MATH 7550-01 INTRODUCTION TO PROBABILITY FALL 2011 Problems. 1 Can you produce an example of a probability space and a sequence of events π΄π in it, ββ such that limπββ π (π΄π ) = 0, π=1 π (π΄π ) = β, π {inο¬nitely many of π΄π occur} = 1? 2 Can you produce an example β βof a probability space and a sequence of events π΄π in it, such that limπββ π (π΄π ) = 0, π=1 π (π΄π ) = β, π {inο¬nitely many of π΄π occur} β (0, 1)? 3 Can you produce an example of a probability space and a sequence of events π΄π in it, ββ such that limπββ π (π΄π ) = 0, π=1 π (π΄π ) = β, π {inο¬nitely many of π΄π occur} = 0? 4 Let β 2 be the class of all rectangles {(π₯, π¦) : π < π₯ β€ π, π < π¦ β€ π}, ο¬nite or inο¬nite, in the plane β2 (if, say, π = β, the inequality π₯ β€ β means the same as π₯ < β: no real number π₯ is equal to β); let πͺ be the class of all open sets in β2 . Prove that π(β 2 ) = π(πͺ). 5 Let π be a Borel set in βπ . Prove that the class of all Borel subsets of π is a π-algebra in π. Prove that this π-algebra is the same as that generated by all subsets of π that are open in π. 6 Prove that if π and π are two random variables (on the same sample space, and taking values in β1 ), then π + π is also a random variable. The deadline for Problems 1 β 6 is September 16. of measures π and π on a space (π, π³ ) such that π(πΆ) = β« 7 Produce an example β« π1 (π₯) π(ππ₯) = π2 (π₯) π(ππ₯), but π{π₯ : π1 (π₯) β= π2 (π₯)} β= 0. Show at which point πΆ πΆ the βproofβ of Theorem 7.1 fails. The deadline for Problem 7 is September 19. 8 Prove or disprove: Let πΉ (π‘), ββ < π‘ < β, be a right-continuous nondecreasing function, πΉ (ββ) = 0, πΉ (+β) = 1. If πΉ has a ο¬nite number of jumps at the points π₯π < π₯π+1 < ... < π₯π , and is constant on the intervals (ββ, π₯π ), (π₯π , π₯π+1 ), ..., (π₯πβ1 , π₯π ), (π₯π , β), then this function is a distribution function of a discrete random variable. 9 Prove or disprove the statement of the previous problem with the ο¬nite sequence of jump points replaced by a countable one that is inο¬nite on one side: π₯π < π₯π+1 < ... < π₯π < π₯π+1 < ..., or ... < π₯π < π₯π+1 < ... < π₯π ; or on both: ... < π₯β1 < π₯0 < π₯1 < π₯2 < ... , assuming that limπβββ π₯π = β β if the sequence is inο¬nite to the left, and that limπββ π₯π = β if it is inο¬nite to the right. 10 Prove or disprove: Let πΉ (π₯), β β < π₯ < β, be a right-continuous nondecreasing function, πΉ (β β) = 0, πΉ (+β) = 1. If β [πΉ (π₯) β πΉ (π₯β )] = 1, π₯ 1 then the function πΉ is a distribution function of a discrete random variable. (The sum has, in appearance, an uncountable number of summands, but in fact all of them except a countable number are equal to 0, because a nondecreasing function can have only countably many discontinuities.) 11 Prove or disprove: Let πΉ (π‘), β β < π‘ < β, be a right-continuous nondecreasing function, πΉ (β β) = 0, πΉ (+ β) = 1. If πΉ has jumps at the points that form a dense set in β1 , then the function πΉ is a distribution function of a discrete random variable. 12 Prove or disprove: Let πΉ (π₯), β β < π₯ < β, be a nondecreasing function, πΉ (β β) = 0, πΉ (+ β) = 1. If πΉ is continuous on β1 , it is a distribution function corresponding to a density. 