Solutions To Mathematics Textbooks/Introduction to Mathematical Statistics (5th edition) (ISBN 8178086301)/Chapter 1/Section 2

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Section 2.1[edit]

1.1[edit]

Find the union A_1\cup A_2 and the intersection A_1\cap A_2 of the two sets A_1 and A_2 where

(a) A_1=\{0,1,2\},A_2=\{2,3,4\} \,

(b) A_1=\{x:0<x<2\},A_2=\{x:1\le x <3\}

(c) A_1=\{(x,y):0<x<2,0<y<2\},A_2=\{(x,y):1<x<3,1<y<3\} \,

Solution (c):

A_1\cup A_2=\{(x,y):0<x<3, 0<y<2 \or 1<x<3,2<y<3\}

A_1\cap A_2=\{(x,y):1<x<2, 0<y<2\}

Rest are similar.


1.3[edit]

List all the possible arrangements of the four letters m,a,r and y. Let A_1 be the collection of the arrangements in which y is in the last position. Let A_2 be the collection of the arrangements in which m is in the first position. Find the union and intersection of A_1 and A_2.

Solution: The arrangements are: mary, mayr, mray, mrya, myra, myar, amry, amyr, aymr, ayrm, arym, army, ryma, ryam, ramy, raym, rmya, rmay, yrma, yram, yamr, yarm, ymra, ymar.

A_1=\{mary,mray,amry,army,ramy,rmay\}

A_2=\{mary, mayr, mray, mrya, myra, myar\}. The rest is obvious.


1.5[edit]

If a sequence of sets A_1,A_2,A_3\cdots is such that A_k\subset A_{k+1}, k=1,2,3,\cdots the sequence is said to be a nondecreasing sequence. Give an example of this kind of sequence of sets.

Solution: Let A_n=(0,n) where n = 1, 2, 3...


1.6[edit]

If a sequence of sets A_1,A_2,A_3\cdots is such that A_k\supset A_{k+1}, k=1,2,3,\cdots the sequence is said to be a nonincreasing sequence. Give an example of this kind of sequence of sets.

Solution: Let A_n=(0,\frac{1}{n}) where n=1,2,3...


1.7[edit]

If A_1,A_2,A_3\cdots are sets such that A_k\subset A_{k+1} k=1,2,3,\cdots, \lim_{k \to \infty}A_k is defined as the union A_1\cup A_2\cup A_3\cup\cdots . Find \lim_{k \to \infty}A_k if

(a) A_k=\{x:1/k\le x\le 3-1/k\}, k=1,2,3\cdots

(b) A_k=\{(x,y):1/k\le x^2+y^2\le 4-1/k\}, k=1,2,3\cdots

Solution: (b) \lim_{k \to \infty}A_k=\{(x,y):1< x^2+y^2<4\}


1.8[edit]

If A_1,A_2,A_3\cdots are sets such that A_k\supset A_{k+1} k=1,2,3,\cdots, \lim_{k \to \infty}A_k is defined as the intersection A_1\cap A_2\cap A_3\cup\cdots . Find \lim_{k \to \infty}A_k if

(a) A_k=\{x:2-1/k< x\le 2\}, k=1,2,3\cdots

(b) A_k=\{x:2<x\le 2+1/k\}, k=1,2,3\cdots

(c) A_k=\{(x,y):0\le x^2+y^2\le 1/k\}, k=1,2,3\cdots

Solution: (c) \lim_{k \to \infty}A_k=\{(0,0)\}


1.9[edit]

For every one dimensional set A let Q(A)=\sum_A f(x) where f(x)=\begin{cases}\frac{2}{3}(\frac{1}{3})^x,  & x=0,1,2,\cdots \\ 0, & otherwise \end{cases}. If A_1=\{x:x=0,1,2,3\} \, and A_2=\{x:x=0,1,2,\cdots\}, find Q(A_1) and Q(A_2).

