/*! 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 167 Let \(Y_{1}\) and \(Y_{2}\) be j... [FREE SOLUTION] | 91Ó°ÊÓ

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Let \(Y_{1}\) and \(Y_{2}\) be jointly distributed random variables with finite variances. a. Show that \(\left[E\left(Y_{1} Y_{2}\right)\right]^{2} \leq E\left(Y_{1}^{2}\right) E\left(Y_{2}^{2}\right) .\) [Hint: Observe that \(E\left[\left(t Y_{1}-Y_{2}\right)^{2}\right] \geq 0\) for any real number t or, equivalently, $$t^{2} E\left(Y_{1}^{2}\right)-2 t E\left(Y_{1} Y_{2}\right)+E\left(Y_{2}^{2}\right) \geq 0$$ This is a quadratic expression of the form \(A t^{2}+B t+C\); and because it is nonnegative, we must have \(B^{2}-4 A C \leq 0 .\) The preceding inequality follows directly.] b. Let \(\rho\) denote the correlation coefficient of \(Y_{1}\) and \(Y_{2} .\) Using the inequality in part (a), show that \(\rho^{2} \leq 1\)

Short Answer

Expert verified
(a) \\( [E(Y_1 Y_2)]^2 \leq E(Y_1^2)E(Y_2^2) \\); (b) \\( \rho^2 \leq 1 \\).

Step by step solution

01

Understand the quadratic expression

We are given the inequality for any real number \(t\): \\[t^{2} E(Y_{1}^{2}) - 2t E(Y_{1}Y_{2}) + E(Y_{2}^{2}) \geq 0.\]\This is a quadratic expression of the form \(At^2 + Bt + C\), where \(A = E(Y_{1}^{2})\), \(B = -2 E(Y_{1}Y_{2})\), and \(C = E(Y_{2}^{2})\). Since this quadratic is nonnegative for all \(t\), the discriminant \(B^2 - 4AC\) must be non-positive, i.e., \(B^2 - 4AC \leq 0\).
02

Compute the discriminant

Substitute the values for \(A\), \(B\), and \(C\) into the discriminant formula: \\[(-2E(Y_{1}Y_{2}))^{2} - 4E(Y_{1}^{2})E(Y_{2}^{2}) \leq 0.\]\Simplifying the left side gives us: \\[4[E(Y_{1}Y_{2})]^{2} - 4E(Y_{1}^{2})E(Y_{2}^{2}) \leq 0.\] \Dividing the entire inequality by 4 results in: \\[[E(Y_{1}Y_{2})]^{2} \leq E(Y_{1}^{2})E(Y_{2}^{2}).\]\This is the required inequality for part (a).
03

Relate inequality to correlation coefficient

Now consider the correlation coefficient \(\rho\) defined as: \\[\rho = \frac{E(Y_{1}Y_{2}) - E(Y_{1})E(Y_{2})}{\sqrt{E(Y_{1}^{2})E(Y_{2}^{2})}}.\]\The numerator represents the covariance, while the denominator is the product of the standard deviations of \(Y_{1}\) and \(Y_{2}\). This ensures the expression is dimensionless and between -1 and 1.
04

Show \(\rho^2 \leq 1\) using the inequality

By the result from step 2, we have: \\[[E(Y_{1}Y_{2})]^{2} \leq E(Y_{1}^{2})E(Y_{2}^{2}).\]\Hence, dividing through by \(E(Y_{1}^{2})E(Y_{2}^{2})\), we get: \\[\left(\frac{E(Y_{1}Y_{2})}{\sqrt{E(Y_{1}^{2})} \cdot \sqrt{E(Y_{2}^{2})}}\right)^{2} \leq 1.\]\Since this expression is exactly \(\rho^2\), we have \(\rho^2 \leq 1\), which completes part (b).

