/*! 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 16 Let \(X_{1}\) and \(X_{2}\) be i... [FREE SOLUTION] | 91影视

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Chapter 4: Problem 16

Let \(X_{1}\) and \(X_{2}\) be independent random variables with nonzero variances. Find the correlation coefficient of \(Y=X_{1} X_{2}\) and \(X_{1}\) in terms of the means and variances of \(X_{1}\) and \(X_{2}\).

Short Answer

Expert verified
The correlation coefficient of \(Y=X_{1} X_{2}\) and \(X_{1}\) in terms of the means and variances of \(X_{1}\) and \(X_{2}\) is given as \(\rho_{Y, X_1} = \frac{E(X_2)}{\sqrt{E(X1^2)E(X2^2) - E(X1)^2E(X2)^2}}\).

Step by step solution

01

Compute Covariance

Calculate the covariance between \(Y\) and \(X_1\). By definition, \(Cov(Y, X_1) = E(YX_1) - E(Y)E(X_1)\). Since Y is defined as \(X_1 X_2\), substitute Y in the equation to get \(Cov(X_1 X_2, X_1) = E(X_1^2 X_2) - E(X_1 X_2)E(X_1)\).
02

Express Covariance in terms of Means and Variances

Go further with the expression obtained: The term \(E(X_1^2 X_2)\) can be expressed as \((E(X_1^2)) (E(X_2))\) since \(X_1\) and \(X_2\) are independent. The term \(E(X_1 X_2)\) is \(E(X_1)E(X_2)\). Substituting these into the equation gives \(Cov(Y, X_1) = (E(X_1^2)) (E(X_2)) - (E(X_1))^2 (E(X_2)) = (E(X_2)) Var(X_1)\), where \(Var(X_1)\) is the variance of \(X_1\).
03

Compute Correlation Coefficient

The correlation coefficient is defined as \(\rho_{Y, X_1} = \frac{Cov(Y, X_1)}{\sqrt{Var(Y)Var(X_1)}}\). Substituting the equation found in Step 2 into the formula, we obtain \(\rho_{Y, X_1} = \frac{(E(X_2)) Var(X_1)}{\sqrt{Var(Y)Var(X_1)}}\).
04

Express \(Var(Y)\)

We realise that \(Var(Y)\) can be broken down as follows: \(Var(Y) = E(Y^2)-E(Y)^2\). Since \(Y=X1X2\), we have \(Var(Y) = E((X1X2)^2) - E(X1X2)^2\). This leads to \(Var(Y) = E(X1^2)E(X2^2) - E(X1)^2E(X2)^2\). So as not to complicate matters, we take this expression as the variance of \(Y\).
05

Conclusion

Insert the previously obtained expressions for our computation back into the correlation coefficient and obtain \(\rho_{Y, X_1} = \frac{(E(X_2)) Var(X_1)}{\sqrt{E(X1^2)E(X2^2) - E(X1)^2E(X2)^2)Var(X_1)}}\). The \(Var(X_1)\) terms in the numerator and the denominator cancel out, leaving \(\rho_{Y, X_1} = \frac{E(X_2)}{\sqrt{E(X1^2)E(X2^2) - E(X1)^2E(X2)^2}}\). This is the final answer.

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Most popular questions from this chapter

Let \(X\) and \(Y\) be independent random variables with means \(\mu_{1}, \mu_{2}\) and variances \(\sigma_{1}^{2}, \sigma_{2}^{2} .\) Determine the correlation coefficient of \(X\) and \(Z=X-Y\) in terms of \(\mu_{1}, \mu_{2}, \sigma_{1}^{2}, \sigma_{2}^{2}\)

For the last exercise, suppose that the sample is drawn from a \(N\left(\mu, \sigma^{2}\right)\) distribution. Recall that \((n-1) S^{2} / \sigma^{2}\) has a \(\chi^{2}(n-1)\) distribution. Use Theorem 3.3.1 to determine an unbiased estimator of \(\sigma\).

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample of size \(n\) from a distribution that is \(N\left(\mu, \sigma^{2}\right)\), where \(\sigma^{2}>0 .\) Show that the sum \(Z_{n}=\sum_{1}^{n} X_{i}\) does not have a limiting distribution.

Suppose \(X\) has a pdf which is symmetric about \(b\); i.e., \(f(b+x)=f(b-x)\), for all \(-\infty

Let \(X\) and \(Y\) have the parameters \(\mu_{1}, \mu_{2}, \sigma_{1}^{2}, \sigma_{2}^{2}\), and \(\rho .\) Show that the correlation coefficient of \(X\) and \(\left[Y-\rho\left(\sigma_{2} / \sigma_{1}\right) X\right]\) is zero.
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