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Chapter 5: Problem 15

Let \(Y_{1}

Short Answer

Expert verified
The expected value of the smallest order statistic, \(Y_{1}\), from a normally distributed random sample is \(-\sigma / \sqrt{\pi}\). The covariance of \(Y_{1}\) and \(Y_{2}\) involves additional steps but is calculated using the formula for covariance.

Step by step solution

01

Joint PDF of Order Statistics

The joint probability density function of \(Y_{1}\) and \(Y_{2}\) given that \(Y_{1} < Y_{2}\) can be written as \[f_{Y_{1}, Y_{2}}(y_{1}, y_{2}) = 2f_{Y}(y_{1})f_{Y}(y_{2})\] where \(f_Y(y)\) is the PDF of a normal distribution. This 2 comes from the number of ways \(Y_{2}\) can be the largest of the two values.
02

Expected Value of \(Y_{1}\)

The expected value of \(Y_{1}\) is given by \[E(Y_{1}) = \int_{-\infty}^{+\infty} y_{1}\int_{y_{1}}^{+\infty} f_{Y_{1}, Y_{2}}(y_{1}, y_{2}) dy_{2} dy_{1}\]. Substituting \(f_{Y_{1}, Y_{2}}(y_{1}, y_{2})\), performing the integration, and simplifying, you get \(E(Y_{1}) = -\sigma / \sqrt{\pi}\).
03

Covariance of \(Y_{1}\) and \(Y_{2}\)

Covariance is given by \(Cov(Y_{1}, Y_{2}) = E(Y_{1} Y_{2}) - E(Y_{1})E(Y_{2})\). First, find \(E(Y_{1} Y_{2}) = \int_{-\infty}^{+\infty} y_{1} y_{2} f_{Y_{1}, Y_{2}}(y_{1}, y_{2}) dy_{1} dy_{2}\). After calculating this, subtract the product of the expectations of the individual distributions (found in a similar way as in step 2) to find the covariance.

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Most popular questions from this chapter

. Let \(Y_{1}

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