binomial and geometric distributions delta on-time performance at hartsfield- jackson atlanta...
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Binomial and Geometric Distributions
Delta On-Time Performance at Hartsfield-Jackson Atlanta International (June, 2003 -
June, 2015)http://www.transtats.bts.gov/OT_Delay/ot_delaycause1.asp?display=data&pn=1
Data / Model
• Total Operations: 2,278,897• On-Time Operations: 1,824,432• Proportion On-Time: 1824432/2278897 = .8006 (.80)• Will consider random samples of various sizes from
this population of operations• Y ≡ # of On-Time operations out of the sample of n• Y ~ Binomial(n , pY = 0.80)• X ≡ # of Flights sampled until the first NOT On-Time
Arrival is selected • X ~ Geometric(pX = 0.20)
Binomial Distribution – Probability Function
2 2 2
3 3 2 2
Binomial Probability Mass Function
!| ~ , 1 1 0,1,..., 0 1
! !
0 since: 0,1,..., 0 1
Binomial Expansion of :
2 : 2
3: 2
n y n yy y
n
n nP Y y Y Bin n p p y p p p p y n p
y y n y
p y y y n p
a b
n a b a ab b
n a b a a b ab
2 2 3 3 2 2 3
0
0 0
2 3 3
General :
1 1 1 1 Thus, a probability distribution
nn n i i
i
n nnn yy n
y y
a b ab b a a b ab b
nn a b a b
i
np y p p p p
y
Geometric Distribution• Used to model the number of Bernoulli trials needed until the
first Success occurs (P(S)=p) First Success on Trial 1 S, y = 1 p(1)=p First Success on Trial 2 FS, y = 2 p(2)=(1-p)p First Success on Trial k F…FS, y = k p(k)=(1-p)k-1 p
1
1 1
1 1 1
* *
*
1 * 0
( ) (1 ) 1,2,...
( ) (1 ) (1 )
Setting 1 and noting that 1, 2,... 0,1,...
1( ) (1 ) 1
1 (1 )
y
y y
y y y
y
y y
p y p p y
p y p p p p
y y y y
pp y p p p
p p
Binomial Distribution – Expected Value
0 0
Binomial Probability Mass Function
!| ~ , 1 1 0,1,..., 0 1
! !
Obtaining the Mean of :
!1 Summand = 0 when 0
! !
!
!
n y n yy y
n nn yy
Yy y
Y
n nP Y y Y Bin n p p y p p p p y n p
y y n y
Y
nE Y yp y y p p y
y n y
nyy n
1
1
1
1 11 1
0 0
1 Pull out ,!
1 !1 Now, set 1 0,1,..., 1 1
1 ! !
1 ! 1 !1 1
! 1 ! ! 1 !
nn yy
y
nn yy
Yy
n nn w n ww w
Yw w
p p n py
nnp p p w y w n y w
y n y
n nnp p p np p p
w n w w n w
0
Now, set 1
!1 1
! !
mmm ww
Yw
m n
mnp p p np p p np
w m w
Geometric Distribution – Expected Value
1
1 1
1
1 1
1
1
Note: 1 and
This interchange is justified due to nature of the convergent series.
y
y y
yy
yy
y y
y
y
E Y yp y y q p
dqq p yq
dq
dq dE Y p p q
dq dq
dE Y p q q
dq
1
1 0
2 2 2
1 since 0 1
1
(1 )(1 )(1) ( 1) 1
1 (1 ) (1 )
y z
y z
q q qq
p q qd q q q pE Y p p
dq q q q p p
Binomial Distribution – Variance and SD
2
0 0
2
Obtaining the Variance of : First obtain 1
!1 1 1 1 Summand = 0 when 0,1
! !
!1 1 1 Pull out
! !
n nn yy
y y
nn yy
y
Y E Y Y E Y E Y
nE Y Y y y p y y y p p y
y n y
nE Y Y y y p p n n
y n y
2
2 2
2
222
0
222
0
2
1 ,
2 !1 1 1
2 ! !
