Tu torial letter 205/1/ 2014Tu Dis ST Sem Dep Examin Solutio toria tribu A26 este artme ation prep...

Tu Dis

ST Sem Dep

Examin

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Statis

ENTS:

torial letter

05/1/

ry II

stics

r

/2014

STA

4

A2603/205//1/2014

2

Dear Student,

This tutorial letter intends to help you prepare for the examination. It contains the following information:

1. About the May/June 2014 examinations.2. Solutions to a previous year’s examination paper.3. Solutions to Assignment 6.

1. ABOUT THE MAY/JUNE 2014 EXAMINATIONS

What you are expected to be able to do in the exam

You are expected to know the following distributions, remember their density functions or probability mass func-tions, recognize them, know when they apply, and be able to calculate probabilities from them:

The Bernoulli distribution The Binomial distribution (the probability mass function will be given in the paper) The Poisson distribution The uniform distribution The normal distribution

You may be asked to work with any other distributions discussed in the study guide as well, in which case thedistribution will be given in the exam paper. In the exam you are given the formulas for the Binomial probabilitymass function, the density function of the general Gamma distribution, the density function of the k-th order statisticfrom a sample of n from a given distribution, and the series expansion of the exponential function. Further formulaswill be given as needed. You will also be given cumulative standard normal distribution tables.

The next page contains a brief summary of what you need to be able to do in the exam.

3 STA2603/205

Random variables Given a discrete or continuous distribution of a random variable (defined via probability mass function/density function or cumulative distribution function),

find: cumulative distribution function from probability mass function,

probability mass function from cumulative distribution function,

probabilities of events distributions of functions of the random variable quartiles, quantiles, medians

Bivariate distributions Given a discrete or continuous joint distribution of two random variables (defined via joint probability mass function/ joint density function or joint cumulative distribution function),

find: joint cumulative distribution function from joint probability mass function and vice versa,

joint density from joint cumulative distribution function and vice versa,

probabilities of events marginal distributions (densities and/or cumulative distribution functions)

conditional distributions Given a joint distribution, test for independence of the random variables Given independence and marginal distributions,

find joint distribution

Work with joint distributions, conditional distributions and marginal distributions (whether deriving density/probability mass functions, cumulative distributions or probabilities of events) Given a joint distribution, find distributions of functions of the random variables,

both by the distribution function method and the transformation method.

Order statistics Use the formula (given in the exam paper) for the density of k‐th order statistics to find distributions for various order statistics Expected values Given distribution or joint distribution,

find expected values of random variables or expected values of functions of the random variables

conditional expectations of random variables and functions of random variables

variances and standard deviations covariances and correlation coefficients moments of the distribution

Moment generating functions Given a distribution (density or probability mass function), find the corresponding moment generating function calculate moments by differentiating the moment generating function Inequalities Apply Chebyshev’s inequality (formula will be given in the exam paper when needed) Limit theorems Apply the central limit theorem Distributions derived from the normal distribution Recognize linear combinations of, sums of squares and quotients of independent normally distributed random variables as coming from normal, chi‐square, F or t distributions, and specify the parameters and degrees of freedom; including the distributions of the sample mean and sample variance.

4

2. SOLUTIONS TO A PREVIOUS YEAR’S EXAMINATION PAPER

Your examination paper for this semester will be approximately similar to previous years’ examination papers. Onesuch paper (October/November 2012) formed the basis for Assignment 6 for this semester. We will give you detailedsolutions to that question paper here. In the following, we give first all the questions, and then the solutions to allthe questions.

More examination papers from previous years are available on the myUnisa website of this module. We will notprovide solutions to them, but you are welcome to contact with your lecturer if you are struggling with a particularquestion. We have also included the tables and the set of equations, exactly as they will be given to you in yourexamination paper. The equations are there just to aid your memory — you will still need to know exactly whichequation to use in which situation! Make sure that you know how to use the given equations and the tables, and howto use your calculator!

We wish you all the best in the examination. Please do contact us for help, if you have any queries!

With regards, Dr E Rapoo

THE QUESTIONS

QUESTION 1

Two discrete random variables X and Y have a joint probability mass function given in the table below.

pXY x y

x1 2 3

1 1814 0

y 2 1818 0

3 14 018

1.1 Calculate the marginal probability mass functions pX x and pY y of X and Y 4

1.2 Are the X and Y independent? Justify your answer! 3

1.3 Calculate the following:

(a) P X 2 Y 2(b) E Y X 2(c) P X 2 Y 1 10

[17]

5 STA2603/205

QUESTION 2

Let

fY Xy x

4x 2y4x 1 0 x 1 0 y 1

0 elsewhere

and

fX x

4x 1

3 0 x 1

0 elsewhere

2.1 Find the marginal density function of Y (6)

2.2 Calculate P

12 X 1 Y 1

(6)