13 Prove or disprove: Let πΉ (π₯), β β < π₯ < β, be a nondecreasing function, πΉ (β β) = 0, πΉ (+ β) = 1. If πΉ is continuous on β1 and piecewise continuously diο¬erentiable (that is, β1 is divided into pieces by points (... <) π₯π < π₯π+1 < ... < π₯π (< ...) so that πΉ is continuously diο¬erentiable on the open intervals between these points, to the left of the smallest π₯π if it exists, and to the right of the greatest of them), it is a distribution function corresponding to a density. 14 Prove or disprove: Let πΉ (π₯), β β < π₯ < β, be a right-continuous nondecreasing function, πΉ (β β) = 0, πΉ (+ β) = 1. If β« β πΉ β² (π₯) ππ₯ = 1, ββ then the function πΉ is a distribution function corresponding to a density. 15 Suppose we take a random permutation of numbers 1, 2, ..., π; that is, the sample space Ξ© consists of all π! sequences π = (π₯1 , π₯2 , ..., π₯π ) such that {π₯1 , π₯2 , ..., π₯π } = {1, 2, ..., π}; (of course, β± = π«(Ξ©)); and the probabilities are taken so that all diο¬erent orders of the natural numbers from 1 to π are equally probable: π {(π₯1 , π₯2 , ..., π₯π )} = 1 . π! The random variable π is equal to the number of numbers π, 1 β€ π β€ π, standing in their own place: for π = (π₯1 , π₯2 , ..., π₯π ), π(π) = π(π₯1 , π₯2 , ..., π₯π ) = #{π : π₯π = π}. It was proved in the lecture that πΈ π = 1, πΈ π 2 = 2. Is πΈ π 3 = 3? The deadline for Problems 8 β 15 is Sep. 30. 16 Let π1 , π2 be two random variables taking the values in measurable spaces (ππ , π³π ), π = 1, 2. Prove that π = (π1 , π2 ) is a random vector with values in the product 2 space (π1 × π2 , π³1 × π³2 ) (that is that the function π : π 7β (β±, π³1 × π³2 )-measurable function from Ξ© to π1 × π2 ). ( ) π1 (π), π2 (π) is an 17 Prove that β¬1 × β¬ 1 = β¬ 2 . 18 disprove that for every sequence π΄1 , π΄2 , ..., π΄π , ... of independent events β© βProve or β β π ( π=1 π΄π ) = π=1 π (π΄π ). 19 Let π (π‘, π) be a (β¬[0, 1] × β±)-measurable function on [0, 1] × Ξ©, taking values 0, 1. Is the following true for all such functions: β« β« πΈ π (π‘, β) #(ππ‘) = πΈ π (π‘, β) #(ππ‘), [0, 1] β [0, 1] β i. e., πΈ π‘β[0, 1] π (π‘, β) = π‘β[0, 1] πΈ π (π‘, β)? (The counting measure # on [0, 1] is not π-ο¬nite, so the positive answer does not follow from Fubiniβs Theorem β but still it could be true.) 20 Prove that if real-valued random variables π1 , π2 have (absolutely) continuous joint distribution with density ππ1 , π2 (π₯1 , π₯2 ), then each of them separately has a continuous one-dimensional distribution, with densities β« β β« β ππ1 (π₯1 ) = ππ1 , π2 (π₯1 , π₯2 ) ππ₯2 , ππ2 (π₯2 ) = ππ1 , π2 (π₯1 , π₯2 ) ππ₯1 . ββ ββ 21 Let π1 , π2 , ..., ππ , ... be an inο¬nite sequence of real-valued random variables. Show that the event {limπββ ππ β€ π₯} belongs to the tail π-algebra β±β₯β for every real π₯ (lim denotes the upper limit; another notation for it is lim sup; and we know that an upper limit, ο¬nite or inο¬nite, always exists). 22 Using the fact that a ο¬nite limπββ π₯π exists if and only if limπββ, πββ (π₯π β π₯π ) = 0, prove that the event {there exists a ο¬nite limit lim ππ } = {π : there exists a ο¬nite limit πββ lim ππ (π)} πββ belongs to the tail π-algebra. 23 Prove or disprove: the event { lim ππ = β β} β β±β₯β . πββ 24 Prove that the random variable limπββ ππ (taking values in the extended real line [ββ, β]) is measurable with respect to the tail π-algebra β±β₯β . 25 Prove or disprove: the event { } π1 + ... + ππ =0 πββ π lim 3 belongs to the tail π-algebra. 26 Prove: the event { the series β β } ππ converges π=1 belongs to β±β₯β . 