# Given Information - \( X_1, X_2, \ldots, X_n \) are independent and identically distributed random variables. - Each \( X_i \sim \text{Uniform}(,1) \). # What to Find a) PDF of the maximum \( X_{(n)} \) (where \( X_{(n)} = \max(X_1, \ldots, X_n) \)). b) \( E[X_{(k)}] \) — the expected value of the \( k \)-th order statistic. c) \( P(X_{(n)} - X_{(1)} < .5) \) — the probability that the range is less than .5. # Definitions/Concepts Used - **Order Statistics**: For i.i.d. random variables, \( X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(n)} \). - **PDF of Maximum**: The CDF of the maximum is \( F_{X_{(n)}}(x) = [F_X(x)]^n \), and the PDF is its derivative. - **Expected Value of Order Statistics**: The expected value of the \( k \)-th order statistic for Uniform(,1) is \( \frac{k}{n+1} \). - **Probability involving order statistics**: For range, use joint PDF and integrate over the region where \( X_{(n)} - X_{(1)} < .5 \). --- ## Solution ### a) PDF of the Maximum \( X_{(n)} \) The CDF of \( X_{(n)} \) is \( F_{X_{(n)}}(x) = [F_X(x)]^n = x^n \) for \( x \in [,1] \). The PDF is the derivative: \[ f_{X_{(n)}}(x) = \frac{d}{dx} x^n = n x^{n-1}, \quad x \in [,1] \] --- ### b) \( E[X_{(k)}] \), the Expected Value of the \( k \)-th Order Statistic The expected value of the \( k \)-th order statistic for \( n \) i.i.d. Uniform(,1) variables is: \[ E[X_{(k)}] = \frac{k}{n+1} \] --- ### c) \( P(X_{(n)} - X_{(1)} < .5) \) Let \( R = X_{(n)} - X_{(1)} \). We want \( P(R < .5) \). The joint PDF of \( X_{(1)} = u \), \( X_{(n)} = v \) (for \( < u < v < 1 \)) is: \[ f_{X_{(1)}, X_{(n)}}(u, v) = n(n-1)(v-u)^{n-2} \] The probability is: \[ P(X_{(n)} - X_{(1)} < .5) = \int_^1 \int_u^{\min(u+.5,1)} n(n-1)(v-u)^{n-2} \, dv \, du \] Change variable \( t = v-u \), so \( v = u+t \) and \( t \in (, \min(.5, 1-u)) \): \[ = \int_^1 \int_^{\min(.5, 1-u)} n(n-1) t^{n-2} \, dt \, du \] Split the integration limits: - For \( u \in [, .5] \), upper limit is .5. - For \( u \in [.5, 1] \), upper limit is \( 1-u \). So, \[ = \int_^{.5} \int_^{.5} n(n-1)t^{n-2} \, dt \, du + \int_{.5}^1 \int_^{1-u} n(n-1)t^{n-2} \, dt \, du \] Calculate the inner integrals: \[ \int_^{a} t^{n-2} dt = \frac{a^{n-1}}{n-1} \] So, \[ = \int_^{.5} n \cdot (.5)^{n-1} du + \int_{.5}^1 n (1-u)^{n-1} du \] \[ = n(.5)^{n-1} \cdot .5 + n \int_{.5}^1 (1-u)^{n-1} du \] \[ = n(.5)^{n} + n \left[ -\frac{(1-u)^n}{n} \right]_{u=.5}^{u=1} \] \[ = n(.5)^n + \left[ (1-.5)^n - (1-1)^n \right] \] \[ = n(.5)^n + (.5)^n - \] \[ = (n+1)(.5)^n \] --- ## Summary - **PDF of the maximum:** \( f_{X_{(n)}}(x) = n x^{n-1} \) for \( x \in [,1] \) - **Expected value of \( k \)-th order statistic:** \( E[X_{(k)}] = \frac{k}{n+1} \) - **Probability that range < .5:** \( P(X_{(n)} - X_{(1)} < .5) = (n+1)(.5)^n \)

# Given Information - \( X_1, X_2, \ldots, X_n \) are independent and identically distributed random variables. - Each \( X_i \sim \text{Uniform}(0, 1) \). # What to Find 1. PDF of the maximum \( X_{(n)} = \max(X_1, \ldots, X_n) \). 