Is - fac.ksu.edu.sa
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Homework I 0th Fsb 438201908 19 jig n Uniform 0,5 Nlr Poisson X fun nix E I f X is of XES n FN n f ni f X DX NIA i Fn co a of j da f DX Is C e to fu o f Ct Es 0.1987 full oTej Is DX f dX fr 1 Is C X E't e Jos 3 Ci se s e 53 0.1919 a Pr N 2 I FN l Pr N 2 l FN O full Pr N 2 l O 1987 0.1919 Pr NZ 2 0.6094 The unconditional probability that N 2 is 0.6094
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Homework I0th Fsb
438201908
19 jig
n Uniform 0,5 Nlr Poisson X
fun nix E I f X is of XESn
FN n f ni f X DXNIA
i Fnco a of j da f DX Is C e to
fu o f Ct Es 0.1987
full oTej Is DX f dX
fr 1 Is C XE't e Jos 3 Ci ses e53 0.1919
a Pr N 2 I FN l
Pr N 2 l FN O fullPr N 2 l O 1987 0.1919
Pr NZ 2 0.6094Theunconditionalprobability that N 2 is 0.6094
Withprobability pPp N n f o S oSPn
GCN 2 Z O5720 0.25
With probability 1 P
Ff N n Y o 5 osn
fp N 2 L O54055 0 375
MixedprobabilityO 25p t 0.375 l P 0.375 O 125P
Homework 20th Fsb
438201908
19 jig
f x O Pcm emr x
9 om
The mean
In f x O Inpcmx Mrc X mlngut
ft 0 Mr'CoX m g C0901
OfCx Mr'CO X Mq O f x O00 geo
Off 0 dx Mr O xFX e dx m9h01 Jfk 0 dx9 O
go.CI i0dxJ mr'CO7Sxfcxio dx m9g 1ffcxiOfdx
ffCx O dx I IX FcxO DX E x
i fmr co E ex Mag I ECN MOIr fog
To obtain thevariance
substitute in
c fCx Mr X Mr COMCO f x O00
OfCaio Mr O X M O f x O00
02 f Xie Mr O X MCOD fCx e Mr401M O f x O00 2
Mr O x Mco of G O00
in
02 fCx 0 Mr O X MCOD fCx e Mr401140 f XieO
Mri 055Ex MC055 f x O
a S 02 fCx 0 dx Mr O f X MCOD fCx 0 dx0
mricoMico f fCxO DXmr COD f Ex MOD fCx dx
off dx ffcx.iodx 0
I
fcx MCOD fcx.io dxSx f x e dx MCO f fCx E dxEG MCO I
M O M O O
0 0 MHMKO j.y cmricoDZfcxMCODTcx.no dx
I Ex MoD fCx dx Mm'f Var X
The Variance is Var X O Mm r
e
osw.ggHomework 3 438201908
19 jig
For Inverse exponentialdistribution
no e
t.name
EIE9oIIieThmIqjl XT
s l O In0 Xj 2 In X
ECO n InG yo 2 In where y
to get let L'Celso
E O no y o
20From data set B n 20 Y E O 10115
I
20 197.7181787 L G 159 780 10115
onThe method of moment estimate
E X 0 F Z
Fromdata set B
C x I f 0.00506
E X 197.72 Same as Max liklihood estimate
For the inverse gammadistr with 2 2
fCx10 02 eOK
3 T Z
fCx10 02 eOK
3
In fCNO 21h0 015 3 In X
l O 21h0 015 31nA
ECE 2h In0 yo 3 1nA wherey
to get let L 07 0
ECO 2h0 y O
from data set B n zo y O101152 395.4363482 L 169.07
The method of moment estimate
ECx 0 T d l Of 2 1 0 01OFx T Z 1
E x I 27 821 1573 1424 420
1424.4 different fromMax liklihood estimate
For inverse gamma distr
f X1 x E 0 e OHdt f x
In f X1 x O L InO 015 att InX In f caa lex 01 l x in 0 015 att InX In T Cx
l ca O an Ino O Xj Gtl InX n In TcaTo get I let x O o LL ca 01 0
to ca ol E XI o
LL x O n th O InCX hq diff Tca 0
After using Matlab140 16 I 0 70888
L Q G 15888
onThe method of moment estimate
E x I 27 821 15743 1424.420
EC 2 02
x l x z132384419
squaring and then deridingby
5 IT ca z 27 43423484444.9 0.15326
I 2 181 8 1424 4 2.181 l 1682 217
different from Max likelihood estimate
Matlab Code
Homework 40th
438201908
19 jig
1 Inverse exponential
f x 01 0 e042
L O f 27 f 82 f 243 ESC250 B
L O 0927 0 ex2
2 05 4243G e 0250313
O e 0100826 l e EsoB
e ECE 7 In O 00.08261 14In x 13 In l e0 50
To get let L'Col 0
L 01 10.0826 t 13e go to e
sso
70
0.052 e0 50 0.0826
I e 01250
189.78 Censored L G 37.85
197 72 without censoring
2 Gamma
fCxa
Acae X O
L X O f 27 f 82 f 243 S 250713
4401 f a I II x e Ii i ma s
Lcd 07 2 1 Is IncxD 7 a Ino 7 In TK TEX 113 InCt Fca
To get 8 I let tf Cx 01 0 LL ca 01 0x Q 7 X 0
21 E ca
diff I FK so
we get 8 295.69 I I 5183 Censored
2561 I I 0.5562 withoutcensorning
3 Inverse gamma
fCx Oh e Ex1 Tca
L X 01 f 271 f 82 f 243 5 2507
LCx I IT on e Ox0 1 9 1 pay
Sx 2501B
Lca o Eq e0 III Tca D
lCa 01 7 1not 7 In Flat 0 att ht t.tt g 13lnFki0g
Toget 8 I let tf Cx 01 0 ELCa 01 0
Cx 01 7 B diff Tca 1 0Cai I
LLca 01 7 Ino diff flat 1 In tyg diff Kai so
we get 8 86.29 I O 4161 Censored
140.16 2 0.7089 without censoring
Homework 50th THE
438201908
19 jig
n 6000 I 15600000 M 16500000
n 6000Z 0.55408 fromFb167Xo Itf 19543 51
Pc _IZ I 2 M
0.55408 15600000 0.44592 16500000
16001328