27 Prove or disprove, for nonnegative ππ : the random variable with respect to the tail π-algebra. ββ π=1 ππ is measurable The deadline for Problems 16 β 27 is October 7. 28 Suppose ππ‘ , 0 β€ π‘ < β, are independent random variables with the same probability density π(π₯). Prove that for no π‘0 β [0, β) does ππ‘ converge in probability to ππ‘0 as π‘ β π‘0 . 29 Let π and π be two random variables taking nonnegative integer values. Prove or disprove: If their generating functions ππ (π ), ππ (π ) coincide for π β [β1, 1], then their distributions ππ and ππ coincide. 30 Give an example of a random variable whose moment generating function is equal to β at all points except at 0. 31 Let a random variable π have the normal distribution with parameters (0, π). Prove that all odd-numbered moments ππ = πΈ π π about zero are equal to 0, and for π = 2π we have: π2π = πΈ π 2π = (2π β 1)!! β ππ , where (2π β 1)!! denotes the product of all odd numbers from 1 to 2π β 1. The deadline for Problems 28 β 31 is October 17. 32 Prove that the normal distribution with parameters (π, π΅), where π΅ is a nonsingular matrix, has the density { β π(π) = const β exp β 12 πππ (π₯π β ππ )(π₯π β ππ )}, π, π where the matrix π = (πππ ) = π΅ β1 . What is the constant in this formula equal to? 33 Let the two-dimensional random vector π have the normal with zero ( distribution ) 1 1 expectation (πΈ π = 0) and the matrix of covariances π΅1 = . Prove that the 1 1 distribution of this random vector is concentrated on some line in the plane (i. e. that almost surely π1 π1 + π2 π2 = const for some (π1 , π2 ) β= (0, 0)). Deduce form this that the normal distribution with parameters (0, π΅1 ) has no density with respect to the two-dimensional Lebesgue measure π2 . 4 34 (Is the)result of the previous problem true if we replace the matrix π΅1 with 1 2 π΅2 = ? 2 4 35* (* means a non-obligatory problem). Prove or disprove: if the characteristic function ππ (π‘) is diο¬erentiable at 0, then πΈβ£πβ£ < β. HINT: A random variable having the standard Cauchy distribution with density π(π₯) = β1 π /(1+π₯2 ) has no expectation; its characteristic function π (π‘) = πββ£π‘β£ is not diο¬erentiable at 0, but this is, so to speak, touch and go: a little more, and it would be: at least the one-sided derivatives are not inο¬nite. Couldnβt we try and consider a density πΛ(π₯) that goes to 0 at ±β a little slower than π(π₯), but still the expectation does not exist? It could be that for this density the characteristic function has zero derivative at π‘ = 0. Better take your density (-ies) symmetric with respect to 0 (even functions): the corresponding characteristic functions will be real-valued. 36 Let a four-dimensional ( ) random vector (π1 , ..., π4 ) have the normal distribution with parameters 0, π΅ = (πππ ) . Find the fourth mixed moment π1111 = πΈ(π1 π2 π3 π4 ). The deadline for Problems 32 β 36 is October 21. 37 Let π1 , π2 , ..., ππ , ..., π be distributions concentrated on nonnegative integers: for all natural π β β β β ππ {π} = π{π} = 1. π=0 π=0 Prove (or disprove) that if lim ππ {π} = π{π} πββ for every π = 0, 1, 2, 3, ..., then ππ βπ€ π (π β β): convergence of the values of the probability mass function implies weak convergence of distributions. 38 Let ππ be the normal distribution with parameters (ππ , ππ ), ππ > 0; suppose ππ β π, ππ β 0 as π β β. Prove that ππ βπ€ πΏπ (π β β), where πΏπ is the distribution concentrated at the point π: πΏπ (πΆ) = 1 if πΆ β π, and = 0 if πΆ ββ π. 39 Prove or disprove that if ππ βπ π, then πππ βπ€ ππ . 