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An Introduction to Measure Theoretic Probability 2nd edition George G. Roussas - Solutions

For any sequence of events {Xn}, n ( 1, show that
The sequence {Xn}, n ³ 1, of r.v.s is said to converge completely to 0, if for every e > 0,P(|Xn|³ e) < ¥.(i) Show that, if {Xn}, n ³ 1, converges completely to 0, then Xn (ii) By means of an example, show that complete convergence is not necessary for a.s. convergence.For
By means of an example, show that Xn X if and only if, for every {n'}( {n}, there exists a further subsequence {n''} ( {n'} such that Xn''X and any two limiting r.v.s are a.s. equal.
By means of an example, show that Xn c ( 0 need not imply that P(Xn ( 0 1 for all n.
If for some en > 0 withen< ¥, it holds thatP(|Xn+1€“Xn| ³ en) < ¥, then show thatXn to a r.v.It suffices to show that {Xn} converges mutually a.s. To this end, set An = (|Xn+1 €“ Xn| ³ en), n ³ 1, and use Exercise 3 in this chapter in order to
For n = 1, 2,..., let Xn and X be (real-valued) r.v.s and let g : ( †’ ( be continuous. Then show that implies g(Xn) g(X) (so the continuity preserves convergence in probability).
Construct an example where r.v.sXnandX, defined on a probability space (W, A, P), are such thatPointwise henceAndBut
Let ( = {-2, - 1, 3, 7}, and let measure μ be defined by: μ ({-2}) = 2, μ({- 1}) = 1, μ({3}) = 3, μ ({7}) = 7. Let X be a r.v. defined by: X(- 2) = X(- 1) = - 1, and X(3) = X(7) = 1. Then, for a = {-2, 3, 7}, compute the integral fA Xd μ.
Let ((, A, Î¼) = ((-5, 5), B(-5,5), ( where ( is the Lebesgue measure, and let the r.v. X be defined as follows:Then, for A = {-1, 4}, compute the integral fA Xd Î¼.
Let ( = {0, 1, 2,....}, let A be the discrete σ-field of subsets of (, and on A, define the function μ by: μ(A) = number of nonnegative integers in A. Then show that μ is a measure, and indeed, a σ-finite measure.
Let g : ( → (0, () be nondecreasing in (0, () and symmetric about 0 (g(-x) = g(x)) x ( (), and let X be a r.v, such that (g(X) < (. Then show that: P(|X| ( c) ( (g (X)/g(c) for every c > 0.
For a r.v. X, show that
LetXbe a simple r.v. defined on (W,A), so thatWhere {Aj, j = 1,€¦., n} is a (measurable) partition of W, and the aj, j = 1,€¦., n, are assumed to be distinct. Next, letBe ay other representation of X. Then show that each Aj, j = 1,€¦., n is the sum of some
IfXandYare two identically distributed integrable r.v.s thenFor any constant c.Now, let g be nonnegative. Then there exist 0 £ gn(x) simple †‘ g(x); i.e.,Which implies that 0 £Simple †‘ g(X) as n†’¥ where Ani = X€“1(Bni). ThenWhereasFor all n, by the previous step. HenceFinally,
If the r.v. X ( 0 and f X d μ = 0, then show that μ(X ( 0) = 0.
If the r.v.Xtakes on the valuesn,n= 1, 2,€¦, with probability 1, so thatShow that
(i) For a r.v. and any two disjoint events A and B, show that(ii) More generally, for any finite collection of pairwise disjoint events Ai, i = 1,€¦., n, show that
If for the pariwise disjoint eventsAi,i= 1,€¦.,n, the integralexists, then the integrals fAi X, i = 1,€¦.., n, exist, and
Let X and Y be two simple r.v.s such f X = f y = ( or f X = f Y = -(. Then show that f (X + Y) exists and f(X + Y) = f X + f Y(= ( or -().
Let μ1 and μ2 be two measures on ((, A) and suppose that f Xd (μ1 + μ2) exists. Then f Xd μi exist, i = 1, 2, and f Xd (μ1 + μ2) = f Xd μ1 + f Xd μ2.
If the r.v.sXi, = 1,€¦.,n, are integrable, then so is the r.v.And
LetXandYbe r.v.s with finite second moments, and set eX= Î¼1, eY = Î¼2, 0 < Var(X) = Ïƒ21, 0 < Var(Y) = Ïƒ22. Then covariance and the correlation coefficient ofXandY, denoted respectively, by Cov(X, Y) and Ï(X, Y), are defined by: Cov(X, Y) =
If the r.v. X is integrable, then show that:By a counterexample, show that the converse need not be true.
