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Give an example of two random variables \(X\) and \(Y\) such that \(E(X Y) \neq E(X) E(Y)\). Here \(X Y\) is the random variable with \((X Y)(s)=X(s) Y(s)\)

Short Answer

Expert verified
Example: Roll a fair die; let \(X\) be the outcome and \(Y\) be 1 if \(X\) is even, else 0. Here, \(E(XY) = 2\) and \(E(X)E(Y) = 1.75\).

Step by step solution

01

Understand the Expectation Definition

Before we dive in, let's understand what we are aiming to demonstrate. We want to show that for two specific random variables, their joint expectation is different from the product of their expectations. This typically implies that the two variables are not independent.
02

Choose Random Variables

Let's take an example involving two discrete random variables: let a fair die be rolled once. Define random variable \(X\) as the outcome of the die roll, and random variable \(Y\) as 1 if the outcome is even and 0 if the outcome is odd. So, \(Y = 1\) if \(X = 2, 4, 6\) and \(Y = 0\) if \(X = 1, 3, 5\).
03

Calculate \(E(X)\) and \(E(Y)\)

Calculate the expected value of \(X\):\[ E(X) = \frac{1}{6}(1 + 2 + 3 + 4 + 5 + 6) = \frac{21}{6} = 3.5 \]Calculate the expected value of \(Y\):\[ E(Y) = \frac{1}{6}(0 + 1 + 0 + 1 + 0 + 1) = \frac{3}{6} = 0.5 \]
04

Calculate \(E(XY)\)

Now, compute the expectation for the product of \(X\) and \(Y\):- \(X \cdot Y = 0\) for \(X = 1, 3, 5\)- \(X \cdot Y = 2\) for \(X = 2\), \(X \cdot Y = 4\) for \(X = 4\), and \(X \cdot Y = 6\) for \(X = 6\)Therefore, \[ E(XY) = \frac{1}{6}(0 + 2 + 0 + 4 + 0 + 6) = \frac{12}{6} = 2 \]
05

Conclude \(E(XY) \neq E(X)E(Y)\)

Finally, compare the two calculated expectations:- \(E(XY) = 2\)- \(E(X)E(Y) = 3.5 \times 0.5 = 1.75\)It is clear that \(E(XY) eq E(X)E(Y)\). This shows that the two random variables \(X\) and \(Y\) are not independent.

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Key Concepts

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

Random Variables
Random variables are fundamental concepts in the world of probability and statistics. They provide a way to quantify uncertain outcomes. A random variable assigns a numerical value to each outcome in a probability experiment, effectively transforming random phenomena into manageable quantities.
For a discrete random variable, the values it can assume are countable, like the outcome of a dice roll. In our example, the random variable \(X\) represents the outcome of rolling a six-sided die. It can take any integer value between 1 and 6.
Recently, we introduced another random variable, \(Y\), which is dependent on \(X\). \(Y\) equals 1 if the die shows an even number (2, 4, or 6) and 0 otherwise (for odds 1, 3, 5). The nature of \(Y\) helps to highlight the concept of dependency, where its value directly hinges on \(X\).
Understanding these variables is crucial as they are used to compute expectations and dependencies, forming the backbone of many statistical analyses.
Independence of Random Variables
When we talk about the independence of random variables, we're referring to a very important property in probability. Two random variables \(X\) and \(Y\) are independent if the occurrence of events related to \(X\) gives no information about events related to \(Y\), and vice versa.
For two independent random variables, the joint expectation \(E(XY)\) should equal the product of their individual expectations \(E(X)E(Y)\).
In our problem, however, \(X\) and \(Y\) are dependent. This is demonstrated because the expectation of \(XY\) is not equal to the product \(E(X)E(Y)\). Specifically, we found \(E(XY) = 2\), while \(E(X)E(Y) = 1.75\). This discrepancy proves that \(X\) and \(Y\) rely on each other instead of being independent.
Independence is a key concept when determining whether results in probability experiments influence each other.
Discrete Probability Distribution
A discrete probability distribution is a list of probabilities associated with each of the possible outcomes of a discrete random variable. It shows how the total 100% probability is spread across all the outcomes.
In the context of a die roll, each number from 1 to 6 has an equal probability of \(\frac{1}{6}\). This forms the probability distribution for \(X\), which assigns a probability to rolling each number.
As for \(Y\), the probability distribution differs as it only distinguishes whether the result of \(X\) is even or odd. Here, both \(Y=0\) and \(Y=1\) still have equal probabilities of \(\frac{1}{2}\) because there are three of each count (evens and odds) on the die.
Discrete probability distributions are crucial as they provide the mathematical foundation needed to compute various probabilities and expectations in practical scenarios.

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