Now, set 2 0,1,..., 2 2
2 !1 1 1
! 2 !
2 !1 1 Now, set 2
! 2 !
!1 1
!
nn yy
y
nn ww
w
nn ww
w
p
nE Y Y n n p p p
y n y
w y w n y w
nE Y Y n n p p p
w n w
nn n p p p m n
w n w
mE Y Y n n p
w m
2 2
0
2 2 2 2 2 2
2 22 2 2 2
1 1 1 1!
1 1
1 1 1
mmm ww
w
Y Y
p p n n p p p n n pw
E Y E Y Y E Y n p np np n p np p
V Y E Y E Y n p np p np np p np p
Geometric Distribution – Variance and SD
21
21 1
2 21
2 21 1
23
32 2 3 2
22 2 2
( 1) ( 1)
1 2 2 22(1 ) ( 1)
1 (1 ) 1
2 1 2(1 ) 2( 1) ( )
( )
yy
y y
y y
y y
d qE Y Y y y q p pq
dq
d dpq q pq q qdq dq
d q d pq pq qpq pq pq qdq q dq q p pq
q p p pE Y E Y Y E Y
p p p p
V Y E Y
2
222 2 2 2
2
2 1 2 1 1( )
p p p qE Y
p p p p p
q
p
Binomial Distribution for On-Time Flights
1 0 1 0 1 1 1
2
0 2 0 1 2 1
1 (Bernoulli Distribution):
10.8 1 0.8 0,1 0 1 0.8 0.2 0.2 1 1 0.8 0.2 0.8
1 0.8 0.8 1 0.8 0.2 0.16 0.40
2
20.8 1 0.8 0,1,2
0 1 0.8 0.2 0.04 1 2 0.8 0.2 0.32
yy
Y
yy
n
p y y p py
E Y V Y
n
p y yy
p p p
2 2 22 1 0.8 0.2 0.64
2 0.8 1.6 2 0.8 0.2 0.32 0.566
In general, what needs to happen for 1 ?
For what value of does that occur for .8?
YE Y V Y
p n p n
n p
Binomial Distributions for n=1,2,3,4,10,25
• In EXCEL: Create a column of values 0,1,2,…,n (Say 0 is in cell
A2) In Cell B2, Type: =BINOM.DIST(A2,n,p,0) Copy and paste that cell alongside 1 (A3),…,n Note that the 0 at the end gives P(Y = y) = p(y) If you use 1 instead, you get P(Y ≤ y) = F(y)
n n n n n n1 2 3 4 10 25
p p p p p p0.8 0.8 0.8 0.8 0.8 0.8
y p(y) y p(y) y p(y) y p(y) y p(y) y p(y)0 0.2 0 0.04 0 0.008 0 0.0016 0 1.02E-07 0 3.36E-181 0.8 1 0.32 1 0.096 1 0.0256 1 4.1E-06 1 3.36E-16
2 0.64 2 0.384 2 0.1536 2 7.37E-05 2 1.61E-143 0.512 3 0.4096 3 0.000786 3 4.94E-13
4 0.4096 4 0.005505 4 1.09E-115 0.026424 5 1.83E-106 0.08808 6 2.43E-097 0.201327 7 2.64E-088 0.30199 8 2.38E-079 0.268435 9 1.8E-06
10 0.107374 10 1.15E-0511 6.27E-0512 0.00029313 0.00117114 0.004015
1 15 0.01177716 0.02944217 0.06234918 0.11084219 0.16334620 0.19601521 0.18668122 0.13576823 0.07083524 0.02361225 0.003778
0.8 0.4 1.6 0.565685 2.4 0.69282 3.2 0.8 8 1.264911 20 2
Several Binomial Distributions with p=0.8
0 1-2.22044604925031E-16
0.0499999999999998
0.0999999999999998
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
0.75
0.8
0.85
0.9
0.95
1
Probability Distribution of On-Time Flights Y ~ Bin(n=1,p=0.80)
p(y)
0 1 2 3
-0.0500000000000001
-6.93889390390723E-17
0.0499999999999999
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
0.5
0.55
0.6
0.65
0.7
Probability Distribution of On-Time Flights Y ~ Bin(n=3,p=0.80)
