Question

Let \(Y_{1}

Short answer

Hence, the pdf of \(Z\) is 2 for \(0<Z<1\), and zero elsewhere.

Step-by-step solution

Understanding the Midrange

The midrange is defined as the average of the smallest and largest values in a data set. We are given the ordered sample \(Y_1\), \(Y_2\), and \(Y_3\). Therefore, the midrange \(Z\) is \((Y_1 + Y_3) / 2\).

Expressing Z in Terms of Y

To find the pdf of \(Z\), we first express it in terms of \(Y\). We have \(Z = \frac{Y_1 + Y_3}{2}\), which equates to \(Z = \frac{1}{2}(Y_1 + Y_3)\). Now, we apply the Jacobian method for transformation of random variables. The Jacobian Transformation matrix, given by the absolute value of the determinant of the transformation, follows the following formula: \[J = \left| \frac{\partial(x,y)}{\partial(u,v)} \right| = \left| \begin{matrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{matrix} \right| \] Using \(Y_2\) as it is and substituting \(Z = \frac{1}{2}(Y_1 + Y_3)\), we can solve determinant.

Applying Jacobian Transformation

Applying the Jacobian transformation, we see that (u,v) = (Y_2, Z) and (x,y) = (Y_1, Y_3). Therefore, the Jacobian matrix becomes:\[ \left| \begin{matrix} \frac{\partial Y_1}{\partial Y_2} & \frac{\partial Y_1}{\partial Z} \ \frac{\partial Y_3}{\partial Y_2} & \frac{\partial Y_3}{\partial Z} \end{matrix} \right| \]After calculations, we get \(J = 2\).

Calculating PDF

With the Jacobian, we calculate the probability density function (pdf). The pdf of Z, \(f_Z(Z)\), is given by the absolute value of the Jacobian multiplied by the pdfs of \(Y_1\), \(Y_2\), and \(Y_3\). As \(Y_1\), \(Y_2\), and \(Y_3\) are from a uniform distribution on (0,1), the pdfs of \(Y_1\), \(Y_2\), and \(Y_3\) are all 1. Therefore, \(f_Z(Z)\) is equal to \(2*1*1*1 = 2\).

Finalizing the PDF

The final pdf for Z must be valid in the given domain. As \(Y_1\), \(Y_2\), and \(Y_3\) are all from a uniform distribution in the interval (0,1) and \(Z = \frac{Y_1 + Y_3}{2}\), the minimum value Z can take is 0, and the maximum is 1. Therefore, the pdf of \(Z\), \(f_Z(Z)\), is 2 for \(0<Z<1\) and zero elsewhere.