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Chapter 4: Problem 15

Let \(Y_{1}

Short Answer

Expert verified
The expectation \(E(Y_{1})\) is \(-\sigma / \sqrt{\pi}\) and the covariance of \(Y_{1}\) and \(Y_{2}\) can be calculated using the formula for covariance and the expectation values of \(Y_{1}\) and \(Y_{2}\).

Step by step solution

01

Calculate \(E(Y_{1})\) using joint pdf

The joint pdf of \(Y_{1}\) and \(Y_{2}\) can be given as follows considering \(N(0, \sigma^2)\): \[f_{Y_{1}, Y_{2}}(y_{1}, y_{2}) = \frac{{2e^{-\frac{{y_{1}^{2} + y_{2}^{2}}}{{2\sigma^{2}}}}}}{\sqrt{2\pi\sigma^{2}}}. \] Using this joint pdf, the expectation \(E(Y_{1})\) can be calculated as: \[E(Y_{1}) = \int_{-\infty}^{y_{2}}y_{1}f_{Y_{1}, Y_{2}}(y_{1}, y_{2})dy_{1}, \] which simplifies to \[E(Y_{1}) = \int_{-\infty}^{y_{2}}\frac{{-2y_{1}e^{-\frac{{y_{1}^{2} + y_{2}^{2}}}{{2\sigma^{2}}}}}}{\sqrt{2\pi\sigma^{2}}}dy_{1}. \] Integrating and simplifying gives \(E(Y_{1}) = -\sigma / \sqrt{\pi}\).
02

Calculate covariance of \(Y_{1}\) and \(Y_{2}\)

Covariance of \(Y_{1}\) and \(Y_{2}\) can be given as \[Cov(Y_{1}, Y_{2}) = E(Y_{1}Y_{2}) - E(Y_{1})E(Y_{2}).\] Since \(Y_{1}\) and \(Y_{2}\) are from \(N(0, \sigma^2)\), it is known that \(E(Y_{1}) = E(Y_{2})\), which makes their covariance to be \[Cov(Y_{1}, Y_{2}) = E(Y_{1}Y_{2}) - [E(Y_{1})]^2.\] Simplifying this gives the covariance of \(Y_{1}\) and \(Y_{2}\).

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Key Concepts

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

Expectation
In probability and statistics, the concept of "expectation" is a fundamental measure which provides the average or mean value of a random variable when it is repeatedly sampled. For the order statistics, specifically for the smallest value in a sample, expectation helps to find out where our central tendency lies when the data is arranged.
  • Order Statistics: In our exercise, we are dealing with order statistics, specifically with the random sample's smallest value, denoted as \( Y_{1} \).
  • Using Joint PDF: To calculate the expectation of the smallest value \( E(Y_{1}) \), the joint probability density function (pdf) of \( Y_{1} \) and \( Y_{2} \) is employed. The joint pdf for two-order statistics from a normal distribution \( N(0, \sigma^{2}) \) is essential because it captures the distribution of two order statistics simultaneously.
  • Integration Process: By integrating over the joint pdf first with respect to \( y_{1} \), the expectation \( E(Y_{1}) \) simplifies to \(-\sigma / \sqrt{\pi}\). This integration takes into account all possible values of \( y_{1} \) when \( y_{1} \leq y_{2} \).
Understanding expectation within the context of order statistics provides insight into the behavior of the smallest sample values under repeated trials.
Covariance
Covariance is a measure of how two random variables change together. It provides insight into the direction and strength of their relationship. If the covariance is positive, the variables tend to increase together; if negative, one tends to increase as the other decreases.
  • Order Statistics Understanding: In our scenario, \( Y_{1} \) and \( Y_{2} \) are order statistics, meaning they represent the smallest and largest values in a sample of two.
  • Formula for Covariance: Given by \( Cov(Y_{1}, Y_{2}) = E(Y_{1}Y_{2}) - E(Y_{1})E(Y_{2}) \), it requires computing the expectations of \( Y_{1} \), \( Y_{2} \), and their product.
  • Symmetry and Simplification: Due to the symmetry of the normal distribution \( N(0, \sigma^2) \), \( E(Y_{1}) = E(Y_{2}) \). This symmetry helps in simplifying the calculation of covariance, focusing on the cross-product expectation \( E(Y_{1}Y_{2}) \).
Covariance in the context of order statistics gives us a quantitative assessment of dependency between ordered values from sampled data.
Joint Probability Density Function
The Joint Probability Density Function (PDF) is a function used to define the likelihood of two random variables occurring simultaneously. For order statistics, especially when derived from continuous distributions, the joint PDF plays a crucial role.
  • Multiple Variables Consideration: Since we deal with \( Y_{1} \) and \( Y_{2} \), the joint PDF allows us to handle both together, reflecting their combined distribution.
  • Calculation in Context: In the exercise, the joint PDF was used to determine various statistics like expectation and covariance. It specifically helped in deriving \( E(Y_{1}) \).
  • Expression of Joint PDF: The joint PDF as provided \( \frac{2e^{-\frac{y_{1}^2 + y_{2}^2}{2\sigma^2}}}{\sqrt{2\pi\sigma^2}} \) indicates how likely pairs \( (y_{1}, y_{2}) \) are to occur given the normal distribution \( N(0, \sigma^2) \).
Understanding the joint PDF is foundational when analyzing the relationship and interaction between random variables, giving a fuller picture of the statistical scenario at hand.

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Most popular questions from this chapter

Let \(X_{1}, X_{2}, \ldots, X_{9}\) be a random sample of size 9 from a distribution that is \(N\left(\mu, \sigma^{2}\right)\) (a) If \(\sigma\) is known, find the length of a \(95 \%\) confidence interval for \(\mu\) if this interval is based on the random variable \(\sqrt{9}(\bar{X}-\mu) / \sigma\) (b) If \(\sigma\) is unknown, find the expected value of the length of a \(95 \%\) confidence interval for \(\mu\) if this interval is based on the random variable \(\sqrt{9}(\bar{X}-\mu) / S\). Hint: \(\quad\) Write \(E(S)=(\sigma / \sqrt{n-1}) E\left[\left((n-1) S^{2} / \sigma^{2}\right)^{1 / 2}\right]\). (c) Compare these two answers.

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For the proof of Theorem 4.8.1, we assumed that the cdf was strictly increasing over its support. Consider a random variable \(X\) with cdf \(F(x)\) that is not strictly increasing. Define as the inverse of \(F(x)\) the function $$ F^{-1}(u)=\inf \\{x: F(x) \geq u\\}, \quad 0

In Exercise \(4.2 .18\) we found a confidence interval for the variance \(\sigma^{2}\) using the variance \(S^{2}\) of a random sample of size \(n\) arising from \(N\left(\mu, \sigma^{2}\right)\), where the mean \(\mu\) is unknown. In testing \(H_{0}: \sigma^{2}=\sigma_{0}^{2}\) against \(H_{1}: \sigma^{2}>\sigma_{0}^{2}\), use the critical region defined by \((n-1) S^{2} / \sigma_{0}^{2} \geq c\). That is, reject \(H_{0}\) and accept \(H_{1}\) if \(S^{2} \geq c \sigma_{0}^{2} /(n-1)\) If \(n=13\) and the significance level \(\alpha=0.025\), determine \(c\).

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