L10.sums2

18:40 01/04/2001 TOPIC. Sums and limits. This section considers the question of when you can interchange the order of summation in a doubly indexed infinite series; the answer is given by Fubini’s Theorem. We also consider the question of when you can bring a limit inside an infinite series; the answer is given in various forms by the Monotone Convergence Theorem, the Dominated Convergence Theorem, and the Sandwich Theorem. The aforementioned theorems apply not only to infinite series, but also to integrals and expectations; we will use them extensively in the rest of the course. Question 1. Suppose x1;1x1;2x1;3¢¢¢ x2;1x2;2x2;3¢¢¢ x3;1x3;2x3;3¢¢¢ ......... is an array of numbers with infinitely many rows and columns. We ask: is it true that X1 m=1³X1 n=1xm;n´ =X1 n=1³X1 m=1xm;n´ ; (1) i.e., can we switch the order of summation? Notice that the expression in ()’s on the LHS is the sum, call it rm, of the elements in the mth row of the array; the LHS is the sum, call it r, of these row sums. Similarly, the expression in ()’s on the RHS is the sum cnof the elements in the nthcolumn, and the RHS itself is the sum cof these column sums. randcare called iterated sums . Example 1. (a) There are cases where r=c. For example, for the array 1 1 1 1 ¢¢¢ 0 1 1 1 ¢¢¢ 0 0 1 1 ¢¢¢ 0 0 0 1 ¢¢¢ ............³ xm;n=n1;ifm·n, 0;otherwise´ 10 – 1one has rm= lim N!1(PN n=1xm;n) =1for each m, so r=1+1+1+¢¢¢=1: Moreover cn= lim M!1(PM m=1xm;n) =nfor each n, so c= 1 + 2 + 3 + ¢¢¢=1=r: (b) Sadly, there are cases where r6=c. For example, for the array 1¡1 0 0 ¢¢¢ 0 1 ¡1 0 ¢¢¢ 0 0 1 ¡1¢¢¢ ............à xm;n=(1;ifn=m, ¡1;ifn=m+ 1, 0;otherwise! one has rm= 0 for each m, so r= 0 + 0 + 0 + ¢¢¢= 0; whereas c1= 1 and cn= 0 for all n¸2, so c= 1 + 0 + 0 + 0 + ¢¢¢= 16=r: ² So far we have tacitly assumed that randcexist. The statement “rexists” means that in the recipe r:= lim M!1·XM m=1³ lim N!1hXN n=1xm;ni´¸ ; all the indicated sums and limits exist (possibly as + 1or¡1); a similar interpretation applies to the statement “ cexists”. There are, of course, cases where randcdon’t exist; see, e.g., Exercise 3. The following theorem says that you can switch the order of sum- mation for a doubly infinite array provided all the array elements are nonnegative (as in Example 1 (a)). 10 – 2 r:=P1 m=1rmwith rm:=P1 n=1xm;n. c:=P1 n=1cnwith cn:=P1 m=1xm;n. Theorem 1 (Fubini I – the nonnegative case). Suppose 0· xm;n· 1 form= 1;2; : : :andn= 1;2; : : :. Let the iterated sums r andcbe defined as above. Then randcboth exist and equal s:= supnX (m;n)2Ixm;n:Iis a finite subset of N£No : (2) Proof ²randcexist:rm:= lim N!1¡PN n=1xm;n¢ exists because the partial sums involved are nondecreasing in N; similarly r:= limM!1¡PM m=1rm¢ exists because the partial sums involved are non- decreasing in M. Parallel statements apply to the columns. ²r·s: Let MandNbe positive integers. Set I=f(m; n)2N£N: 1·m·M;1·n·Ng. Then XM m=1³XN n=1xm;n´ =X (m;n)2Ixm;n·s: ²²²²² ²²²²² ²²²²² ²²²²² ²²²²² ²²²²² 11 22 MN AsN! 1 , the LHS above tends toPM m=1rm(see Exercise 1), so XM m=1rm·s: AsM! 1 , the LHS above tends to r, sor·s. ²r¸s: Let Ibe a finite subset of N£N. There exist positive integers MandNsuch that I½ f(m; n)2N£N: 1·m·M;1·n·Ng =)X (m;n)2Ixm;n·XM m=1³XN n=1xm;n´ ·XM m=1rm·r: Since this is true for each I, we have s·r. ²r=s=c: We have just shown r=s. A similar