40 Prove or disprove: if πππ βπ€ ππ , then ππ βπ π. 41 For every π, let the random variables ππ , ππ be independent; and let ππ βa. s. π, ππ βa. s. π. Prove that then π and π are independent. The deadline for Problems 37 β 41 is October 28. 42 Check that { π (π‘) = 1 β β£π‘β£, 0, β£π‘β£ β€ 1, β£π‘β£ > 1, 5 is a characteristic function of a continuous distribution on the real line. Find its density. 43 For a random variable π having the density obtained in the previous problem, prove that πΈ β£πβ£ = β. 44 Let π1 , π2 , ..., ππ , ... be a sequence of independent random variables having the distribution with the density found in Problem 42 (such a sequence exists by Theoπ1 + ... + ππ rem 13.5). Let ππ = . Does the sequence ππ converge in probability to a π constant as π β β? If yes, what is this constant equal to? 45 For the sequence of random variables of the previous problem, does the weak limit of the distribution of ππ (π€) lim πππ πββ exist? If yes, is the limiting distribution discrete or continuous, and what is its probability mass function or its density? We call a Markov chain time-homogeneous (or just homogeneous) if its transition matrix is the same at all steps: π1 = π2 = ... = ππ = ... = π . For homogeneous Markov chains the transition matrix π ππ = π πβπ (the (πβπ)-th power of the one-step transition matrix π ) depends only on the diο¬erence π β π = π, and its entries πππ π₯π¦ also depend (π) (π) (π) π on this diο¬erence only: πππ π₯π¦ = ππ₯π¦ , where ππ₯π¦ are the entries of the power π (ππ₯π¦ are not the powers of ππ₯π¦ , the (π₯, π¦)-th entry of the matrix π , this is why we use (π) with parentheses in the subscript). A substantial part of the theory of Markov chains deals with the behavior of the π(π) step transition probabilities ππ₯π¦ = π {ππ+π = π¦β£ππ = π₯} as the number of steps π β β. It turns out sometimes (under some conditions) that these conditional probabilities have the limit as π β β that does not depend on the number π₯ of the matrix row. This can be interpreted as follows: under these conditions, the events {ππ+π = π¦} and {ππ = π₯} (or: the random variables ππ+π and ππ ) become independent in the limit as π β β. Or: under these conditions the events or random variables separated by a growing number of steps (by a growing time interval) become independent in the limit. I want you to solve several problems about this phenomenon (in the more intuitive words, disappearance of dependence with lapse of time; in a more prosaic and more precise (π) words, existence, for each pair π₯, π¦ β π, of the limit limπββ ππ₯π¦ not depending on π₯ β or existence of the limit limπββ π π being a matrix all of whose rows are identical to one another). ( ) 0 1 46 Let π = . Find the limit limπββ π π . 1/2 1/2 HINT: Represent the matrix π in the form π = π΄ β π· β π΄β1 , where π΄ is a non-singular matrix, and π· a diagonal one. 6 47 Let β β 0 0 1 0 0 0 1 β β 0 π =β β . 1/2 0 1/2 0 0 1/2 0 1/2 Does the limit limπββ π π exist? Are the rows of this limiting matrix identical? 48 The same question for the matrix β 0 β 2/3 π =β 0 1/3 1/3 0 0 1/3 2/3 0 0 2/3 β 2/3 0 β β 1/3 0 (which is the same as that in the example with formula (25.5), only with 5 replaced with 4 β which makes all the diο¬erence). 49* The same question for the matrix π given by formula (25.5). The deadline for Problems 42 β 48 is November 4. 