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Show that if \(X\) and \(Y\) are independent rv's, then \(E(X Y)=\) \(E(X) \cdot E(Y)\). Then apply this in Exercise 25.

Short Answer

Expert verified
If \(X\) and \(Y\) are independent, then \(E(X Y) = E(X) \cdot E(Y)\). Apply this directly to compute the product of expectations in Exercise 25.

Step by step solution

01

Understanding Independence of Random Variables

Two random variables \(X\) and \(Y\) are independent if the occurrence of one does not affect the occurrence of the other. Mathematically, this is expressed as \(P(X \leq x, Y \leq y) = P(X \leq x) \cdot P(Y \leq y)\).
02

Definition of Expected Value

The expected value of a function of random variables is a weighted average, where the weights are the probabilities. Given two random variables, the expected value of their product is \(E(XY) = \int_{-fty}^{fty} \int_{-fty}^{fty} xy \, f_{XY}(x, y) \, dx \, dy\), where \(f_{XY}(x, y)\) is the joint density function.
03

Applying Independence to Joint Density

If \(X\) and \(Y\) are independent, their joint density function \(f_{XY}(x, y)\) can be expressed as the product of their marginal densities, \(f_X(x)\) and \(f_Y(y)\). Hence, \(f_{XY}(x, y) = f_X(x) \cdot f_Y(y)\).
04

Calculating the Expectation

Substituting the joint density into the expected value formula, we have \(E(XY) = \int_{-fty}^{fty} \int_{-fty}^{fty} xy \, f_X(x) f_Y(y) \, dx \, dy\). Separating the integrals due to the independence gives \(\int_{-fty}^{fty} x \, f_X(x) \, dx \cdot \int_{-fty}^{fty} y \, f_Y(y) \, dy\).
05

Identifying as Product of Expectations

Recognize that \(\int_{-fty}^{fty} x \, f_X(x) \, dx = E(X)\) and \(\int_{-fty}^{fty} y \, f_Y(y) \, dy = E(Y)\). Thus, \(E(XY) = E(X) \cdot E(Y)\), proving the result for independent random variables.
06

Applying to Exercise 25

To apply this result, any context in Exercise 25 involving independent random variables \(X\) and \(Y\) and requiring the computation of \(E(XY)\), can directly use the relationship \(E(XY) = E(X) \cdot E(Y)\). No further calculations are needed if the expectations \(E(X)\) and \(E(Y)\) are known.

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Key Concepts

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

Independence of Random Variables
Independence is a fundamental concept in probability theory that simplifies the analysis of random variables. Two random variables, say \(X\) and \(Y\), are considered independent if knowing the outcome of one does not change the probability distribution of the other. In simpler terms, they do not influence each other.

Mathematically, this is expressed using the joint probability: \[ P(X \leq x, Y \leq y) = P(X \leq x) \cdot P(Y \leq y). \] This equation tells us that for independent variables, the joint probability density is simply the product of their individual probabilities.

Understanding independence is crucial because it allows us to decompose problems into simpler parts, making calculations easier and more intuitive. Independence lets us apply certain rules like direct multiplication when dealing with joint probabilities or expectations.
Joint Density Function
A joint density function describes the probability distribution of two or more random variables. It is a function that provides the likelihood of different outcomes occurring together in a bivariate or multivariate space.

For two random variables \(X\) and \(Y\), their joint density function \(f_{XY}(x, y)\) is used to determine the probability that \(X\) is at a certain value while \(Y\) is at another value simultaneously. The equation is generally defined as:\[ f_{XY}(x, y) \] which represents the probability per unit area in the \(xy\) plane.

When \(X\) and \(Y\) are independent, the joint density simplifies to the product of their marginal densities. This transformation significantly simplifies the calculations involved in determining the probabilities or expectations of their joint behavior.
Marginal Density
Marginal density is concerned with the probability distribution of a subset of a collection of random variables. It is called 'marginal' because it is obtained by integrating the joint density function over the unwanted variables.

For a pair of random variables \(X\) and \(Y\), the marginal density for \(X\), \(f_X(x)\), is obtained by:\[ f_X(x) = \int_{-\infty}^{\infty} f_{XY}(x, y) \, dy. \]Similarly, the marginal density for \(Y\) can be found by integrating out \(x\).
  • Makes it possible to consider individual effects of variables.
  • Eliminates the unnecessary complexity of dealing with higher dimensionality.

