Probability Distributions – Finite RV’s

• Random variables first introduced in Expected Value

• def. A finite random variable is a random variable that can assume only a finite number of distinct values

• Example: Experiment-Toss a fair coin twice X( random variable)- number of

heads X can assume only 0, 1, 2

Probability mass function

def. T he p rob ab ility m ass function (p .m .f.) o f a fin ite random variab le , X , is g iven b y

)()( xXPxf X . 1)(0 xf X .

xX xf

all

1)( .

Probability mass function(p.m.f)-Small f

E x a m p le 1 : B o x c o n ta in s fo u r $ 1 c h ip s , th re e $ 5 c h ip s , tw o $ 2 5 c h ip s , a n d o n e $ 1 0 0 c h ip . L e t X b e th e d e n o m in a t io n o f a c h ip s e le c te d a t r a n d o m . T h e p .m .f . o f X i s d is p la y e d b e lo w .

x $ 1 $ 5 $ 2 5 $ 1 0 0

)( xf X 0 .4 0 .3 0 .2 0 .1

Probability Mass Function

00.050.10.150.20.250.30.350.40.45

$1 $5 $25 $100

x

f X(x)

Cumulative distribution function(c.d.f)

d e f . T h e c u m u l a t i v e d i s t r i b u t i o n f u n c t i o n ( c . d . f . ) o f a n y r a n d o m v a r i a b l e , X , i s g i v e n b y )()( xXPxF X .

D o m a i n c o n s i s t s o f a l l r e a l n u m b e r s xxF X as 0)( xxF X as 1)(

Cumulative distribution function(c.d.f)- Big F

Example 1 (continued): The c.d.f. of X is displayed below.

100 0.1

10052 9.0

255 7.0

51 4.0

1 0

)(

xif

xif

xif

xif

xif

xFX

Cumulative Distribution Function

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-20 0 20 40 60 80 100 120

x

FX(x)

Calculating Probabilities-Using p.m.f & c.d.f

2.0)25($)25$( XfXP

9.0

2.03.04.0

)25($)5($)1($

)25$()5$()1$()25$(

XXX fff

XPXPXPXP

9.0)25($)25$( XFXP

6.0

1.02.03.0

)100($)25($)5($)5$(

XXX fffXP

6.0

4.01

)1($1

)1$(1

)5$(1)5$(

XF

XP

XPXP

Example 2: The c.d.f. of a random variable X is given below.

16 0.1

169 9.0

94 0.7

41 3.0

10 1.0

0 0

)(

xif

xif

xif

xif

xif

xif

xFX

Find the p.m.f. of X.

p.m.f

x 0 1 4 9 16 )(xfX 0.1 0.2 0.4 0.2 0.1

Expected value –Finite R.V

T h e e x p e c t e d v a l u e o r m e a n o f a f i n i t e r a n d o m v a r i a b l e i s g i v e n b y

x

XX xfxμXE all

)()( .

Example 1 (continued):

x )(xfX )(xfx X $ 1 0.4 0.40 $ 5 0.3 1.50 $ 25 0.2 5.00 $ 100 0.1 10.00 Sum 1.0 X $16.90

90.16$

1.0100$2.025$3.05$4.01$

)(

XXE

Example 3: A recent Gallup Poll showed that 55% of Americans own a cell phone. Let X be the number of Americans in a sample of size three who own a cell phone. Let C be the event than an individual owns a cell phone and let D be the event that an individual does not own a cell phone.

DDDDDCDCDDCC

CDDCDCCCDCCCS

45.055.01)( and 55.0)( DPCP

136125.0

)45.0()55.0(

)()()(

)()(

2

DPCPCP

DCCPCCDP

136125.0)()()( DCCPCDCPCCDP

4083750

136125013612501361250

22

.

...

)DCC(P)CDC(P)CCD(P

)DCCCDCCCD(P

)(f)X(P X

The p.m.f. of X is given by x 0 1 2 3

)(xfX 0.091125 0.334125 0.408375 0.166375

The c.d.f. is given by

3 1

32 833625.0

21 425250.0

10 091125.0

0 0

)(

xif

xif

xif

xif

xif

xFX

The random variable in this example is a special kind of finite random variable. A Bernoulli trial is an experiment that has exactly two outcomes, “success” and “failure”.

A binomial random variable gives the number of “successes” in n independent Bernoulli trials, where the probability of “success” on each trial is equal to p. BINOMDIST-show excel

T h e e x p e c t e d v a l u e o f a b i n o m i a l r a n d o m v a r i a b l e , X , w i t h p a r a m e t e r s n a n d p i s g i v e n b y pnXE X )( . E x a m p l e 3 ( c o n t i n u e d - c a l l p r o b l e m ) :

65.155.03)( XXE p e o p l e

Probability Distributions – Continuous RV’s

d e f. A c o n tin u o u s ra n d o m v a r ia b le is a ra n d o m v a ria b le th a n c a n a s su m e a n y v a lu e in so m e in te rv a l o f n u m b e rs . E x a m p le s : T -th e le n g th o f tim e , m e a su re d in m in u te s , a n d p a r ts o f m in u te s ,b e tw e e n a rr iv a ls o f p h o n e c a lls a t a c o m p a n y s w itc h b o a rd In th e o ry -T c a n a ssu m e a n y p o s itiv e re a l n u m b e r a s a v a lu e

H e re th e in te rv a l - ),[ 0

Probability density function-p.d.f

d e f . T h e p ro b a b ili ty d e n s ity fu n c tio n (p .d .f .) o f a c o n tin u o u s ra n d o m v a ria b le , X , is g iv e n b y )( xf X . 0)( xf X . T h e to ta l a re a b e tw e e n th e g ra p h o f

)( xf X a n d th e h o riz o n ta l a x is m u s t b e e q u a l to 1 .

p.d.f

Example 4: The p.d.f. of T, the weekly CPU time (in hours) used by an accounting firm, is given below.

