91Ó°ÊÓ

To find the joint probability density function of \(X_1\) and \(X_2\), we make use of the given probability density function for a single observation. Since the problem states that the sample is drawn from the same distribution, we have: \(f_{X_1}(x_1) = 2 x_1\) for \(0 < x_1 < 1\) and \(f_{X_2}(x_2) = 2x_2\) for \(0 < x_2 < 1\) Since \(X_1\) and \(X_2\) are independent, the joint probability density function is the product of their individual probability density functions: \(f_{X_1, X_2}(x_1, x_2) = f_{X_1}(x_1) f_{X_2}(x_2) = (2x_1)(2x_2) = 4x_1 x_2\) This is valid in the region \(0 < x_1 < 1\) and \(0 < x_2 < 1\).

Let's define two new variables for the range and the sum of the sample: \(R = X_{max} - X_{min}\) (range) \(S = X_1 + X_2\) (sum) Now, we will find the relationships between \(R\), \(S\), \(x_1\), and \(x_2\). The range and sum can be represented as follow: Case 1: \(x_1 \ge x_2\) \(R = x_1 - x_2\) \(S = x_1 + x_2\) Case 2: \(x_2 \ge x_1\) \(R = x_2 - x_1\) \(S = x_1 + x_2\) Now, we will find the Jacobian matrix and its determinant for the transformation from \((x_1, x_2)\) to \((R, S)\): \(\frac{\partial(x_1, x_2)}{\partial(R, S)} = \begin{vmatrix} \frac{\partial x_1}{\partial R} & \frac{\partial x_1}{\partial S} \\ \frac{\partial x_2}{\partial R} & \frac{\partial x_2}{\partial S} \\ \end{vmatrix}\) For Case 1: \(\frac{\partial(x_1, x_2)}{\partial(R, S)} = \begin{vmatrix} 1 & 1 \\ -1 & 1 \\ \end{vmatrix} = 2\) For Case 2: \(\frac{\partial(x_1, x_2)}{\partial(R, S)} = \begin{vmatrix} -1 & 1 \\ 1 & 1 \\ \end{vmatrix} = 2\) In both cases, the determinant of the Jacobian matrix is 2.

With the joint probability density function of \(X_1\) and \(X_2\) and the determinant of the Jacobian matrix from the transformation, we can find the joint probability density function of \(R\) and \(S\): \(f_{R, S}(r, s) = f_{X_1, X_2}(x_1, x_2) \frac{\partial(x_1, x_2)}{\partial(R, S)} = 4x_1 x_2 (2) = 8x_1 x_2\)

We are only interested in the range \(R\), which is the difference between the two observations. To find the distribution of the range, we will integrate the joint probability density function of R and S with respect to S: \(f_R(r) = \int_{-\infty}^{\infty} f_{R, S}(r, s) ds\) For Case 1: \(f_R(r) = \int_{r}^{1+r} 8(x_1 - r)(x_1 - r + r) dx_1 = \int_{r}^{1+r} 8(x_1 - r)(x_1) dx_1\) For Case 2: \(f_R(r) = \int_{-r}^{1-r} 8(x_2 + r)(x_2) dx_2\) Since the distribution is symmetric, we can just calculate one of these integrals and multiply it by 2: \(f_R(r) = 2 \left(\int_{r}^{1+r} 8(x_1 - r)(x_1) dx_1\right)\) Given the density function is only valid from 0<x<1, the range of R can be 0<r<1. Therefore \(f_R(r)\) : For \(0 < r < 1\): \(f_R(r) = 16 \left(-\frac{1}{3} r^3 +\frac{1}{2}r^2 \right)\) So the derived distribution of the range for a sample of size 2 from a distribution with density function \(f(x)=2x, 0<x<1\) is given by: \(f_R(r) = 16 \left(-\frac{1}{3} r^3 +\frac{1}{2}r^2 \right)\) for \(0 < r < 1\).