50* For a Markov chain with the transition matrix of Problem 46 , prove that πΌβ (β±β€π , β±β₯π ) β 0 as π β β. At what rate does it go to 0 (that is, e. g.: as some power of π β or of π β π; or exponentially fast; or: faster than any exponential function)? 51 Let π0 , π1 , π2 , ..., ππ , ... be independent random variables having the normal distribution with parameters (0, 1) (the standard normal distribution). Let π1 = (π1 β π0 )2 , π2 = π1 + π2 + ... + ππ (π2 β π1 )2 , ..., ππ = (ππ β ππβ1 )2 , ... . Prove that there exists limπββ (π ) . π What is this limit? We cannot use the laws of large numbers that we had for independent random variables: ππ are deο¬nitely dependent. And we cannot use Theorem 26.4 β even if we check that πΌβ (β±ππ , πβ€π , β±ππ , πβ₯π ) β€ π½(π β π), π½(π) β 0 (π β β), because the random variables ππ are unbounded (just as normal random variables). 52* For the random variables of the previous problem prove that the strong law of large π1 + π2 + ... + ππ numbers takes place: the sequence converges almost surely. π HINT: Try to copy the proof of Theorems 16.3, 16.4: in the sum π β πΈ[(ππ β πΈ ππ )(ππ β πΈ ππ )(ππ β πΈ ππ )(ππ β πΈ ππ )] π, π, π, π=1 not all summands with π, π, π, π diο¬erent are equal to 0 as for independent random variables, but the number of nonzero summands can be counted and estimated. 7 53 Let π be a continuous random variable with probability density β§ β¨ 4 β π₯, ππ (π₯) = 8 β© 0, π₯ β [0, 4], π₯β / [0, 4]; π = (π β 1)2 . Find the conditional expectation πΈ{πβ£π = π¦}. 54 Let random variables π, π be independent and have normal distributions with parameters, respectively, (0, 1) and (0, 2); and π = π + π. Find the conditional distribution of π under the condition π = π§. Find the conditional expectation πΈ(πβ₯π) and the conditional (( )2 ° ) variance πΈ π β πΈ(πβ₯π) ° π . 55 Let π1 , π2 , ..., ππ , ... be independent random variables having each the same normal distribution with parameters (0, 1). Find (a version of) πΈ(π12 β₯β±β₯β ). 56 Let π and π be independent random variables with densities ππ (π₯), ππ (π¦); π = π + π. Find the conditional density ππβ£π=π₯ (π§). The deadline for Problems 51 β 56 is Nov. 11. ( ( ) 57 Prove that if the random variable π is π-measurable, πΈβ£πβ£ < β, πΈ β£πβ£ β πΈ β£πβ£β₯π) < β, then πΈβ£π πβ£ < β. 58 Let π1 , π2 , ..., ππ , ... be independent random variables with πΈ ππ = 0, πΈ ππ2 = ππ2 ; ππ = π1 + ... + ππ . 2 β₯π1 , ..., ππ ). Find πΈ(ππ+1 59 Let π = β€+ = {0, 1, 2, 3, ...}; [π(π β π‘)]π¦βπ₯ πβπ(π βπ‘) π(π‘, π₯, π , π¦) = (π¦ β π₯)! if π₯, π¦ β β€+ , π¦ β₯ π₯, and 0 for π¦ < π₯, where π β₯ 0 (you see that this is, in fact, up to a shift, the Poisson distribution with parameter π β (π β π‘)). Check that the conditions β 1 , β 3 , β 4 are satisο¬ed. The Markov process with transition probabilities (30.5) and with right-continuous trajectories πβ (π) is called the Poisson process with parameter (the rate) π. 60 Prove: If ππ‘ , π‘ β₯ 0, is a Markov process on π = β€+ with transition probabilities (30.5) (supposing that such a process exists), then for π β₯ π‘ the random variable ππ β ππ‘ has the Poisson distribution with parameter π β (π β π‘). 61 For the same process prove that for 0 β€ π‘0 β€ π‘1 β€ ... β€ π‘π the random variables ππ‘1 β ππ‘0 , ππ‘2 β ππ‘1 , ..., ππ‘π β ππ‘πβ1 are independent. 