Let's tackle each part of your questions step-by-step, with clear explanations. --- ## **1. PDF of the Range \( R = U_{(n)} - U_{(1)} \) for a Sample from \( U(0,1) \)** Let \( U_1, U_2, ..., U_n \) be i.i.d. from \( U(0,1) \). Let \( U_{(1)} \) and \( U_{(n)} \) be the minimum and maximum. Let \( R = U_{(n)} - U_{(1)} \). **Step 1: Joint PDF of \( U_{(1)} \) and \( U_{(n)} \):** The joint PDF is: \[ f_{U_{(1)},U_{(n)}}(x,y) = n(n-1)(y - x)^{n-2}, \quad 0 < x < y < 1 \] **Step 2: Change of Variables** Let \( R = y - x \), \( X = x \). The Jacobian is 1. So \( y = x + r \), and the bounds are: - \( 0 < x < 1 - r \) - \( 0 < r < 1 \) **Step 3: Find PDF of \( R \)** The joint PDF in new variables is: \[ f_{X,R}(x, r) = n(n-1) r^{n-2}, \quad 0 < x < 1 - r,\ 0 < r < 1 \] To get the marginal PDF of \( R \): \[ f_R(r) = \int_{x=0}^{1-r} n(n-1) r^{n-2} dx = n(n-1) r^{n-2} (1 - r) \] for \( 0 < r < 1 \). **Final Answer:** \[ \boxed{ f_R(r) = n(n-1) (1 - r) r^{n-2}, \quad 0 < r < 1 } \] --- ## **2. Joint PDF of Spacings for Uniform Order Statistics** Let \( U_{(1)} < U_{(2)} < ... < U_{(n)} \) be the order statistics. Define: - \( X_1 = U_{(1)} \) - \( X_2 = U_{(2)} - U_{(1)} \) - ... - \( X_n = U_{(n)} - U_{(n-1)} \) - \( X_{n+1} = 1 - U_{(n)} \) So \( X_1 + X_2 + ... + X_{n+1} = 1 \), and \( X_i > 0 \). **Step 1: Joint PDF of Order Statistics** The joint PDF of the order statistics for U(0,1) is: \[ f_{U_{(1)},...,U_{(n)}}(u_1,...,u_n) = n!, \quad 0 < u_1 < u_2 < ... < u_n < 1 \] **Step 2: Change of Variables to Spacings** Let: - \( x_1 = u_1 \) - \( x_2 = u_2 - u_1 \) - ... - \( x_n = u_n - u_{n-1} \) - \( x_{n+1} = 1 - u_n \) Then: - \( u_1 = x_1 \) - \( u_2 = x_1 + x_2 \) - ... - \( u_n = x_1 + ... + x_n \) The Jacobian of this transformation is 1 (since it is a lower triangular matrix with 1's on the diagonal). The region is \( x_i > 0 \), \( x_1 + ... + x_{n+1} = 1 \). **Step 3: The Joint PDF** Thus, the joint PDF is: \[ f_{X_1,...,X_{n+1}}(x_1, ..., x_{n+1}) = n! \] for \( x_i > 0, \ x_1 + ... + x_{n+1} = 1 \). This is the **Dirichlet distribution** with all parameters 1. **Final Answer:** \[ \boxed{ f_{X_1,...,X_{n+1}}(x_1, ..., x_{n+1}) = n! \cdot \mathbf{1}\left(\sum_{i=1}^{n+1} x_i = 1, \ x_i > 0\right) } \] --- ## **3. Order Statistics from General Continuous Distribution** Let \( X_1,...,X_n \) be i.i.d. from a continuous and strictly increasing cdf \( F \). Let \( Y_1 = \min X_i \), \( Y_n = \max X_i \). Define \( Z_1 = F(Y_1) \), \( Z_n = F(Y_n) \). ### (a) **Find the joint and marginal distributions of \( Z_1, Z_n \).** **Step 1: Distribution of \( Z_1 = F(Y_1) \)** The cdf of \( Y_1 \) is: \[ P(Y_1 \leq y) = 1 - P(\text{all } X_i > y) = 1 - (1 - F(y))^n \] So the cdf of \( Z_1 \) at \( z \) is: \[ P(Z_1 \leq z) = P(F(Y_1) \leq z) = P(Y_1 \leq F^{-1}(z)) = 1 - (1-z)^n,\quad 0 < z < 1 \] So the pdf is: \[ f_{Z_1}(z) = n(1-z)^{n-1},\quad 0 < z < 1 \] Similarly, for \( Z_n = F(Y_n) \): \[ P(Y_n \leq y) = P(\text{all } X_i \leq y) = F(y)^n \] So \[ P(Z_n \leq z) = P(F(Y_n) \leq z) = P(Y_n \leq F^{-1}(z)) = z^n \] Thus, \[ f_{Z_n}(z) = n z^{n-1},\quad 0 < z < 1 \] **Step 2: Joint Distribution** The joint cdf for \( Y_1 \leq y_1 \), \( Y_n \leq y_n \) (\( y_1 < y_n \)) is: \[ P(Y_1 \leq y_1, Y_n \leq y_n) = P(\text{all } X_i \leq y_n) - P(\text{all } X_i < y_1) = F(y_n)^n - F(y_1)^n \] But we want the joint pdf of \( Z_1, Z_n \): Let \( z_1 < z_n \), then \[ P(Z_1 \leq z_1, Z_n \leq z_n) = P(Y_1 \leq F^{-1}(z_1), Y_n \leq F^{-1}(z_n)) \] But for \( F \) strictly increasing, \( F^{-1}(z_1) < F^{-1}(z_n) \), so \[ = F(F^{-1}(z_n))^n - F(F^{-1}(z_1))^n = z_n^n - z_1^n \] The joint pdf is the mixed partial derivative: \[ f_{Z_1,Z_n}(z_1, z_n) = \frac{\partial^2}{\partial z_1 \partial z_n} (z_n^n - z_1^n) = n(n-1)(z_n - z_1)^{n-2} \] But this is not quite right—let's do it carefully. The joint pdf of the minimum and maximum of i.i.d. variables with pdf \( f(x) \) and cdf \( F(x) \) is: \[ f_{Y_1, Y_n}(y_1, y_n) = n(n-1) [F(y_n) - F(y_1)]^{n-2} f(y_1) f(y_n),\quad y_1 < y_n \] Now, \( Z_1 = F(y_1), Z_n = F(y_n) \), so the Jacobian is: \[ \left| \frac{\partial(y_1, y_n)}{\partial(z_1, z_n)} \right| = \frac{1}{f(F^{-1}(z_1)) f(F^{-1}(z_n))} \] So, \[ f_{Z_1, Z_n}(z_1, z_n) = n(n-1) (z_n - z_1)^{n-2},\quad 0 < z_1 < z_n < 1 \] **Final Answer:** \[ \boxed{ f_{Z_1, Z_n}(z_1, z_n) = n(n-1) (z_n - z_1)^{n-2},\quad 0 < z_1 < z_n < 1 } \] and the marginals are: \[ \boxed{ f_{Z_1}(z_1) = n (1 - z_1)^{n-1},\quad f_{Z_n}(z_n) = n z_n^{n-1},\quad 0 < z_1, z_n < 1 } \] --- ### (b) **Prove \( Z_1 / Z_n \) and \( Z_n \) are Independent** Let \( W = Z_1 / Z_n \), \( V = Z_n \). Let’s find the joint pdf of \( (W, V) \): Set \( Z_1 = w v \), \( Z_n = v \), for \( 0 < w < 1 \), \( 0 < v < 1 \), and \( z_1 = w v < z_n = v \). The Jacobian is: \[ \left| \frac{\partial(z_1, z_n)}{\partial(w, v)} \right| = \begin{vmatrix} v & w \\ 0 & 1 \end{vmatrix} = v \] So, the joint pdf is: \[ f_{W,V}(w, v) = f_{Z_1, Z_n}(w v, v) \cdot v \] Recall \( f_{Z_1, Z_n}(z_1, z_n) = n(n-1) (z_n - z_1)^{n-2} \), so \[ f_{W,V}(w, v) = n(n-1) (v - w v)^{n-2} v = n(n-1) v (1 - w)^{n-2} v^{n-2} = n(n-1) (1-w)^{n-2} v^{n-1} \] for \( 0 < w < 1, 0 < v < 1 \). Now, let's compute the marginals: - Marginal of \( W \): \[ \int_0^1 f_{W,V}(w, v) dv = n(n-1) (1-w)^{n-2} \int_0^1 v^{n-1} dv = n(n-1) (1-w)^{n-2} \cdot \frac{1}{n} = (n-1)(1-w)^{n-2} \] - Marginal of \( V \): \[ \int_0^1 f_{W,V}(w, v) dw = n(n-1) v^{n-1} \int_0^1 (1-w)^{n-2} dw = n(n-1) v^{n-1} \cdot \frac{1}{n-1} = n v^{n-1} \] So, \[ f_{W,V}(w,v) = f_W(w) f_V(v) \] Thus, **\( W \) and \( V \) are independent**. **Final Answer:** \[ \boxed{ Z_1/Z_n \quad \text{and} \quad Z_n \quad \text{are independent.} } \] --- **Summary Table** | Question | Final Answer | |:---|:---| | 1. PDF of Range \( R \) | \( f_R(r) = n(n-1)(1-r) r^{n-2}, \ 0 < r < 1 \) | | 2. Joint PDF of Spacings | \( f_{X_1,...,X_{n+1}}(x_1, ..., x_{n+1}) = n! \) if \( \sum x_i = 1, x_i > 0 \) | | 3a. Joint, Marginal of \( Z_1, Z_n \) | \( f_{Z_1, Z_n}(z_1, z_n) = n(n-1)(z_n - z_1)^{n-2}, 0 < z_1 < z_n < 1 \) <br> \( f_{Z_1}(z_1) = n(1-z_1)^{n-1} \), \( f_{Z_n}(z_n) = n z_n^{n-1} \) |

