/*! 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 1 If \(A_{1}, A_{2}, \ldots, A_{k}... [FREE SOLUTION] | 91影视

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Chapter 9: Problem 1

If \(A_{1}, A_{2}, \ldots, A_{k}\) are events, prove, by induction, Boole's inequality $$ P\left(A_{1} \cup A_{2} \cup \cdots \cup A_{k}\right) \leq \sum_{1}^{k} P\left(A_{i}\right) $$ Then show that $$ P\left(A_{1}^{c} \cap A_{2}^{c} \cap \cdots \cap A_{k}^{c}\right) \geq 1-\sum_{1}^{b} P\left(A_{i}\right) $$

Short Answer

Expert verified
The proof for the Boole's inequality is done by the method of induction by considering the base case where it holds for k=1 and then we assume it holds for k=n and prove for k=n+1. Then we prove that \(P(A_{1}^{c} \cap A_{2}^{c} \cap \cdots \cap A_{k}^{c}) \geq 1 - \sum_{1}^{k} P(A_{i})\) by using the complement rule and the result obtained from Boole's inequality.

Step by step solution

01

Prove the Base Case

For the base case, let k=1. Then, the inequality becomes \(P(A_{1}) \leq P(A_{1})\), which is clearly true, because the probability of any event cannot exceed itself.
02

Prove the Induction Step

Assume that the inequality holds for k = n (i.e., \(P(A_{1} \cup A_{2} \cup \ldots \cup A_{n}) \leq \sum_{1}^{n} P(A_{i})\)). We have to prove that it holds for k = n+1. Using probability rules, we can express \(P(A_{1} \cup A_{2} \cup \ldots \cup A_{n+1})\) as \(P((A_{1} \cup A_{2} \cup \ldots \cup A_{n}) \cup A_{n+1}) \leq P(A_{1} \cup A_{2} \cup \ldots \cup A_{n}) + P(A_{n+1})\). By our induction assumption, \(P(A_{1} \cup A_{2} \cup \ldots \cup A_{n}) \leq \sum_{i=1}^{n} P(A_{i})\), hence, \(P(A_{1} \cup A_{2} \cup \ldots \cup A_{n+1}) \leq \sum_{i=1}^{n+1} P(A_{i})\).
03

Show the Second Inequality

We know that \(P(A_{1}^{c} \cap A_{2}^{c} \cap \ldots \cap A_{k}^{c})\) is the probability of the complement of the union \(A_{1} \cup A_{2} \cup \ldots \cup A_{k}\). From Boole's inequality, we obtain that \(P(A_{1} \cup A_{2} \cup \ldots \cup A_{k}) \leq \sum_{1}^{k} P(A_{i})\), and thus \(1 - P(A_{1} \cup A_{2} \cup \ldots \cup A_{k}) \geq 1 - \sum_{1}^{k} P(A_{i})\). Hence, \(P(A_{1}^{c} \cap A_{2}^{c} \cap \ldots \cap A_{k}^{c}) \geq 1 - \sum_{1}^{k} P(A_{i})\).

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Most popular questions from this chapter

Given the following observations associated with a two-way classification with \(a=3\) and \(b=4\), compute the \(F\) -statistic used to test the equality of the column means \(\left(\beta_{1}=\beta_{2}=\beta_{3}=\beta_{4}=0\right)\) and the equality of the row means \(\left(\alpha_{1}=\alpha_{2}=\alpha_{3}=0\right)\), respectively. $$ \begin{array}{ccccc} \hline \text { Row/Column } & 1 & 2 & 3 & 4 \\ \hline 1 & 3.1 & 4.2 & 2.7 & 4.9 \\ 2 & 2.7 & 2.9 & 1.8 & 3.0 \\ 3 & 4.0 & 4.6 & 3.0 & 3.9 \\ \hline \end{array} $$

