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Chapter 5: Problem 95

Let the discrete random variables \(Y_{1}\) and \(Y_{2}\) have the joint probability function $$p\left(y_{1}, y_{2}\right)=1 / 3, \quad \text { for }\left(y_{1}, y_{2}\right)=(-1,0),(0,1),(1,0)$$ Find \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right)\). Notice that \(Y_{1}\) and \(Y_{2}\) are dependent. (Why?) This is another example of uncorrelated random variables that are not independent.

Short Answer

Expert verified
The covariance \(\operatorname{Cov}(Y_1, Y_2)\) is 0.

Step by step solution

01

Mean Calculation

Start by finding the mean of each random variable. The mean of a discrete random variable \(Y\) is calculated as \(E(Y) = \sum y \, P(Y=y)\).For \(Y_1\), we have:\[E(Y_1) = (-1)(1/3) + 0(1/3) + 1(1/3) = 0.\]For \(Y_2\), we have:\[E(Y_2) = 0(1/3) + 1(1/3) + 0(1/3) = 1/3.\]
02

Expected Value of Product

Next, find the expected value of the product of \(Y_1\) and \(Y_2\). This is calculated as \(E(Y_1 Y_2) = \sum (y_1 y_2) \, P(Y_1=y_1, Y_2=y_2)\).\[E(Y_1 Y_2) = (-1 \times 0)(1/3) + (0 \times 1)(1/3) + (1 \times 0)(1/3) = 0.\]
03

Covariance Calculation

Use the covariance formula \(\operatorname{Cov}(Y_1, Y_2) = E(Y_1 Y_2) - E(Y_1)E(Y_2)\).Substitute the values calculated in the previous steps:\[\operatorname{Cov}(Y_1, Y_2) = 0 - (0)(1/3) = 0.\]

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Key Concepts

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

Joint Probability Function
The joint probability function is a way to describe the probability of multiple random variables occurring at the same time. In the context of discrete random variables, it's essential to understand how each potential outcome of the variables affects the overall probability. When dealing with two variables, such as \(Y_1\) and \(Y_2\), the joint probability function \(p(y_1, y_2)\) gives you the probability that \(Y_1\) equals \(y_1\) and \(Y_2\) equals \(y_2\) simultaneously.
For the given exercise, we have the joint probability function:
  • \(p(-1, 0) = 1/3\)
  • \(p(0, 1) = 1/3\)
  • \(p(1, 0) = 1/3\)
This distribution demonstrates how each pair of \((y_1, y_2)\) occurs with equal likelihood. Joint probabilities are foundational for calculating other statistical measures, such as covariance, by allowing us to determine combined behavior of two discrete random variables.
Discrete Random Variables
Discrete random variables are types of variables that can take on a countable number of distinct values. They are essential in probability and statistics because they neatly lend themselves to counting processes and probability distributions.
For example, \(Y_1\) and \(Y_2\) in the exercise are discrete because they take on specific values from a set: \(-1, 0, 1\).
The key features of discrete random variables include:
  • They have a list of possible outcomes.
  • The probability of each outcome can be calculated.
  • The sum of all probabilities equals 1.
Understanding discrete random variables helps in easily calculating other statistics like mean and variance, and is crucial for grasping more complex topics such as joint probability functions and covariance.
Uncorrelated Random Variables
Uncorrelated random variables refer to pairs of variables where their covariance is zero. This concept indicates that there's no linear relationship between the two variables. In statistics, this is a straightforward way to assess the relationship between two random variables without assuming they are independent.
The exercise in question shows that \(Y_1\) and \(Y_2\) have a covariance of 0, as calculated via:\[\operatorname{Cov}(Y_1, Y_2) = E(Y_1 Y_2) - E(Y_1)E(Y_2) = 0 - (0)(1/3) = 0\]It's important to note that uncorrelated does not mean independent. Independence requires that the occurrence of one event doesn't affect the probability of the other, a stronger condition than being uncorrelated. In this exercise, \(Y_1\) and \(Y_2\) are dependent despite being uncorrelated because the definition of their distribution relies on both variables.

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Most popular questions from this chapter

If \(Y_{1}\) and \(Y_{2}\) are independent random variables, each having a normal distribution with mean 0 and variance 1, find the moment-generating function of \(U=Y_{1} Y_{2} .\) Use this moment-generating function to find \(E(U)\) and \(V(U)\). Check the result by evaluating \(E(U)\) and \(V(U)\) directly from the density functions for \(Y_{1}\) and \(Y_{2}\)

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.

Suppose that the number of eggs laid by a certain insect has a Poisson distribution with mean \(\lambda\). The probability that any one egg hatches is \(p .\) Assume that the eggs hatch independently of one another. Find the a. expected value of \(Y\), the total number of eggs that hatch. b. variance of \(Y\).

If \(c\) is any constant and \(Y\) is a random variable such that \(E(Y)\) exists, show that \(\operatorname{Cov}(c, Y)=0\).

In Exercise \(5.3,\) we determined that the joint probability distribution of \(Y_{1}\), the number of married executives, and \(Y_{2},\) the number of never- married executives, is given by $$p\left(y_{1}, y_{2}\right)=\frac{\left(\begin{array}{c}4 \\\y_{1}\end{array}\right)\left(\begin{array}{c}3 \\ y_{2}\end{array}\right)\left(\begin{array}{c}2 \\\3-y_{1}-y_{2}\end{array}\right)}{\left(\begin{array}{l}9 \\\3 \end{array}\right)}$$ where \(y_{1}\) and \(y_{2}\) are integers, \(0 \leq y_{1} \leq 3,0 \leq y_{2} \leq 3,\) and \(1 \leq y_{1}+y_{2} \leq 3 .\) Find \(\operatorname{Cov}\left(Y_{1}, Y_{2}\right)\).
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