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Chapter 2: Problem 7

Let \(X_{1}, X_{2}, X_{3}, X_{4}\) have the joint pdf \(f\left(x_{1}, x_{2}, x_{3}, x_{4}\right)=24,0

Short Answer

Expert verified
The joint pdf of \(Y_{1}, Y_{2}, Y_{3}, Y_{4}\) is given by \(g(y_{1}, y_{2}, y_{3}, y_{4}) = 24*Y_{4}^{3}\). The variables \(Y_{1}, Y_{2}, Y_{3}, Y_{4}\) are mutually independent as the joint pdf factors into product of marginal pdf's for each variable.

Step by step solution

01

Transform Variables

Define the variables \(Y_{1}, Y_{2}, Y_{3}, Y_{4}\) as follows: \(Y_{1} = X_{1} / X_{2}\), \(Y_{2} = X_{2} / X_{3}\), \(Y_{3} = X_{3} / X_{4}\), and \(Y_{4} = X_{4}\). Then rewrite the \(X_i\) variables in terms of \(Y_i\): \(X_{1} = Y_{1}Y_{2}Y_{3}Y_{4}\), \(X_{2} = Y_{2}Y_{3}Y_{4}\), \(X_{3} = Y_{3}Y_{4}\), and \(X_{4} = Y_{4}\).
02

Compute Jacobian Determinant

To transform the joint pdf from \(X_{i}\) to \(Y_{i}\), we have to compute the Jacobian determinant. The Jacobian matrix is a square matrix of all first-order partial derivatives of a vector-valued function. The Jacobian determinant for this transformation is given by \(|J| = | \partial(X_{1}, X_{2}, X_{3}, X_{4})/ \partial(Y_{1}, Y_{2}, Y_{3}, Y_{4})|\), which simplifies to \(|J| = Y_{4}^{3}\).
03

Obtain the Joint pdf of Y's

Now, we can find the joint pdf of \(Y_{1}, Y_{2}, Y_{3}, Y_{4}\) by multiplying the original joint pdf (which is 24 within the specified range) by the absolute value of the Jacobian determinant |J|. Thus, the joint pdf \(g(y_{1}, y_{2}, y_{3}, y_{4}) = 24*Y_{4}^{3}\).
04

Verify Independence

Variables are mutually independent if the joint pdf factors into a product of marginal pdf's for each variable. So, if we can write \( g(y_{1}, y_{2}, y_{3}, y_{4}) = h_{1}(y_{1})h_{2}(y_{2})h_{3}(y_{3})h_{4}(y_{4}) \), then the variables are mutually independent. It can be seen that \( g(y_{1}, y_{2}, y_{3}, y_{4}) = 1 * 1 * 1 * 24*Y_{4}^{3} \), which is factored with respect to each variable. Therefore, the variables \(Y_{1}, Y_{2}, Y_{3}, Y_{4}\) are mutually independent.

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Most popular questions from this chapter

Let \(X_{1}\) and \(X_{2}\) have the joint pmf \(p\left(x_{1}, x_{2}\right)=x_{1} x_{2} / 36, x_{1}=1,2,3\) and \(x_{2}=1,2,3\), zero elsewhere. Find first the joint pmf of \(Y_{1}=X_{1} X_{2}\) and \(Y_{2}=X_{2}\), and then find the marginal pmf of \(Y_{1}\).

Let \(X_{1}, X_{2}, X_{3}\), and \(X_{4}\) be four independent random variables, each with pdf \(f(x)=3(1-x)^{2}, 0y)=P\left(X_{i}>y, i=1, \ldots, 4\right)\).

Let \(\mathbf{X}=\left(X_{1}, \ldots, X_{n}\right)^{\prime}\) be an \(n\) -dimensional random vector, with the variancecovariance matrix given in display (2.6.13). Show that the \(i\) th diagonal entry of \(\operatorname{Cov}(\mathbf{X})\) is \(\sigma_{i}^{2}=\operatorname{Var}\left(X_{i}\right)\) and that the \((i, j)\) th off diagonal entry is \(\operatorname{Cov}\left(X_{i}, X_{j}\right)\).

Let \(X\) and \(Y\) have the joint pdf \(f(x, y)=6(1-x-y), x+y<1,0

Let \(X_{1}\) and \(X_{2}\) have the joint pdf \(f\left(x_{1}, x_{2}\right)=15 x_{1}^{2} x_{2}, 0
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