argument shows c=s. 10 – 3r:=P1 m=1rmwith rm:=P1 n=1xm;n. c:=P1 n=1cnwith cn:=P1 m=1xm;n. (2):s:= sup©P (m;n)2Ixm;n:Iis a finite subset of N£Nª . For nonnegative summands, the supremum in (2) is called the double sum of the xm;n’s, denotedP m;nxm;n. For summands of arbitrary sign, the double sum is defined as s=X m;nxm;n:=hX m;nx+ m;ni ¡hX m;nx¡ m;ni =s+¡s¡;(3) provided at least one of the two double sums s+ands¡on the RHS is finite; otherwise sis said not to exist. In (3) x+ m;nandx¡ m;ndenote respectively the positive and negative parts of xm;n. (Note though thats+ands¡may not be the positive and negative parts of s.) Here is the main theorem. It implies that you can switch the order of summation for the xm;n’s if the iterated sum — taken in either order — of the jxm;nj’s is finite. Theorem 2 (Fubini II – the quasi-integrable case). Suppose ¡1 · xm;n· 1 form= 1;2; : : :andn= 1;2; : : :. The double sumsof the xm;n’s exists if and only if at least one of the following iterated sums is finite: r¡:=X1 m=1³X1 n=1x¡ m;n´ ; r +:=X1 m=1³X1 n=1x+ m;n´ ; c¡:=X1 n=1³X1 m=1x¡ m;n´ ; c +:=X1 n=1³X1 m=1x+ m;n´ : If the double sum exists, then so do the iterated sums randc, and r=s=c. According to Fubini I, one has r¡=s¡=c¡, and r+=s+=c+. This proves the first assertion of the theorem and implies that in prin- ciple it doesn’t matter whether you verify one of the row conditions, or one of the column conditions. However, it may be a little easier to work with the rows in one application, but with the columns in 10 – 4 18:40 01/04/2001 r¡:=P1 m=1¡P1 n=1x¡ m;n¢ : r +:=P1 m=1¡P1 n=1x+ m;n¢ another. There are cases (see, e.g., Exercise 4) where randcexist and are equal, but sdoes not exist; hence the conditions of the the- orem are sufficient, but not necessary, for being able to interchange the order of summation. Proof I will do the case where r¡<1, which impliesP1 n=1x¡ m;n<1for all m, and x¡ m;n<1for all mandn. (4) I need to show that randcexist and equal s. Since xm;n=x+ m;n¡ x¡ m;n, we have rm;N:=XN n=1xm;n=XN n=1x+ m;n¡XN n=1x¡ m;n; note that the difference on the RHS is well defined since the second term is finite by (4). As N! 1 , XN n=1x+ m;n!(r+)m:=X1 n=1x+ m;n XN n=1x¡ m;n!(r¡)m:=X1 n=1x¡ m;n<1 so rm:= lim N!1rm;Nexists and equals ( r+)m¡(r¡)m. (5) Adding (5) for m= 1; : : : ; M gives XM m=1rm=XM m=1(r+)m¡XM m=1(r¡)m: AsM! 1 , XM m=1(r+)m!r+=X1 m=1(r+)m XM m=1(r¡)m!r¡=X1 m=1(r¡)m<1 so r= lim N!1PM m=1rmexists and equals r+¡r¡=s+¡s¡=s. Since c¡=r¡<1, a similar argument shows cexists and equals s. 10 – 5Generalizations. Using measure theory, one can show that Fubini I and II hold not just for sums, but also for integrals and expectations, and combinations of such. For example ZhZ f(x; y)dxi dy=ZhZ f(x; y)dyi dx=ZZ f(x; y)dx dy (6) provided fis nonnegative, or ZhZ jf(x; y)jdxi dy <1;orZhZ jf(x; y)jdyi dx <1:(7) Similarly for a random variable X, EhZ f(t; X)dti =Z E£ f(t; X)¤ dt (8) provided fis nonnegative, or EhZ jf(t; X)jdti <1;orZ E£ jf(t; X)j¤ dt <1: (9) There are some additional technical conditions that are needed for these results, namely, the function fmust be jointly measurable in its two arguments and the integrations have to be taken with respect to¾-finite measures. We’ll ignore these condition in this course. Example 2. Suppose A1,A2,: : :is an infinite sequence of events. LetNbe the random variable which records how many of these events occur: N(!) =X1 n=1IAn(!) for each !2Ω. Since IAn¸0 for each n, we have E(N)=E³X1 n=1IAn´ = Fub IX1 n=1E(IAn)=X1 n=1P[An]:(10) If the sum on the RHS here is finite, then we must have N(!)