/*! 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 84 We considered two individuals wh... [FREE SOLUTION] | 91Ó°ÊÓ

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We considered two individuals who each tossed a coin until the first head appears. Let \(Y_{1}\) and \(Y_{2}\) denote the number of times that persons \(A\) and \(B\) toss the coin, respectively. If heads occurs with probability \(p\) and tails occurs with probability \(q=1-p,\) it is reasonable to conclude that \(Y_{1}\) and \(Y_{2}\) are independent and that each has a geometric distribution with parameter p. Consider \(Y_{1}-Y_{2}\), the difference in the number of tosses required by the two individuals. a. Find \(E\left(Y_{1}\right), E\left(Y_{2}\right),\) and \(E\left(Y_{1}-Y_{2}\right)\) b. Find \(E\left(Y_{1}^{2}\right), E\left(Y_{2}^{2}\right),\) and \(E\left(Y_{1} Y_{2}\right)\) (recall that \(Y_{1}\) and \(Y_{2}\) are independent). c. Find \(E\left(Y_{1}-Y_{2}\right)^{2}\) and \(V\left(Y_{1}-Y_{2}\right)\) d. Give an interval that will contain \(Y_{1}-Y_{2}\) with probability at least \(8 / 9\)

Short Answer

Expert verified
\(E(Y_1)=E(Y_2)=\frac{1}{p}\), \(E(Y_1-Y_2)=0\); \(V(Y_1-Y_2)=\frac{4-2p}{p^2}\); Interval: \((-3\sqrt{\frac{4-2p}{p^2}}, 3\sqrt{\frac{4-2p}{p^2}})\).

Step by step solution

01

Determine Expectation of Geometric Distribution

The expectation or mean of a geometric random variable with parameter \(p\) is given by \(E(Y) = \frac{1}{p}\). Thus, \(E(Y_1) = \frac{1}{p}\) and \(E(Y_2) = \frac{1}{p}\) due to symmetry and independence.
02

Compute Expectation of the Difference

Since \(Y_1\) and \(Y_2\) are independent, the expectation of the difference is the difference of the expectations: \[E(Y_1 - Y_2) = E(Y_1) - E(Y_2) = \frac{1}{p} - \frac{1}{p} = 0.\]
03

Compute Second Moments of Geometric Distribution

The second moment of a geometric random variable is \(E(Y^2) = \frac{2-p}{p^2}\). Therefore, \[E(Y_1^2) = \frac{2-p}{p^2}\] and \(E(Y_2^2) = \frac{2-p}{p^2}\).
04

Compute Expectation of Product of Independent Variables

For independent variables, \(E(Y_1 Y_2) = E(Y_1)E(Y_2)\). Therefore, \[E(Y_1 Y_2) = \frac{1}{p}\times\frac{1}{p} = \frac{1}{p^2}.\]
05

Compute the Second Moment of the Difference

Using the formula \((Y_1-Y_2)^2 = Y_1^2 + Y_2^2 - 2Y_1Y_2\), the expectation is:\[(E(Y_1 - Y_2)^2 = E(Y_1^2) + E(Y_2^2) - 2E(Y_1Y_2)\]Substitute the values from previous steps:\[= \frac{2-p}{p^2} + \frac{2-p}{p^2} - 2\cdot \frac{1}{p^2} = \frac{4-2p}{p^2}.\]
06

Compute Variance of the Difference

Variance is the second moment minus the square of the expectation.\[V(Y_1-Y_2) = E((Y_1-Y_2)^2) - (E(Y_1-Y_2))^2\]\[= \frac{4-2p}{p^2} - 0^2 = \frac{4-2p}{p^2}.\]
07

Find the Confidence Interval

To find an interval that contains \(Y_1 - Y_2\) with probability at least \(8/9\), you can use Chebyshev's inequality. For any random variable \(Z\) with mean \(\mu\) and variance \(\sigma^2\), \[P(|Z-\mu| \geq k\sigma) \leq \frac{1}{k^2}\].Setting \(\frac{1}{k^2} = \frac{1}{9}\) gives \(k = 3\). Therefore, \[Y_1 - Y_2 \in \left(\mu - 3\sigma, \mu + 3\sigma\right)\] with probability at least \(8/9\).Thus, since \(\mu = 0\) and \(\sigma^2 = \frac{4-2p}{p^2}\), \(\sigma = \sqrt{\frac{4-2p}{p^2}}\), we have:\[Y_1 - Y_2 \in \left(-3\sqrt{\frac{4-2p}{p^2}}, 3\sqrt{\frac{4-2p}{p^2}}\right)\].