8 62 Let π = [0, β), π = β1 (and, of course, π³ = β¬ 1 ). Check that the density 2 1 π(π‘, π₯, π , π¦) = β πβ(π¦βπ₯) /2(π‘βπ ) 2π(π‘ β π ) (as a function of its density-argument π¦ this is the normal density with parameters (π₯, π β π‘)) satisο¬es the Chapman β Kolmogorov equation (30.7). 63 For a Markov process ππ‘ , π‘ β [0, β), with the transition function given by (30.6) with the density (30.8), and π (π‘, π₯, π‘, πΆ) = πΏπ₯ (πΆ) (assuming such a process exists) prove that the diο¬erence ππ β ππ‘ for π β₯ π‘ has the normal distribution with parameters (0, π β π‘) and that for 0 β€ π‘0 β€ π‘1 β€ ... β€ π‘π the diο¬erences ππ‘1 β ππ‘0 , ππ‘2 β ππ‘2 , ..., ππ‘π β ππ‘πβ1 are independent. 64 Let the initial distribution π at time π‘0 = 0 be concentrated at the point π₯0 : π(πΆ) = πΏπ₯0 (πΆ), and the ο¬nite-dimensional distributions be given by formula (31.1) with π (π‘, π₯, π , πΆ) of the previous problem. Check that the joint distribution of ππ‘π βs (if the stochastic process ππ‘ , π‘ β₯ 0, exists) is a normal one. With what parameters? The deadline for Problems 57 β 64 is Nov. 18. 65 Find ππ₯ for the Poisson process (see Problem 59 ). 66 Find the matrix π΄ = (ππ₯π¦ ) for the Poisson process. Find the distribution of the position of the process after the ο¬rst jump if it starts from the point π₯ = 5 at time 0. 67 Let ππ₯ > 0. Prove that the time π1 of the ο¬rst jump and the position ππ1 of the process at this time are independent with respect to the probability measure ππ₯ . β« ππ₯π¦ HINT: Prove that ππ₯ {π1 β πΆ, ππ1 = π¦} = ππ₯ πβππ₯ π‘ ππ‘ β for every Borel set ππ₯ πΆ πΆ β [0, β) and π¦ β= π₯. To do this, start with πΆ = [π, π), 0 β€ π β€ π <β« β): such sets for a semi-algebra, and if two ο¬nite measures ππ₯ {π1 β πΆ, ππ1 = π¦} and πβππ₯ π‘ ππ‘ β ππ₯π¦ πΆ coincide on a semi-algebra, they coincide on the π-algebra β¬0, β generated by it. Then use the equality ππ₯ {π1 β [π, π), ππ1 = π¦} = limββ0+ ππ₯ {π1β β [π, π), ππ1β = π¦}, where the probability depending on β is expressed as a sum of some products: a sum of a ο¬nite geometric progression. 68 Let a function π(π‘), π‘ β₯ 0, be continuous, let it have, for every π‘ β [0, β), a rightπ+ hand derivative β(π‘) = π(π‘), which is also continuous on [0, β). Prove that then the ππ‘ π two-sided derivative π(π‘) exists for all π‘ > 0. β« π‘ ππ‘ HINT: Take πΛ(π‘) = π(π‘) β β(π ) ππ . The function πΛ(π‘) is continuous and has a zero 0 right-hand derivative at every point; we have to prove that it is a constant. Now you take any proof of the corresponding fact for the two-sided derivative, and try to adapt it to the one-sided one. 9 69 Check that both kinds of Kolmogorov equations are satisο¬ed for the transition probabilities of the Poisson process (and this is a way how these probabilities can be found starting with the matrix π΄). ( ) ( ) β1 1 70 Let π = {0, 1}, π΄ = . Find the solution π π‘ = π(π‘, π₯, π¦) π₯, π¦βπ of the 2 β2 π π‘ π π‘ matrix equation π = π΄π π‘ or π = π π‘ π΄ with the initial condition π 0 = πΌ. Find the ππ‘ ππ‘ limits lim π‘ββ π(π‘, π₯, π¦). β β β2 2 0 71 Let π = {0, 1, 2}, π΄ = β 1 β3 2 β . Do the limits lim π‘ββ π(π‘, π₯, π¦) exist? If 0 1 β1 they do, ο¬nd them by solving the system π β π΄ = 0, π β 1 = 1 (π and 0 are row vectors, 1 the column vector with all components equal to 1). 72 Let π = {0, 1, 2, ..., π, ...