Ch3_FunctionsRandomVariables_torresgarcia.pdf

8/16/2019 Ch3_FunctionsRandomVariables_torresgarcia.pdf

http://slidepdf.com/reader/full/ch3functionsrandomvariablestorresgarciapdf 1/38



Reference: Most slides adapted from Montgomery et al. (2011)

RANDOM VARIABLES AND PROBABILITYDISTRIBUTIONS1. Random Numbers, Random Variables (RVs), Random Experiment (3-1 & 3-2)

2. Probability Laws (3-3)

3. Continuous RVs (3-4 & 3-5) CRVs Distributions: Normal, Lognormal, Gamma, Weibull, Beta

4. Probability Plots (3-6)

5. Discrete RVs (3-7 & 3-8) DRVs Distributions: Binomial and others

6. Poisson Process (3-9)

7. Normal Approximation to the Binomial and Poisson Distributions (3-10)

8. More than one RV and Independence (3-11)

9. Functions of RVs (3-12)

10. Statistics and the Central Limit Theorem (3-13)




INTRODUCTION

Sometimes, we are interested in analyzing situationsthat are characterized by many random variablesthat are “tied” to one another

Examples: Physiology/Human Factors: a subject’s height and weight

Manufacturing: defect types

Epidemiology: disease risks (high blood pressure, high LDL cholesterol, smoking, etc.)

3




3-11 More Than One Random Variable

and Independence

3-11.1 Joint Distributions




JOINT PROBABILITY MASS FUNCTIONLet X1 be the number of times that a circuit board has been reworked.

Let X2 be the number of critical defects in the circuit board.

The joint probability mass function for X1 and X2

5

),(),( 221121 x X x X P x x p

X1

0 1 2

X2 0 1/10 4/10 1/10

1 2/10 2/10 0

x y

y x p

y x p

1),(

1),(0Is p(x1 , x2) a valid joint probability mass function?




MARGINAL DISTRIBUTIONS

We can derive separate distributions for both X1 and X2 from the jointdistribution, p(x1 , x2)

These separate distributions are called Marginal DistributionsMarginal distribution of X1 :

Marginal distribution of X2 :

6

),()(

),()(

212

211

1

2

x x p x p

x x p x p

x

x

10

10

10

1)2(

10

6

10

2

10

4)1(

... 10

3

10

2

10

1

10000

hence,

210

1

1

21211

2

1

0

111

2

x p

x p

similarly

) ,x p(x ) ,x p(x ) p(x

, , ) for k ,x p(x ) p(xk) p(x x

10

40

10

2

10

2)1(

... 10

6

10

1

10

4

10

1

)2,0(0100

0

hence...

10

2

122121

2

2

2

0

122

1

x p

similarly

x x p ) ,x p(x ) ,x p(x

) p(x

, ) for k ,x p(x ) p(xk) p(x x

X1

0 1 2

X2 0 1/10 4/10 1/10

1 2/10 2/10 0

p(x1, x2):


http://slidepdf.com/reader/full/ch3functionsrandomvariablestorresgarciapdf 7/38 Reference: Most slides adapted from Montgomery et al. (2011)

SUMMARIZING MARGINAL DISTRIBUTIONS

7

2 x10

1

1 x10

6

0 x10

3) p(x

1

1

11

1 x10

4

0 x10

6) p(x

2

22

p(x1, x2): X1

0 1 2

X2 0 1/10 4/10 1/10

1 2/10 2/10 0


http://slidepdf.com/reader/full/ch3functionsrandomvariablestorresgarciapdf 8/38 Reference: Most slides adapted from Montgomery et al. (2011)

EXPECTATIONSGiven:

What is E(X1)?

What is E(X2)?

