econ118 practice quiz

8/9/2019 ECON118 Practice Quiz

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Practice Quiz Problems for ECON 118

Tomoki FUJII

1. b 0.072727272 . . . can be considered as a sum of geometric sequencewith initial terma and common ratior. What area andr? Write b ina fraction in the lowest terms.

2. b 1.010101010 . . . can be considered as a sum of geometric sequencewith initial terma and common ratior. What area andr? Write b ina fraction in the lowest terms.

3. Letf(x) = e2x + 2ex1, wheree is the base of the natural logarithm.Findf(x). At what value ofx is f(x) maximized? Find the maximumvalue off(x).

4. Letf(x) =

x2

2 + ln xforx >0. Find f(x) and f(x). At what value

ofxisf(x) maximized? Find the maximum value off(x).

5. You throw two balanced dice. LetX be the sum of the numbers thedice are displaying after the throw. Find P(X = 2). Answer in afraction in the lowest terms.

6. (Continued from previous question) What are E[X] and E[max(X, 10)]?Answer in a fraction in the lowest terms.

7. Write 1.1111 in the form of a fraction in the lowest terms. Noticethat this number can be considered as a sum of a geometric sequence.

8. a= 0.1010101 . . . can be considered as a sum of a geometric sequence.What are the initial value and common ratio of the sequence? Writeain a fraction in the lowest terms.

9. Consider a sequence 1, 11+i

, 1(1+i)2

, . . . , for i >0. What is the sum Softhis sequence?

These problems are provided for you to practice on your own. You do not need to

submit what you did. In the actual quiz, you should expect about five questions.

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10. SupposeXis a random variable that takes 0 with probability 12

, 2 with

probability 14 and 6 with probability 14 . Find E[X], and E[X2].

11. Let f(x) =ex + ln x 3x2. Find f(x) and f(x).12. Let f(x) =ex

2

. Find f(x) and f(x).

13. You have a small die and a big die, both of which are balanced. Youthrow them together. Let Xand Ybe the numbers the big and smalldice are displaying. Find E[X], E[X2], and E[max(X, Y)].

14. Letf(x) =

x(b x), whereb >0 andx (0, b). Findf(x), and themaximum off(x) on (0, b).

15. Suppose thatg(x) =ex2ex3x. Find the maximum and minimumofg(x) when xsatisfies ln 2 x ln 2. You may use ln 2 0.693.

16. Let f(x) = [x(b x)]c, where b, c >0 and x (0, b). Findf(x), andthe maximum off(x) on (0, b).

17. Suppose there are two goodXandY. Your utility function is given byU(X, Y) = ln X+(1) ln Y with (0, 1). The prices ofXandYare pand q. Suppose your income is I. Find the optimal consumptionschedule (X, Y ) that maximizes your utility U(X, Y) subject to the

budget constraint pX+qY =I.

18. Let X be the number displaying after a throw of an unbalanced die.The random variableXfollows the probability distribution given in thetable below. Find the probability that X is an even number? Answerin a fraction in the lowest terms. Also write E[X] in a fraction in thelowest terms.

x 1 2 3 4 5 6P(X=x) 1/21 2/21 3/21 4/21 5/21 6/21

19. Writea= 0.45454545 . . . in a fraction in the lowest terms.

20. Let x > 0 and f(x) = ex

+ex, where e is the base of the naturallogarithm or the Napiers constant. Find f(x).

21. Find the minimum off(x) in the previous problem.

22. Let X be the number displaying after a throw of an unbalanced die.The random variableXfollows the probability distribution given in thetable below. Find the probability that X is an odd number? Answerin a fraction in the lowest terms.

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x 1 2 3 4 5 6

P(X=x) 0 1/4 1/3 1/4 0 1/6

23. What are E[X] and E[X2]? Answer in a fraction in the lowest terms.

24. Suppose you throw two balanced dice. Let X be the product of thenumbers the dice are displaying after the throw. Find P(X= 4) andE[X]. Answer in a fraction in the lowest terms.

