Avd. Matematisk statistik

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1 Avd. Matematisk statistik TENTAMEN I SF940 SANNOLIKHETSTEORI/EXAM IN SF940 PROBABILITY THE- ORY TUESDAY THE 9 th OF OCTOBER a.m p.m. Examinator : Timo Koski, tel , tjtkoski@kth.se Tillåtna hjälpmedel Means of assistance permitted : Appendix in A.Gut: An Intermediate Course in Probability. Formulas for probability theory SF940. L. Råde & B. Westergren: Mathematics Handbook for Science and Engineering. Pocket calculator. You should define and explain your notation. Your computations and your line of reasoning should be written down so that they are easy to follow. Numerical values should be given with the precision of two decimal points. You may apply results stated in a part of an exam question to another part of the exam question even if you have not solved the first part. The number of exam questions Uppgift is six 6. Solutions written in Swedish are, of course, welcome. Each question gives maximum ten 0 points. 30 points will guarantee a passing result. The grade Fx the exam can completed by extra examination for those with 7 9 points. Solutions to the exam questions will be available at starting from Tuesday 9 th of October 03 at.5 p.m.. The exam results will be announced at the latest on Tuesday the th of November, 03. Your exam paper will be retainable at elevexpeditionen during a period of seven weeks after the date of the exam. Lycka till!

2 forts tentamen i sf Uppgift a Let A, B, C be events in a probability space Ω, F, P. Assume that A B C. Show that P C c P A c + P B c 3 p b Let {X n } n be a sequence of r.v. s in a probability space and X and Y be two random variables in the same space. Take ɛ > 0 and let C { X Y ɛ}, A { X n X ɛ/} and B { X n Y ɛ/}. Check that A B C. 3 p c Assume next that X n P X, as n and that X n P Y, as n. What do you now know about X and Y in view of a and b? 4 p Uppgift The random variables X, X,... are independent and N0,. Set N def min{k X k 0} a Show that the probability density function of X N is { φx, x 0, f XN x 0, x < 0. Aid: It may be helpful to compute the probability P X N x N n. 5 p b Find E [X N ] and Var [X N ].. 5 p Uppgift 3 We say that a random variable X is Linnikα-distributed, α > 0, if its characteristic function is given by ϕ X t + t. α Let X, X,... be independent and identically Linnikα-distributed, Let N FSp, and let N be independent of X, X,.... Set S N def X + X X N. Find the distribution of p /α S N. 0 p

3 forts tentamen i sf Uppgift 4 Use a central limit theorem for suitably chosen Poisson random variables to prove that n n k lim n e n k!. k0 0 p Uppgift 5 X {Xt < t < } is a Gaussian stochastic process. Its mean function is µt for all t and its autocovariance function is Cov [Xt, Xs] C X h max 0, h, h t s. a Find the distribution of the bivariate random variable Xt Xt 0., Xt 0. T. 3 p b Why does it hold that Xt Xt 0., Xt 0. T d Xt + Xt + 0.8, Xt T? p c Find the probability P Xt > Xt 0. + Xt 0.. p d We sample the process X in time so that our samples are Show that X k Xk, k,,.... n n X P k a, k as n + and determine the limiting value a. Be so kind and justify Your steps of solution very carefully. 4 p Uppgift 6 W {W t t 0} is a Wiener process, hu is a function such that h udu <. Let 0 for any t > 0 Y t e t 0 hudw u. Find the autocorrelation function of the process Y {Y t t 0} as { e t s R Y t, s h udu+ s 0 h udu t > s e s t h udu+ t 0 h udu t s. Aid: The rule of double expectation may turn out to be useful to start with. 0 p

4 Avd. Matematisk statistik SOLUTIONS TO THE EXAM TUESDAY THE 9 th OF OCTOBER 03. Uppgift a If A B C, then it holds for the complement sets that C c A B c. By De Morgan s law Beta we get A B c A c B c. Thus by properties of any probability measure the first inequality in Collection of formulas P C c P A c B c and by Boole s inequality Beta and the Collection of Formulas which establishes the assertion as claimed. P A c B c P A c + P B c, b For given ɛ > 0 we have C { X Y ɛ}, A { X n X ɛ/} and B { X n Y ɛ/}. We need to check that We get by the triangle inequality that A B C, X Y X X n + X n Y X X n + X n Y But A B is the event that both A { X n X ɛ/} and B { X n Y ɛ/} hold. Hence if the event A B holds, i.e., if the event A B holds, then Thus it holds that A B C. X X n + X n Y ɛ/ + ɛ/ ɛ, X Y ɛ. c X n P X, as n and that X n P Y, as n mean that and P A c P { X n X > ɛ/} 0 P B c P { X n Y > ɛ/} 0 as n. Hence the inequality in a, which can be used in view of b, means that for any ɛ > 0. Hence we have shown that P C c P { X Y > ɛ} 0 P X Y 0 i.e., that the limiting random variable for convergence in probability is almost surely unique.