2. \( E[X_{(k)}] \) — the expected value of the \( k \)-th order statistic. 3. \( P(X_{(n)} - X_{(1)} < 0.5) \) — the probability that the range is less than 0.5. # Definitions/Concepts Used - **Order Statistics**: For i.i.d. random variables, \( X_{(1)} \leq X_{(2)} \leq \cdots \leq X_{(n)} \). - **PDF of Maximum**: The cumulative distribution function (CDF) of the maximum is \( F_{X_{(n)}}(x) = [F_X(x)]^n \), where \( F_X(x) \) is the CDF of the uniform distribution. - **Expected Value of Order Statistics**: The expected value of the \( k \)-th order statistic for Uniform(0,1) is \( E[X_{(k)}] = \frac{k}{n+1} \). - **Probability involving Order Statistics**: For the range, use the joint PDF and integrate over the defined region. --- ## Solution ### a) PDF of the Maximum \( X_{(n)} \) 1. The CDF of \( X_{(n)} \) is given by: \[ F_{X_{(n)}}(x) = [F_X(x)]^n = x^n \quad \text{for } x \in [0, 1]. \] 2. The PDF is the derivative of the CDF: \[ f_{X_{(n)}}(x) = \frac{d}{dx} x^n = n x^{n-1}, \quad x \in [0, 1]. \] --- ### b) Expected Value of the \( k \)-th Order Statistic The expected value of the \( k \)-th order statistic for \( n \) i.i.d. Uniform(0,1) variables is: \[ E[X_{(k)}] = \frac{k}{n+1}. \] --- ### c) Probability \( P(X_{(n)} - X_{(1)} < 0.5) \) 1. Let \( R = X_{(n)} - X_{(1)} \). We want: \[ P(R < 0.5). \] 2. The joint PDF of \( X_{(1)} = u \) and \( X_{(n)} = v \) (for \( 0 < u < v < 1 \)) is: \[ f_{X_{(1)}, X_{(n)}}(u, v) = n(n-1)(v-u)^{n-2}. \] 3. The probability can be expressed as: \[ P(X_{(n)} - X_{(1)} < 0.5) = \int_0^1 \int_u^{\min(u+0.5, 1)} n(n-1)(v-u)^{n-2} \, dv \, du. \] 4. Change variables with \( t = v-u \): \[ = \int_0^1 \int_0^{\min(0.5, 1-u)} n(n-1) t^{n-2} \, dt \, du. \] 5. Split the integration limits based on \( u \): - For \( u \in [0, 0.5] \): upper limit is \( 0.5 \). - For \( u \in [0.5, 1] \): upper limit is \( 1-u \). 6. Calculate the inner integrals: \[ \int_0^a t^{n-2} dt = \frac{a^{n-1}}{n-1}. \] 7. Combine the results: \[ = \int_0^{0.5} n \cdot (0.5)^{n-1} \, du + \int_{0.5}^1 n (1-u)^{n-1} \, du. \] 8. Final simplification yields: \[ = n(0.5)^n + (0.5)^n = (n+1)(0.5)^n. \] --- ## Summary - **PDF of the maximum**: \( f_{X_{(n)}}(x) = n x^{n-1} \) for \( x \in [0, 1] \) - **Expected value of \( k \)-th order statistic**: \( E[X_{(k)}] = \frac{k}{n+1} \) - **Probability that range < 0.5**: \( P(X_{(n)} - X_{(1)} < 0.5) = (n+1)(0.5)^n \)