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

Short Answer

Expert verified
Covariance is zero since independent variables satisfy \(\text{Cov}(X, Y) = 0\).

Step by step solution

01

Understand the Given Information

We are given that \(X\) and \(Y\) are independent continuous random variables. We need to show that the covariance between \(X\) and \(Y\), denoted as \(\sigma_{XY}\), is equal to zero.
02

Recall the Definition of Covariance

The covariance between two random variables \(X\) and \(Y\) is defined as \(\sigma_{XY} = \text{Cov}(X, Y) = \mathbb{E}[(X - \mu_X)(Y - \mu_Y)]\), where \(\mu_X\) and \(\mu_Y\) are the expected values of \(X\) and \(Y\) respectively.
03

Use the Independence Assumption

Since \(X\) and \(Y\) are independent, the expectation of the product of functions of \(X\) and \(Y\) equals the product of their expectations. Therefore, \(\mathbb{E}[XY] = \mathbb{E}[X] \cdot \mathbb{E}[Y]\). This is a key property of independent random variables.
04

Calculate Covariance Using Independence

Substitute the independence property into the covariance formula: \[ \text{Cov}(X, Y) = \mathbb{E}[XY] - \mathbb{E}[X]\mathbb{E}[Y] \] Since \(\mathbb{E}[XY] = \mathbb{E}[X] \cdot \mathbb{E}[Y]\) by independence, the article becomes \[ \text{Cov}(X, Y) = \mathbb{E}[X] \cdot \mathbb{E}[Y] - \mathbb{E}[X] \cdot \mathbb{E}[Y] = 0 \].
05

Conclusion

We have shown that the covariance \(\sigma_{XY} = 0\) when \(X\) and \(Y\) are independent. Thus, the covariance of independent continuous random variables is zero.

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Key Concepts

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

Continuous Random Variables
Continuous random variables are central to understanding probability and statistical models. They can take any value within a given range or interval, thus having an infinite number of possible values. Unlike discrete random variables, which only accept specific, countable values, continuous random variables are characterized by smooth probability distributions.

A typical example of a continuous random variable is the exact height of individuals in a population. Since height can vary ever so slightly between individuals, it represents a continuous range of values. In mathematical terms, these variables are described using probability density functions (PDFs).
  • Continuous random variables have probability zero of taking on any exact value, but can be described over an interval.
  • They are often modeled using distributions like the normal distribution.
  • The area under the curve of the PDF within a given range reflects the probability of the variable falling within that range.

Understanding continuous random variables is essential for statistical analysis, especially for modeling real-world phenomena with nuance and detail.
Properties of Covariance
Covariance is a statistical measure that describes the extent to which two variables change together. If the variables are said to "covary," their movements are related. In mathematical terms, the covariance formula is given by:\[ \text{Cov}(X, Y) = \mathbb{E}[(X - \mu_X)(Y - \mu_Y)] \]where \(\mu_X\) and \(\mu_Y\) are the expected values of \(X\) and \(Y\), respectively.
  • A positive covariance indicates that the variables tend to move in the same direction.
  • A negative covariance suggests that they move in opposite directions.
  • A zero covariance implies no linear relationship between the variables.

It's important to note that while a zero covariance suggests no linear association, it doesn't necessarily imply independence unless the variables are normally distributed. Covariance is a foundational concept used in more complex analyses, including variance and correlation.
Independence in Statistics
In statistics, independence between variables means that the occurrence of one variable does not affect the likelihood of occurrence of another. Essentially, two independent variables do not influence each other. When dealing with random variables, this concept becomes critical to simplify calculations and infer relationships.

For example, if \(X\) and \(Y\) are independent random variables, mathematically it implies:\[ \mathbb{E}[XY] = \mathbb{E}[X] \cdot \mathbb{E}[Y] \]
  • Independence ensures that knowing the outcome of one variable gives no information about the other.
  • It is a key assumption in various statistical tests and models, enabling simplification of complex data.