If X is a simple r.v. (not necessarily nonnegative), defined on the (product) space ((1 ( (2, A1 ( A2, ( = μ1 ( μ2) with μ1 and μ2 σ-finite, for which ( Xd ( exists, then show directly that ((( X ((1, (2) d μ1 μ2 = ((( X ((1, (2) d μ2 d μ1 = ((( X ((1, (2) d( = ( X d(.
Let X be a r.v. defined on the product space ((1 ( (1, A1 ( A2, ( = μ1 ( μ2) and suppose that ( X d( exists. Then show that the Fubini theorem holds true.
If the r.v.sX1,€¦.,Xnare i.i.d. with eX1 = Î¼ ÃŽ Ã‚ and Var (X1) = Ïƒ2ÃŽ (0, ¥), then by the CLT,WhereIs the sample mean of the Xjs. Show that the CLT implies the Weak Law of Large Numbers (WLLN);
Let ((, A, Î¼) = ((0, 1], B(0, 1], () where ( is the Lebesgue measure, and consider the function f : (0, 1] †’ ( defined byThen investigate whether or not the integral ((0, 1] f d( exists. Also, compute it if it exits, as well as the ((0, 1) | f | d(.
For n = 1, 2, ..., consider the r.v.s X1, X2,.... and X, and show that
Take (W, A, P) = ((0, 1), B(0, 1), l), l being the Lebesgue measure, and for n = 0, 1,€¦, define the r.v.sXnby:Then, as n†’¥, show that:(as it should be, by Theorem 2, since 0 £ Xn £ 1 for all n, and the bounds are integrable).
Define the r.v. X by X(() = 1/2( and then show that (( Xd μ = 2.
Forn³ 1, letXn,Yn, andXbe r.v.s defined on the probability space (W, A, P), and suppose thatAs n†’¥. Then, as n†’¥:(i)A constant, implies(ii)Implies
Consider the following situation. Let ((, A, P) be a probability space and let X be an integrable r.v. On A, define the (finite) measure v by v (A) = (A X dP. Then show that v 0 there exists ( = ( (() (> 0) such that P (A) < ( implies v(A) < (.
Let X be an integrable r.v. Then, for every ( > 0, there is a simple r.v. X( such that f|X - X(| < (.
Let X be a nonnegative r.v. Then:(i) Use the Fubini Theorem to show that(ii) Apply this result in case the d.f. of X is F(x) = 1 -e-(x for x ( 0 and 0 for x 0); or F(x) = 0 for x
Establish the following generalized version of part (iii) of Theorem 2 (Fatou Lebesgue Theorem). Namely, forn= 1, 2,€¦., letXn,Un,VnandX, U, Vbe r.v.s such thatUn£Xn£Vna.e.,n³ 1, and asn†’¥,Ã² Un †’ Ã² U finite and
Establish the following generalized (Dominated Convergence Theorem). Namely, for n = 1, 2, ...., let Xn, Un, X, U be r.v.s such that finite. Then, as n→(: (i) ( Xn → X finite; (ii) ( |Xn - X| → 0; (iii) (A Xn → (A X uniformly in A( A.
Leave the setup and the assumptions intact except that the assumptionsAndAre replaced byAndAs n†’¥. Then conclusions (i)€“(iii) hold.
Let μ be a σ-finite measure and let φ be a nonnegative σ-additive and finite function, both defined on the measurable space ((, A). Then show that φ 0, there exists ( = ( (() > 0 such that μ (A) < ( implies φ (A) < (.
All sets figuring below are subsets of the product space W1´ W2. Then show that(i) E Ã F implies Ew1 Ã Ew1 and Ew2 Ã Ew2, w1 ÃŽ W1, w2 ÃŽ, W2.(ii) E Ã‡ F Ã˜ implies
(i) Consider the measure space ((, A, μ) and let C be a fixed set in A. On A, define the function μo by: μo (A) = μ (A ( C). Then show that μo is a measure. (ii) Let X a r.v. defined on ((, A, μ) and suppose that ( Xd μ exists. Then show that ( Xd μo also exists and that ( Xd μo = (C Xd μ.
