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Chapter 5: Problem 33

Show that if \(X, Y\), and \(Z\) are rv's and \(a\) and \(b\) are constants, then \(\operatorname{Cov}(a X+b Y, Z)=a \operatorname{Cov}(X, Z)+\) \(b \operatorname{Cov}(Y, Z)\)

Short Answer

Expert verified
\(\operatorname{Cov}(aX + bY, Z) = a \operatorname{Cov}(X, Z) + b \operatorname{Cov}(Y, Z)\).

Step by step solution

01

Recall the Definition of Covariance

The covariance between two random variables \(U\) and \(V\) is defined as \(\operatorname{Cov}(U, V) = E[(U - E[U])(V - E[V])]\). We will use this formula to explore the expression \(\operatorname{Cov}(aX + bY, Z)\).
02

Substitute for Covariance

To show \(\operatorname{Cov}(aX + bY, Z)\), we substitute into the definition: \[\operatorname{Cov}(aX+bY, Z) = E[((aX + bY) - E[aX + bY])(Z - E[Z])]\].
03

Expand the Expectation

Inside the expectation, expand and rearrange: \[E[((aX + bY) - aE[X] - bE[Y])(Z - E[Z])]\]. Use distributive property to expand it:\[aE[(X - E[X])(Z - E[Z])] + bE[(Y - E[Y])(Z - E[Z])]\].
04

Factor Out Constants

Since expectation is a linear operator, factor out the constants \(a\) and \(b\):\[aE[(X - E[X])(Z - E[Z])] + bE[(Y - E[Y])(Z - E[Z])]\].This yields the expression: \(a \operatorname{Cov}(X, Z) + b \operatorname{Cov}(Y, Z)\).
05

Conclusion

We have shown that by the properties of expectation and the definition of covariance, \[\operatorname{Cov}(aX + bY, Z) = a \operatorname{Cov}(X, Z) + b \operatorname{Cov}(Y, Z)\]. This proves the required result.

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Key Concepts

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

Random Variables
Random variables are foundational elements in probability and statistics. A random variable, often denoted by symbols like \(X\), \(Y\), or \(Z\), is a variable whose possible values are numerical outcomes of a random phenomenon. Unlike deterministic variables, random variables can take on different values depending on the outcome of an uncertain event.

There are two primary types of random variables:
  • Discrete Random Variables: These have a countable number of distinct outcomes, such as rolling a die (1 through 6).
  • Continuous Random Variables: These have an infinite number of possible values within a given range, like the exact height of individuals.
In our context, the random variables \(X\), \(Y\), and \(Z\) represent such numerical outcomes within probabilistic experiments. Understanding how these variables interact using operations like covariance helps us measure and interpret their relationships.
Expectation
Expectation, also known as expected value, is a crucial concept in the study of random variables. It provides a measure of the "central tendency" of a random variable by calculating the average value it might take, over a large number of trials or scenarios.

Mathematically, the expectation of a random variable \(X\), denoted as \(E[X]\), is calculated as:
  • For Discrete Random Variables: The sum of the products of each possible value of \(X\) and their respective probabilities.
  • For Continuous Random Variables: The integral of the product of the value and its probability density function.
In the covariance problem, expectation is used to derive expressions like \(E[(U - E[U])(V - E[V])]\) which help in determining the covariance between two random variables.
Linear Operator
The term "linear operator" is central to mathematical transformations involving random variables. In simple terms, a linear operator is a mathematical object that satisfies two properties: additivity and homogeneity.

These properties imply:
  • Additivity: Applying the operator to a sum of two functions \(f\) and \(g\) results in the sum of their operator-applied results, i.e., \(L(f + g) = L(f) + L(g)\).
  • Homogeneity: Applying the operator to a scaled function results in the scale multiplied by the operator-applied result, i.e., \(L(af) = aL(f)\) where \(a\) is a constant.
The expectation \(E\) is an example of a linear operator because linearity allows us to factor out constants from within it. This feature is used to simplify expressions during the calculation of covariance, such as in the step where constants \(a\) and \(b\) are factored out.
Properties of Expectation
The properties of expectation empower us to approach complex problems involving random variables with greater ease. Some important properties are:
  • Linearity: The expectation of a sum is the sum of expectations, \(E[aX + bY] = aE[X] + bE[Y]\).
  • Constants: The expectation of a constant is just the constant itself, \(E[c] = c\).
  • Expectation of a Constant Times a Random Variable: You can factor out a constant, \(E[aX] = aE[X]\).
These properties, particularly linearity, are essential in calculating the covariance in the exercise, allowing us to simplify expressions and break down complicated terms into manageable parts. They underpin the factorial steps in the solution, making complex expressions clearer and easier to compute.

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Most popular questions from this chapter

A health-food store stocks two different brands of a type of grain. Let \(X=\) the amount (lb) of brand A on hand and \(Y=\) the amount of brand B on hand. Suppose the joint pdf of \(X\) and \(Y\) is \(f(x, y)\) \(=\left\\{\begin{array}{cl}k x y & x \geq 0, y \geq 0,20 \leq x+y \leq 30 \\\ 0 & \text { otherwise }\end{array}\right.\) a. Draw the region of positive density and determine the value of \(k\). b. Are \(X\) and \(Y\) independent? Answer by first deriving the marginal pdf of each variable. c. Compute \(P(X+Y \leq 25)\). d. What is the expected total amount of this grain on hand? e. Compute \(\operatorname{Cov}(X, Y)\) and \(\operatorname{Corr}(X, Y)\). f. What is the variance of the total amount of grain on hand?

A pizza place has two phones. On each phone the waiting time until the first call is exponentially distributed with mean one minute. Each phone is not influenced by the other. Let \(X\) be the shorter of the two waiting times and let \(Y\) be the longer. It can be shown that the joint pdf of \(X\) and \(Y\) is \(f(x, y)=2 e^{-(x+y)}, 0

This week the number \(X\) of claims coming into an insurance office is Poisson with mean 100 . The probability that any particular claim relates to automobile insurance is \(.6\), independent of any other claim. If \(Y\) is the number of automobile claims, then \(Y\) is binomial with \(X\) trials, each with "success" probability .6. a. Determine \(E(Y \mid X=x)\) and \(V(Y \mid X=x)\). b. Use part (a) to find \(E(Y)\). c. Use part (a) to find \(V(Y)\).

Show that if \(Y=a X+b(a \neq 0)\), then \(\operatorname{Corr}(X, Y)=\) \(+1\) or \(-1\). Under what conditions will \(\rho=+1\) ?

Suppose that \(X\) is uniformly distributed between 0 and 1. Given \(X=x, Y\) is uniformly distributed between 0 and \(x^{2}\) a. Determine \(E(Y \mid X=x)\) and then \(V(Y \mid X=x)\). Is \(E(Y \mid X=x)\) a linear function of \(x\) ? b. Determine \(f(x, y)\) using \(f_{X}(x)\) and \(f_{Y \mid X}(y \mid x)\). c. Determine \(f_{y}(y)\).
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