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1 Av Matematisk statistik TENTAMEN I SF940 SANNOLIKHETSTEORI/EXAM IN SF940 PROBABILITY THE- ORY, WEDNESDAY OCTOBER 5, 07, Examinator : Boualem Djehiche, tel , boualem@kthse Tillåtna hjälpmeel/permitte means of assistance: Appenix in A Gut: An Intermeiate Course in Probability, Formulas for probability theory SF940, L Råe & B Westergren: Mathematics Hanbook for Science an Engineering an pocket calculator All use notation must be explaine an efine Reasoning an the calculations must be so etaile that they are easy to follow Each problem yiels max 0 p You may apply results state in a part of an exam question to another part of the exam question even if you have not solve the first part A preliminarily lower boun of 5 points will guarantee a passing result If you have receive 5 bonus points from the home assignments, you may skip Problem a) If you have receive 0 bonus points, you may skip the whole Problem Solutions to the exam questions will be available at starting from Monay October 30, 07 Goo luck! Problem Let W t)) t 0 be a Wiener process Set, for t > 0, X n := n j + W tj + )) W tj)), n a) Determine the istribution of X n 5 p) b) Show that X n converges in istribution to some ranom variable X, as n Fin the istribution of the limit ranom variable X 5 p) Hint: k= k = π 6 ) Problem Suppose that the joint characteristic function of X an Y equals ϕ X,Y s, t) = E[e isx+ity ] = exp αe is ) + βe it ) + γe is+t) ) ), with α > 0, β > 0, γ > 0 a) Show that X an Y both have a Poisson istribution, but that X + Y oes not 6 p)

2 cont exam sf940 Fall 06 b) Are X an Y inepenent? 4 p) Problem 3 Let X, X, be inepenent, U0, )-istribute ranom variables, an let N Poλ) be inepenent of X, X, Set Y N = 0 when N = 0) Y N := max{x, X,, X N } a) Determine the istribution function an the characteristic function of Y N 4 p) b) Show that E[Y N ] as λ 3 p) c) Show that λ Y N ) converges in istribution as λ, an etermine the limit istribution 3 p) Problem 4 Let X an Y be ranom variables such that E[X] = 0, VarX) = σ > 0 an Y X = x Nx, a ) ) X a) Determine a istribution of X such that the vector becomes normally istribute with mean µ = an covariance matrix Λ Determine µ, µ an Λ 6 ) Y µ p) µ b) Compute E[e X Y = y] 4 p) Problem 5 Let X, X, be inepenent, U, )-istribute ranom variables, a) Show that the limit in probability of max j n X j is, an of min j n X j is -, as n 6 p) b) Set Y n := n j= X j n max X j j n Show that Y n converges in istribution as n a etermine the limiting istribution 4 p)

3 Av Matematisk statistik Suggeste solutions to the exam Wenesay October 5, 07 The problems can be solve using other methos than the suggeste below Problem a) The Wiener process has the inepenent increments property This implies that the ranom variables Y j := W tj) W tj )), j =, n, are inepenent Moreover, Y j Y N0, t) j N0, t ) Thus, j+ j+) X n = n Y j j + N n 0, t b) We may use the characteristic function to see that ) X n N 0, t = N j + ) ) j + ) ) 0, π t 6 Problem a) We have X Poα + γ) an Y Poβ + γ), since ϕ X s) = ϕ X,Y s, 0) = exp {α + γ)e is )}, ϕ Y t) = ϕ X,Y 0, t) = exp {β + γ)e it )} To fin the istribution of X + Y, we compute its characteristic function: ϕ X+Y s) = ϕ X,Y s, s) = exp {α + β)e is ) + γe is )} Clearly, it is not of the form exp {θe is )}, for some θ > 0 X +Y is not Poisson istribute b) X an Y are not inepenent, since ϕ X,Y s, t) ϕ X s)ϕ Y t) Problem 3 First note that if X U0, ) then its istribution function is F X x) = x, x 0, ) Moreover, the probability generation function of N Poλ) is g N t) = E[t N ] = e λt ) Since the X, X, are ii U0, ) an are inepenent of N, by conitioning on N, we have F YN x) = P max{x, X,, X N } x) = E[F X x)) N ] = g N F X x)) = e λx )

