Avd. Matematisk statistik

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1 Avd Maemaisk saisik TENTAMEN I SF294 SANNOLIKHETSTEORI/EXAM IN SF294 PROBABILITY THE- ORY WEDNESDAY THE 9 h OF JANUARY 23 2 pm 7 pm Examinaor : Timo Koski, el , jkoski@khse Tillåna hjälpmedel Means of assisance permied : Appendix 2 in AGu: An Inermediae Course in Probabiliy Formulas for probabiliy heory SF294 L Råde & B Wesergren: Mahemaics Handbook for Science and Engineering Pocke calculaor You should define and explain your noaion Your compuaions and your line of reasoning should be wrien down so ha hey are easy o follow Numerical values should be given wih he precision of wo decimal poin You may apply resul saed in a par of an exam quesion o anoher par of he exam quesion even if you have no solved he firs par The number of exam quesions (Uppgif) is six (6) Soluions wrien in Swedish are, of course, welcome Each quesion gives maximum en () poin 3 poin will guaranee a passing resul The grade Fx (he exam can compleed by exra examinaion) for hose wih poin Soluions o he exam quesions will be available a hp://wwwmahkhse/maa/gru/sf294/ saring from Wednesday 9 h of January 23 a 75 pm The exam resul will be announced a he laes on Friday he 25 h of January 23 Your exam paper will be reainable a elevexpediionen during a period of seven weeks afer he dae of he exam Lycka ill!

2 for enamen i sf Uppgif Le (Ω, F) be a measurable space and le (R, B (R)) designae he real line and i Borel sigma field, respecively a) Give he definiion of a real valued random variable X in (Ω, F) ( p) b) Give he definiion of a Borel funcion f (2 p) c) Le X be a random variable, and f be a Borel funcion Se Y f(x) Prove ha Y is a random variable in (Ω, F) Aid: Use he definiions a) and b) (7 p) Uppgif 2 X Exp(), Y Exp() X and Y are indepeden Find he probabiliy P (X Y ) Please show your compuaions ( p) Uppgif 3 X and Y are wo independen coninuous random variables Their probabiliy densiy funcions are f X (x) and f Y (y), respecively a) Find he probabiliy densiy funcion f U (u) of he raio U X Aid: The rick is o Y consider he change of variables U X and V Y, ie, V is an auxiliary variable (5 p) Y b) Show ha if X N(, ) and Y N(, ) and X and Y are independen, hen C(, ) (5 p) X Y Uppgif 4 Consider he Wiener process {W () } Inroduce Y () W () W (/2) > a) Find he disribuion of Y () (4 p) b) Find he auocorrelaion funcion of he process {Y () > } (4 p) c) Is he process {Y () > } weakly saionary? Is he process {Y () > } a Wiener process? Jusify your answers (2 p)

3 for enamen i sf ( p) Uppgif 5 N {N() } is a Poisson process wih inensiy λ > Show ha for > s > here is he condiional disribuion ( N(s) N() y Bin y, s ) Aid: Compue firs P (N(s) x, N() y) wih y x ( p) Uppgif 6 {X n } n is a sequence of indepeden and idenically disribued rv s Their common mean is µ and heir common variance is < σ 2 < + Se S n X + + X n, n Show ha n S n nµ S n + nµ d N (, a 2), as n +, and deermine a 2 Please jusify your soluion in deail ( p)

4 Avd Maemaisk saisik SOLUTIONS TO THE EXAM SATURDAY THE 9 h OF JANUARY 23 Uppgif a) A real valued random variable is a real valued funcion X : Ω R such ha for every se A B, he Borel σ algebra over R, X (A) {ω : X(ω) A} F b) A funcion f : R R is called a Borel funcion, if for every se A in B, he Borel σ algebra, we have ha f (A) {x R f(x) A} belongs o he Borel σ algebra, ie, f (A) B c) Le A be a Borel se, ie, A B We consider Y (A) {ω Ω Y (ω) A} By consrucion we have Y (ω) f(x(ω)), and hus Y (A) {ω Ω f(x(ω)) A} {ω Ω X(ω) f (A)}, where he inverse image is f (A) {x R f(x) A} Since f is a Borel funcion, we have by definiion ha f (A) B, since A B Bu hen {ω Ω X(ω) f (A)} F, since X is a random variable Bu hereby we have esablished ha Y (A) F for any A in B, which by definiion means ha Y is a random variable We wrie Uppgif 2 P (X Y ) P (X Y ), and deermine he disribuion of X Y using he characerisic funcion As X and Y are assumed independen we ge ϕ X Y () ϕ X () ϕ Y () ϕ X () ϕ Y ( ),

