# Given Information - Random variables \( X_1, X_2, ..., X_n \) are i.i.d. and follow the uniform distribution: \[ X_i \sim \text{Uniform}(, 1) \] # What Have to Find a) PDF of the maximum: \( X_{(n)} = \max\{X_1, X_2, ..., X_n\} \) b) Expected value of the \( k \)-th order statistic: \( E[X_{(k)}] \) c) Probability \( P(X_{(n)} - X_{(1)} < .5) \), where \( X_{(1)} = \min\{X_1, ..., X_n\} \) # Definition or Concept Used - **Order statistics:** For i.i.d. random variables, the \( k \)-th order statistic is the \( k \)-th smallest value. - **PDF of order statistics:** For \( X_i \sim \text{Uniform}(,1) \), the PDF of the \( k \)-th order statistic is: \[ f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!}x^{k-1}(1-x)^{n-k}, \quad < x < 1 \] - **CDF of maximum:** \( F_{X_{(n)}}(x) = [F_X(x)]^n \) - **PDF of maximum:** Derivative of the CDF. - **Probability calculation:** Use integration of joint PDF or CDF properties. # Solution **a) PDF of the maximum \( X_{(n)} \):** The CDF of the maximum is: \[ F_{X_{(n)}}(x) = P(X_{(n)} \leq x) = P(X_1 \leq x, ..., X_n \leq x) = [F_X(x)]^n = x^n, \quad < x < 1 \] The PDF is: \[ f_{X_{(n)}}(x) = \frac{d}{dx}F_{X_{(n)}}(x) = n x^{n-1}, \quad < x < 1 \] --- **b) \( E[X_{(k)}] \):** The PDF of the \( k \)-th order statistic is: \[ f_{X_{(k)}}(x) = \frac{n!}{(k-1)!(n-k)!} x^{k-1}(1-x)^{n-k}, \quad < x < 1 \] The expected value is: \[ E[X_{(k)}] = \int_^1 x f_{X_{(k)}}(x) dx \] \[ = \frac{n!}{(k-1)!(n-k)!} \int_^1 x \cdot x^{k-1} (1-x)^{n-k} dx \] \[ = \frac{n!}{(k-1)!(n-k)!} \int_^1 x^k (1-x)^{n-k} dx \] \[ = \frac{n!}{(k-1)!(n-k)!} \cdot \text{B}(k+1, n-k+1) \] where \(\text{B}(a, b)\) is the Beta function: \(\text{B}(a, b) = \frac{\Gamma(a)\Gamma(b)}{\Gamma(a+b)}\). For integers, \(\Gamma(m) = (m-1)!\). Thus, \[ E[X_{(k)}] = \frac{k}{n+1} \] --- **c) \( P(X_{(n)} - X_{(1)} < .5) \):** Let \( X_{(n)} = y \), \( X_{(1)} = x \), \( < x < y < 1 \). We require \( y - x < .5 \implies x > y - .5 \) and \( y < x + .5 \). Joint PDF: \[ f_{X_{(1)}, X_{(n)}}(x, y) = n(n-1)(y-x)^{n-2}, \quad < x < y < 1 \] Probability: \[ P(X_{(n)} - X_{(1)} < .5) = \int_^1 \int_{x}^{\min(x+.5,\,1)} n(n-1)(y-x)^{n-2} dy\, dx \] Change order for easier computation: \[ = \int_^1 \int_{\max(, y-.5)}^{y} n(n-1)(y-x)^{n-2} dx\, dy \] \[ = \int_^1 n(n-1) \left[ \int_{\max(, y-.5)}^{y} (y-x)^{n-2} dx \right] dy \] Let \( u = y - x \), \( du = -dx \), bounds: \( x = y \to u= \), \( x = \max(, y-.5) \to u = y-\max(, y-.5) = \min(y, .5) \): So, \[ = \int_^1 n(n-1) \int_{}^{\min(y, .5)} u^{n-2} du\, dy \] \[ = \int_^{.5} n(n-1) \int_{}^{y} u^{n-2} du\, dy + \int_{.5}^{1} n(n-1) \int_{}^{.5} u^{n-2} du\, dy \] Evaluating the inner integrals: \[ \int_{}^{a} u^{n-2} du = \frac{a^{n-1}}{n-1} \] So, \[ = \int_^{.5} n y^{n-1} dy + \int_{.5}^{1} n [.5]^{n-1} dy \] \[ = n \left[ \frac{y^n}{n} \right]_^{.5} + n [.5]^{n-1} (1 - .5) \] \[ = (.5)^n + .5 n (.5)^{n-1} \] \[ = (.5)^n (1 + n) \] --- # Summary - **PDF of the maximum:** \( f_{X_{(n)}}(x) = n x^{n-1}, \quad < x < 1 \) - **Expected value of the \(k\)-th order statistic:** \( E[X_{(k)}] = \frac{k}{n+1} \) - **Probability:** \( P(X_{(n)} - X_{(1)} < .5) = (1 + n) \cdot (.5)^n \)