}, ππ₯π₯ = β π, ππ₯, π₯+1 = π, where π is a positive constant, all other ππ₯π¦ = 0. Find all solutions of the equation ππ΄ = 0. Does the discrete distribution on π with the probability mass function π(π‘, π₯, β) converge as π‘ β β to some probability distribution on π? 73 Let π , π be independent random variables with exponential distributions with pa( ) rameters π and π. Find the distribution of the random variable π (π) = min π (π), π (π) . 74 Let π = {0, 1, 2, ..., π, ...}, π00 = βπ, π01 = π, ππ, πβ1 = π, πππ = βπ β π, ππ, π+1 = π for π > 0, all other ππ₯π¦ = 0, where π and π are positive constants (negative numbers on the main diagonal of the inο¬nite matrix π΄, positive numbers on the diagonal one row above the main one and one row below it, and the rest zeros; the Markov process describes a system with one service and unrestricted queue). Find all solutions of the equation π β π΄ = 0. For π = π = 2, does the limiting distribution for π(π‘, π₯, β) as π‘ β β exist; and if it does, what is this limiting distribution? The deadline for Problems 65 β 74 is Nov. 30. 75 Do the tail π-algebras associated with the Markov chains of Problems 47 , 48 consist only of events of probability 0 and 1? β 76 Assuming that ππ‘ = ππ‘ / π‘, 0 < π‘ < β, where ππ‘ is the Markov process of Problem 62 , is a Markov process, ο¬nd its transition density ππ (π‘, π₯, π , π¦) (it seems obvious that it ought to have one). β of the random variable ππ = βHINT: ππ (π‘, π₯, π , π¦) should be the conditional density ππ / π under the condition ππ‘ = π₯, that is,βππ‘ = π₯ β π‘. We know the conditional distribution of ππ under the condition ππ‘ = π₯ β π‘: itβs a normal distribution with such and such parameters; we should be able to ο¬nd the conditional distribution of ππ‘ being a linear function of ππ‘ too. 77 Check that ππ (π‘, π₯, π , π¦) satisο¬es the Chapman β Kolmogorov equation (all other necessary properties are quite obvious). 10 78 Check that ππ‘ , 0 < π‘ < β, is a Markov process with the transition density found in Problem 76 . 79 Let π(π₯) be a probability density on the real line. Does there exist a family of random variables ππ‘ , π‘ β₯ 0, that are independent and have the distribution with the density π(π₯)? 80 Let the joint distribution of random variables π, π, π be normal with parameters (0, π΅), where β β 2 β1 β1 π΅ = β β1 2 β1 β . β1 β1 2 Is this distribution continuous? That is: does this distribution have a density ππππ (it is not important whether this density is a continuous function or not)? 81 For the random variables of Problem 80 , ο¬nd the conditional distribution of the random variable π under the condition that π is ο¬xed (with respect to π). That is, answer the following questions: Is the conditional distribution a discrete one, i. e., β π {π β πΆβ₯π} = ππ (π) ? π: π₯π βπΆ If yes, ο¬nd π π (π) = π {π = π₯π β₯π}. Is the conditional distribution a continuous one, i. e., β« π {π β πΆβ₯π} = π πβ₯π (π₯) ππ₯ ? πΆ If yes, ο¬nd the conditional density ππβ₯π (π₯) (depending on π = π(π)). Is the conditional distribution neither discrete nor continuous? If so, ο¬nd the conditional distribution function πΉπβ₯π (π‘) = π {π β€ π‘β₯π}. Note that, in all these cases, we can try to ο¬nd not the things (conditional probabilities, or conditional densities) under the condition that π is known (ο¬xed) (which is denoted with β₯π ), but rather under the condition that π = π¦, for an arbitrary real π¦. 82 Let (π, π) have the uniform distribution on the square [0, 1]2 . Does the random variable π = π 2 + π 2 have a continuous distribution? 