5 for enamen i sf and by he Appendix B of Gu i i( ) i + i + 2 Here a reference o Appendix B of Gu gives ha X Y L () However, since ϕ X Y () is a real funcion, we ge X Y d Y X In oher words P (X Y ) P (Y X ) 2 ANSWER a): P (X Y ) 2 Uppgif 3 a) We find he join pdf, here f U,V (u, v), and marginalize o U from ha he desired pdf The inverse map is found as Then he Jacobian is Then Hence he disribuion of he raio U X Y This can be wrien as f U (u) X h (U, V ) UV, Y h 2 (U, V ) V J v u v f U,V (u, v) f X (uv) f Y (v) v f U (u) f X (uv) f Y (v) vdv is given by he marginal densiy f U,V (u, v) dv f X (uv) f Y (v) vdv () ANSWER a): f U (u) f X (uv) f Y (v) vdv f X (uv) f Y (v) vdv b) If X N(, ) and Y N(, ) and X and Y are independen, hen () gives for any u (, ) f U (u) e 2 u2 v 2 e 2 v2 vdv e 2 u2 v 2 e 2 v2 vdv We compue he firs inegral in he righ hand side e 2 u2 v 2 e 2 v2 vdv e 2 v2 (+u 2 ) vdv

6 for enamen i sf In he same manner we ge Thus [ + u 2 e ] 2 v2 (+u 2 ) v + u 2 e 2 u2 v 2 e 2 v2 vdv + u 2 e 2 u2 v 2 e 2 v2 vdv e 2 u2 v 2 e 2 v2 vdv + u 2 We have found f U (u), < u < π + u2 This is he probabiliy densiy funcion of C(, ), as was o be shown + u 2 π + u 2 Uppgif 4 a) Since {W () } is a Wiener process Y () W () W (/2) > is a linear combinaion of wo Gaussian random variables and hus a Gaussian random variable Since E [W ()], we ge E [Y ()] By properies of he Wiener process (Colleion of Formulas), Z W () W (/2) N(, /2) Then Y () Z and Var [Y ()] Var [Z] 2 2 ANSWER a): Y () N (, 2), > b) The auocorrelaion funcion of he process {Y () > } is given by R Y (, s) E [Y ()Y (s)] E [(W () W (/2))(W (s) W (s/2))] (E [(W ()W (s)] E [(W ()W (s/2)] E [(W (/2)W (s)] + E [(W (/2)W (s/2)]) (min (, s) min (, s/2) min (/2, s) + min (/2, s/2)) Here we mus disinguish beween four differen cases > s > s > /2 Then R Y (, s) (s s/2 /2 + s/2) (s /2)

7 for enamen i sf > /2 > s Then R Y (, s) (s s/2 s + s/2) 2 s > 2 s > > s/2 Then R Y (, s) ( s/2) 22 s > s/2 > Then R Y (, s) ( /2 + /2) ANSWER b): 2 & 2 22 above c) The process {Y () > } is no weakly saionary, as i auocorrelaion funcion R Y (, s) is no a funcion of s The process {Y () > } is no a Wiener process, eg, i has no almos surely he value a Also, Y () N (, 2) N (, ) Uppgif 5 Since N {N() } is a Poisson process we ge for > s and y x P (N(s) x, N() y) P (N(s) x, N() N(s) y x) P (N(s) x) P (N() N(s) y x), since he incremen are independen As N is a Poisson process wih inensiy λ > we have λs (λs)x P (N(s) x) e, x! and λ( s) (λ( s))y x P (N() N(s) y x) e (y x)! Then Then P (N(s) x, N() y) (λs)x (λ( s)) y x e λ x!(y x)! P (N(s) x N() y) P (N(s) x, N() y) P (N() y) y! (λs) x (λ) x (λ( x!(y x)! (λ) x s))y x (λ) y y! x!(y x)! ( s y! x!(y x)! ( s ) x (λ) x (λ( s))y x (λ) y ) x (λ( s)) y x (λ) y x (λs) x (λ( s)) y x e λ x!(y x)! e λ (λ)y y!

8 for enamen i sf y! ( s x!(y x)! ( ) y (s x ) x ( s ) y x ) x ( s ) y x This is he probabiliy mass funcion p X (x) of X Bin ( y, s ) Uppgif 6 We wrie S n nµ n i (X i µ) Then S n nµ n S n + nµ n i n (X i µ) S n + nµ n n n i (X i µ) S n + nµ We ake firs a look a he numeraor n n i By he cenral limi heorem as n + In he denominaor σ n (X i µ) σ σ n n i (X i µ) n (S n n + nµ) n (X i µ) i n (X i µ) d N (, ), i n (S n + nµ) n S n + µ P µ + µ 2µ, n +, since he weak law of large numbers enails S n n also applied one of he cases of Cramér - Sluzky Then he Cramér - Sluzky heorem yields S n nµ n S n + nµ n i (X i µ) n (S n n + nµ) as n +, where Z N(, ) Hence, if a 2 σ2 4µ 2, we have n S n nµ S n + nµ P N(, a 2 ), P σ 2µ Z, P µ, as n +, where we as n + ANSWER: a 2 σ2 4µ 2