/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 Suppose that \(X\) and \(Y\) are... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose that \(X\) and \(Y\) are independent continuous random variables. Show that \(\sigma_{X Y}=0 .\)

Short Answer

Expert verified
The covariance \(\sigma_{XY}\) of independent random variables \(X\) and \(Y\) is zero.

Step by step solution

01

Understand the Concept of Independence

When two random variables, \(X\) and \(Y\), are independent, it means that the occurrence of an event related to \(X\) does not affect the occurrence of an event related to \(Y\). Mathematically, this is represented by the joint probability density function: \(f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y)\) where \(f_X(x)\) and \(f_Y(y)\) are the marginal probability density functions of \(X\) and \(Y\) respectively.
02

Define Covariance

Covariance measures the degree to which two variables change together. For two random variables \(X\) and \(Y\), the covariance \(\sigma_{XY}\) is defined as \(\sigma_{XY} = E[(X - \mu_X)(Y - \mu_Y)]\), where \(\mu_X\) and \(\mu_Y\) are the expected values of \(X\) and \(Y\), respectively.
03

Substitute Independence into Covariance

Because \(X\) and \(Y\) are independent, the expected value of the product of \(g(X)\) and \(h(Y)\) can be simplified using the product of expectations: \(E[g(X)h(Y)] = E[g(X)]E[h(Y)]\). Applying this property to the covariance formula, we have \(E[(X-\mu_X)(Y-\mu_Y)] = E[(X-\mu_X)]E[(Y-\mu_Y)]\).
04

Simplify the Covariance Expression

From Step 3, we now have \(\sigma_{XY} = E[X - \mu_X]E[Y - \mu_Y]\). Notice \(E[X - \mu_X] = 0\) and \(E[Y - \mu_Y] = 0\) because \(\mu_X\) and \(\mu_Y\) are the means. Thus, \(\sigma_{XY} = 0 \cdot 0 = 0\).
05

Conclusion: Interpret the Result

The covariance \(\sigma_{XY} = 0\) indicates that there is no linear relationship between the independent random variables \(X\) and \(Y\). Therefore, when \(X\) and \(Y\) are independent, their covariance is zero.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Covariance
Covariance is a statistical measure that describes how two random variables change together. If one variable tends to increase when the other increases, they have positive covariance. Conversely, if one decreases when the other increases, they exhibit negative covariance. Thus, covariance can be a measure of the strength of a linear relationship between two variables.

For random variables \(X\) and \(Y\), the covariance \(\sigma_{XY}\) is calculated using the formula:
  • \(\sigma_{XY} = E[(X - \mu_X)(Y - \mu_Y)]\)
Here, \(\mu_X\) and \(\mu_Y\) represent the expected values or means of \(X\) and \(Y\) respectively.

When \(X\) and \(Y\) are independent, their covariance is zero. This follows from the fact that the measure of dependency between the two variables is nonexistent, as they do not affect each other's outcomes. Hence, there is no trend or direction of change that both variables follow together.
Probability Density Function
In probability theory, a probability density function (PDF) is a function that describes the likelihood of a continuous random variable to take on a particular value. It is a vital tool in evaluating probabilities for continuous random variables, and its integral over a range gives the probability that the variable falls within that range.

To understand this, consider a continuous random variable \(X\) with PDF \(f_X(x)\). Then the probability that \(X\) falls between two values \(a\) and \(b\) is:
  • \( P(a \leq X \leq b) = \int_{a}^{b} f_X(x) \, dx \)
The function \(f_X(x)\) itself does not give probabilities directly, but rather densities. For two independent random variables \(X\) and \(Y\), their joint PDF is simply the product of their individual PDFs:

  • \(f_{X,Y}(x,y) = f_X(x) \cdot f_Y(y)\)
This property forms an essential understanding that underpins the independence of \(X\) and \(Y\), in essence implying that knowing the value of \(X\) does not provide information about the value of \(Y\), and vice versa.
Expected Value
The expected value is a fundamental concept in probability, representing the average or mean value of a random variable based on its probability distribution. It essentially gives a long-term average if you were to repeat an experiment or observation infinite times.

For a continuous random variable \(X\) with probability density function \(f_X(x)\), the expected value \(E[X]\) is defined as:
  • \( E[X] = \int_{-\infty}^{\infty} x \, f_X(x) \, dx \)
This formula means you multiply each possible value of the random variable by its probability density and sum across all possible values.

Understanding expected value is crucial, especially when examining independence and covariance. For example, if random variables \(X\) and \(Y\) are independent, the expected value of their product equals the product of their expected values:
  • \(E[X \cdot Y] = E[X] \cdot E[Y]\)
This property is extensively used in proving that the covariance of two independent random variables is zero, emphasizing the non-existence of a linear relationship between them as they do not influence each other's expected outcomes.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The width of a casing for a door is normally distributed with a mean of 24 inches and a standard deviation of \(1 / 8\) inch. The width of a door is normally distributed with a mean of \(237 / 8\) inches and a standard deviation of \(1 / 16\) inch. Assume independence. (a) Determine the mean and standard deviation of the difference between the width of the casing and the width of the door. (b) What is the probability that the width of the casing minus the width of the door exceeds \(1 / 4\) inch? (c) What is the probability that the door does not fit in the casing?

\(X\) and \(Y\) are independent, normal random variables with \(E(X)=2, V(X)=5, E(Y)=6,\) and \(V(Y)=8 .\) Determine the following: (a) \(E(3 X+2 Y)\) (b) \(V(3 X+2 Y)\) (c) \(P(3 X+2 Y<18)\) (d) \(P(3 X+2 Y<28)\)

An aircraft is flying at a constant altitude with velocity magnitude \(r_{1}\) (relative to the air) and angle \(\theta_{1}\) (in a twodimensional coordinate system). The magnitude and direction of the wind are \(r_{2}\) and \(\theta_{2},\) respectively. Suppose that the wind angle is uniformly distributed between 10 and 20 degrees and all other parameters are constant. Determine the probability density function of the magnitude of the resultant vector \(r=\left[r_{1}^{2}+r_{2}^{2}+r_{1} r_{2}\left(\cos \theta_{1}-\cos \theta_{2}\right)\right]^{0.5}\)

Determine the value of \(c\) such that the function \(f(x, y)=c x y\) for \(01.8,1

In the transmission of digital information, the probability that a bit has high, moderate, or low distortion is 0.01 , \(0.04,\) and \(0.95,\) respectively. Suppose that three bits are transmitted and that the amount of distortion of each bit is assumed to be independent. Let \(X\) and \(Y\) denote the number of bits with high and moderate distortion of the three transmitted, respectively. Determine the following: (a) Probability that two bits have high distortion and one has moderate distortion (b) Probability that all three bits have low distortion (c) Probability distribution, mean, and variance of \(X\) (d) Conditional probability distribution, conditional mean, and conditional variance of \(X\) given that \(Y=2\)

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.