th m m m m central moment : E[( X X) ] ( X X) ( x X) f ( x)
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1 1 Trasform Techiques h m m m m mome : E[ ] x f ( x) dx h m m m m ceral mome : E[( ) ] ( ) ( x) f ( x) dx A coveie wa of fidig he momes of a radom variable is he mome geeraig fucio (MGF). Oher rasform echiques are characerisic fucios, z-rasform, ad Laplace rasform. Mome Geeraig Fucio For a real, he MGF of he radom variable is M () E[ e ] e k x k e p ( x ) discree x k e f ( x) dx coiuous Example - Beroulli p p, x 1 ( x) 1 p, x M () e e (1 p) e p pe 1 p is a real variable Example - Expoeial M () e e f ( x) dx 0 x x 0 x e e dx e x dx
2 Properies of MGF 1) Fid momes easil from he MGF Recall 3 3 e 1! 3!, Takig expecaio, 3 3 M () e 1! 3! Differeiaig m imes, m ( m) d m M (0) M ( ) m d 0 ) Ca show wo radom variables have he same probabili disribuio M() MY() f( u) fy( u) If wo radom variables ad Y have he same MGF, ad Y have he same probabili disribuio. 3) Covergece Cosider a sequece of radom variables,, wih cdf F ( x), F ( x), ad heir mome geeraig fucios M ( ), M ( ),. F ( x) F( x) iff M ( ) M ( )
3 3 4) Sum of Idepede RVs Cosider Z Y. If ad Y are idepede radom variables wih M ( ) ad M ( ) respecivel, he Y Proof: M () M () M () Z Y M () e Z Z e ( Y) e e Y if ad Y are idepede, e Y M () M () e Y 5) Ohers If Y a b, he. b M () e M ( a) Y
4 4 Beroulli p p, x 1 ( x) 1 p, x 0 M () pe 1 p M (0) 1, which mus be rue for a. (1) M () pe (1) M (0) p () M () pe () M (0) p I fac, M () pe 1 p 3 1 p p 1! 3! 3 1 p p p! 3! Recall for a radom variable, Therefore 3 3 M () 1! 3! p for 1,,
5 5 Poisso k e is a Poisso radom variable wih P k for k 0,1,,. k! The MGF is M () e (1 e ) Homework. Derive Poisso MGF Derive he MGF of a Poisso radom variable. Proper - Sum of idepede Poisso is Poisso Whe ad Y are idepede Poisso wih arrival raes ad respecivel, heir MGF are Thus 1(1 e ) (1 e ) () ad (). M e M e M () M () M () Y Y e ( )(1 e ) 1 Y is Poisso wih rae. 1 Y 1
6 6 Expoeial Le ~ Exp. M () M () ! 3! 1 1!! 3 3!! Proper Sum of Idepede Expoeial Wha is he sum of idepede expoeial? Suppose M(), M (). M Y () 1 Y Y is o expoeial, bu becomes -Erlag whe : M Y () 1
7 7 Proper Expoeial ad Gamma Fucio Relaio bewee he momes of a expoeial radom variable ad he gamma fucio. Le be a expoeial radom variable wih 1. we kow which meas Eq, e1 shows! x x e dx! e1 0 ( 1) x e dx! 0 x
8 8 Gaussia Le ~ N m,. M () e m Homework. Derive he MGF of Gaussia Proper - Gaussia Sum of idepede Gaussia is Gaussia.* Proof. e Y Y m m Y Le M ( ) e ad M ( ) e. If ad Y are idepede, M () M () M (). Y Y Y m my my Y which is he MGF of N m,. Also oe ha M () e Y Y m my Noe I fac, a liear combiaio of joil Gaussia radom variables is Gaussia. Suppose ad Y are joil Gaussia radom variables. For a cosas a, b ad c, defie V a by c V is a Gaussia radom variable. ad Y do o have o be idepede for V o be Gaussia.