2.3 Determine PY X (6)

[18]

QUESTION 3

Suppose X and Y are two independent random variables of the exponential distribution with parameter 2

3.1 Find the distribution function of the random variable U X Y (8)3.2 Find the moment generating function of X (6)

3.3 Calculate EX5

by differentiating the moment generating function of X (4)

3.4 Find the moment generating function of the random variable W 2XY (6)[24]

QUESTION 4

4.1 Let X1 and X2 be two independent random variables so that the variances of X1 and X2 are 21 k and 22 2 respectively. Given that the variance of Y 3X1 2X2 is 35 find the value of k (4)

4.2 Find the mean and the variance of the sum Y of the items of a random sample of size 5 from the distributionwith probability density function

fX x 6x1 x 0 x 1

0 elsewhere

(8)[12]

6

QUESTION 5

5.1 Suppose that the number of insurance claims (N ) filed during a period of one year is Poisson distributed withEN 10000 Use the normal approximation to the Poisson distribution to approximate PN EN 500 (8)

5.2 State if each of the following statements is true or false. If the statement is false, explain why it is false.

(a) If two random variables come from the same distribution with the same parameters, then they cannot beindependent.

(b) Two discrete random variables are independent if their marginal probability mass functions both sum upto 1.

(c) If fY y fX Y x y for all possible values x and y then X and Y are independent.(d) Only continuous random variables have moment generating functions.

(e) If the joint distribution function FXY x y is known then the marginal distribution of X can be foundby integrating FXY over all values of Y (10)

[18]

QUESTION 6Let X1 X2 X10 be a sample from a N 2 25 distribution, independent of the sample Y1Y2 Y20 comingfrom a N2 100 distribution. Let

X 110

10i1

Xi Y 12020j1

Yj

S2X 19

10i1Xi X2 S2Y

119

20j1Yj Y 2

2X

110

10i1Xi 22

2Y

120

20j1Yj 22

6.1 Explain why Q 6

j1

Yj 2

10

2does not follow a 210 distribution. (3)

6.2 Write down the name of the distribution of U 19S2Y

102and give the value(s) of the parameter(s).

(3)

6.3 Define V A

2X

2Y

For what value of the constant A does V follow an F distribution? What are the degrees

of freedom of this F distribution? Justify your answer! (5)

[11]

TOTAL: [100]

7 STA2603/205

FORMULAE

pXk n

kpk1 pnk

fX x

x1ex

fnx n!k 1!n k! [FXx]

k1 [1 FXx]nk fXx

etx 1 t x t x2

2! t x

3

3!

8

Table 1: Cumulative Standardized Normal Probabilities

9 STA2603/205

Table 1 (continued)

10

THE SOLUTIONS

QUESTION 1

1.1 The marginal probability mass function for X is found by summing up over all y-values, over each column;and the marginal probability mass function for Y is found by summing up over all x-values, over rows. Weget:

pXx

18 18 14 48 12 x 1

14 18 0 38 x 20 0 18 18 x 3

pY x

18 14 0 38 y 1

18 18 0 28 14 y 2

14 0 18 38 y 3

1.2 The random variables are not independent. For instance

pXY 3 1 0 pX3 pY 1 18 38

1.3 (a)

PX 2 Y 2 P X 2 and Y 2P Y 2

pXY 1 2 pXY 2 2pY 2

18 18

14

1

(b) We need to first find the conditional probability distribution of Y given that X 2. We get

PY yX 2 P X 2 Y yP X 2

pXY 2 ypX 2

14

38 23 y 1

18

38 13 y 2

0 y 3It follows that

EY X 2 1 PY 1X 2 2 PY 2X 2 3 PY 3X 2

1 23 2 1

3 3 0 4

3

(c)

PX 2 Y 1 PX 2 since Y 1 is always true

PX 2 or X 3 pX2 pX3 12

11 STA2603/205

QUESTION 2

2.1 The marginal density function of Y is found by integrating the joint density function over all x-values. Herewe are not given the joint density function directly, but it can always be found from the given function asfollows:

fXY x y fY Xyx fXxTherefore, we have

fY y

fXY x ydx

fY X yx fX x dx

1

0

4x 2y4x 1

4x 13

dx 0 y 1

1

0

4x 2y3

dx

46

x2 23

yx1

0

46 2

3y 2 2y

3 for 0 y 1

and 0 elsewhere.

2.2 To find P 1

2 x 1 Y 1

we need to first find the conditional density of X given Y . This is given by

fX Y x y fXY x yfY y

4x2y3

22y3 4x2y22y 2xy1y 0 x 1

0 elsewhere.