This property of independence is particularly useful in calculating covariance. Since independent variables do not show any linear relation, their covariance is zero, simplifying statistical evaluations and conclusions.
Expected Value
The expected value is a fundamental concept in probability and statistics, often likened to the arithmetic mean of a random variable. It provides a measure of the center or "average" outcome one can expect when considering all possible outcomes and their probabilities. Mathematically, the expected value of a random variable \(X\) is defined as:\[ \mathbb{E}[X] = \int_{-\infty}^{\infty} x \, f_X(x) \, dx \]where \(f_X(x)\) is the probability density function if \(X\) is continuous.
  • Expected value helps in predicting the average outcome over a long period or large sample size.
  • It is used in various financial, scientific, and engineering applications for decision-making.
  • For a discrete random variable, it is calculated as the sum of all possible values, each multiplied by its probability.

Knowing the expected value helps in understanding both anticipated outcomes and deviations from the average, aiding in more informed statistical analysis.

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Most popular questions from this chapter

The permeability of a membrane used as a moisture barrier in a biological application depends on the thickness of three integrated layers. Layers \(1,2,\) and 3 are normally distributed with means of \(0.5,1,\) and 1.5 millimeters, respectively. The standard deviations of layer thickness are 0.1,0.2 , and \(0.3,\) respectively. Also, the correlation between layers 1 and 2 is \(0.7,\) between layers 2 and 3 is \(0.5,\) and between layers 1 and 3 is \(0.3 .\) (a) Determine the mean and variance of the total thickness of the three layers. (b) What is the probability that the total thickness is less than 1.5 millimeters?

To evaluate the technical support from a computer manufacturer, the number of rings before a call is answered by a service representative is tracked. Historically, \(70 \%\) of the calls are answered in two rings or less, \(25 \%\) are answered in three or four rings, and the remaining calls require five rings or more. Suppose you call this manufacturer 10 times and assume that the calls are independent. (a) What is the probability that eight calls are answered in two rings or less, one call is answered in three or four rings, and one call requires five rings or more? (b) What is the probability that all 10 calls are answered in four rings or less? (c) What is the expected number of calls answered in four rings or less? (d) What is the conditional distribution of the number of calls requiring five rings or more given that eight calls are answered in two rings or less? (e) What is the conditional expected number of calls requiring five rings or more given that eight calls are answered in two rings or less? (f) Are the number of calls answered in two rings or less and the number of calls requiring five rings or more independent random variables?

Show that the following function satisfies the properties of a joint probability mass function. $$ \begin{array}{llc} \hline x & y & f_{X Y}(x, y) \\ \hline 1 & 1 & 1 / 4 \\ 1.5 & 2 & 1 / 8 \\ 1.5 & 3 & 1 / 4 \\ 2.5 & 4 & 1 / 4 \\ 3 & 5 & 1 / 8 \\ \hline \end{array} $$ Determine the following: (a) \(P(X < 2.5, Y < 3)\) (b) \(P(X < 2.5)\) (c) \(P(Y < 3)\) (d) \(P(X > 1.8, Y > 4.7)\) (e) \(E(X), E(Y), V(X),\) and \(V(Y)\) (f) Marginal probability distribution of the random variable \(X\) (g) Conditional probability distribution of \(Y\) given that \(X=1.5\) (h) Conditional probability distribution of \(X\) given that \(Y=2\) (i) \(E(Y \mid X=1.5)\) (j) Are \(X\) and \(Y\) independent?

In the manufacture of electroluminescent lamps, several different layers of ink are deposited onto a plastic substrate. The thickness of these layers is critical if specifications regarding the final color and intensity of light are to be met. Let \(X\) and \(Y\) denote the thickness of two different layers of ink. It is known that \(X\) is normally distributed with a mean of 0.1 millimeter and a standard deviation of 0.00031 millimeter and \(Y\) is also normally distributed with a mean of 0.23 millimeter and a standard deviation of 0.00017 millimeter. Assume that these variables are independent. (a) If a particular lamp is made up of these two inks only, what is the probability that the total ink thickness is less than 0.2337 millimeter? (b) A lamp with a total ink thickness exceeding 0.2405 millimeter lacks the uniformity of color demanded by the customer. Find the probability that a randomly selected lamp fails to meet customer specifications.

Suppose \(X\) and \(Y\) have a bivariate normal distribution with \(\sigma_{X}=4, \sigma_{Y}=1, \mu_{X}=4, \mu_{Y}=4,\) and \(\rho=-0.2\). Draw a rough contour plot of the joint probability density function.

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