stat 400 2.3 – moment generating function stepanov uiuc nguyen€¦ · nguyen the k th moment of...
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STAT 400 UIUC 2.3 – Moment Generating Function Stepanov
Dalpiaz Nguyen
The k th moment of X (the k th moment of X about the origin), µ k , is given by
µ k = E ( X k ) =
The k th central moment of X (the k th moment of X about the mean), µ k' , is given by
µ k' = E ( ( X – µ ) k ) =
The moment-generating function of X, M X ( t ), is given by
M X ( t ) = E ( e t X ) =
Theorem 1: M X' ( 0 ) = E ( X ) M X
" ( 0 ) = E ( X 2 )
M X( k )
( 0 ) = E ( X k )
Theorem 2: M X 1 ( t ) = M X 2
( t ) for some interval containing 0
Þ f X 1 ( x ) = f X 2
( x ) Theorem 3: Let Y = a X + b. Then M Y ( t ) = e b t M X ( a t ) Example 1: Suppose a random variable X has the following probability distribution: x f ( x ) Find the moment-generating function of X, M X ( t ). 10 0.20 11 0.40 12 0.30 13 0.10
( )å ×x
k xfx all
( ) ( )å ×-x
k xfx all
μ
( )å ×x
xt xfe all
the 1st moment of X
M ,= E ( X
')
0To define a RV,we can use on ofthe followings:• pmf if X is discrete• adf.mgf
tf X is discrete
first oftenvotive of M× Ct) evaluated at t -- OT 2nd derivative
t
Kth derivative
-
MxCt) = I e'tx.f (x)
allx
= et lo f ( Io) t et"
f ( Il) t et"
f (iz) t etbf ( 13)=O.de/OttO.4e'tttO.3ekttO.t
2
Example 2: Suppose the moment-generating function of a random variable X is M X ( t ) = 0.10 + 0.15 e
t + 0.20 e 2 t + 0.25 e
– 3 t + 0.30 e 5 t.
Find the expected value of X, E(X). Example 3: Suppose a discrete random variable X has the following probability distribution: f ( 0 ) = P( X = 0 ) = , f ( k ) = P( X = k ) = , k = 1, 2, 3, …
a) Find the moment-generating function of X, M X ( t ).
21 2 e-!2
1
kk ×
Mi (t) = ¥Mx (t) = Of 0.15et t O.2 (2) e"t 0.25 f-3) e-Stt 0.3 (5) est
= 0.15 et t 0.4 e"- 0.75 e
-Stt 1.5est
E ( X) = M'
×(O) = 0.15 e° t 0.4 e° - 0.75e
° t 1.5e° =⑤
OR
÷f÷÷÷t÷÷:÷ ::c:::*::::*.-0.75 =⑦I.5
M× ( t) = ↳ etx . f (x) = et -o . flo) tett.fclltettfqe.tn#a,
= (2 - e'"
) t IE,etkzk.TK
.
K"
as e.tk/zk as (ett)K
= (2 - e") t Ey= ( 2 - e") t E-
K-- I k !
= Is - e"4tfEol - ELY]""
teeth y= 2 -e
"' teeth - ,=eet/2-e42
3
b) Find the expected value of X, E ( X ), and the variance of X, Var ( X ). Example 4: Let X be a Binomial ( n, p ) random variable. Find the moment-generating function of X. Example 5: Let X be a geometric random variable with probability of “success” p. a) Find the moment-generating function of X.
et et
Mx ( t) - et - e"'t I offer eet" (dat et) = yet . Leth-
Milt) = tzeteetk ⇒ E (x) = Mi co) = I e°ee%=tVar ( X ) = E ( X
') - ( E Cx)] ' d
-47M "xLt) -- I [ date'- EE] -- I [ et . EE + Iet.EE. et ]→
F- CX) - M "×Co) -- {Ceo . e + I eo.EE?teoJ='zfe'" t Ie"']=¥e"
var ( x) -- Elk) - [ ECXD-
= fee "' - ( Ie'")! 'fe"
X N Binom ( n, p) ffix)
pmf of X : ICX ( F) p" ( l - p)"-×
,x-- O,I. . .. . n .
Mx (t) = Eetx . fcx) = ,¥oet× - ( Tx ) pix ( i - p)"- X
aux
=
,!Z ( g) ( p - et ICI - p )
"
= [ ( I - p) t pet] "↳ Binomial Theorem:
Xu Geom ( p,
( at b)" II akbn
-k
x-I( don't need to
know
Mxlt) -- off et"- f (x) = ,⇐,
et ? ( I - p ) .
p etx this for exam)
-
=p,E?(et)? ( t - p )" = pet¥7 Cet )"( t -pi
"
-- pet Eilert- p't! get E. let " -PD
?'÷÷i
4
b) Use the moment-generating function of X to find E ( X ). Example 6: a) Find the moment-generating function of a Poisson random variable. Consider ln M X ( t ). ( cumulant generating function )
( ln M X ( t ) ) ' = ( ln M X ( t ) )
" =
Since M X ( 0 ) = 1, M X' ( 0 ) = E ( X ), M X" ( 0 ) = E ( X 2 ),
( ln M X ( t ) ) ' | t = 0 = E ( X ) = µ X
( ln M X ( t ) ) " | t = 0 = E ( X 2 ) – [ E ( X ) ] 2 = s X
2 b) Find E ( X ) and Var ( X ), where X is a Poisson random variable.
( )( )
X
X
M
M ' tt ( ) ( ) ( )
( )[ ] 2 X
2 XXX
M
MMM
' "
t
ttt úûù
êëé-×
p - et
Mx (t) -- ftp.ji th - incl - p)←
For the geometric seriesto converge, lbasel LI .
¥ ( p -et) path - etc '-PD ⇒ let Ci - p>Is I-7 always
µ ;ftp.etll- et ( I - p)] - teth -p)] pet
⇒ et ( I - p) ( l positive
# ⇒ ttlncl- p) so[ I - ( I - p) et ]'
⇒ t C - incl- p)
p.et-F.pe , t C - incl - p) .F- ( x) = MICO) =%÷,eoyT -
-⑦Mdt) -↳etx.flxl-x.IE?dTxe--e-dx..Eoldet5xl.=e-d.edet==D
.Tied
In Mxlt) = Net - D( tnMx Ct))
'= det ⇒ E (x) = ( inMxlt))
'
It.- o
-- dei -④
( hM×Ct))"= det ⇒ Var (X) = ( lnM×CtD" It .- o -- dei --④