Consider the (Ïƒ-finite) measure space ((, A, Î¼), and let {Ai, i = 1, 2,....} be a (measurable) partition of (. For each i, define the measure Î¼i by: Î¼i (A) = Î¼ (A ( Ai). Then, if X is a r.v. defined on ((, A, Î¼) for
Establish the following generalized version of Theorem 2 (The Holder inequality). Namely, show that for any n (( 2) r.v.s X1,...., Xn and any positive numbers r1, ...., rn with n/r1 + ... + 1/rn = 1, it holds that
If ε |X| < (, show that ( |X| is absolutely continuous; that is (A |X| → 0 when P(A) → 0.
Forn= 1, 2,€¦, letXn,YnandX,Ybe r.v.s such thatP(Xn³Yn³ 0) = 1.AndFinite. Then show that
For any r.v. X defined on the measure space ((, A, µ), show directly that the Markov inequality holds; namely, µ (|X| ( c) ( c-r ( |X|r d µ for any r > 0 and any c > 0?
For a r.v X and some r > 0, show that ε |X|r < ( if and only if ε|X - c|r < ( for every (finite) constant c?
For a r.v. X with ε X = µ ( ( and Var (X) = (2 = 0, show that P (X = µ) = 1?
Forn= 1,2,€¦., consider the r.v.sXnandXand show that(i)If and only if(ii)If and only if
(i) Forr> 1 andxjÃŽ Ã‚, j = 1, €¦, n show that 1 and xj ( (, j" class="fr-fic fr-dii">(ii) From part (i), deduce that(On the other hand, by the Minkowski inequality),
Forn= 1, 2,€¦, suppose thatXnandXare r.v.s defined on the probability space (W,A,P), and suppose thatAnd {Xn} are uniformly integrable. Then show thatUniformly in A ÃŽ A.
If for n = 1, 2,..., the r.v.s {Xn} and {Yn} are uniformly integrable, show that {Xn + Yn} are also uniformly integrable?
For n = 1, 2,..., let Xn be identically distributed with Îµ Xn ( (. The show that {Xn} are uniformly integrable, whereIs the sample mean of X1,..., Xn.
Let g : I open interval in ( → ( be convex; i.e., for every x, x' ( I and every a ( [0, 1], it holds g(ax + (1 - a)x') ( ag (x) + (1 - a) (x) + (1 - a) g (x'). Then show that(i) g is continuous.(ii) For every x0 ( I. there exits ( (x0) ( ( such that g(x) - g (x0) ( ( (x0) ( (x - x0), x ( I.
LetWith ÎµYr < ¥ (for some r > 0) andThen show that Îµ|X|r < ¥ and
Let the r.v.sXn,n³ 1, be defined as follows :Xn(w) =2nif w ÃŽ (0, 1/n), andXn(w) = 0 otherwise, where (W,A,P) = ((0, 1], B(0, 1], l) and l is the Lebesgue measure. Then show thatpointwise, butTo any finite number for any r > 0; in factFor any r > 0.
Forn= 1, 2,€¦., let the r.v.Xnbe defined bywhere c is a positive constant.Then show that(i)(ii)(iii)
LetFor all n. Then, ifShow that
Forn= 1, 2,€¦., letXn,YnandX,Ybe r.v.s defined on the probability space (W,A,P), and suppose thatWhere r, s > 1 with 1/r + 1/s = 1. Then show that
Let (W,A,P) = ((0, 1), B(0, 1), l), where l is the Lebesgue measure, and letXnbe a r.v. defined byThen show thatFor any r > 0, andIndeed,Pointwise.
