/*! 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} Q8E In Example 12.6.7, let \(\left( ... [FREE SOLUTION] | 91影视

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Chapter 12: Q8E (page 849)

In Example 12.6.7, let \(\left( {{X^*},{Y^*}} \right)\) be a random draw from the sample distribution\({F_n}\) . Prove that the correlation \({X^*} and {Y^*}\) is R in Eq. (12.6.2).

Short Answer

Expert verified

\(R = \frac{{\sum\limits_{i = 1}^n {\left( {{X_i} - \bar X} \right)} \left( {{Y_i} - \bar Y} \right)}}{{\sqrt {\sum\limits_{i = 1}^n {{{\left( {{X_i} - \bar X} \right)}^2}} \sum\limits_{i = 1}^n {{{\left( {{Y_i} - \bar Y} \right)}^2}} } }}\)

Step by step solution

01

To Prove that the correlation between \({X^*} and {Y^*}\) is R 

The correlation coefficient of X and Y is

\({\rho _{X,Y}} = \frac{{Cov(X,Y)}}{{{\sigma _X} \times {\sigma _Y}}}\)

Notice first that for such a random sample from the sample distribution,\({F_n}\)the expected values are

\(\begin{aligned}{l}E\left( {{X^*}} \right) &= \bar X\\E\left( {{Y^*}} \right) &= \bar Y\end{aligned}\)

and the variances are given by

\(\begin{aligned}{l}V\left( {{X^*}} \right) &= \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{X_i} - \bar X} \right)}^2}} \\\\V\left( {{Y^*}} \right) &= \frac{1}{n}\sum\limits_{i = 1}^n {{{\left( {{Y_i} - \bar Y} \right)}^2}} \end{aligned}\)

Finally, to get the sample correlation coefficient of

\(R = \frac{{\sum\limits_{i = 1}^n {\left( {{X_i} - \bar X} \right)} \left( {{Y_i} - \bar Y} \right)}}{{\sqrt {\sum\limits_{i = 1}^n {{{\left( {{X_i} - \bar X} \right)}^2}} \sum\limits_{i = 1}^n {{{\left( {{Y_i} - \bar Y} \right)}^2}} } }}\)

it notices that the covariance\(is Cov(X,Y)\)is

\(\begin{aligned}{l}Cov(X,Y) &= E\left( {\left( {{X^*} - \bar X} \right)\left( {{Y^*} - \bar Y} \right)} \right)\\\\ &= \sum\limits_{i = 1}^n {\left( {{X_i} - \bar X} \right)} \left( {{Y_i} - \bar Y} \right)\end{aligned}\)

Because the sample distribution\({F_n}\)is discrete. Finally, by substituting all values and using the fact that\({\sigma _{{X^*}}} = \sqrt {V\left( {{X^*}} \right)} \)it is true that

\(\begin{aligned}{l}{\rho _{X,Y}} &= \frac{{Cov(X,Y)}}{{{\sigma _X} \times {\sigma _Y}}}\\\\ &= \frac{{\sum\limits_{i = 1}^n {\left( {{X_i} - \bar X} \right)} \left( {{Y_i} - \bar Y} \right)}}{{\sqrt {\sum\limits_{i = 1}^n {{{\left( {{X_i} - \bar X} \right)}^2}} \sum\limits_{i = 1}^n {{{\left( {{Y_i} - \bar Y} \right)}^2}} } }}\quad \quad \end{aligned}\)

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Most popular questions from this chapter

Use the blood pressure data in Table 9.2 that was described in Exercise 10 of Sec. 9.6. Suppose now that we are not confident that the variances are the same for the two treatment groups. Perform a parametric bootstrap analysis of the sort done in Example 12.6.10. Use v=10,000 bootstrap simulations.

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b. Correct the level of the two-sample t-test by computing the appropriate quantile of the bootstrap distribution of \(\left| {{U^{(i)}}} \right|.\)

c. Compute the standard simulation error for the quantile in part (b).

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