8

p(x1, x2): X1

0 1 2X2 0 1/10 4/10 1/10

1 2/10 2/10 0

2 x10

1

1 x10

6

0 x10

3) p(x

1

1

11

1 x10

4

0 x10

6) p(x

2

22

5

4

10

12

10

61

10

30)(

)()(

1

2

0

111

1

X E

x p x X E x

2

0

1

02111

1 2),()(

x x x x p x X E or

5

2

10

41

10

60)(

)()(

2

1

0

222

2

X E

x p x X E x

1

0

2

0

2122

2 1

),()( x x

x x p x X E or




CONDITIONAL PROBABILITY

9

)(

),()|(

2

212211

x p

x x p x X x X p

)(

),()|(

1

211122

x p

x x p x X x X p

p(x1, x2): X1

0 1 2

X2 0 1/10 4/10 1/10

1 2/10 2/10 0

2 x10

1

1 x10

6

0 x10

3) p(x

1

1

11

1 x10

4

0 x10

6) p(x

2

22

Joint probability mass function

Marginal of x1 Marginal of x2

p(X1 = 0| X2= 0) = (1/10)/(6/10) = 1/6

p(X1 = 1| X2= 0) = (4/10)/(6/10) = 4/6

p(X1 = 2| X2= 0) = (1/10)/(6/10) = 1/6

p(X1 = 0| X2= 1) = (2/10)/(4/10) = 2/4 p(X1 =

1| X2= 1) = (2/10)/(4/10) = 2/4

p(X1 = 2| X2= 1) = (0/10)/(4/10) = 0

p(X2 = 0| X1= 0) = (1/10)/(3/10) = 1/3

p(X2 = 1| X1= 0) = (2/10)/(3/10) = 2/3

p(X2 = 0| X1= 1) = (4/10)/(6/10) = 4/6

p(X2 = 1| X1= 1) = (2/10)/(6/10) = 2/6

p(X2 = 0| X1= 2) = (1/10)/(1/10) = 1

p(X2 = 1| X1= 2) = (0)/(1/10) = 0




CONDITIONAL EXPECTATION

What is the expected value of X1, given that X2 = 0?That is, find E(X1| X2 = 0).

Using similar logic, the following quantities can becomputed:E(X1| X2 = 0) = 1 (from above), E(X1| X2 = 1) = 2/4

E(X2 | X1= 0) = 2/3, E(X2 | X1= 1) = 2/6, E(X2 | X1= 2) = 0

Pick one of the above and verify

10

16

12

6

41

6

10

)0|()0|( 21

2

0

121

1

x x p x X X E

x




JOINT DENSITY FUNCTION:

Example:

11

10 and 1x0for)2(3

2),( 212121 x x x x x f

1),( 1221

1

1

2

2

dxdx x x f

high x

low x

high x

low x

),( 21 x x f

113

2

13

2

1)2(3

2

1

1

0

1

1

1

0

1

0

2

21

12

1

0

1

0

21

1

1

2

1 2

dx x

dx x x x

dxdx x x

x

x

x x

TRUE 12

3

3

2

112

1

3

2

123

21

0

1

2

1

x

x




MARGINAL DENSITY FUNCTIONSJust as in the discrete case, we can derive the marginal density functionsfor X1, X2, etc. from the joint density function

Mathematical definitions:Marginal density of X1:

Marginal density of X2:

12

2211 ),()(

2

dx x x f x f x

1212 ),()(1

dx x x f x f x

10 13

2

6

4

3

2

3

4

3

2),()(

11

1

0

2

221

2

1

0

2122

1

0

11

2

x for x x x x

dx x xdx x x f x f x

10 1431

34

31

3

4

3

2),()(

22

1

0

12

2

1

1

1

0

2112

1

0

12

1

x for x x x x

dx x xdx x x f x f x




CRITICAL THINKING

What is and

They are both equal to 1.

How can you find F(x1), F(x2)?

How can you find E(X1), E(X2), VAR(X1), VAR(X2)?

13

1

1

0

1)(

1

dx x f x

2

1

0

2 )(

2

dx x f x

Same as always.




ELEMENTARY DEFINITION

14

Xof Marginal

YandXof Jointx)|(YYof lConditiona

,

Yof Marginal

YandXof Jointy)|(XXof lConditiona

Similarly

Holds for both discrete and continuous.





and Independence

3-11.2 Independence




CONDITIONAL PROBABILITY AND INDEPENDENCE

If X and Y are independent and have densityfunctions f(x) and f(y)...then their Joint Density Function, f(x,y) = f(x)f(y)

Example: Let () = 1/3 − 1 ≤ ≤ 2

Let () = 1/8 2 ≤ ≤ 10

Then (, ) = 1/24 − 1 ≤ ≤ 2 2 ≤ ≤ 10

You can verify the above by doubly integratingf(x,y) over the ranges of x and y.

17




RANDOM VARIABLES AND PROBABILITYDISTRIBUTIONS1. Random Numbers, Random Variables (RVs), Random Experiment (3-1 & 3-2)

2. Probability Laws (3-3)

3. Continuous RVs (3-4 & 3-5) CRVs Distributions: Normal, Lognormal, Gamma, Weibull, Beta

4. Probability Plots (3-6)

5. Discrete RVs (3-7 & 3-8) DRVs Distributions: Binomial and others

6. Poisson Process (3-9)

7. Normal Approximation to the Binomial and Poisson Distributions (3-10)

8. More than one RV and Independence (3-11)

9. Functions of RVs (3-12)

10. Statistics and the Central Limit Theorem (3-13)




FUNCTION TO RANDOM VARIABLES

Image Source: onlinecourses.science.psu.edu




ALGORITHM #1: DISCRETE CASE

HOW TO DEVELOP THE PMF AND CDF OF A DISCRETE

RANDOM VARIABLE (Y) THAT IS A FUNCTION OF ANOTHERDISCRETE RANDOM VARIABLE (X)

1. Find the values of y.

2. Sum the pmf values of x that correspond to each y toobtain p(y)

3. Compute F(y) by using the relationship: F(y) = P(Y < y)

20

From 1 - 3 above we can get what we need to find E(Y) and V(Y)