Let \(X_{1}, X_{2}, \ldots, X_{n}\) be a random sample from a normal distribution \(N\left(\mu, \sigma^{2}\right)\). Show that $$ \sum_{i=1}^{n}\left(X_{i}-\bar{X}\right)^{2}=\sum_{i=2}^{n}\left(X_{i}-\bar{X}^{\prime}\right)^{2}+\frac{n-1}{n}\left(X_{1}-\bar{X}^{\prime}\right)^{2}, $$ where \(\bar{X}=\sum_{i=1}^{n} X_{i} / n\) and \(\bar{X}^{\prime}=\sum_{i=2}^{n} X_{i} /(n-1)\). Hint: \(\quad\) Replace \(X_{i}-\bar{X}\) by \(\left(X_{i}-\bar{X}^{\prime}\right)-\left(X_{1}-\bar{X}^{\prime}\right) / n\). Show that \(\sum_{i=2}^{n}\left(X_{i}-\bar{X}^{\prime}\right)^{2} / \sigma^{2}\) has a chi-square distribution with \(n-2\) degrees of freedom. Prove that the two terms in the right-hand member are independent. What then is the distribution of $$ \frac{[(n-1) / n]\left(X_{1}-\bar{X}^{\prime}\right)^{2}}{\sigma^{2}} ? $$

Let \(\mathbf{X}^{\prime}=\left[X_{1}, X_{2}\right]\) be bivariate normal with matrix of means \(\boldsymbol{\mu}^{\prime}=\left[\mu_{1}, \mu_{2}\right]\) and positive definite covariance matrix \(\mathbf{\Sigma}\). Let $$ Q_{1}=\frac{X_{1}^{2}}{\sigma_{1}^{2}\left(1-\rho^{2}\right)}-2 \rho \frac{X_{1} X_{2}}{\sigma_{1} \sigma_{2}\left(1-\rho^{2}\right)}+\frac{X_{2}^{2}}{\sigma_{2}^{2}\left(1-\rho^{2}\right)} $$ Show that \(Q_{1}\) is \(\chi^{2}(r, \theta)\) and find \(r\) and \(\theta\). When and only when does \(Q_{1}\) have a central chi-square distribution?

Let \(Q_{1}\) and \(Q_{2}\) be two nonnegative quadratic forms in the observations of a random sample from a distribution which is \(N\left(0, \sigma^{2}\right) .\) Show that another quadratic form \(Q\) is independent of \(Q_{1}+Q_{2}\) if and only if \(Q\) is independent of each of \(Q_{1}\) and \(Q_{2}\) Hint: \(\quad\) Consider the orthogonal transformation that diagonalizes the matrix of \(Q_{1}+Q_{2}\). After this transformation, what are the forms of the matrices \(Q, Q_{1}\) and \(Q_{2}\) if \(Q\) and \(Q_{1}+Q_{2}\) are independent?

Let the \(4 \times 1\) matrix \(\boldsymbol{Y}\) be multivariate normal \(N\left(\boldsymbol{X} \boldsymbol{\beta}, \sigma^{2} \boldsymbol{I}\right)\), where the \(4 \times 3\) matrix \(\boldsymbol{X}\) equals $$ \boldsymbol{X}=\left[\begin{array}{rrr} 1 & 1 & 2 \\ 1 & -1 & 2 \\ 1 & 0 & -3 \\ 1 & 0 & -1 \end{array}\right] $$ and \(\beta\) is the \(3 \times 1\) regression coeffient matrix. (a) Find the mean matrix and the covariance matrix of \(\hat{\boldsymbol{\beta}}=\left(\boldsymbol{X}^{\prime} \boldsymbol{X}\right)^{-1} \boldsymbol{X}^{\prime} \boldsymbol{Y}\). (b) If we observe \(\boldsymbol{Y}^{\prime}\) to be equal to \((6,1,11,3)\), compute \(\hat{\boldsymbol{\beta}}\).
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