Moment Generating Function

Transcription

1 1 Mome Geeraig Fucio m h mome m m m E[ ] x f ( x) dx m h ceral mome m m m E[( ) ] ( ) ( x ) f ( x) dx Mome Geeraig Fucio For a real, 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 0 M () e 0 1 e (1 p) e p pe 1 p is ay real cosa Example - Expoeial M () e e f ( x) dx x x x x e e dx e dx 0 0 1

2 Properies of MGF 1) Fid momes easily from he MGF Recall 3 3 e 1! 3!, Takig expecaio, 3 3 M () E[ e ] 1! 3! Differeiaig m imes, m ( m) d m M (0) M ( ) m d 0 ) Ca show wo radom variables have he same probabiliy disribuio M() MY() f( u) fy( u) If wo radom variables ad Y have he same mome geeraig fucio, ad Y have he same probabiliy disribuio. 3) Covergece Cosider a sequece of radom variables,, wih cdf F ( x), F ( x), ad mome geeraig fucios M ( ), M ( ),. 1 F ( x) F( x) iff M ( ) M( ) ) Sum of Idepede RVs Cosider Z Y. If ad Y are idepede radom variables wih M ( ) ad M ( ) respecively, he M () M () M () Z Y Y

3 3 Proof: 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 3

4 4 Beroulli p p, x 1 ( x) 1 p, x 0 M () e 0 1 e (1 p) e p pe 1 p M (0) 1, which mus be rue for ay. (1) M () pe (1) M (0) p () M () pe () M (0) p I fac, Therefore M () pe 1 p p p 1! 3! 1 p p p! 3! p for 1,, 4

5 5 Poisso is a Poisso radom variable wih k e P k for k 0,1,,. k! The MGF is M () e (1 e ) Propery - Poisso Sum of idepede Poisso is Poisso. Whe ad Y are idepede Poisso wih arrival raes ad respecively, Thus 1 e (1 ) (1 e ) Y M () e, M () e M () M () M () e Y Y ( )(1 e ) 1 1 5

6 6 Expoeial Le ~ Exp. M () M ()! ! 3! 1 1!! 3 3! Propery - Expoeial Wha is he sum of idepede expoeial? M (), M () M Y () 1 Y Y is o expoeial, bu becomes -Erlag whe 1 6

7 7 Propery - Expoeial 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! x x e dx! 0 We have show ( 1) x e dx! 0 x 7

8 8 Gaussia Le ~ N m,. M () e m proof : ( xm) x 1 M () e e dx 1 e x ( m ) xm m 1 m x( m ) dx e e dx e Homework Fid higher momes of a zero mea Gaussia radom variable. Assume ~ N(0, ). Fid. Hi. Use he Taylor series of he mome geeraig fucio. Aswer. 0, for odd!, for eve / /! 4! For example, 3! 4 4 8

9 9 Propery - 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,. Therefore Y is Gaussia. Also oe ha M () e Y Y m my Noe I fac, ay liear combiaio of joily Gaussia radom variables is Gaussia. Suppose ad Y are joily Gaussia radom variables. For ay 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

10 10 Gamma Le ~ Gamma(, ). 1 ( x) x f ( x) e for x0. ( ) M ( ) for. Proof: M () e 0 1 x ( x) x e e dx ( ) 1 x x e d x ( ) 0 provided 0 ( ) 0 1 y y e dy for. Noe o Gamma Fucio z 1 x ( z) x e dx 0 We ca show ( z1) z( z) z! (1)

11 11 Propery Relaio bewee Gamma ad Erlag Disribuios -Erlag is a special case of Gamma, where is a posiive ieger. Le be a posiive ieger, ad assume ha 1,,, are idepede expoeial radom variables wih arrival rae. Defie 1. is he -h arrival ime i a Poisso arrival process wih arrival rae. Sice is Exp, Sice j M M j j (). are idepede, (). The MGF of is he MGF of Gamma,. is referred o as a -Erlag radom variable, ad is a special case of Gamma, where is a posiive ieger. 11

12 1 Propery Sum of idepede gamma is gamma. x y Le ~ Gamma, ad Y ~ Gamma,. Assume ad Y are idepede. Y ~ Gamma, x y Proof: x y M(), MY() x MY() Y ~ Gamma, x y y Suppose ad are posiive iegers. x y 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. y 1

13 13 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/ MY (). Special case of Gamma, 1/ 1 1 Sice Chi(1) Gamma,, 1 ( y) fy ( y) e ( ) 1 e y y y 1 1 ad y 0 Homework Fid he MGF of Chi(1). 13

14 14 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. Y 1 k is Gamma,. k / 1/ 1 MY (). 1/ 1 k Sice Chi k Gamma,, 1 ( y) fy ( y) e ( ) y 1 k ad 14

15 15 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 Y N P N 1,, Y M e Y e N P N e P N e log 1 P N M P N e M Defie u log M. 15

16 16 Y M P N e e 1 u N N u 1 M u e Differeiaig e1, MY MN u u u MN u u log M MN u 1 M u M e Noe u 0 logm 0 log10. Therefore from e, MY MN u 1 M u M which saes Y N 0 0 u0 0 16

17 17 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 Y Y Y N N N N N N N N 17

18 18 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 ). 18