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Answers

1. b= 0.0727272 = 0.072+0.00072+0.0000072 + . Thus, it can beconsidered as a sum of a geometric sequence with initial valuea= 0.072and common ratio r= 0.01. b= 0.072

10.01= 4

55.

2. b = 1.0101010101 = 1 + 0.01 + 0.0001 + . Thus, it can be con-sidered as a sum of a geometric sequence with initial value a= 1 andcommon ratio r= 0.01. r= 1

10.01= 100

99.

3. f(x) = 2e2x + 2ex. Solving f(x) = 0, we haveex = 1 or x= 0. Themaximum is f(0) = 0.

4. f(x) =x+ 1x

and f(x) =1 1x2

. f(x) = 0 when x = 1. Themaximum is f(1) = 1/2.

5. X= 2 occurs only when both dice are displaying 1. Therefore, P(X=2) = 1/36.

6. First, notice P(X = 2) = P(X= 12) = 1/36, P(X = 3) = P(X =11) = 2/36,P(X= 4) =P(X= 10) = 3/36,P(X= 5) =P(X= 9) =4/36, P(X= 6) = P(X= 8) = 5/36, and P(X= 7) = 6/36. Thus,E[X] = (2+12)1/36+(3+11)2/36+(4+10)3/36+(5+9)4/36+(6+8) 5/3 6 + 7 6/36 = 7 (This is a sure way to do, but there is a simplerway. You can calculate E[X] as twice the expectation of a single diethrow). E[max(X, 10)] = P(X 10) 10 + P(X= 11) 11 + P(X =12) 12 = 33/36 10 + 2/36 11 + 1/36 12 = 91/9.

7. Notice that 1.1111 = 1 + 0.1 + 0.0 1 + 0.001 + . Thus, it is a sumof a geometric sequence with a unit initial value and the common ratioof 0.1. Hence,

1.1111 =t=0

10t

= 1

1 101

= 10

10 1=

10

9

8. a= 0.1010101 . . . can be considered as a sum of a geometric sequencewith initial value 0.1 and common ratio 0.01. a= 0.1

10.01= 10

99.

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9. This is a geometric sequence. The initial value is unity, and the common

ratio is 11+i . Notice that 11+i

< 1. Therefore, the sum is S=i=0(1 +i)i = 1

1 11+i

= 1+ii

.

10. E[X] = 0 12

+ 2 14

+ 6 14

= 2. E[X2] = 02 12

+ 22 14

+ 62 14

= 10.

11. f(x) =ex +x1 6x, f(x) =ex x2 6.

12. You use the chain rule. For f(x), you also need to use the productrule.

f(x) = 2xex2

f(x) = (2x) ex2

+ 2x

ex2

= 2ex2

+ (2x)2ex2

= (4x2 + 2)ex2

13. Note that P(X = 1) = P(X = 2) = P(X = 6) = 16

. If you arenot comfortable with this notation, read the slides carefully and makeyourself familiar with it.

E[X] =6

x=1

xP(X=x)

= 1 P(X= 1) + 2 P(X= 2) + + 6 P(X= 6)= (1 + 2 + + 6) 1

6=

7

2

E[X2] =6

x=1

x2P(X=x)

= 12 P(X= 1) + 22 P(X= 2) + + 62 P(X= 6)= (12 + 22 + + 62) 1

6=91

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Now, let Z = max(X, Y). Z = 1 when X = 1 and Y = 1. ThusP(Z= 1) = P(X = 1&Y = 1) = 1

36. Likewise, P(Z= 2) = P(X =

1&Y = 2) + P(X = 2&Y = 2) + P(X = 2&Y = 1 ) = 336

, andP(Z= 3) = P(X= 1&Y= 3) +P(X= 2&Y= 3) +P(X= 3&Y =3) +P(X= 3&Y= 2) +P(X= 3&Y= 1) = 5

36.