4 cont exam sf940 Fall 06 More precisely, 0 if x 0, F YN x) = e λx ) if 0 < x <, if x Thus, F YN has two parts: a egenerate part, 0, which correspons to the case where Y N = 0 which occurs when N = 0) an a smooth part, e λx ) when x > 0 To obtain the characteristic function, we first erive the ifferential of the smooth part of F YN x) This is We have ϕ YN t) = E[e ity N ] = b) E[Y N ] = i ϕ Y N 0) = λ c) We have e λx ) ) x = e itx F YN x) = P N = 0)+ { λe λx ) if 0 < x <, 0 elsewhere 0 e itx λe λx ) x = e λ + λ e it e λ) it + λ e λ ) E[Y N ] as λ ϕ λ YN )t) = E[e itλ Y N ) ] = e itλ E[e itλy N ] = e itλ ϕ YN tλ) = ) e λ it) it Since e λ it) = e λ 0, as λ, it hols that e λ it) 0, as λ Thus, λ Y N ) Exp) ϕ λ YN )t) it = ϕ Exp)t) Problem 4 a) We compute the characteristic function of X, Y ) in two ways: First, we use conitioning to obtain ϕ X,Y s, t) := E[e isx+ity ] = E[E[e isx+ity X]] = E[e isx E[e ity X]] = Y X = x Nx, a )) That is = E[e isx e itx a t ] = e a t E[e is+t)x ] E[e is+t)x ] = e a t ϕx,y s, t) Next, we use the assumption ) that X, Y ) Nµ, Λ), where µ = µ, µ ), with µ = E[X] = σ c 0, an Λ = c b, where σ = VarX) an c := CovX, Y ), b = VarY ) to be etermine together with µ = E[Y ] We have ϕ X,Y s, t) = exp { isµ + tµ ) σ s + b t + cst) } = µ = 0) = exp { itµ σ s + b t + cst) }

5 cont exam sf940 Fall 06 3 { E[e is+t)x ] = exp itµ } σ s + b a )t + cst) The assumption that X, Y ) Nµ, Λ) implies that X NE[X], VarX)) = N0, σ ) E[e is+t)x ] = exp { } σ s + t) Thus, we shoul have e σ s+t) = exp { itµ } σ s + b a )t + cst), s, t Ientifying the coefficients, we obtain that µ = 0, c = σ an b = a +σ Hence, X, Y ) Nµ, Λ), where ) ) 0 σ σ µ =, Λ = 0 σ a + σ b) To compute E[e X Y = y], we note that X Y = y Nµ, σ ), where µ = 0 + σ y 0) = σ a + σ E[e X Y = y] = e µ + σ = exp σ a + σ y, σ = σ σ4 a + σ a + σ y + σ ) σ4 a + σ ) Problem 5 Before, answering a) an b), we note that if X U, ) then F X x) = P X x) = x +, < x <, E[X] = 0, VarX) = /3 a) Pick ɛ > 0 an set X ) := min j n X j an X n) := max j n we have P X n) > ɛ) = P X n) > + ɛ) + P X n) < ɛ) But, since all X j, ), then X n), ) This implies that P X n) > + ɛ) = 0 Thus, ) n ɛ P X n) > ɛ) = P X n) < ɛ) = X js are ii) = F X ɛ)) n = n + ɛ) n = This, in turn suggest that 0 < ɛ < otherwise we get negative probabilities), which implies P that P X n) > ɛ) 0 as n, ie X n) We have P X ) + > ɛ) = P X ) > ɛ ) + P X ) < ɛ + ))

6 cont exam sf940 Fall 06 4 But, P X ) < ɛ + )) = 0, as all X j, ) Moreover, ie X ) P X ) > ɛ ) = F X ɛ )) n = P + ɛ ) n = ) n ɛ P X ) + > ɛ) = 0, n, b) By the Central Limit Theorem, it hols that n n j= X j X n) P Use Cramèr-Slutsky s theorem to obtain Y n N 0, ) 3 ) n ɛ N0, ) Moreover, by a) 3