Therefore,fX Y x y 1 2x 11 1

2x 12

x 12 0 x 1

and

P

12 X 1 Y 1

1

12

x 1

2

dx

12

x2 12

x11

2

12 1

2

12 1

4 1

2 1

2

1

18 1

4

5

8

2.3

P Y X yx

fXY x y dxdy 1

0

x0

4x 2y3

dy

dx

1

0

134xy y2x

0

dx

10

134x2 x2 dx 1

0

13 5x2dx

13 5

3x31

0 1

3 5

3 5

9

12

QUESTION 3

X and Y are both from the exponential distribution with parameter 2, which means that their density functionsare

fXx ex x 0fY y ey y 0

and the joint density function, since X and Y are independent, is

fXY x y ex ey

3.1 Let U X Y . Then the distribution function of U is

FU u PU u PX Y u xyu

fXY x y dxdy

u

0

ux0

eydyexdx for u 0

u

0

1 eux exdx u

0

ex eu dx

1 eu ueu u 0 zero elsewhere

where 2

Alternatively, we can use the transformation method here. If we take

U X Y V X

then

u g1x y x y g2x y x

with inverse transformationsx y u

The Jacobian is

J x y

g1x

g1y

g2x

g2y

1 11 0

1Therefore,

fUV u fXY u e eu 2eu for u 0

13 STA2603/205

and thereforefU u

fUV u d u

02eudu u2eu u 0

andFU u

u

fU a da u

0a2eada u 0

which gives the same result as before, but the calculations are much longer!

3.2

MX t Eet X

0etxexdx

0etxdx

t etx

0

t 0 1

t (assuming t

3.3 FromMt

t t1

We get

M t 11 t2 t2M t 21 t3 2 t3M t 312 t4 6 t4Mit 416 t4 24 t5Mt 5124 t6 120 t6

and thereforeEX5 M 0 1205

3.4 For W 2XY , the moment generating function isMW t E

etW E et2XY

e2txy fXY x y dxdy

0

0

e2txyexeydxdy

QUESTION 4

4.1 Since X1 and X2 are independent,

V arY V ar3X1 2X2 V ar3X1 V ar 2X2 9V arX1 4V arX2 9 k 4 2 9k 8

so to have V arY 35, we must have9k 8 35∴ 9k 35 8 27∴ k 3

14

4.2 For a random variable X from the distribution

fXx

6x 1 x 0 x 10 elsewhere

we get

EX 1

0x 6x 1 x dx 1

2

EX2 1

0x2 6x 1 x dx 3

10

and therefore

V arX EX2 EX2 310

12

2 1

20

Now,Y X1 X2 X3 X4 X5

where the Xi are all independent of each other and have the same distribution as X above, i 1 5Therefore

EY 5 Ex 5 12 5

2

V arY 5V arX 5 120 1

4

QUESTION 5

5.1 The normal approximation to the Poisson distribution states that if N has the Poisson distribution, then

Z N E N V ar N

is approximately a standard normal random variable when the Poisson parameter is large enough.

Here, EN V arN 10000. Therefore, we get

P N E N 500 P N E N

V ar N

500V arN

PN E NV ar N

500V arN P

Z 500

10000

P Z 5 0

5.2 (a) False. The variables may or may not be independent.

(b) False. The marginal distribution must always sum up to 1, whether the random variables are independentor not.

15 STA2603/205

(c) False. Random variables X and Y would be independent if we had fY y fY Xyx for all x yvalues.

(d) False. Moment generating function is defined also for discrete random variables.

(e) False. What is true is that the marginal density function of X can be found by integrating fXY , the jointdensity function, over all values of y.

QUESTION 6

6.1 Q is the sum of six independent squared standard normal variables, so it has the 2 6 distribution, not the210 distribution.

6.2 Since each Yj N2 100, we haveYj 2 N0 100

andYj 2

10 N 0 1

Yj 2

10

2 21

20j1

y j 2

10

2 2 20

But20j1

Yj 2

10

2 1

100

20j1

Yj 2

2 19

100

119

20j1

Yj 2

2 19102

U R

Therefore, R has the 2 20 distribution.

6.3 We haveX j 2

5 N 01 Yj 2

10 N 0 1

so20

i1

Xi 2

5

2 2 10

20j1

Yj 2

10

2 2 20

and

W

10

i1

Xi 2

5

210

20j1

X j 2

10

220

F10 20Here,

V A 25100

W

and therefore, if we haveA 25

100 1 ∴ A 100

25 4

then V has the F1020 distribution.

16

3. SOLUTIONS TO ASSIGNMENT 6

The correct answers to the multiple choice questions are given below. Please see Section 2 in this tutorialletter for the correct solutions to the examination questions that these questions are based on!

1. The correct answer is 2. 11. The correct answer is 1.








9. The correct answer is 4. 19 The correct answer is 4.

10. The correct answer is 3. 20 The correct answer is 4.

Tu torial letter 205/1/ 2014Tu Dis ST Sem Dep Examin Solutio toria tribu A26 este artme ation prep...

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