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Verify P(Z = 4) = 736

, P(Z = 5) = 936

and P(Z = 6) = 1136

. There-

fore,E[max(X, Y)] =E[Z] = 1 136+2 336+3 536+4 736+5 936+6 1136 = 16136.

14. f(x) = b2x2

x(bx). f(x) = 0 when x = b

2. It is straightforward to see

that f(x) > 0 if and only if 0 < x < b2

and f(x) < 0 if and onlyif b

2 < x < b. Thus, f(x) takes a maximum at x = b

2. Thus, the

maximum off(x) isf( b2

) = b2

.

15. Taking the derivative ofx, we have g(x) = ex + 2ex 3 =ex(ex 1)(ex 2). Since ex >0, g(x) = 0 if and only ifex is equal to 1 or 2.g

(x)< 0 if and only if 1 < ex

0 if and only ifx > ln 22 . So, g(x) is locally concave around

x= 0, and thusg(x) takes a local maximum around x = 0, and a localminimum at x= ln 2.

To find the minimum, we compare the values of g(x) at x = ln 2and x= ln 2. Since g( ln2) = 7/2 + 3 ln 2 1.421, and g(ln2) =13 l n 2 1.079, the minimum occurs atx= ln 2 and the minimumis7/2 + 3 ln2.

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-2

-1.5

-1

-0.5

-1 -0.5 0 0.5 1

-7/2+3ln 2

1-3ln 2

-ln 2ln 2

16. f(x) =c[x(b x)]c1(b 2x). Obviously, f(x) = 0 when x= b2

. It isstraightforward to show that f(x) > 0 if and only if 0 < x < b

2

andf(x) < 0 if and only if b

2< x < b. Thus, f(x) takes a maximum at

x= b2

. Thus, the maximum off(x) isf( b2

) =b2

2c.

17. The maximization you face is as follows:

maxX,Y

ln X+ (1 ) ln Y s.t. pX+qY =I

Plugging the constraint in the objective function, we have

maxX

ln X+ (1 )ln(IpX)q1.

Now, let us define f(X) = ln X+ (1)ln(I pX)q1. Then,f(X) = 1

X+ (1 ) p

IpX. Setting this equal to zero and solving for

X, we have X = Ip

and Y = IpX

q = (1)I

q .

18. P(Xis an even number) = P(X= 2) +P(X= 4) +P(X= 6) = 221

+421

+ 621

= 47

. E[X] = 1P(X= 1)+2P(X= 2)+3P(X= 3)+4P(X=4)+5 P(X= 5)+ 6 P(X= 6) = 1

21+ 4

21+ 9

21+ 16

21+ 25

21+ 36

21= 91

21= 13

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19. a= 0.45454545 . . . can be considered as a sum of a geometric sequence

with initial value 0.45 and common ratio 0.01. Thus, a= 0.4510.01 = 511 .

20. f(x) = ex2

+ex

21. f(x) = 0 when ex = ex2

. This occurs when x= 1. Thus, the minimumisf(1) = 2e.

22. X is an odd number when X = 1, X = 3 or X = 5. Thus, P(X =1) +P(X= 3) +P(X= 5) = 0 + 1/3 + 0 = 1/3.

23. E[X] = 10+2 1/4+3 1/3+4 1/4+5 0+6 1/6 = 1/2+1+1+1 = 7/2.

E[X2

] = 12

0+22

1/4+32

1/3+42

1/4+52

0+62

1/6 = 1+3+4+6 = 14.24. X = 4 occurs when both are two or one of the dice is 1 and the

other is 4. Therefore, P(X = 4) = 3/36 = 1/12. To find E[X], wedistinguish the two dice by arbitrarily calling the first die and the sec-ond die. Let X1 and X2 be the numbers that the first and seconddice are displaying, respectively. By definition X = X1X2. There-fore, E[X] = E[X1X2] =

6i=1

6j=1i j P(X1 = i, X2 = j) =

136

6i=1

6j=1i j = 136

6i=1i

6j=1j

= 2121

36 = 49

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econ118 practice quiz

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