9 9 Gamma Le ~ Gamma(, ). Prrof. M () e 1 ( x) x f ( x) e for x0. ( ) ( ) M for. 0 1 x ( x) x e e dx ( ) 1 x x e d x ( ) 0 provided 0 ( ) 0 1 e d for. Noe o Gamma Fucio ( z1) x e dx 0 z x We ca show ( z1) z( z) z! (1) 1 1
10 10 Proper Relaio bewee Gamma ad Erlag Disribuios m-erlag is a special case of Gamma, where m is a posiive ieger. Le m be a posiive ieger, ad assume ha,,, are idepede expoeial radom variables wih arrival rae. 1 m Defie Y 1 m. Y is he m-h arrival ime i a Poisso arrival process wih arrival rae. Y is referred o as a m-erlag radom variable. Sice is Exp, Sice (). are idepede, MY (). The MGF of Y is ideical o he MGF of Gamma, wih m. A m-erlag radom variable is a special case of Gamma, where j M j j m is a posiive ieger m.
11 11 Proper Sum of idepede gamma is gamma. x Le ~ Gamma, ad Y ~ Gamma,. Assume ad Y are idepede. Y ~ Gamma, x Proof: x M(), MY() x MY() Y ~ Gamma, x Example. m-erlag Suppose ad are posiive iegers. x For example, suppose 3 ad 4. x is he ime whe he 3rd arrival occurs, Y is he ime whe he 4h arrival occurs, ad Y is he ime whe he 7h arrival accurs.
12 1 Chi-square wih 1 degree of freedom Le ~ N(0,1) Defie Y. Y ~ Chi 1. Chi - square disribuio of 1 degree of freedom 1/ 1/ 1 MY (). 1/ Homework. MGF of Chi Derive he MGF of Chi(1). Proper. Chi is a special case of gamma. 1 1 MY () Special case of Gamma, Fid pdf of Y: 1 1 Sice Chi(1) Gamma,, 1 ( ) fy ( ) e ( ) 1 e 1 1 ad 0
13 13 Chi-square wih k degrees of freedom Le,, be idepede N 0, k Defie Y. k k / Y ~ Chi k, Chi - square disribuio wih k degrees of freedom. 1/ 1 1 k MY (). special case of Gamma, 1/ Proof. Sice is N 0,1, j 1 1 j is Chi 1 ad hus is Gamma,. j j Sice are idepede, are idepede. Also sum of idepede gamma is gamma. 1 k Therefore Y is Gamma,. M Y 1/ () 1/ k / 1. Fid pdf of Y: 1 k Sice Chi k Gamma,, 1 ( ) fy ( ) e ( ) 1 k ad
14 14 Sum of a Radom Number of Radom Variables Le,, be iid radom variables wih mea ad. 1 Defie a ew radom variables as Y 1 where N is a radom variable wih N ad. Y N N N VAR Y N N Proof. Le M, M, M deoe MGF of Y,, N. Y N N is a posiive radom variale wih P N, 1,,. The MGF of Y is Y M e e Y N P N e P N e log 1 P N M P N e Defie u log M. M P N e Y e 1 u N N u M 1 M u e
15 15 Differeiaig e1, MY MN u u u MN u u log M MN u 1 M u M e Noe ha as 0, u u 0 log M 0 log1 0. Therefore from e, MY MN u 1 M u M which saes Y N 0 0 u0 0 Wrie e as M MY MN u e3 u M Differeiaig e3 wih respec o, M M MY MN u u MN u M u u M N u 1 M Recallig, M M M M M M M M M u MN u u M u M Subsiuig 0, ad hus u 0, Y N N N N
16 16 Y Y Y N N N N N N N N Example. Queue Legh A queue coais N packes. Each packe coais bis. 1 N N N ad are radom variables. Assume N,, ad are kow. Le Y idicae he umber of bis i he queue: Y Y N N VAR( Y ) N. If 0, ha is, packes are of a fixed legh, he Y N, N ad VAR( Y ).
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