Probability Distributions – Finite RV’s
description
Transcript of Probability Distributions – Finite RV’s
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 .