91Ó°ÊÓ

We need to find the CDF \(P(R \le r, M \le m)\) for the given random variables. Since \(R \le r\) and \(M \le m\), we have: - \(X_{(1)} \ge m - \frac{1}{2}r\) - \(X_{(n)} \le m + \frac{1}{2}r\) Now, we consider the probabilities of the minimum and the maximum of \((-X_1, ..., X_n)\) satisfying the inequalities above: \(P\left(X_{(1)} \ge m - \frac{1}{2}r\right) = \left(1-\left(m - \frac{1}{2}r\right)\right)^n\) \(P\left(X_{(n)} \le m + \frac{1}{2}r\right) = \left(m + \frac{1}{2}r\right)^n\) Now, we consider the probability of both these events occurring simultaneously: \(P\left(R \le r, M \le m\right) = P\left(X_{(1)} \ge m - \frac{1}{2}r, X_{(n)} \le m + \frac{1}{2}r\right)\) Since \((-X_1, ..., X_n)\) are independent and identically distributed, their joint CDF can be factorized as a product: \(P\left(R \le r, M \le m\right) = \left[\left(m+\frac{1}{2}r\right)^n - \left(1-\left(m-\frac{1}{2}r\right)\right)^n\right]^n\)

To find the joint PDF, we need to differentiate the CDF \(P(R \le r, M \le m)\) with respect to \(r\) and \(m\). So, we have: \(f_{R,M}(r, m) = \frac{\partial^2}{\partial r \partial m} P\left(R \le r, M \le m\right)\) Differentiating the CDF with respect to \(r\) and \(m\): \(f_{R,M}(r, m) = \frac{\partial^2}{\partial r \partial m} \left[\left(m+\frac{1}{2}r\right)^n - \left(1-\left(m-\frac{1}{2}r\right)\right)^n\right]^n\) After differentiating, we get: \(f_{R,M}(r, m) = \frac{n^2}{2^{n-1}} \left[ \left(m+\frac{1}{2}r\right)^{n-1} - \left(m-\frac{1}{2}r\right)^{n-1} \right] \left[ \left(m+\frac{1}{2}r\right)^{n-1} + \left(m-\frac{1}{2}r\right)^{n-1} \right]\) This is the joint density function of \(R\) and \(M\).