Forn= 1, 2,€¦., let the r.v.Xnbe defined by
Forn= 1, 2,€¦., let the r.vXnbe defined by(i) Show that(ii) Determine the condition c and r (> 0) must satisfy, so that
Establish the following generalized version of Theorem 3 (the Minkowski inequality). Namely, show that for any n (( 2) r.v.s X1,..., Xn and any r ( 1. It holds that
(i) For any r.v. X, show that P(|X| ( c) ( ε |X|/c (c > 0), by using the obvious inequality |X| ( c). In particular, if X ( 0, then P(X ( c) ( εX/c? (ii) For any r.v. X and any c > 0, show that P(X ( c) ( e-tc εetX (t > 0), P(X ( c) ( e-tc εetX (t < 0). (iii) If εX2 < (, the show that P(|X| >
For n ( 1, let Xn be r.v.s with ÎµXn = µn, (2 (Xn) = (2n, and suppose that, as n †’ (, µn †’ µ ( (, and (2n †’ 0. Then show that Îµ(Xn - µ)2 †’ 0 and
Forn ‰¥1, letXnandYnbe r.v.s defined on the probability space (Î©,A,P), and suppose that |Xn| and |Yn|,n ‰¥ 1, are uniformly integrable and as(i) |Xn €“ Yn|, n ‰¥ 1, are uniformly integrable.(ii) Îµ |Xn €“
Consider the probability space (W,A,P), and let Q be an open subset of Ã‚. Let g(Ã—; Ã—) : W ´ Q †’ Ã‚ be (A´BQ) €“ measurable, where BQis the s-field of Borel subsets of Q, we say that g(Ã—;q) is
All r.v.s appearing below are defined on the probability space ((, A, P). (i) If X ( 0, then ε X ( 0 and ε X = 0 only if P (X = 0) = 1. (ii) Let X ( Y with finite expectations. Then ε X ( εY and εX = εY only if P(X = Y) = 1. (iii) Let X > Y with finite expectations. Then εX > εY. (iv) Let g
IfFor some r > 0, then
IfAndFor some r > 0, it follows that P(X ¹ Y) = 0.
Construct r.v.sXn,n³ 1, andXon some probability space (Î©,A,P) such thatButFor any r > 0.
Construct r.v.sXn,n‰¥ 1, andXon some probability space (Î©,A,P) such thatButFor any r > 0.
Let Xn, n = 1,2,..., be r.v.s such that ε |xn| ( ( M (< (), n ( 1, for some ( > 0. Then show that |xn| a, n ( 1, are uniformly integrable for all a with 0 < a < (?
For n = 1, 2,..., let Xn be r.v.s such that P (Xn = cn) = 1/n P (Xn = 0) = 1 - 1/n, for some c > 0. Investigate whether or not these r.v.s are uniformly integrable?
Verify the relationUsed in the proof of Lemma 1; namely, if {An} n ³ 1, from a non-increasing sequence andThen for any n0 ³ 1, the given relation holds. In particular,If
Given the finite measures µn, n = 1, 2,...., on the measurable space ((, A), define a probability measure µ such that µn
Let C be the class of all probability measure defined on the measurable space ((, A), and letd(P, Q) = || P - Q || = 2 sup {|P(A) - Q (A)|; A ( A}.(i) Then show that d is a distance in C; i.e., show that d(P, Q) ( and d(P, Q) = 0 only if P = Q, d(P, Q) = d(Q, P), and d(P, Q) ( d(P, R) + d(R, Q)
Let U (a, () be the uniform distribution over the interval [a, (] (a < (), and for n ( 1, let Pn and Qn be the (probability) measures corresponding to U(- 1/n, 1) and U(0, 1 + 1/n). Then show that ||Pn - Qn|| → 0 as n → (, by using part (ii) of Exercise 3?
Let P and Q be two probability measures defined on the measurable space ((, A), and suppose that P ( Q (i.e., P 0.
Let P, Q and f, g be as in Exercise 3 and, without loss of generality, we suppose that the dominating measure µ is a probability measure (e.g,. µ = (P + Q)/2). Define ( by ( = ( ( (fg)1/2 d µ, and show that: (i) ( ( 1. (ii) 2(1 - () ( d(P, Q) ( 2(1 - (2)1/2, where (by Exercise 3), d(P, Q) = ((|
Forn‰¥ 1, letfnbe real-valued measurable functions defined on an open setEŠ‚ R containing 0, and suppose that |fn(x)| £M(< ¥) for allnand allxÃŽE, and thatFor x ÃŽ E. Then show that, for any sequenceThere exists a subsequence {xm} Ã
For n ( 1, let fn be real-valued measurable functions defined on (, and suppose that |fn (x)| ( M ( for x ( (. Then for any sequence {xn} with it holds ( 0 a.e. [(], where ( is the Lebesgue measure?
Let F : ( → [0, 1] be non-decreasing, right continuous, F(- () = 0 and F(() = 1 (i.e., a d.f. of a r.v.), and let F-1 be defined by F-1 (y) = inf {x ( (; F(x) ( y}, y ( [0, 1]. Next, consider the probability space ((, A, P) = ([0, 1], B[0, 1], P), where P = ( is the Lebesgue measure, and on (,
Let X be a r.v. with continuous d.f. FX, and define the r.v. Y as follows: Y = X If X ( C, and Y = C if X > C, where C is a constant. Then (i) Determine the d.f. of the r.v. Y, FY. (ii) Show that FY = F1 + F2, where F1 is a continuous d.f. and F2 is a step function?
