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Chapter 4: Problem 15

Let \(Y_{1}

Short Answer

Expert verified
The expected value of the first order statistic \( Y_{1} \) is \(-\sigma / \sqrt{\pi}\) and the covariance of \(Y_{1}\) and \(Y_{2}\) can be attained by evaluating \( Cov(Y_{1}, Y_{2}) = E[Y_{1}.Y_{2}] - E[Y_{1}]E[Y_{2}] \).

Step by step solution

01

Determining the joint pdf

The density function of a normally distributed random variable is given by, \( f(y) = \frac{1}{\sqrt{2\pi\sigma^{2}}}e^{-y^{2}/2\sigma^{2}} \). Given two independently distributed variables, the joint pdf will be the product of their individual pdfs. So, \( f(Y_{1}, Y_{2}) = f(Y_{1}) \cdot f(Y_{2}) =\frac{1}{2\pi\sigma^{2}}e^{-(Y_{1}^{2}+Y_{2}^{2})/2\sigma^{2}} \).
02

Evaluating E(Y1)

Evaluation of \( E\left(Y_{1}\right) \) is based on the formula, \( E[X] = \int{x.f(x)}dx \). Here \( f(X) \) is the joint pdf. Considering the symmetry of normal distribution and properties of odd functions, the expected value of \( Y_{1} \) becomes, \( E\left(Y_{1}\right) = - \int_{0}^{\infty}{y_{1}f(y_{1})}dy_{2} = -\sigma / \sqrt{\pi} \).
03

Calculating Covariance

The covariance of two random variables is given by \( Cov(X, Y) = E[X.Y] - E[X]E[Y] \). Before calculating the covariance, we need to know the expected value \( E[Y_{2}] \). By symmetry of normal distribution, \( E[Y_{2}] = +\sigma / \sqrt{\pi} \). Now, we can use these to find the covariance of \(Y_{1}\) and \(Y_{2}\).

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Most popular questions from this chapter

In the baseball data set discussed in the last exercise, it was found that out of the 59 baseball players, 15 were left-handed. Is this odd, since the proportion of left-handed males in America is about \(11 \% ?\) Answer by using \((4.2 .7)\) to construct a \(95 \%\) approximate confidence interval for \(p\), the proportion of left-handed baseball players.

Let the result of a random experiment be classified as one of the mutually exclusive and exhaustive ways \(A_{1}, A_{2}, A_{3}\) and also as one of the mutually exhaustive ways \(B_{1}, B_{2}, B_{3}, B_{4}\). Say that 180 independent trials of the experiment result in the following frequencies: \begin{tabular}{|c|c|c|c|c|} \hline & \(B_{1}\) & \(B_{2}\) & \(B_{3}\) & \(B_{4}\) \\ \hline\(A_{1}\) & \(15-3 k\) & \(15-k\) & \(15+k\) & \(15+3 k\) \\ \hline\(A_{2}\) & 15 & 15 & 15 & 15 \\ \hline\(A_{3}\) & \(15+3 k\) & \(15+k\) & \(15-k\) & \(15-3 k\) \\ \hline \end{tabular} where \(k\) is one of the integers \(0,1,2,3,4,5\). What is the smallest value of \(k\) that leads to the rejection of the independence of the \(A\) attribute and the \(B\) attribute at the \(\alpha=0.05\) significance level?

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