<1, i.e.,!2Anfor at most finitely many n, for almost all sample points !. This result is called the first Borel-Cantelli Lemma (see page 7-16). 10 – 6 Question 2. Suppose x1= (x1(k))1 k=1,x2= (x2(k))1 k=1,: : :is an infinite sequence of infinite sequences such that x(k) := lim n!1xn(k) exists for each k. We ask: is it true that lim n!1³X1 k=1xn(k)´ =X1 k=1x(k); (11) i.e., can we bring the limit on ninside the sum? The answer is — not without some conditions. For example, suppose the xn(k)’s are given by the array 1 0 0 0 ¢¢¢ 0 1 0 0 ¢¢¢ 0 0 1 0 ¢¢¢ 0 0 0 1 ¢¢¢ ............xn(k) =n1;ifk=n 0;otherwise, where nis the row index and kis the column index. Then X1 k=1xn(k) = 1 for each n=)lim n!1³X1 k=1xn(k)´ = 1: However x(k) = lim n!1xn(k) = 0 for each k=)X1 k=1x(k) = 0 : We are going to present some conditions under which (11) does hold. But first, here is some terminology. An infinite sequence x= (x(k))1 k=1of extended real numbers (i.e., finite numbers or §1) is said to beintegrable , written x2 L, ifP1 k=1jx(k)j<1, i.e., if the seriesP1 k=1x(k) is absolutely convergent; the integral of an integrable x is Z x:=X1 k=1x(k) = lim K!1³XK k=1x(k)´ : (12) xis said to be quasi-integrable from below , written x2 Q¡, if the sequence x¡= ((x(k))¡)1 k=1is integrable, or, equivalently, if there is 10 – 7x= (x(k))1 k=12 L ()P1 k=1jx(k)j<1 x= (x(k))1 k=12 Q¡()P1 k=1x¡(k)<1 an integrable sequence `= (`(k))1 k=1such that `·xin the sense that`(k)·x(k) for all k. Similarly, xis said to be quasi-integrable from above , written x2 Q +, if the sequence x+= ((x(k))+)1 k=1 is integrable, or, equivalently, if there is an integrable sequence u= (u(k))1 k=1such that x·u. Finally, xis said to be quasi-integrable , written x2 Q, if it is quasi-integrable from below or above (or both). The integral of a quasi-integrable xis taken to beZ x:=Z x+¡Z x¡=X1 k=1x+(k)¡X1 k=1x¡(k) =X1 k=1x(k) := lim K!1XK k=1x(k): (13) Note that the collection Qof quasi-integrable sequences is Q¡[ Q+, whereas the collection Lof integrable sequences is Q¡\ Q +. The theorems below only apply to integrable or quasi-integrable sequences. Example 3. (a) The sequence x= (1=(k(k+ 1)))1 k=1is integrable, with integralZ x=X1 k=1x(k) = lim K!1XK k=1³1 k¡1 k+ 1´ = 1: (b) The sequence x= (1)1 k=1is not integrable, but it is quasi-integrable from below with integralZ x=X1 k=11 = lim K!1XK k=11 =1: (c) The infinite seriesP1 k=1(¡1)k=kis convergent in the usual sense, i.e.,c:= lim K!1¡PK k=1(¡1)k=k¢ exists and is finite (by calculus, c=¡log(2)). However the sequence x= ((¡1)k=k)1 k=1is not quasi- integrable, sinceP1 k=1x¡(k) =P kodd1=k=1=P keven1=k=P1 k=1x+(k). The theorems that follow don’t apply to this sequence. ² 10 – 8 18:40 01/04/2001 x= (x(k))1 k=12 Q¡()P1 k=1x¡(k)<1. Forx2 Q¡,R x=P1 k=1x(k) = lim K!1PK k=1x(k). For extended real numbers ´1,´2,´3,: : :, and ´, the notation ´n"´means ´1·´2·´3<¢¢¢and´= lim n´n.