EXPONENTIAL PROBABILITY DISTRIBUTION

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1 MTH/STA 56 EXPONENTIAL PROBABILITY DISTRIBUTION As discussed in Exaple (of Secion of Unifor Probabili Disribuion), in a Poisson process, evens are occurring independenl a rando and a a unifor rae per uni of ie. Suppose ha we label a ie zero he insan a which we begin observing he Poisson process. Now le T denoe he ie a which he rs even occurs. Consider he even ft > g ha he ie of he rs even is greaer han. This even occurs if and onl if here are zero evens in he xed inerval (0; ]. Neverheless, i follows fro Secion of Poisson Probabili Disribuion ha he probabili of zero evens occurring in an inerval of xed lengh is p (0; ) ()0 e e 0! Therefore, P ft > g e fro which we nd he disribuion funcion for T ; ha is, and is densi funcion is F (; ) e for > 0 f (; ) d d F (; ) e for > 0 This is called he exponenial densi funcion. Forall, we have he following disribuion. De niion. The coninuous rando variable Y has an exponenial probabili disribuion wih paraeer > 0 if is probabili densi funcion is given b e for > 0 f (; ) 0 elsewhere. An coninuous rando variable ha follows an exponenial disribuion wih paraeer > 0 is referred o as he exponenial rando variable wih paraeer > 0. The disribuion funcion for he exponenial rando variable Y is given b e for > 0 F (; ) 0 elsewhere. The gures below plo he densi and disribuion funcions for he exponenial rando variable wih paraeer.

2 Theore. paraeers are The ean and variance of he exponenial rando variable Y wih E (Y ) and V ar (Y ) ; respecivel. The oen-generaing funcion is Thus, and Proof. Y () for < The oen-generaing funcion is given b Y () E e Y Z 0 e e d for < : E (Y ) d d Y () 0 E Y d d Y () 0 Z 0 e ( ( ) 0 ( ) 3 ( ) ( ) e ) ( d : 3 )! 0 Hence, V ar (Y ) E Y [E (Y )] :

3 Exaple. Suppose ha a sse conains a cerain pe of coponen whose ie (in ears) o failure is given b he rando variable T, disribued exponeniall wih paraeer 5. If ve of hese coponens are insalled in di eren sses, wha is he probabili ha a leas wo are sill funcioning a he end of 8 ears? Soluion. equal o We rs observe ha a given coponen is sill funcioning afer 8 ears is P ft > 8g e 85 0: Now le Y represens he nuber of coponens funcioning afer 8 ears which is clearl a binoial rando variable wih paraeer n 5 and p 0:. Therefore, he desired probabili equals P fy g P fy g X 0 Also, he ean and variance of he survival ie T are 5 (0:) (0:8) 5 0:7373 0:67: E (T ) 5 5 and V ar (T ) (5) 5: Meorless Proper. The exponenial disribuion has he sae eorless proper ha we found for he geoeric disribuion. Theore (Meorless Proper). Suppose Y has an exponenial disribuion wih paraeer. For an posiive nubers a and b, P (Y > a + b j Y > a) P (Y > b) : Proof For an posiive nubers a and b, P (Y > a + b j Y > a) P (Y > a + b and Y > a) P (Y > a + b) P (Y > a) P (Y > a) P (Y a + b) F (a + b) P (Y a) F (a) e (a+b) e a e b F (b) P (Y > b) : Thus, if in Exaple we have observed a 4 ears wih no failures, he probabili ha i will be a leas b ears unil he rs failure is unchanged fro he original value for his probabili when we begin observaion. The exponenial disribuion is he onl coninuous probabili disribuion wih he eorless proper. This proper ells us ha a used exponenial coponen is esseniall as good as new. I a be equivalenl saed as P (Y > a + b) P (Y > a) P (Y > b) : (4.) 3

4 Then he converse of Theore can be shown o be also rue as follows. Theore 3. Le Y be a non-degenerae nonnegaive rando variable. If (4:) holds for an nubers a > 0 and b > 0, hen Y has an exponenial disribuion wih soe paraeer > 0. Proof De ne F () F () P (Y > ) for all. Le c > 0 and and n be posiive inegers. Then, b (4:), i is eas o see ha F (nc) F (c) h n c i and F (c) F : Now we clai 0 < F () <. If F (), hen F (n) F () n for all n, which conradics F (+) 0. If F () 0, hen F F () 0 an so F 0 for all, which conradics F ( ). Since 0 < F () <, wrie F () e / for soe appropriae > 0. I follows fro he second equali of he above ha F F () e / : Thus, b he rs equali of he above, we have n n F F e (n)/ ; ha is, F () e / for an posiive raional nuber. B he righ coninui of F (), i follows ha F () e / for an posiive real nuber. Therefore, Y has an exponenial disribuion wih paraeer > 0. The reason ha he geoeric and exponenial disribuion should boh have he eorless proper can be seen b reebering ha he coninuous ie Poisson process can be derived as a lii of a sequence of independen Bernoulli rials. A geoeric rando variable Y is he nuber of Bernoulli rials unil he rs success occurs and he exponenial rando variable T is he ie of occurrence o he rs even (success). In fac, as seen in Secion??, if Y is a geoeric rando variable wih paraeer p, hen P fy > ng q n ( p) n where q p. In deriving he Poisson process in Secion??, we se p n and divide he inerval (0; ] ino n subinervals of lengh n. Then he evens fy > ng and ft > g are equivalen and n P ft > g li P fy > ng li ( p) n li e n! n! n! n so he exponenial disribuion funcion is he lii of he geoeric disribuion funcion; he exponenial rando variable inheris he eorless proper fro he geoeric rando variable. 4

5 Suppose ha we begin observing a Poisson process a ie zero and le T r be he ie o occurrence of he rh even (r ). his rando variable associaes wih he negaive binoial rando variable de ned on independen Bernoulli rials. Le be an xed posiive nuber and consider he even ft r > g ha he ie o he rh even is greaer han. he even ft r > g is equivalen o he even fy r g, where Y is he nuber of evens ha occurs in (0; ], because he ie o he rh even can exceed if and onl if here are r or fewer evens in (0; ]. Since Y is a Poisson rando variable wih paraeer, we have P ft r > g P fy r g Xr k0 () k e and he disribuion funcion for T r, he ie o he rh occurrence, is F (; r; ) P ft r g P ft r > g k! Xr k0 () k e The rando variable T r is called he Erlang rando variable wih paraeers r and. The densi funcion of T r is f (; r; ) d F (; r; ) d " d d e e () e! e e + e e + 3 e r r e + r r e (r )! (r )! r (r )! r e for > 0:! k! () r e (r )! # Erlang Disribuion. Le T r be he rando variable represening he waiing ie o occurrence of he rh even in a Poisson process wih paraeer. hen he densi funcion of T r is r f (; r; ) (r )! r e for > 0 0 elsewhere As will be derived laer in he chaper, he ean and variance of he Erlang rando variable are r and r and he oen-generaing funcion is Y () ( ) for < : 5

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