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Key Concepts

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

Expectation
Expectation, often referred to as the mean, is a fundamental concept in probability and statistics. For a random variable, expectation gives us the average or expected value it would take on upon many repetitions of the same experiment. In the context of a geometric distribution, which measures how many trials it takes to get the first success, the expected value can be easily calculated using the formula:\[E(Y) = \frac{1}{p}\]where \(p\) is the probability of success on a single trial.
  • If \(p = 0.5\), meaning the probability of heads is 50%, then the expectation is \(E(Y) = 2\).
  • A smaller \(p\) implies higher expectation since it might take more tries to get the first head.
When dealing with two independent random variables, like \(Y_1\) and \(Y_2\) in the given example, it becomes straightforward to evaluate the expectation of their difference:The expectation of the difference, \(E(Y_1 - Y_2)\), is calculated as:\[E(Y_1 - Y_2) = E(Y_1) - E(Y_2) = \frac{1}{p} - \frac{1}{p} = 0\]This shows that, on average, there is no difference in the number of tosses required by both individuals to get the first head.
Variance
Variance provides a measure of how spread out the values of a random variable can be. Specifically, variance tells us the average of the squared differences from the mean; it answers the question, "how much do values deviate from expectation?"For a geometric distribution, variance is calculated with:\[V(Y) = \frac{1-p}{p^2}\]This represents how much variability exists in the number of trials before a success, due to potential variations in the number of failures before a success is achieved. If \(p\) is small, the variance becomes large because it might take many more trials to get a first success.In the case of two independent geometric variables, \(Y_1\) and \(Y_2\), we often are interested in the variance of their difference, \(V(Y_1 - Y_2)\).The formula utilized:\[V(Y_1-Y_2) = \frac{4-2p}{p^2}\]This formula indicates that variance will depend on the probability \(p\), with smaller values of \(p\) leading to greater uncertainty or spread in the difference between the trials for each person.
Chebyshev's Inequality
Chebyshev's Inequality is a key tool in probability and statistics that provides a bound on how likely a random variable's value is to be significantly distant from its mean. This inequality is very useful for determining "how much" data lies within a specific range from the mean, regardless of the distribution type.For any random variable \(Z\), with expectation \(\mu\) and variance \(\sigma^2\), Chebyshev's Inequality states:\[P(|Z - \mu| \geq k\sigma) \leq \frac{1}{k^2}\]This tells us that, at least \(1 - \frac{1}{k^2}\) of the values are within \(k\) standard deviations of the mean.
  • To achieve a probability of at least \(\frac{8}{9}\), we set \(\frac{1}{k^2} = \frac{1}{9}\), giving \(k = 3\).
  • This implies that, for \(Y_1 - Y_2\), at least \(\frac{8}{9}\) of values should lie within three standard deviations from the mean, \(0\).
This makes Chebyshev's Inequality a powerful, non-specific tool to estimate probabilities of intervals containing the random variable.
Independent Random Variables
Understanding independent random variables is essential in probability, as they describe variables whose occurrences do not influence or affect each other. In practice, this implies that knowing the outcome of one variable provides no information about the outcome of the other.When working with independent variables \(Y_1\) and \(Y_2\) in our example, many simplifications emerge:
  • The expectation of their product is simply the product of their expectations, \(E(Y_1 Y_2) = E(Y_1)E(Y_2)\).
  • This means \(E(Y_1 Y_2) = \frac{1}{p^2}\) in our case.
  • The variance of the difference \(V(Y_1 - Y_2)\) also benefits from this independence, as it is calculated by treating the two variables separately before combining their variances.
This independence simplifies many of the calculations, allowing us to directly apply results from individual variables to the combined or derived variables like \(Y_1 - Y_2\). This principle is why independence is a significant and frequently leveraged assumption in statistical problems and analyses.

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Most popular questions from this chapter

A quality control plan calls for randomly selecting three items from the daily production (assumed large) of a certain machine and observing the number of defectives. However, the proportion \(p\) of defectives produced by the machine varies from day to day and is assumed to have a uniform distribution on the interval (0,1) . For a randomly chosen day, find the unconditional probability that exactly two defectives are observed in the sample.

In Exercise 5.9 we determined that $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll} 6\left(1-y_{2}\right), & 0 \leq y_{1} \leq y_{2} \leq 1 \\ 0, & \text { elsewhere } \end{array}\right.$$ is a valid joint probability density function. Find $$\text { a. } E\left(Y_{1}\right) \text { and } E\left(Y_{2}\right)$$ $$\text { b. } V\left(Y_{1}\right) \text { and } V\left(Y_{2}\right)$$ $$\text { c. } E\left(Y_{1}-3 Y_{2}\right)$$

When commercial aircraft are inspected, wing cracks are reported as nonexistent, detectable, or critical. The history of a particular fleet indicates that \(70 \%\) of the planes inspected have no wing cracks, \(25 \%\) have detectable wing cracks, and \(5 \%\) have critical wing cracks. Five planes are randomly selected. Find the probability that a. one has a critical crack, two have detectable cracks, and two have no cracks. b. at least one plane has critical cracks.

If \(Y_{1}\) is the total time between a customer's arrival in the store and leaving the service window and if \(Y_{2}\) is the time spent in line before reaching the window, the joint density of these variables, according to Exercise 5.15 , is $$f\left(y_{1}, y_{2}\right)=\left\\{\begin{array}{ll}e^{-y_{1}}, & 0 \leq y_{2} \leq y_{1} \leq \infty \\\0, & \text { elsewhere }\end{array}\right.$$ Are \(Y_{1}\) and \(Y_{2}\) independent?

The number of defects per yard \(Y\) for a certain fabric is known to have a Poisson distribution with parameter \lambda. However, \lambda itself is a random variable with probability density function given by $$f(\lambda)=\left\\{\begin{array}{ll} e^{-\lambda}, & \lambda \geq 0 \\ 0, & \text { elsewhere } \end{array}\right.$$ Find the unconditional probability function for \(Y\)

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