p(y)
0 1 2 3 40
0.05
0.1
0.15
0.2
0.25
0.3
0.35
0.4
0.45
Probability Distribution of On-Time Flights Y ~ Bin(n=4,p=0.80)
p(y)
0 1 2 3 4 5 6 7 8 9 10
-0.04
-2.77555756156289E-17
0.04
0.08
0.12
0.16
0.2
0.24
0.28
0.32
0.36
0.4
Probability Distribution of On-Time Flights Y ~ Bin(n=10,p=0.80)
p(y)
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
-0.025
-3.46944695195361E-17
0.025
0.05
0.075
0.1
0.125
0.15
0.175
0.2
0.225
0.25
Probability Distribution of On-Time Flights Y ~ Bin(n=25,p=0.80)
p(y)
0 3 6 9 12 15 18 21 24 27 30 33 36 39 42 45 48 51 54 57 60 63 66 69 72 75 78 81 84 87 90 93 96 990
0.02
0.04
0.06
0.08
0.1
0.12
Probability Distribution of On-Time Flights Y ~ Bin(n=100,p=0.80)
p(y)
Geometric Distribution Probabilities
1 1
2 1
11
2 2
1 1
1 1 1
1 1 0.20
2 2 0.20 0.80 0.16
0.20 0.80
11 1 1 0.20 .805.00 20.00
0.20 .20 .04
20 4.47
11
1
X X X
X X X X
xxX X
X
X X
X
xx x xi i X
X X X X Xi i i X
P X p p q p
P X p p q p q
P X x p x p q
pE X V X
p p
qF x P X x p x p q p q p
q
1 .80
In general, what is the smallest such that for fixed 0 1
For the Airline data, what is the smallest x such that 0.95
xxXq
x P X x c c
P X x
Geometric Distribution
• In EXCEL: Create a column of values 1,2,…,Y* for some large
value of Y* (Say 1 is in cell A2) In Cell B2, Type: =NEGBINOM.DIST(A2-1,1,p,0) Copy and paste that cell alongside 1 (A3),…,Y* Note that the 0 at the end gives P(Y = y) = p(y) If you use 1 instead, you get P(Y ≤ y) = F(y)
Geometric Distribution Probabilities and CDFGeometric (p=.20)y p(y) F(y)
1 0.2 0.22 0.16 0.363 0.128 0.4884 0.1024 0.59045 0.08192 0.672326 0.065536 0.7378567 0.052429 0.7902858 0.041943 0.8322289 0.033554 0.865782
10 0.026844 0.89262611 0.021475 0.91410112 0.01718 0.93128113 0.013744 0.94502414 0.010995 0.9560215 0.008796 0.96481616 0.007037 0.97185317 0.005629 0.97748218 0.004504 0.98198619 0.003603 0.98558820 0.002882 0.98847121 0.002306 0.99077722 0.001845 0.99262123 0.001476 0.99409724 0.001181 0.99527825 0.000944 0.996222
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 250
0.05
0.1
0.15
0.2
0.25
Geometric Distribution for Probability of y Trials until 1st NON On-Time Flight
p(y)
Moment-Generating Function
2 3 4
2 3 4
2 3 42 3 4
'Define:
1 ...2! 3! 4!
'Assuming for 1,2,3,.... :
1 ...2! 3! 4!
12! 3! 4!
kk
ty
k
tY ty
y y
y y y y
E Y
ty ty tye ty
k
ty ty tym t E e e p y ty p y
t t tp y t yp y y p y y p y y p
2 3 4
1 2 3 4
0
2 3
1 2 3 4 1
2
2 3 4 2
...
' ' ' '1 ...2! 3! 4!
'If exists: 0
2 3 4' ' ' ' 'Note: ' 0 ... ' 02! 3! 4!
