/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Let \(X_{1}, \ldots, X_{n}\) be ... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 2

Let \(X_{1}, \ldots, X_{n}\) be a random sample from a uniform \((a, b)\) distribution. Let \(Y_{1}=\min X_{i}\) and let \(Y_{2}=\max X_{i} .\) Show that \(\left(Y_{1}, Y_{2}\right)^{\prime}\) converges in probability to the vector \((a, b)^{\prime}\).

Short Answer

Expert verified
\((Y_{1}, Y_{2})^{\prime}\) converges in probability to the vector \((a, b)^{\prime}\).

Step by step solution

01

Define \(Y_1\) and \(Y_2\)

Let \(Y_1 = \min(X_1, \ldots, X_n)\) and \(Y_2 = \max(X_1, \ldots, X_n)\), and \(X_1, \ldots, X_n\) are a random sample of size \(n\) from a uniform distribution on \((a, b)\). Ignore the case where \(a=b\).
02

Find the distribution of \(Y_1\) and \(Y_2\)

The cumulative distribution function (CDF) \(F_{Y_1}(y)\) of \(Y_1\) is 1 when \(y > a\), hence \(F_{Y_1}(y)=0\) for \(y \leq a\). For \(a<y<b\), \(F_{Y_1}(y) = 1 - (b - y)^n / (b - a)^n\). Similarly, the CDF \(F_{Y_2}(y)\) of \(Y_2\) is 0 when \(y < b\), hence \(F_{Y_2}(y) = 1\) for \(y \geq b\). For \(a<y<b\), \(F_{Y_2}(y) = (y - a)^n / (b - a)^n\).
03

Apply Probability Convergence

The definition of \(Y_1\) and \(Y_2\) converging in probability to \(a\) and \(b\) respectively means as \(n\) approaches infinity, \(F_{Y_1}(y)\) approaches 1 for all \(y > a\) and \(F_{Y_2}(y)\) approaches 1 for all \(y \geq b\). With the CDF of \(Y_1\) and \(Y_2\), it is clear that as \(n \rightarrow \infty\), the limits of \(F_{Y_1}(y)\) and \(F_{Y_2}(y)\) for \(y > a\) and \(y \geq b\) respectively are 1. Therefore, \(Y_1\) and \(Y_2\) converge in probability to \(a\) and \(b\) respectively.

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