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Recall that two events \(A\) and \(B\) are called independent if \(p(A B)=p(A) p(B) .\) Similarly two random variables \(x\) and \(y\) are called independent if the joint probability function \(f(x, y)=g(x) h(y) .\) Show that if \(x\) and \(y\) are independent, then the expectation or average of \(x y\) is \(E(x y)=E(x) E(y)=\mu_{x} \mu_{y}\).

Short Answer

Expert verified
If X and Y are independent, then \(E(XY) = E(X)E(Y) = \mu_x \mu_y\).

Step by step solution

01

Define Independence of Random Variables

Given two random variables, X and Y, they are independent if their joint probability function is the product of their individual probability functions: \(f(x, y) = g(x) h(y)\).
02

Define Expectation of a Function of Random Variables

The expectation of the product of X and Y is defined as \(E(XY)\). By definition, this means integrating the product \(x * y\) over the joint probability density function \(f(x, y)\):\[E(XY) = \int \int xy f(x, y) \, dx \, dy\]
03

Substitute Independence Condition

Given that X and Y are independent, substitute \(f(x, y) = g(x) h(y)\) into the expectation formula:\[E(XY) = \int \int xy g(x) h(y) \, dx \, dy\]
04

Separate the Integrals

Separate the double integral into the product of two single integrals: \[E(XY) = \int x g(x) \, dx \int y h(y) \, dy\]
05

Recognize Each Integral as an Expectation

Recognize that \(\int x g(x) \, dx\) is the definition of the expected value of X, \(E(X) = \mu_x\), and \(\int y h(y) \, dy\) is the expected value of Y, \(E(Y) = \mu_y\). Thus,\[E(XY) = E(X) \, E(Y) = \mu_x \mu_y\]

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Key Concepts

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

Joint Probability Function
The concept of a joint probability function is essential when dealing with multiple random variables. Simply put, it's a function that gives us the probability that each of the variables falls within a specific range or takes on certain values simultaneously. For two random variables, X and Y, their joint probability function is denoted as \(f(x, y)\). If X and Y are independent, the joint probability function can be expressed as the product of their individual probability functions: \(f(x, y) = g(x) h(y)\). This means that the probability of X and Y occurring together is just the product of their separate probabilities. The joint probability function is crucial for understanding interactions between variables, particularly when determining their combined behavior.
Expectation of a Function
The expectation of a function of random variables provides us with the long-term average or mean value you would expect if an experiment were repeated many times. For example, the expectation of the product of X and Y, denoted as \(E(XY)\), is simply the average value of the product over all possible values of X and Y. Mathematically, this is expressed as \int \int xy f(x, y) \, dx \, dy\
.
This integral sums up the product \(xy\) weighted by their joint probabilities over the possible values of X and Y. For independent variables, substituting \(f(x, y) = g(x)h(y)\) allows breaking down the process into simpler, separate calculations for each variable.
Integration of Probability Density
The integration of a probability density function helps us find the expectation of a random variable or any function involving random variables. If you want to calculate the expectation of the product of independent random variables X and Y, you'd integrate their product over the joint density function:

\(E(XY) = \int \int xy f(x, y) \, dx \, dy\).

Substituting \(f(x, y) = g(x)h(y)\) since X and Y are independent, we get:

\(E(XY) = \int \int xy g(x)h(y) \, dx \, dy\).

The integration separates into:

\(E(XY) = ( \int x g(x) \, dx ) ( \int y h(y) \, dy )\).

Essentially, this breaks the problem into finding the expectations of X and Y separately and then multiplying these results. Integration in this context simplifies computing expectations for independent variables.
Expected Value
The expected value or expectation of a random variable is a fundamental concept in probability and statistics. It gives you a measure of the 'central' tendency or the mean value you'd anticipate over many trials. For any random variable X, its expected value, \(E(X)\), is computed as: \(E(X) = \int x g(x) \, dx\).

Similarly, for Y, the expected value is: \(E(Y) = \int y h(y) \, dy \).

If X and Y are independent random variables, the expected value of their product is simply the product of their individual expected values:

\(E(XY) = E(X)E(Y) = \mu_x \mu_y\).

This result is very useful when dealing with complex systems as it allows for simplifying calculations assuming independence of the involved variables. Understanding expected values is key to making predictions and informed decisions based on probabilistic models.

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Most popular questions from this chapter

A thick coin has probability \(\frac{3}{7}\) of falling heads, \(\frac{3}{7}\) of falling tails, and \(\frac{1}{7}\) of standing on edge. Show that if it is tossed repeatedly it has probability 1 of eventually standing on edge.

Two people are taking turns tossing a pair of coins; the first person to toss two alike wins. What are the probabilities of winning for the first player and for the second player? Hint: Although there are an infinite number of possibilities here (win on first turn, second turn, third turn, etc.), the sum of the probabilities is a geometric series which can be summed; see Chapter 1 if necessary.

Given a family of two children (assume boys and girls equally likely, that is, probability 1/2 for each), what is the probability that both are boys? That at least one is a girl? Given that at least one is a girl, what is the probability that both are girls? Given that the first two are girls, what is the probability that an expected third child will be a boy?

Let \(m_{1}, m_{2}, \cdots, m_{n}\) be a set of measurements, and define the values of \(x_{i}\) by \(x_{1}=\) \(m_{1}-a, x_{2}=m_{2}-a, \cdots, x_{n}=m_{n}-a,\) where \(a\) is some number (as yet unspecified, but the same for all \(x_{i}\) ). Show that in order to minimize \(\sum_{i=1}^{n} x_{i}^{2},\) we should choose \(a=(1 / n) \sum_{i=1}^{n} m_{i} .\) Hint: Differentiate \(\sum_{i=1}^{n} x_{i}^{2}\) with respect to \(a .\) You have shown that the arithmetic mean is the "best" average in the least squares sense, that is, that if the sum of the squares of the deviations of the measurements from their "average" is a minimum, the "average" is the arithmetic mean (rather than, say, the median or mode).

A basketball player succeeds in making a basket 3 tries out of 4. How many tries are necessary in order to have probability \(>0.99\) of at least one basket?

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