6 12' ' ' ''' 0 ... '' 03! 4!
y
kk k
kk
t
y
t t tt
d m tm t m E Y
dt
t t tm t m
t tm t m
Moment-Generating Function – Binomial Distribution
0 0 0
0
1 1
2 2 1
Binomial Distribution:
1 1
1 1
' 1 ' 0 1
'' 1 1 1
n n n yn y n ytY ty ty y t y
y y y
n y nn yt t
y
n nt t
n nt t t t
n nm t E e e p y e p p e p p
y y
npe p p pe
y
m t n p pe pe m n p p p np
m t n n p pe pe n p pe pe
2 2 2
222 2 2 2 2
1 2 2 1
'' 0 1 1
' ' ' '1 1 1
Airline Values:
1 0.80 0.80 0.20 0.80n nt t
m n n p np n p np p
E Y np E Y n p np p V Y n p np p np np p
m t e e
Geometric Distribution – MGF
1
1
1
1 1 1
2 2
2 2
2 2 2
( )
1 1 (1 )
1 (1 ) (1 ) (1 ) (1 )'
1 (1 ) 1 (1 )
1' 0
1 (1 )1 (1 )
tY ty y
y
t t ty yty y t t
t ty y y
t t t t t t t
t t
t
t
m t E e e q p
p p pqe pe pee q qe qe
q q q qe p e
p e pe pe p e pe p p e p p em t
p e p e
pe p pm E Y
p ppp e
m
2
4
2 2 2 2 2 2
4 4
22
4 3 2 2 2
1 (1 ) 2 1 (1 ) (1 )''
1 (1 )
1 2(1 ) (1 ) 2(1 ) 2(1 ) 1 (1 )
1 (1 ) 1 (1 )
1 1 1 1 1 1'' 0
t t t t t
t
t t t t t t t
t t
p e pe pe p e p et
p e
pe p e p e p e p e pe p e
p e p e
p q q q q q qm V Y
p p p p p p
Probability-Generating Functions
0 1 2 3
0
2 3
2
Define: 1 ... 1 where is a positive integer
is a Random Variable that takes on integer values: 0,1,2,...
0 1 2 3 ...
1 1 2 3 ...
' 0 1 2 2 3
k
Y y
y
E Y Y Y k k
Y y
P t E t t p y t p t p t p t p
tp t p t p
P t p tp t
0
0
1
3 ...
' 1 0 1 2 2 3 3 ...
'' 0 0 2 1 2 3 2 3 ...
'' 1 0 0 2 1 2 3 2 3 ...
1 1
1 ... 1
y
y
kk k
kk t
p
P p p p
yp y E Y
P t p tp
P p p
y y p y E Y Y
d P tP t P t E Y Y Y k
dt
Probability-Generating Functions - Binomial
0 0
0
1
1
2 2
2 2
Binomial Distribution:
Note: 1 2 ... 0
1
1 1
' 1
' 1 1
'' 1 1
'' 1 1 1 1
nn yY y y y
y y
nny n y
y
n
n
n
n
p n p n p
nP t E t t p y t p p
y
npt p p pt
y
P t n p pt p
P n p p p np E Y
P t n n p pt p
P n n p p p n n
2 21 1
For Airline Data: 1 0.80 0.80 0.20 0.80
n
n n
p p p E Y Y
P t t t
Geometric Distribution – PGF
1
1 1 1
1
1
2 2 2
2 2
3
( )
1 1 (1 )
1 (1 ) (1 )1 (1 ) (1 )'
1 (1 ) 1 (1 ) 1 (1 )
1' 1
1 (1 )
2 (1'' 2 1 (1 ) (1 )
yY y y y y
y y y
y
y
p pP t E t t q p t q tq
q q
ptq pt pttq
q tq p t
p p t p tp t p pt p pP t
p t p t p t
p pP E Y
p pp
pP t p p t p
3
3 2
2
2 2 2
)
1 (1 )
2 12 (1 )'' 1 1
2 1 1 1 2 2 1 1
p
p t
pp pP E Y Y
p p
p p p pV Y
p p p p p