4 if0

40 if)4(

0 if0

)( 264

3

t

ttt

t

tfT

-0.1

0

0.1

0.2

0.3

0.4

0.5

-4 -2 0 2 4 6

t

f T (t )

Relationship between Probability & Area of p.d.f - for Continuous

R.V

)( bXaP is equal to the area between the graph of )(xfX and the horizontal axis over the interval ],[ ba .

E x a m p le 4 (c o n t in u e d ) : )21( TP i s e q u a l to th e a re a b e tw e e n th e g ra p h o f )( tfT a n d th e t -a x is o v e r th e in te rv a l ]2,1[ .

- 0 .1

0

0 .1

0 .2

0 .3

0 .4

0 .5

- 4 - 2 0 2 4 6

t

f T ( t )

Important

For any continuous random X, 0)( xXP .

E x a m p le 4 ( c o n t in u e d ) : T h e c .d . f . o f T i s g iv e n b e lo w .

4 1

40 )316(

0 0

)( 3256

1

tif

tiftt

tif

tFT

c.d.f

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-4 -2 0 2 4 6

t

F T (t )

Use the c.d.f. to find )21( TP .

26170

13161256

123162

256

112

12

1221

33

.

))(())((

)(F)(F

)T(P)T(P

)T(P)T(P)T(P

TT

Additional tools are needed to compute the expected value of a continuous random variable.

Uniform random variable

def. A continuous uniform random variable is a random variable defined on an interval

],[ ba such that every subinterval of ],[ ba having the same length has the same probability.

p.d.f for uniform random variable

If X is a continuous uniform random variable on the interval ]u,[0 , then

.uxif

uxifu

xif

)x(fX

0

0 1

0 0

c.d.f for uniform random variable

. 1

0

0 0

)(

uxif

uxifu

xxif

xFX

Expected value for uniform random variable

2)(

uXE X

Example for Uniform random variable

E x a m p le 5 : A b u s a rr iv e s a t a b u s s to p e v e ry 1 0 m in u te s . L e t W b e th e w a itin g tim e ( in m in u te s ) u n til th e n e x t b u s . T h e p .d .f . a n d c .d .f . o f W a re g iv e n b e lo w .

10 0

100 101

0 0

)(

wif

wif

wif

wfW

Graph of p.d.f for uniform

-0.020

0.000

0.020

0.040

0.060

0.080

0.100

0.120

-5 0 5 10 15

w

f W (w )

Graph of c.d.f for uniform

0

10 1

10 10

0 0

)(

wif

wifw

wif

wFW

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

1.2

-5 0 5 10 15

w

F W (w )

Find )64( WP .

-0.020

0.000

0.020

0.040

0.060

0.080

0.100

0.120

-5 0 5 10 15

w

f W (w )

2.010.0)46()64( WP

2.0104

106

)4()6(

)4()6(

)4()6()64(

WW FF

WPWP

WPWPWP

Example 5 (continued): The expected value

of W is given by 5

2

10)( WWE

.

Exponential random variable

def. An exponential random variable may be used to model the length of time between consecutive occurrences of some event in a fixed unit of space or time.

p.d.f/c.d.f/ expected value –Exponential random variable

I f X i s a n e x p o n e n t i a l r a n d o m v a r i a b l e w i t h p a r a m e t e r , t h e n

0.

10 0

)( / xife

xifxf xX

0. 1

0 0)( / xife

xifxF xX

XXE )( .

E x a m p l e 6 : O n a v e r a g e , t h r e e c u s t o m e r s p e r h o u r u s e t h e A T M i n a l o c a l g r o c e r y s t o r e . L e t T b e t h e t i m e ( i n m i n u t e s ) b e t w e e n c o n s e c u t i v e c u s t o m e r s . T h e p . d . f . a n d c . d . f . o f T a r e g i v e n b e l o w .

0

201

0 0)( 20/ tife

tiftf tT

-0.01

0.00

0.01

0.02

0.03

0.04

0.05

0.06

-20 0 20 40 60 80 100 120

t

f T (t )

0 1

0 0)( 20/ tife

tiftF tT

- 0 .2

0 .0

0 .2

0 .4

0 .6

0 .8

1 .0

1 .2

- 2 0 0 2 0 4 0 6 0 8 0 1 0 0 1 2 0

t

F T ( t )

F i n d )15( TP .

- 0 . 0 1

0 . 0 0

0 . 0 1

0 . 0 2

0 . 0 3

0 . 0 4

0 . 0 5

0 . 0 6

- 2 0 0 2 0 4 0 6 0 8 0 1 0 0 1 2 0

t

f T ( t )

4724.0

)1(1

)15(1

)15(1)15(

20/15

e

TP

TPTP

E x a m p le 6 (c o n tin u e d ): T h e e x p e c te d v a lu e o f T is g iv e n b y 20)( TTE .