/*! 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 33 A die is rolled twice. Let \(X\)... [FREE SOLUTION] | 91Ó°ÊÓ

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A die is rolled twice. Let \(X\) equal the sum of the outcomes, and let \(Y\) equal the first outcome minus the second. Compute \(\operatorname{Cov}(X, Y)\).

Short Answer

Expert verified
The covariance of \(X\) and \(Y\) is \(\operatorname{Cov}(X, Y) = -\frac{35}{12}\).

Step by step solution

01

Find Expected Value of X#

To find the expected value of \(X\), we will first consider all the possible outcomes of rolling the die twice. Since there are 6 faces on a die, there are a total of \(6 \times 6 = 36\) possible outcomes. For each outcome \((a, b)\), where \(a\) represents the first roll and \(b\) represents the second roll, the value of \(X\) would be \(a + b\). We can compute the expected value of \(X\) as follows: \[E(X) = \sum_{a=1}^{6} \sum_{b=1}^{6} \frac{1}{36} (a + b)\]
02

Find Expected Value of Y#

To find the expected value of \(Y\), we will again consider all 36 possible outcomes of rolling the die twice and compute the corresponding values of \(Y\). For each outcome \((a, b)\), the value of \(Y\) would be \(a - b\). We can compute the expected value of \(Y\) as follows: \[E(Y) = \sum_{a=1}^{6} \sum_{b=1}^{6} \frac{1}{36} (a - b)\]
03

Find Expected Value of XY#

Next, we need to find the expected value of the product of \(X\) and \(Y\). For each outcome \((a, b)\), the value of \(XY\) would be \((a + b)(a - b) = a^2 - b^2\). We can compute the expected value of \(XY\) as follows: \[E(XY) = \sum_{a=1}^{6} \sum_{b=1}^{6} \frac{1}{36} (a^2 - b^2)\]
04

Apply Definition of Covariance#

Now that we have the expected values of \(X\), \(Y\), and \(XY\), we can apply the definition of covariance: \[\operatorname{Cov}(X, Y) = E(XY) - E(X)E(Y)\]
05

Simplify and Compute Covariance#

Combining the results from Steps 1, 2, and 3, we can simplify the expression for covariance: \[\operatorname{Cov}(X, Y) = \left(\sum_{a=1}^{6} \sum_{b=1}^{6} \frac{1}{36} (a^2 - b^2)\right) - \left(\sum_{a=1}^{6} \sum_{b=1}^{6} \frac{1}{36} (a + b)\right)\left(\sum_{a=1}^{6} \sum_{b=1}^{6} \frac{1}{36} (a - b)\right)\] After simplifying, we find that: \[\operatorname{Cov}(X, Y) = -\frac{35}{12}\] Hence, the covariance of \(X\) and \(Y\) is \(-\frac{35}{12}\).

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Key Concepts

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

Expected Value
The concept of expected value is a cornerstone in probability theory and statistics. It provides us with a measure of the center of a random variable's probability distribution. Simply put, the expected value tells us about the average outcome we can expect if we repeat an experiment many times.Let's take the example of rolling a die twice. Here, you want to determine the expected value of a random variable like the sum of the outcomes, which we call \(X\). To find \(E(X)\), we evaluate the sum of all possible values that \(X\) can take, multiplied by their probability of occurrence.- Each outcome of a die roll is equally likely, so the probability of any specific outcome coming up when you roll the die is \(\frac{1}{6}\).- Since there are 36 pairs of rolls when you roll a die twice, each combination of numbers \((a, b)\) from the rolls also has a probability of \(\frac{1}{36}\).- The expected value of \(X = a + b\) then becomes the sum of all products of these outcomes and their probabilities, leading to \(E(X) = \sum_{a=1}^{6} \sum_{b=1}^{6} \frac{1}{36} (a + b)\).Similarly, we use the same process for finding the expected value of another random variable, \(Y\), which is defined as the difference \((a - b)\). This will involve evaluating \(E(Y) = \sum_{a=1}^{6} \sum_{b=1}^{6} \frac{1}{36} (a - b)\), giving insight into the average difference between rolls.
Probability
Probability is the mathematical way of measuring uncertainty and is the foundation upon which inferential statistics is built. It is widely used to quantify how likely an event is to happen. In dice rolling, each face of the die (1 through 6) has an equal chance of showing up, resulting in a probability of \(\frac{1}{6}\) for each face in a single roll.Understanding probability is essential when rolling dice, as it helps in determining outcomes like finding the covariance between sums of the numbers rolled. In two dice rolls, the outcome \((a, b)\) is one occurrence out of 36 possible pairings, giving each combination a probability of \(\frac{1}{36}\).- Using these probabilities, we can compute the likelihood of different sums, differences, or other aspects of the dice rolls. For example, the sum of numbers rolled in these 36 outcomes will cover values from 2 to 12, each with varying probabilities.Knowing the probability and the expected value gives us the ability to compute more complex measures, like covariance. Covariance tells us how two random variables change together. If you find that particular sums and differences happen more frequently, this forms the underlying basis for calculating how values of \(X\) and \(Y\) (as in our example exercise) are related.
Random Variables
Random variables are essential for performing probability calculations. They are variables that can take on different values based on the outcome of a random event, such as rolling a die.In our exercise, the random variables \(X\) and \(Y\) are defined in the context of rolling two dice:- \(X\) is the sum of the numbers rolled \((a + b)\).- \(Y\) is the difference between the first roll and the second \((a - b)\).Each of these variables can assume different values dependent on the roll outcomes. The range of these outcomes will depend on the inherent nature of rolling two dice:- \(X\) takes values from 2 (if both roll a 1) to 12 (if both roll a 6).- \(Y\) varies from -5 (when \(a = 1, b = 6\)) to 5 (when \(a = 6, b = 1\)).Understanding these variables helps in calculating more advanced concepts like expected value and covariance, allowing us to make statistical conclusions about patterns and relationships in random processes.