When dealing with independent random variables, the joint density dependency simplifies to this product form, as the densities act independently of each other.
Product of Expectations
The expected value, or expectation, of a random variable is a measure of its central tendency. When dealing with the product of two independent random variables \(X\) and \(Y\), you can calculate the expectation of their product as the product of their individual expectations.

This relationship is expressed mathematically as:\[ E(XY) = E(X) \cdot E(Y) \]when \(X\) and \(Y\) are independent. This simplifies the computation significantly because you independently calculate each expectation and then multiply them together.

It's powerful because it allows us to bypass more complex integration of the joint density function directly, utilizing the factorization property of independence:
  • Reduces computational burden in multi-variable scenarios.
  • Leverages the simplicity of separate expectation evaluations.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be random variables denoting \(n\) independent bids for an item that is for sale. Suppose each \(X_{i}\) is uniformly distributed on the interval [100, 200]. If the seller sells to the highest bidder, how much can he expect to earn on the sale? [Hint: Let \(Y=\max \left(X_{1}, X_{2}, \ldots, X_{n}\right)\). First find \(F_{Y}(y)\) by noting that \(Y \leq y\) iff each \(X_{i}\) is \(\leq y\). Then obtain the pdf and \(E(Y)\).]

A binary communication channel transmits a sequence of "bits" ( 0 s and \(1 \mathrm{~s})\). Suppose that for any particular bit transmitted, there is a \(10 \%\) chance of a transmission error (a 0 becoming a 1 or a 1 becoming a 0 ). Assume that bit errors occur independently of one another. a. Consider transmitting 1000 bits. What is the approximate probability that at most 125 transmission errors occur? b. Suppose the same 1000-bit message is sent two different times independently of one another. What is the approximate probability that the number of errors in the first transmission is within 50 of the number of errors in the second?

a. Use the rules of expected value to show that \(\operatorname{Cov}(a X+b\), \(c Y+d)=a c \operatorname{Cov}(X, Y)\). b. Use part (a) along with the rules of variance and standard deviation to show that \(\operatorname{Corr}(a X+b, c Y+d)=\operatorname{Corr}(X, Y)\) when \(a\) and \(c\) have the same sign. c. What happens if \(a\) and \(c\) have opposite signs?

A certain market has both an express checkout line and a superexpress checkout line. Let \(X_{1}\) denote the number of customers in line at the express checkout at a particular time of day, and let \(X_{2}\) denote the number of customers in line at the superexpress checkout at the same time. Suppose the joint pmf of \(X_{1}\) and \(X_{2}\) is as given in the accompanying table. $$ \begin{array}{cc|cccc} & & 0 & 1 & 2 & 3 \\ \hline {}{}{x_{1}} & 0 & .08 & .07 & .04 & .00 \\ & 1 & .06 & .15 & .05 & .04 \\ & 2 & .05 & .04 & .10 & .06 \\ & 3 & .00 & .03 & .04 & .07 \\ & 4 & .00 & .01 & .05 & .06 \end{array} $$ a. What is \(P\left(X_{1}=1, X_{2}=1\right)\), that is, the probability that there is exactly one customer in each line? b. What is \(P\left(X_{1}=X_{2}\right)\), that is, the probability that the numbers of customers in the two lines are identical? c. Let \(A\) denote the event that there are at least two more customers in one line than in the other line. Express \(A\) in terms of \(X_{1}\) and \(X_{2}\), and calculate the probability of this event. d. What is the probability that the total number of customers in the two lines is exactly four? At least four?

Suppose the sediment density \((\mathrm{g} / \mathrm{cm})\) of a randomly selected specimen from a certain region is normally distributed with mean \(2.65\) and standard deviation \(.85\) (suggested in "Modeling Sediment and Water Column Interactions for Hydrophobic Pollutants," Water Research, 1984: 1169–1174). a. If a random sample of 25 specimens is selected, what is the probability that the sample average sediment density is at most 3.00? Between \(2.65\) and 3.00? b. How large a sample size would be required to ensure that the first probability in part (a) is at least .99?

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