By a simple example demonstrate that it is possible that Fn n ( 1, are d.f.s of r.v.s withA d.f., but not that of a r.v.
For n ( 1, let Xn and X be r.v.s defined on the probability space ((, A, P), and suppose that, as n †’ (, Îµ|Xn| †’ Îµ|X|Then show that |Xn|, n ( 1, are uniformly integrable?
Forn‰¥1, letFnandFbe d.f.s of r.v.s let fnbe real-valued measurable functions defined on be continuous, and letg: R †’ [0, ˆž] be continuous, Assume that:(i) Fn †’ F as n †’ ˆž.(ii) (iii) uniformly on finite
(i) IfThen show that(ii) By an example, show that the converse statement in part (i) need not be true.
Let Fn, n ( 1, be d.f.s of r.v.s and suppose thatWhere F is a d.f. with Var F = 1. Then show that F is the d.f. of a r.v.
Show that the Weak Compactness Theorem (Theorem 5) still holds, if condition 1 in Definition 1 is replaced by: (0 () F(x) ( B, x ( (, n ( 1, for some B < (?
Show that Theorem 6 - 8 hold true if, for all x ( ( 0 ( F(x) ( B and 0 ( Fn (x) ( B, n = 1,2,..., for some B > 0, as indicated in Remark 6?
Forn= 1, 2,€¦, consider the r.v.sXn,XandYsuch that |Xn| £Y, ÎµYr< ¥ (for somer> 0), andThen show that
Let X be a r.v. with d.f. F; i.e., F(x) = P(X ( x), x ( (. Then show that (i) P(X < x) = F(x -) (the left-hand side limit). (ii) F(x) is continuous at x if and only if P(X = x) = 0.
In the following expression, determine with constants a and (, so that the function F defined is the d.f. of a r.v. F(x) = 0, x ( 0, and F(x) = + (e-x2/2, x > 0.
Forn= 1, 2,€¦, letFnandFbe d.f.s such thatAnd let F be continuous. Then show thatUniformly in x ÃŽ Ã‚.For Îµ > 0, choose a and b sufficiently small and sufficiently large, respectively, so that F(a) < Îµ/3, F (¥) €“ F(b) <
If X is an integrable r.v. and {An, n = 1, 2,....} are pairwise disjoint events, show thatSplit X into X+ and X- and use Corollary 1 (ii) to the Lebesgue Monotone Convergence Theorem (Theorem 1 in Chapter 5).
Let X and Y be discrete r.v.s and recall (from the application following Theorem 11) that B'Y is the (-field of subsets of ( defined by B'Y = {B' ( (; Y-1(B') = A for some A ( A}. For x, y ( ( with P(Y = y) > 0, consider the conditional probability P(X = x|Y = y) = P(X = x, Y = y) / P(Y = y) And
The Dominated Convergence Theorem in its classical from states: If |Xn| ³Ya.s., ³ 1, ÎµY< ¥, and eitherOrThenFinite.In the framework of conditional expectations, we have shown that: If |Xn| £ Y a.s., n ³1, ÎµY < 1, ÎµY < ¥, andWith ÎµX finite, thenFor any s-field B Ã A.By
Let the r.v.sXandYhave the Bivariate normal distribution with parameters Î¼1, Î¼2in Ã‚, 0 < s1, s2< ¥, andrÃŽ [€“ 1, 1] so that their joint probability density function (p.d.f.) is given byWhere(i) Show that the exponent may be written
In reference to Exercise 12 (ii), and without any actual operations, conclude that the conditional distribution of the r.v. Y, give X = x, is N(b, (22 (1 - (2)), where b = (2 + ((2/a1 (x - (1); and by symmetry, the conditional distribution of X, given Y = y, is N(c, (21 (1 - (2)), where c = (1 +
(i) In reference to Exercise 12 and 13, and by writing ε(XY) = ε[ε(XY)|X], show that ε(XY) = (1(2 + ((1(2. (ii) Use the definition of the conversance of two r.v.s X and Y (with finite second moments), Cov (X, Y) = ε[(X - εX)(Y - εY)] = ε(XY) - (εX) (εY), in order to conclude that, in the

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