´n#´is defined similarly. For infinite sequences x1,x2,: : :, and x, xn"xmeans xn(k)"x(k) for k= 1, 2, : : :, while xn#xmeans xn(k)#x(k) for k= 1, 2, : : :. Theorem 3 (The Monotone Convergence Theorem (MCT)). Suppose x1,x2,: : :andxare infinite sequences of extended real numbers. MCT ¡: Ifxn"xandx12 Q¡, then xn2 Q¡for all n,x2 Q¡, andR xn"R x: (14) MCT +: Ifxn#xandx12 Q +, then xn2 Q +for all n,x2 Q +, andR xn#R x: (15) Example 4. (a) Suppose x1= (1;0;0;0; : : :) x2= (1;1;0;0; : : :)³ xn(k) =n1;ifk·n 0;otherwise´ x3= (1;1;1;0; : : :) etc. Then xn"x= (1;1;1; : : :) and x12 L ½ Q ¡.MCT ¡asserts thatR xn"R x. This is correct, sinceR xn=n" 1=R x. (b) Suppose x1= (1;1;1;1; : : :) x2= (0;1;1;1; : : :)³ xn(k) =n1;ifk¸n 0;otherwise´ x3= (0;0;1;1; : : :) etc. Then xn#x= (0;0;0;0; : : :). The xn’s and xare all quasi- integrable. HoweverR xn=1doesn’t tend down toR x= 0. This doesn’t contradict MCT +, because no xnis inQ+. ² 10 – 9MCT ¡:xn"xandx12 Q¡=)xn2 Q¡,x2 Q¡, andR xn"R x. Proof I’ll prove MCT ¡;MCT +then follows by changing signs. To begin with, consider the case where x1is nonnegative (i.e., x1(k)¸0 for all k). Then the xn’s and xare nonnegative, and hence trivially quasi-integrable. For indices m < n we have xm(k)·xn(k)·x(k) for each k; adding over k= 1;2; : : :givesR xm·R xn·R x. Thus L:= lim nR xnexists and L·R x: To get the opposite inequality, let Kbe a positive integer. Then PK k=1x(k) =PK k=1limnxn(k) = lim nPK k=1xn(k) ·limnP1 k=1xn(k) = lim nR xn=L; Letting K! 1 givesR x=P1 k=1x(k)·L. Now consider the general case, where x¡ 1is assumed to be in- tegrable. Since x¡ 1(k)¸x¡ n(k)¸x¡(k) for all k, this implies 1>R x¡ 1¸R x¡ n¸R x¡, and hence that xn2 Q¡andx2 Q¡. Moreover x¡ 1(k) is finite for each k. Define infinite sequences y1,y2,: : :, and y by setting yn(k) =xn(k) +x¡ 1(k) and y(k) =x(k) +x¡ 1(k) for each k. Since the yn’s are nonnegative and tend up to y, we have R yn"R y by the nonnegative case treated above. But R yn=R (xn+x¡ 1) =R xn+R x¡ 1;and R y=R (x+x¡ 1) =R x+R x¡ 1: SinceR x¡ 1is finite, it follows thatR xn"R x. 10 – 10 MCT ¡:xn"xandx12 Q¡=)xn2 Q¡,x2 Q¡, andR xn"R x. Given an infinite sequence x1,x2,: : :of infinite sequences, one defines inf nxnand lim inf nxnelement by element, i.e., (infnxn)(k) := inf nxn(k) and (lim inf nxn)(k) = lim inf nxn(k) fork= 1, 2, : : :. Similarly for sup’s and limsup’s. Theorem 4 (Fatou’s Lemma (FL)). Letx1,x2,: : :be an infinite sequence of infinite sequences of extended real numbers. FL¡:If there exists an integrable sequence `such that `·xnfor alln, then xn2 Q¡for all n,lim inf nxn2 Q¡, and R lim inf nxn·lim inf nR xn: (16) FL+:If there exists an integrable sequence usuch that xn·ufor alln, then xn2 Q +for all n,lim supnxn2 Q +, and lim supnR xn·R lim supnxn: (17) Note that the hypothesis of the lower half ( FL¡) of Fatou’s Lemma is trivially satisfied if xn¸0 for all n. Proof of FL¡.Putyn= inf p¸nxpandy= lim inf nxn. Then yn"y(by definition) and y12 Q¡(since `·y1): Hence by MCT ¡,yn2 Q¡for all n,y2 Q¡, and R (lim inf xn) =R y= by (14)limnR yn= lim nR (infp¸nxp) ·limn(infp¸nR xp) = lim inf nR xn: Example 5. Reconsider the sequences x1,x2,: : :in Example 4b. Notice that 0 ·xnfor all n, and lim inf nxn=x= lim supnxn, where x= (0;0;0; : : :).FL¡applies and correctly asserts that 0 =R x· lim inf nR xn= lim inf n1=1. The conclusion (17) to FL+would be1 · 0, which is obviously false. This doesn’t contradict FL+, since there is no integrable sequence usuch that xn·ufor all n. ² 10 – 11FL¡: 0·xnfor all n=)R lim inf nxn·lim inf nR xn. Theorem 5 (The Sandwich Theorem). Let`1,`2,: : :,`,x1,x2, : : :,x,u1,u2,: : :, and ube infinite sequences of real numbers such that S1`n!