coin at random. If it is a. nickel, what is the probability that it came from your right pocket?

Using the normal approximation to the binomial distribution, and tables [or calculator for \(\phi(t)\) ], find the approximate probability of each of the following: Between 195 and 205 tails in 400 tosses of a coin.

Define the sample variance by \(s^{2}=(1 / n) \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} .\) Show that the expected value of \(s^{2}\) is \([(n-1) / n] \sigma^{2} .\) Hints: Write $$ \begin{aligned} \left(x_{i}-\bar{x}\right)^{2} &=\left[\left(x_{i}-\mu\right)-(\bar{x}-\mu)\right]^{2} \\ &=\left(x_{i}-\mu\right)^{2}-2\left(x_{i}-\mu\right)(\bar{x}-\mu)+(\bar{x}-\mu)^{2} \end{aligned} $$ Find the average value of the first term from the definition of \(\sigma^{2}\) and the average value of the third term from Problem 2, To find the average value of the middle term write $$ (\bar{x}-\mu)=\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}-\mu\right)=\frac{1}{n}\left[\left(x_{1}-\mu\right)+\left(x_{2}-\mu\right)+\cdots+\left(x_{n}-\mu\right)\right] $$ show by Problem \(7.12\) that $$ E\left[\left(x_{i}-\mu\right)\left(x_{j}-\mu\right)\right]=E\left(x_{i}-\mu\right) E\left(x_{j}-\mu\right)=0 \quad \text { for } \quad t \neq j $$ and evaluate \(E\left[\left(x_{i}-\mu\right)^{2}\right]\) (same as the first term). Collect terms to find $$ E\left(s^{2}\right)=\frac{n-1}{n} \sigma^{2} $$

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