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Chapter 1: Problem 10

For every one-dimensional set \(C\), define the function \(Q(C)=\sum_{C} f(x)\), where \(f(x)=\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)^{x}, x=0,1,2, \ldots\), zero elsewhere. If \(C_{1}=\\{x: x=0,1,2,3\\}\) and \(C_{2}=\\{x: x=0,1,2, \ldots\\}\), find \(Q\left(C_{1}\right)\) and \(Q\left(C_{2}\right)\). Hint: Recall that \(S_{n}=a+a r+\cdots+a r^{n-1}=a\left(1-r^{n}\right) /(1-r)\) and, hence, it follows that \(\lim _{n \rightarrow \infty} S_{n}=a /(1-r)\) provided that \(|r|<1\).

Short Answer

Expert verified
The results are \(Q(C_{1}) = \frac{80}{81}\) and \(Q(C_{2}) = 1\).

Step by step solution

01

Calculate Q(C_1)

In this step, we will calculate \(Q(C_{1})\) by summing up the values of \(f(x)\) for all \(x\) in \(C_{1}\), i.e. \(x = 0, 1, 2, 3\). Hence, \(Q(C_{1}) = f(0) + f(1) + f(2) + f(3)\). Substituting \(f(x) =\left(\frac{2}{3}\right)\left(\frac{1}{3}\right)^{x}\) for all \(x\) in \(C_{1}\) and adding up the results, we get \(Q(C_{1}) = \frac{2}{3} + \frac{2}{9}+ \frac{2}{27}+ \frac{2}{81}\)
02

Simplify Q(C_1)

In this step, a common denominator (81) is used to add the numbers. Hence, the simplified form of \(Q(C_{1})\) will be \(Q(C_{1})= \frac{54 +18 +6 +2}{81} = \frac{80}{81}\)
03

Calculate Q(C_2)

To calculate \(Q(C_{2})\), we will need to sum up the values of function \(f(x)\) for all \(x\) in \(C_{2}\); however, \(C_{2}\) is an infinite set. Here the geometric series sum formula comes into play. The formula is: \(S = \frac{a}{1-r}\) provided \(|r|<1\), where \(a\) is the first term and \(r\) is the common ratio. Here, \(a = \frac{2}{3}\) (the first term) and \(r = \frac{1}{3}\) (the common ratio). Therefore, \(Q(C_{2}) = S = \frac{\frac{2}{3}}{1 - \frac{1}{3}}\)
04

Simplify Q(C_2)

In this step, simply the expression to get the final value of \(Q(C_{2})\). Hence, \(Q(C_{2}) = \frac{2}{3} \times \frac{3}{2} = 1\)

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Most popular questions from this chapter

A professor of statistics has two boxes of computer disks: box \(C_{1}\) contains seven Verbatim disks and three Control Data disks and box \(C_{2}\) contains two Verbatim disks and eight Control Data disks. She selects a box at random with probabilities \(P\left(C_{1}\right)=\frac{2}{3}\) and \(P\left(C_{2}\right)=\frac{1}{3}\) because of their respective locations. A disk is then selected at random and the event \(C\) occurs if it is from Control Data. Using an equally likely assumption for each disk in the selected box, compute \(P\left(C_{1} \mid C\right)\) and \(P\left(C_{2} \mid C\right)\).

Players \(A\) and \(B\) play a sequence of independent games. Player \(A\) throws a die first and wins on a "six." If he fails, \(B\) throws and wins on a "five" or "six." If he fails, \(A\) throws and wins on a "four," "five," or "six." And so on. Find the probability of each player winning the sequence.

Hunters \(\mathrm{A}\) and \(\mathrm{B}\) shoot at a target; the probabilities of hitting the target are \(p_{1}\) and \(p_{2}\), respectively. Assuming independence, can \(p_{1}\) and \(p_{2}\) be selected so that \(P(\) zero hits \()=P(\) one hit \()=P(\) two hits \() ?\)

Consider \(k\) continuous-type distributions with the following characteristics: pdf \(f_{i}(x)\), mean \(\mu_{i}\), and variance \(\sigma_{i}^{2}, i=1,2, \ldots, k .\) If \(c_{i} \geq 0, i=1,2, \ldots, k\), and \(c_{1}+c_{2}+\cdots+c_{k}=1\), show that the mean and the variance of the distribution having pdf \(c_{1} f_{1}(x)+\cdots+c_{k} f_{k}(x)\) are \(\mu=\sum_{i=1}^{k} c_{i} \mu_{i}\) and \(\sigma^{2}=\sum_{i=1}^{k} c_{i}\left[\sigma_{i}^{2}+\left(\mu_{i}-\mu\right)^{2}\right]\) respectively.

Let \(X\) have the pdf \(f(x)=x^{2} / 9,0
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