`,xn!x, and un!uasn! 1 (element by element), S2`n·xn·unfor each n, and S3the`n’s,un’s,`, and uare integrable and limn!1R `n=R `and limn!1R un=R u: Then the xn’s and xare integrable and S4limn!1R xn=R x. Simply put, the xn’s can be integrated to the limit provided they are “sandwiched” between lower bounds `n’s and upper bounds un’s which can be integrated to the limit. Proof xnis integrable because `n·xn·unand`nandunare integrable; similarly, xis integrable because `·x·uand`andu are integrable. Since 0·xn¡`n!x¡`and 0 ·un¡xn!u¡x Fatou’s Lemma implies that R x¡R `=R (x¡`) =R lim inf n(xn¡`n) · (byFL¡)lim inf nR (xn¡`n) = (byS3)lim inf nR xn¡R `; (18) R u¡R x=R (u¡x) =R lim inf n(un¡xn) ·lim inf nR (un¡xn) =R u¡lim supnR xn: (19) Hence lim supnR xn· (by (19))R x· (by (18))lim inf nR xn: 10 – 12 18:40 01/04/2001 Sandwich Theorem. If ( S1)`n!l,xn!x, and un!u, (S2)`n·xn·unfor each n, and (S3)R `n!R lfinite, andR un!R ufinite, then ( S4)R xn!R xfinite. Example 6. LetP1,P2,: : :andPbe probability measures on the set Nof positive integers, and let f1,f2,: : :andfbe the corresponding probability mass functions. Thus Pn[B] =X k2Bfn(k) and P[B] =X k2Bf(k) for each subset BofN. Suppose f(k) = lim n!1fn(k) for each k2N: For each B, ¯¯Pn[B]¡P[B]¯¯=¯¯¯X k2Bfn(k)¡X k2Bf(k)¯¯¯ =¯¯¯X k2B¡ fn(k)¡f(k)¢¯¯¯·X k2B¯¯fn(k)¡f(k)¯¯ ·X1 k=1¯¯fn(k)¡f(k)¯¯=Z¯¯fn¡f¯¯:=vn: The Sandwich Theorem implies that the bound vnhere tends to 0 as n! 1 . Indeed since `n:= 0·xn:=jfn¡fj ·un:=fn+f; `n!`:= 0; x n!x:= 0; u n!u:= 2f;R `n= 0!0 =R `andR un=R fn+R f= 2!2 =R u; we have vn=Z xn!Z x= 0: We’ve shown that if Pn[B]!P[B] for each one-point set B, then Pn[B]!P[B] uniformly for all subsets BofN. ² 10 – 13Sandwich Theorem. If ( S1)`n!l,xn!x, and un!u, (S2)`n·xn·unfor each n, and (S3)R `n!R lfinite, andR un!R ufinite, then ( S4)R xn!R xfinite. The following theorem is used over and over. Theorem 6 (The Dominated Convergence Theorem (DCT)). Letx1,x2,: : :andxanddbe infinite sequences of real numbers such that D1limn!1xn(k) =x(k)for each k, D2jxn(k)j ·d(k)for all nandk, and D3dis integrable. Then the xn’s and xare integrable and D4R x= lim n!1R xn. Proof Apply the Sandwich Theorem with `n=¡d=`andun= d=u. The sequence dabove is called a dominator . It is of course essential that the dominator be integrable and that it dominate ev- eryxn. The conditions for the DCT are stronger than those of the Sandwich Theorem. For example, consider the sequences x1= (1;¡1;0;0;0; : : :); x2= (0;1;¡1;0;0; : : :);à xn(k) =(1;ifk=n, ¡1;ifk=n+ 1, 0;otherwise! x3= (0;0;1;¡1;0; : : :); etc. Here xn!x= (0;0;0; : : : ) andR xn= 0!0 =R x. This conclusion can be deduced (somewhat artificially) from the Sandwich Theorem by taking `n=xn=unand`=x=u. However, it can not be deduced from the DCT because there is no integrable dominator in this situation; indeed, any sequence dsatisfying D2hasd(k)¸1 for all k, and so can’t be integrable. 