EXAMPLE FOR DISCRETE Random variable has pmf

Random variable = 202

21

otherwise 0

2xif .1

1xif .3

0xif .6 p(x)

otherwise 0

80yif .1

20yif .3 0yif .6 p(y)

80yif 1

80y20if .9

20y0if .6

0yif 0F(y)




ANOTHER EXAMPLE FOR DISCRETE Let

Find p(y) if Y = |X|+1

Find F(y)

Graph F(y)

23

1xif .3

0xif .2

1-xif .1

2-xif .4 p(x)

21

10

21

32

Y X

3yif 0.4-2)P(X

2yif 0.40.30.11) p(X-1) p(X

1yif 0.20) p(X p(y)

31.0

3y20.6

2y10.2

1y0

F(y)

y




ALGORITHM #2:HOW TO DEVELOP THE CDF AND PDF OF A CONTINUOUS

RANDOM VARIABLE (Y) THAT IS A FUNCTION OFANOTHER CONTINUOUS RANDOM VARIABLE (X)

1. Find the range of Y.

2. Obtain the CDF, F(y) = P(Y < y)

3. Differentiate F(y) with respect to y to get the probability density, f(y).

or

24

From 1 - 3 above we can get what we need to find E(Y) and V(Y)

dy

dx x f y f )()(




EXAMPLE is a random variable with probability density,

() = 0.5( − 1) 1 ≤ ≤ 3

If = 2 – 3, find (), (), (), and ().

25

f(x)

X1 3

1

() = 7/3 () = 17/3 – (7/3)^2 = 2/9




EXAMPLE FOR CONTINUOUS is a random variable with probability density,

() = 0.5( − 1) 1 ≤ ≤ 3

If = 2 – 3 = ( + 3)/2

1. Range of W: Find in lower and upper ranges of X

Lower W: 2(1) – 3 = -1 (x = 1)

Upper W: 2(3) – 3 = 3 (x = 3)

2. CDF of W: F(w) = P(W < w) = P(2X-3 < w) = P(X < (w+3)/2)

3. pdf of w:

26

1612

12

3

24)1(5.)(

2

2

3

1

2

3

1

2

ww

w x x

dx xdx x f

w w

3w1-for8

1)()(

w

w f w F dw

d

8

1

4

1

8

3

2

11

2

35.0

)2

3(

12

35.0)()(

www

dw

wd

w

dw

dx x f w f

or




3-12 Functions of Random Variables




Now we have the pdf of Y and can determine E(Y) and

VAR(Y):

Because W is a nice, linear function of X, we could havedetermined E(W) and VAR(W) this way also: () = (2 − 3) = (2) − (3) = 2() – (3) = 2([7/3]) − 3 = 5/3

() = (2 − 3) = 22() = 4(2/9) = 8/9

28

3

5

8

1w)(w)(

3

1

3

1

dww

dww f W E

9

8

3

5

3

11)]([)()(

3

11 8

1w)(w)(

2

22

3

1

2

3

1

22

W E W E W VAR

dww

dww f W E





3-12.1 Linear Combinations of Independent

Random Variables






Random Variables





3-12.2 What If the Random Variables Are NotIndependent?

If = + , then

() = ( + ) = () + () () = ( + ) = 2() + 2() + (, )

Note: If X and Y are independent, then

() = 2() + 2() because (,) = 0




DISCRETE CASE EXAMPLE

33

y

0 1

p(x,y) = x 2 0.2 0.4

4 0.3 0.1

E(X) = 2.8

VAR(X) = .96

E(Y) = .5

VAR(Y) = .25

E(XY) = 1.2

(,) = −0.2

= −0.408

What is COV(X,Y) and r?

)()()(),( Y E X E Y X E Y X COV XY

4082.0489.0

2.025.096.0

2.0)()(

),(

Y X

XY

Y VAR X VAR

Y X COV

r

2.04.12.15.08.22.1),( XY

Y X COV

Y X

XY

Y VAR X VAR

Y X COV

r

)()(

),(




CONTINUOUS CASE EXAMPLE

(, ) = 1/24 for -1 < x < 2 and 2 < y < 10

It can be shown that

f(x) = 1/3 for -1 < x < 2, E(X) = 1/2, VAR(X) = 3/4

f(y) = 1/8 for 2 < y < 10, E(Y) = 6, VAR(Y) = 16/3

COV(X,Y) = ?, r = ?

34

396

288

96

12

96

300

248

3

48

3

48

)1(

48

4

4824)(

10

2

210

2

10

2

210

2

2

1

210

2

2

1

ydy

y

dy y y

dy y x

dxdy xy

Y X E

06

2

13)()()(),(

21211221

X E X E X X E X X COV

)()()(),( 21211221 X E X E X X E X X COV





3-12.3 What If the Function Is Nonlinear?





3-12.3 What If the Function

Is Nonlinear?




3-12.3 What If the Function Is Nonlinear?

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