Solution: Q(A_1)=\frac{80}{81} and Q(A_2)=1 using the formulae of sum of a geometric series.


1.10[edit]

For every one dimensional set A for which the integral exists, let Q(A)=\int_A f(x) where f(x)=\begin{cases}6x(1-x),  & 0<x<1 \\ 0, & elsewhere \end{cases}, otherwise let Q(A) be undefined. If A_1=\{x:\frac{1}{4}<x<\frac{3}{4}\}, A_2=\{\frac{1}{2}\} and A_3=\{x:0<x<10\}, find Q(A_1), Q(A_2) and Q(A_3).

Solution: Q(A_1) can be found by integrating f(x) from \frac{1}{4} to \frac{3}{4}, Q(A_2)=0 as integral over a set of measure zero is 0 and Q(A_3) can be found by integrating f(x) from 0 to 1.


1.11[edit]

Let Q(A)=\int_A\int(x^2+y^2)dxdy for every two dimensional set A for which the integral exists; otherwise let Q(A) be undefined. If A_1=\{(x,y):-1\le x\le 1, -1\le y\le 1\}, A_2=\{(x,y):-1\le x=y\le 1\}, and A_3=\{(x,y):x^2+y^2\le 1\}, find Q(A_1), Q(A_2) and Q(A_3).

Solution: Q(A_1)= \int_{-1}^1\int_{-1}^1(x^2+y^2)dxdy=\frac{8}{3}, Q(A_2) is zero since it represents volume of a sheet and Q(A_3)=\int_{0}^{2\pi}\int_{0}^1(r^2)rdrd\theta if we introduce polar coordinates. This evaluates to \frac{\pi}{2}.


1.15[edit]

To join a certain club, a person must be either a statistician or a mathematician or both. Of the 25 members in this club, 19 are statisticians and 16 are mathematicians. How many persons in the club are both a statistician and a mathematician?

Soluion: The number of persons which are statisticians or mathematicians = number of statisticians + number of mathematicians - number of persons which are both statisticians and mathematicians. This can be proved directly using properties of the counting measure. The general proof is as follows: Let A and B be two finite sets. Endow the natural numbers with the counting measure which we shall denote by n. We shall denote union by + and intersection by . and the context will make it clear whether we mean actual addition, multiplication or union, intersection. Now A-B (Here - represents the relative complement, AB and B-A are disjoint and as n is sigma additive so

  1. n(A+B)=n(A-B)+n(B-A)+n(AB)
  2. n(A)=n(A-B)+n(AB) which implies n(A-B)=n(A)-n(AB)
  3. n(B)=n(B-A)+n(AB) which implies n(B-A)=n(B)-n(AB)

Substituting n(A-B) and n(B-A) from (2) and (3) into (1) gives us n(A+B)=n(A)+n(B)-n(AB). So the answer is 10.


1.16[edit]

After a hard fought football game, it was reported that of the 11 starting players, 8 hurt a hip, 6 hurt an arm, 5 hurt a knee, 3 hurt both a hip and an arm, 2 hurt both a hip and a knee, 1 hurt both an arm and a knee, and no one hurt all three. Comment on the accuracy of the report.

Solution: We shall first establish that in the language of the previous problem:


n(A+B+C)=n(A)+n(B)+n(C)-n(AB)-n(BC)-n(CA)+n(ABC) \,.


Now,


n(A+B+C)=n(A+B)+n(C)-n((A+B)C)\, =n(A)+n(B)+n(C)-n(AB)-n(AC+BC)\, =n(A)+n(B)+n(C)-n(AB)-{n(AC)+n(BC)-n(ACBC)}\, =n(A)+n(B)+n(C)-n(AB)-n(BC)-n(CA)+n(ABC)\,


Now if we let A, B and C to be the set of players with injured hips, knees and arms respectively, we would find the contradiction on substituting the appropriate terms in the above formula. Hence the report is not accurate.