83 Let the random variables π and π be independent, π having a discrete distribution with π {π = 0} = π {π = 2} = π {π = 4} = 1/3, and π having the uniform distribution on the interval [0, 3]. Find the distribution of the random variable π = π + π. Is this distribution discrete? continuous? neither discrete nor continuous? According to which of this cases takes place, ο¬nd the probability mass function of π, or its density, or its distribution function πΉπ (π‘). The deadline for Problems 75 β 83 is Dec. 9. 11 84 Let Ξπ for every π > 0 be a Poissonian random variable with parameter π. Prove that Ξπ converges in probability to +β as π β β (the notation: Ξπ βπ β); that is, that for every πΆ we have lim πββ π {Ξπ > πΆ} = 1. 85 Let π1 , π2 , ..., ππ , ... be independent random variables, ππ having the Poisson distribution with parameter 1/π(π + 1). Does the event {π1 + π2 + ... + ππ + ... < β} have probability equal to 1? equal to 0? strictly between 0 and 1? In the ο¬rst case, what is the distribution of the random variable π = π1 +π2 +...+ππ +...? To be concrete: ο¬nd the probability π {π = 1} with accuracy up to 0.01. In the second and in the third case: does the sum π1 +π2 +...+ππ βπ +β (convergence in probability to inο¬nity, see the previous problem)? 86 Let π have a uniform distribution on the interval (0, 2); π = ln π. Find πΈ π. Is πΈβ£πβ£π ο¬nite for all natural π? 87 Let π1 , π2 , ..., ππ , ... be independent random variables with uniform distribution on the interval (0, 2). βπ Find πΈ π=1 ππ . Which of the following events have probability 1, and which 0: { lim πββ π β } { ππ = 0 ; lim πββ π=1 π β } ππ = +β ; π=1 π β { } there exists a ο¬nite positive limit lim ππ ; πββ { lim π β πββ π=1 ππ does not exist, lim π β πββ π=1 { lim πββ π=1 ππ = 0, lim πββ π β π β } ππ = π3/2 ; π=1 } ππ = + β ? π=1 βπ HINT: Take ln π=1 ππ ; use the previous problem and the Strong Law of Large Numbers (a concrete theorem of this group). 88 Prove or disprove: If a random variable π has a ο¬nite ο¬fth moment πΈ π 5 , it also has a ο¬nite second moment πΈ π 2 . 89 Suppose the random variable( π has a ο¬nite π1 , π2 are two arbitrary ° expectation; ) ° random variables. We know that πΈ πΈ(πβ₯π1 , π2 ) π1 = πΈ(πβ₯π1 ) (almost surely, of course; or: one of the versions of one of these conditional expectations is the other conditional expectation). Is the iterated conditional expectation ° ( ) πΈ πΈ(πβ₯π1 )° π1 , π2 necessarily equal to πΈ(πβ₯π1 ) (a. s.)? Is it necessarily equal to πΈ(πβ₯π1 , π2 )? Or it may be not equal (a. s.) to either of these two conditional expectations, but equal to some quite diο¬erent random variable? 12 90 Let π and π be independent and have Poisson distribution with parameters, correspondingly, π and π. Find the conditional distribution of the random variable π if π + π is known; i. e., ο¬nd π {π = πβ₯π + π}. 91 Prove or disprove: Let ππ , π be random variables with ο¬nite expectations. Then if the distribution of ππ converges weakly to that of π as π β β, then also the expectations converge: πΈ ππ β πΈ π. The deadline for Problems 84 β 91 is Dec. 16, noon. 92* Let ππ‘ , π‘ β₯ 0, be a Wiener process starting from a non-random point π₯0 β₯ 0 at Λ π‘ = β£ππ‘ β£. time 0: π0 β‘ π₯0 . Let π Λ π‘. For π‘ > 0, ο¬nd the probability density of the random variable π Λ π‘. 93* Find the ο¬nite-dimensional densities of the stochastic process π Λ π‘ a Markov process? 94* Is π 13
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