5 forts tentamen i sf Uppgift a We express as and by independence we get and as the r.v. s are I.I.D., P X N x N n P X n x X < 0,..., X n < 0, X n 0, P X n x X n 0, P X y X 0 and by definition of conditional probability Thus X N is independent of N and we get P 0 X y P X 0 Φx Φ0 Φx Φ0 x 0. f XN x d dx P X N x N n d dx P X N x φxi x 0. ANSWER a: f XN x φx, x 0, f XN x 0, x < 0. b E [X N ] + 0 Then by the symmetry property of x φx x e x / dx ] [ e x / π π 0 π π. E [ ] + XN x φxdx 0 + x φxdx + 0 x φxdx. Hence VarX N E [ X N] E [XN ] π. as was to be found. Uppgift 3 We use the composition formula to represent the characteristic function of the sum S N as ϕ SN t g N ϕ X t,

6 forts tentamen i sf where g N t is the probability generating function of N. Since N FSp we have by definition and by Appendix B, p p p p g N t t k p N k k0 where we require that pt <. t k p p k k t k p k p t k p k p k k0 pt p pt p pt pt pt, This p.g.f. is given directly in Råde- Westergren: Beta, chapter 7., if one observes that FSp is called geometric Gp in loc.cit, Then the composition formula gives ϕ SN t g N ϕ X t p + t α p + t α. Then we get the characteristic function of p /α S N as ϕ p /α S N t ϕ SN p /α t p +p t α p + p t α +p t α pϕ X t pϕ X t p +p t α p +p t α p p + p t α + t. α By uniqueness of the characteristic function we get that p /α S N is Linnikα-distributed. ANSWER : p /α S N Linnik α. Uppgift 4 Let U i Po be independent, for i,,...,. Set X n U +... U n. Then it follows by identical Po -distributions that E [U i ], Var [U i ],

7 forts tentamen i sf and in addition we know see Beta that and thus P X n k e e n n nk k! n k0 X n Pon.. We get thereby that n k k! P X n n P X n n 0 P Xn n 0. n We write X n n n n i0 U i n Then we get by the central limit theorem, that n i0 U i n d N0,, as n. By definition, the convergence in distribution means that Xn n n i0 P 0 P U i 0 F n U n n i0 i 0 Φ0 n. This is as asserted. 0 p Uppgift 5 a We set and write this as with a vector notation where and Y Xt Xt 0., Y Xt 0. Y def X def B Y Y BX Xt Xt 0. Xt Since X is a vector with a multivariate normal distribution as the process X is Gaussian, the distribution of Y is the multivariate normal distribution Collection of Formulas N Bµ, BCB T.

8 forts tentamen i sf where µ is the mean vector of X and C is the covariance matrix of X. From the information about the process X we get that µ and C Then matrix computations give 0 Bµ 9/0 8/0 9/0 9/0 8/0 9/ , BCB T 0. ANSWER : Xt Xt 0. Xt 0. N , 0. b Since the process X is Gaussian and weakly stationary constant mean function, and the a.c.f. depends only on the difference between time instants, it is also strictly stationary. Therefore Xt Xt 0. Xt 0. d Xt + Xt + 0. Xt + 0. Xt + Xt Xt Since the desired conclusion follows. Xt + Xt Xt B Xt + Xt Xt c When we use the notations from part a, the probability P Xt > Xt 0. + Xt 0. can be written as P Y > Y However, the covariance matrix in a shows that Y and Y are independent as they are non-correlated Gaussians. We see also that Y N0, 0.. Hence we get Y P Y > Y P Y > P > Φ, as Y 0. N0,. ANSWER : P Xt > Xt 0. + Xt 0. Φ 0..

9 forts tentamen i sf d If the samples of the process X in time are X k Xk, k,,...,. then for every k and E [X k ] Thus Cov X k, X l Cov [Xk, Xl] C X k l max 0, k l. Cov X k, X l { if k l 0 if k l. Thus X k s are Gaussian and non correlated, therefore independent r.v. s and the standard weak law of large numbers shows that n as n +, since E [X k ] for all k. n X P k, k By definition we have Uppgift 6 R Y t, s E [Y t Y s] E [e t 0 hudw u e s 0 hudw u ] We assume t > s > 0 and set Fs W equal to the sigma field generated by W up to time s. Then double expectation gives [ E E [e t 0 hudw u e ]] s 0 hudw u Fs W Then we use the linearity properties by definition of the Wiener integral [ E E [e t s hudw u+ s 0 hudw u e ]] s 0 hudw u Fs W E [e s 0 hudw u E [e t s hudw u Fs W where we took out what is known w.r.t. Fs W. But the random variable e t s hudw u is a function of the increments of W after s and hence independent of Fs W. Therefore E [e s 0 hudw u E [e ]] t s hudw u Now we know by Collection of Formulas that Z t hudw u N0, t s s h udu. Hence we see first that E [ e Z] ψ Z ]]

10 forts tentamen i sf is the momentgenerating function ψ Z t of Z evaluated at t. But both the Collection of Formulas and Beta give unanimously that ts h udu v ψ Z v e we change from t to v as the argument in order to avoid confusion with the time variable and thus E [e ] ts t s hudw u h udu e. But in the same manner, for Y N0, s 0 h udu. E Therefore we have found for t > s that [e ] ts s 0 hudw u h udu ψ Y e e s 0 h udu R Y t, s e t s hudu+ s 0 h udu The result for t s is found analogously. We can write compactly this as R Y t, s e maxs,t mins,t hudu+ mint,s 0 h udu