TMA 4265 Stochastic Processes

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1 TMA 4265 Sochasic Processes Norges eknisk-naurvienskapelige universie Insiu for maemaiske fag Soluion - Exercise 8 Exercises from he ex book 5.2 The expeced service ime for one cusomer is 1/µ, due o he exponenial disribuion. As he exponenial variables are assumed no o have any memory, he expeced service ime for he cusomer being served as you ener he bank is also 1/µ. The expeced amoun of ime you wai unil i is your urn is herefore 5/µ. You are also expeced o use 1/µ before you are done, hence he expeced ime you will spend in he bank is 6/µ. 5.4 T i Service ime for cusomer i Find: Pr(A is he las person leaving) Pr(T B + T C < T A ) p Service ime T i 1 min: p The disribuion for T i is Pr(T i ) { 1 3 1,2,3 ellers 2. okober 28 Side 1

2 p Pr(T B + T C < T A ) Pr(T B + T C 2, T A 3) Pr(T B T C 1, T A 3) c) T i exp(µ) iid. Alernaively p Pr(T B + T C < T A ) Pr(T B + T C < T A, T B < T A ) Pr(T B + T C < T A T B < T A ) Pr(T B < T A ) Pr(T C < T A ) Pr(T B < T A ) (T A memoryless) p Pr(C leaves before A, B leaves before A) Pr (C leaves before A B leaves before A) Pr (B leaves before A) }{{}}{{} exp disr memoryless + symmerical symmerical 5.6 p Pr(Smih is NOT las) Pr(Jones is las) + Pr(Brown is las) Pr(T B < T J ) Pr(T S < T J ) + Pr(T J < T B ) Pr(T S < T B ) (see 5.2) λ 2 λ 2 λ 1 + λ 1 λ 1 + λ 2 λ 1 + λ 2 λ 1 + λ 2 λ 1 + λ 2 ( ) 2 ( ) 2 λ2 λ1 + λ 1 + λ 2 λ 1 + λ See he book a page 748. (Page 662 in 8 h ediion) X exp(λ) 2. okober 28 Side 2

3 We need: i) P(X < c) 1 e λc ii) f X X<c (x) Now we ge he definiion, see page 97 { fx (x) < x < c oherwise E[X X < c] xf X X<c (x)dx 1 c xλe λx dx c 1 [ xe λx 1 λ e λx ] c 1 [ 1 λ (1 e λc ) ce λc ] x f X(x) dx (1 e λc ) 1 λ c 1 We have E[X X > c] c + E[X] c + 1 λ Therefore we ge he ideniy Exercises from exams E[X X < c] 1 [E[X] (1 ) E[X X > c]] 1 [ 1 λ (1 )(c + 1 ] λ ) 1 λ c 1 Augus 4, Oppg. 2 Le X be he number of ype A errors and Y be he number of ype B errors, where X is Poisson disribued wih inensiy λ 1 and Y is Poisson disribued wih inensiy λ okober 28 Side 3

4 We wan o show ha X + Y is Poisson disribued wih inensiy λ 1 + λ 2. P(X + Y n) P(X k,y n k) k P(X k)p(y n k) k k λ k 1 e λ 1 k! e (λ 1+λ 2 ) λ n k 2 e λ 2 (n k)! k (2) e (λ 1+λ 2 ) (λ 1 + λ 2 ) n k!(n k)! λk 1λ2 n k We recognize his as a Poisson disribuion wih inensiy λ 1 + λ 2. The ransiion dendoed by is due o he fac ha X and Y are independen. The ransiion denoed by (2) is due o he fac ha he Binomial expansion of (λ 1 + λ 2 ) n is (λ 1 + λ 2 ) n k k!(n k)! λk 1 λn k 2 A naural objecion agains modeling he number of errors on he componen as a Poisson process wih consan inensiy is ha he probabiliy of an error will increase wih increasing ime in mos cases. We le Z X + Y. Then X is he number of errors of ype A, Y is he number of errors of ype B, and Z is he oal number of errors. We wan o find P(X 1 Z 1) (2) P(X 1,Z 1) P(Z 1) P(X 1)P(Y ) P(Z 1) P(X 1,Y ) P(Z 1) λ 1 e λ 1 e λ 2 (λ 1 + λ 2 )e (λ 1+λ 2 ) λ 1 λ 1 + λ 2 is due o ha fac he if X 1 and Z 1, hen Y. (2) is due o he fac ha X and Y are independen. (X and Z are no independen.) c) We le X(u) be he number of errors in he inerval (,u]. Similarly, X() is he number of errors in he inerval (,], wih u. Given ha X() n, we wan o find he disribuion of X(u). 2. okober 28 Side 4

5 P(X(u) k X() n) P(X(u) k,x() n) P(X() n) P(X(u) k,x() X(u) n k) P(X() n) e λu (λu) k k! k!(n k)! e λ( u) (λ( u)) n k (n k)! e λ (λ) n (u ) k ( 1 u ) n k, which we recognize as a Binomial disribuion wih parameer u/. Commen o : The number of errors on he inerval (, u] is no independen of he number of errors on he inerval (,]. The number of errors on he inerval (u,] is, hough, independen of he number of errors on he inerval (,u]. We wan o find he probabiliy of exacly k(< n) errors in he inerval (u,] given ha here occurs n errors in he inerval (,]. P(X() X(u) k X() n) P(X(u) n k X() n) (2) (n k)! (u ) n k ( u) n, 1 which we recognize as a Binomial disribuion wih parameer 1 u/. Commen o (2): We use k n k in he disribuion derived above. 2. okober 28 Side 5