To prove \( Z_1/Z_n \) and \( Z_n \) are independent, we show their joint pdf factors into the product of marginals by transforming variables and computing the Jacobian. Since the joint factors, they are independent, confirming the independence of the ratio and the maximum transformed variables.

To prove \( Z_1/Z_n \) and \( Z_n \) are independent, we transform variables and compute their joint pdf, which factors into the product of marginals, showing independence. This confirms that the ratio \( Z_1/Z_n \) and \( Z_n \) are statistically independent due to the factorization of their joint distribution.

To show independence of \( Z_1/Z_n \) and \( Z_n \), we perform a variable transformation and compute their joint density, which factors into the product of marginals. This factorization confirms the independence, establishing that the ratio and the maximum are statistically independent.

To find the PDF of the range \( R = U_{(n)} - U_{(1)} \), we derive the joint distribution of the minimum and maximum of uniform samples, then change variables to \( R \) and \( X \). Integrating over \( X \), the resulting marginal PDF is \( f_R(r) = n(n-1)(1 - r) r^{n-2} \) for \( 0<r<1 \).

1. The PDF of the range \( R = U_{(n)} - U_{(1)} \) is: \[ \boxed{ f_R(r) = n(n-1)(1 - r) r^{n-2}, \quad 0 < r < 1 } \] 2. The joint pdf of spacings \(\{X_i\}\) from order statistics of U(0,1) is: \[ \boxed{ f_{X_1,\dots,X_{n+1}} = n! \quad \text{on} \quad \left\{\sum_{i=1}^{n+1} X_i = 1, X_i > 0 \right\} } \] 3. For \( Z_1 = F(Y_1) \), \( Z_n = F(Y_n) \), the joint density is: \[ \boxed{ f_{Z_1,Z_n}(z_1,z_n) = n(n-1)(z_n - z_1)^{n-2}, \quad 0<z_1<z_n<1 } \] and \( Z_1/Z_n \) is independent of \( Z_n \).