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Key Concepts

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

Covariance
The covariance is a measure that tells us how much two random variables vary together. If you have two random variables, say, one representing the height of students and another representing their weight, covariance can indicate whether taller students tend to weigh more or less. It's calculated as the expected value of the product of the deviations of each random variable from their respective means. In formula terms, if we consider two random variables, \(Y_1\) and \(Y_2\), the covariance is given by \(E((Y_1 - E(Y_1))(Y_2 - E(Y_2)))\).
Covariance can be positive, negative, or zero:
  • A positive covariance indicates that as one variable increases, the other tends to increase as well.
  • A negative covariance suggests that as one variable increases, the other tends to decrease.
  • A covariance of zero implies no linear relationship between the variables.
However, one limitation of covariance is that it is scale-dependent and doesn't give insight into the strength of the relationship, which is where the correlation coefficient comes in.
Correlation Coefficient
The correlation coefficient, often denoted by \(\rho\), takes the concept of covariance a step further by standardizing it, making it easier to interpret. It provides a measure of the strength and direction of a linear relationship between two random variables, scaled to lie between -1 and 1.
The formula for the correlation coefficient is \(\rho = \frac{E((Y_1 - E(Y_1))(Y_2 - E(Y_2)))}{\sqrt{E(Y_1^2) - [E(Y_1)]^2} \cdot \sqrt{E(Y_2^2) - [E(Y_2)]^2}}\).
Here's what the values of \(\rho\) indicate:
  • \(\rho = 1\) means a perfect positive linear relationship.
  • \(\rho = -1\) denotes a perfect negative linear relationship.
  • \(\rho = 0\) suggests no linear correlation.
In part (b) of the problem, we used the Cauchy-Schwarz inequality derived earlier to show that \(\rho^2 \leq 1\), which affirms that the correlation coefficient indeed fits within its -1 to 1 range.
Quadratic Expression
A quadratic expression is a polynomial of degree two, generally written in the form \(Ax^2 + Bx + C\). Such an expression can describe a parabola and might intersect the x-axis, sit entirely above it, or be tangent to it.
In the problem, the expression \(t^2 E(Y_1^2) - 2t E(Y_1Y_2) + E(Y_2^2)\) is a quadratic expression in \(t\). If we visualize this as a parabola, needing it to be non-negative for all \(t\) suggests the parabola never dips below the t-axis. This condition is equivalent to ensuring its discriminant \(B^2 - 4AC \leq 0\). Rephrased in our specific context, this condition leads directly to the Cauchy-Schwarz inequality. Quadratics are crucial in deriving this relationship between expectations of the products and the expectations of squares.
Random Variables
Random variables represent quantities that have possible outcomes determined by chance. They are foundational in probability and statistics and can be either discrete or continuous.
In this exercise, we deal with random variables \(Y_1\) and \(Y_2\), which are jointly distributed. This means their outcomes are not independent, but their behaviors are linked or correlated. Each random variable has expectations (mean), variances (spread), and possibly, covariances when considered together with others.
Understanding random variables involves knowing:
  • Their expected value \(E(Y)\), giving an average or mean by weighing outcomes by their probabilities.
  • Their variance \(V(Y) = E(Y^2) - [E(Y)]^2\), a measure of the spread or dispersion of their values.
These concepts are key when applying inequalities like Cauchy-Schwarz in dealing with expectations and variances. Such insights are pivotal in drawing conclusions about underlying relationships in data and probabilistic models.

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Most popular questions from this chapter

In Exercise \(5.16, Y_{1}\) and \(Y_{2}\) denoted the proportions of time that employees I and II actually spent working on their assigned tasks during a workday. The joint density of \(Y_{1}\) and \(Y_{2}\) is given by $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} y_{1}+y_{2}, & 0 \leq y_{1} \leq 1,0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ Employee I has a higher productivity rating than employee II and a measure of the total productivity of the pair of employees is \(30 Y_{1}+25 Y_{2}\). Find the expected value of this measure of productivity.

In the production of a certain type of copper, two types of copper powder (types A and B) are mixed together and sintered (heated) for a certain length of time. For a fixed volume of sintered copper, the producer measures the proportion \(Y_{1}\) of the volume due to solid copper (some pores will have to be filled with air) and the proportion \(Y_{2}\) of the solid mass due to type A crystals. Assume that appropriate probability densities for \(Y_{1}\) and \(Y_{2}\) are $$\begin{array}{l} f_{1}\left(y_{1}\right)=\left\\{\begin{array}{ll} 6 y_{1}\left(1-y_{1}\right), & 0 \leq y_{1} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right. \\ f_{2}\left(y_{2}\right)=\left\\{\begin{array}{ll} 3 y_{2}^{2}, & 0 \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right. \end{array}$$ The proportion of the sample volume due to type A crystals is then \(Y_{1} Y_{2} .\) Assuming that \(Y_{1}\) and \(Y_{2}\) are independent, find \(P\left(Y_{1} Y_{2} \leq .5\right)\)

Suppose that \(Y_{1}\) and \(Y_{2}\) are independent Poisson distributed random variables with means \(\lambda_{1}\) and \(\lambda_{2},\) respectively. Let \(W=Y_{1}+Y_{2} .\) In Chapter 6 you will show that \(W\) has a Poisson distribution with mean \(\lambda_{1}+\lambda_{2} .\) Use this result to show that the conditional distribution of \(Y_{1},\) given that \(W=w\), is a binomial distribution with \(n=w\) and \(p=\lambda_{1} /\left(\lambda_{1}+\lambda_{2}\right) \cdot^{\star}\)

A technician starts a job at a time \(Y_{1}\) that is uniformly distributed between 8: 00 A.M. and 8: 15 A.M. The amount of time to complete the job, \(Y_{2}\), is an independent random variable that is uniformly distributed between 20 and 30 minutes. What is the probability that the job will be completed before 8:30 A.M.?

The weights of a population of mice fed on a certain diet since birth are assumed to be normally distributed with \(\mu=100\) and \(\sigma=20\) (measurement in grams). Suppose that a random sample of \(n=4\) mice is taken from this population. Find the probability that a. exactly two weigh between 80 and 100 grams and exactly one weighs more than 100 grams. b. all four mice weigh more than 100 grams.

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