1

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