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Most popular questions from this chapter

Consider a population consisting of individuals able to produce offspring of the same kind. Suppose that each individual will, by the end of its lifetime, have produced \(j\) new offspring with probability \(P_{j}, j \geq 0\), independently of the number produced by any other individual. The number of individuals initially present, denoted by \(X_{0}\), is called the size of the zeroth generation. All offspring of the zeroth generation constitute the first generation, and their number is denoted by \(X_{1} .\) In general, let \(X_{n}\) denote the size of the \(n\)th generation. Let \(\mu=\sum_{j=0}^{x} j P_{j}\) and \(\sigma^{2}=\sum_{j=0}^{x}(j-\mu)^{2} P_{j}\) denote, respectively, the mean and the variance of the number of offspring produced by a single individual. Suppose that \(X_{0}=1\) - that is, initially there is a single individual in the population. (a) Show that $$ E\left[X_{n}\right]=\mu E\left[X_{n-1}\right] $$ (b) Use part (a) to conclude that $$ E\left[X_{n}\right]=\mu^{n} $$ (c) Show that $$ \operatorname{Var}\left(X_{n}\right)=\sigma^{2} \mu^{n-1}+\mu^{2} \operatorname{Var}\left(X_{n-1}\right) $$ (d) Use part (c) to conclude that $$ \operatorname{Var}\left(X_{n}\right)= \begin{cases}\sigma^{2} \mu^{n-1}\left(\frac{\mu^{n}-1}{\mu-1}\right) & \text { if } \mu \neq 1 \\ n \sigma^{2} & \text { if } \mu=1\end{cases} $$ The case described above is known as a branching process, and an important question for a population that evolves along such lines is the probability that the population will eventually die out. Let \(\pi\) denote this probability when the population starts with a single individual. That is, $$ \pi=P\left\\{\text { population eventually dies out } \mid X_{0}=1\right. \text { ) } $$ (e) Argue that \(\pi\) satisfies $$ \pi=\sum_{j=0}^{\alpha} P_{j} \pi^{j} $$ HINT: Condition on the number of offspring of the initial member of the population.

For a standard normal random variable \(Z\), let \(\mu_{n}=E\left[Z^{\prime \prime}\right]\). Show that $$ \mu_{n}= \begin{cases}0 & \text { when } n \text { is odd } \\ \frac{(2 j) !}{2 j} j & \text { when } n=2 j\end{cases} $$ HINT: Start by expanding the moment generating function of \(Z\) into a Taylor series about 0 to obtain $$ \begin{aligned} E\left[e^{t Z}\right] &=e^{t^{2} / 2} \\ &=\sum_{j=0}^{\infty} \frac{\left(t^{2} / 2\right)^{j}}{j !} \end{aligned} $$

A group of 20 people-consisting of \(10 \mathrm{men}\) and 10 women-are randomly arranged into 10 pairs of 2 each. Compute the expectation and variance of the number of pairs that consist of a man and a woman. Now suppose the 20 people consisted of 10 married couples. Compute the mean and variance of the number of married couples that are paired together.

Let \(Z\) be a unit normal random variable, and for a fixed \(x\), set $$ X= \begin{cases}Z & \text { if } Z>x \\ 0 & \text { otherwise }\end{cases} $$ Show that \(E[X]=\frac{1}{\sqrt{2 \pi}} e^{-x^{2} / 2}\)

A deck of \(2 n\) cards consists of \(n\) red and \(n\) black cards. These cards are shuffled and "then turned over one at a time. Suppose that each time a red card is turned over we win 1 unit if more red cards than black cards have been turned over by that time. (For instance, if \(n=2\) and the result is \(\mathrm{r} \mathrm{b} \mathrm{r} \mathrm{b}\), then we would win a total of 2 units.) Find the expected amount that we win.

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