10 – 14 Generalizations. Using measure theory, one can show that the Monotone Convergence Theorem, Fatou’s Lemma, the Sandwich The- orem, and the Dominated Convergence Theorem hold not just for sums of sequences, but also for integrals of (measurable) functions and expectations of random variables. For example, suppose X1,X2, : : :andXare random variables, all defined on some common proba- bility space endowed with a probability measure P. The expectation version of MCT ¡says that if Xn(!)"X(!) for ( P-almost) all sample points !2Ω and X12 Q¡¡ meaning E(X¡ 1)<1¢ , then XnandXare in Q¡(and so have expectations) and E(Xn)"E(X); (20) according to the hypotheses E(Xn) and E(X) can’t be ¡1; they can however be + 1. The expectation version of Fatou’s Lemma says in part that if the Xn’s are nonnegative, then E(lim inf nXn)·lim inf nE(Xn): (21) The expectation version of the DCT says that if Xn(!)!X(!) for ( P-almost) all sample points !2Ω and there is an integrable random variable Dsuch that jXn(!)j ·D(!) for all nand for ( P-almost) all !2Ω, then XnandXhave finite expectations and E(Xn)!E(X): (22) 10 – 15The following definition and exercise cover some issues that the text assumes you are familiar with. Let x1,x2,: : :andxbe extended real-numbers. One says that xnconverges to xasn! 1 , and writes x= limnxnorxn!x, if for each real number w < x one has w·xnfor all sufficiently large n, and, similarly, for each real number y > x one has xn·yfor all sufficiently large n. Ifx=1only the “ w- condition” is required; if x=¡1only the “ y-condition” is required. One has xn!x() lim inf nxn=x= lim supnxn. For example, limn¡ 1 + (¡1)n=n¢ = 1, lim npn=1, and lim nlog(1=n) =¡1. Exercise 1. Suppose x1,x2,: : :andy1,y2,: : :are extended real numbers such that xn+ynis defined (i.e., not of the form ( §1) + (¨1)) for each n. (i) Suppose yn!yand (lim inf nxn)+yis defined. Show that lim inf n(xn+yn) = (lim inf nxn) +y. (ii) Suppose xn!x andyn!yandx+yis defined. Show that xn+yn!x+y. ¦ Exercise 2. Show that Fubini I is a special case of Fubini II. ¦ Exercise 3. Form= 1, 2, : : :andn= 1, 2, : : :, put xm;n=½(¡1)n¡1;ifm·n, 0; otherwise. Show that the double sum sand the iterated sums randcdo not exist. ¦ Exercise 4. For (m; n)2N£N, set xm;n=8 >>< >>:1;ifm=n, ¡1;if (m; n) = (2 k¡1;2k) or (2 k;2k¡1) for some k2N, 0;otherwise. Show that the iterated sums randcexist and are equal, but the double sumP m;nxm;ndoes not exist. ¦ Fubini’s theorem was used implicitly in the argument leading up to Theorem 7.3. The following exercise makes the use explicit. 10 – 16 18:40 01/04/2001 Exercise 5. LetXbe a random variable with distribution function F and left-continuous representing function R. (a) Show that for each u2(0;1), R+(u) =Z1 x=0IfR(u)>xgdx and use Fubini I to show thatZ1 0R+(u)du=Z1 0¡ 1¡F(x)¢ dx: (b) Similarly, use Fubini I to show that Z1 0R¡(u)du=Z0 ¡1F(x)dx: [Hint: use the switching formula (1.5).] ¦ Exercise 6. Use Fubini’s theorem and induction on kto show that the volume of the unit ball B:=f(x1; : : : ; x k) :x2 1+¢¢¢+x2 k·1g inRkequals Vk:=¼k=2 Γ(k=2 + 1): (23)¦ Exercise 7. LetFbe a distribution function and let cbe a positive number. Show thatR1 ¡1¡ F(x+c)¡F(x)¢ dx=c. ¦ Exercise 8. LetXandYbe two random variables such that for each x2R, the conditional distribution of Ygiven X=xhas a density, say, fx; thus P[Y2BjX=x] =R Bfx(y)dyfor each (Borel) subset BofR. Let f(y) be the result of averaging fx(y) with respect to the distribution ¹ofX, i.e., f(y) :=Z1 ¡1fx(y)¹(dx) =E¡ fX(y)¢ : (24) Show that Yhas density f. ¦ 10 – 17Exercise 9. Deduce the MCT from FL. ¦ The text gave a very simple proof of the DCT, but that came at the end of long series of arguments. The next exercise asks you to prove the DCT various ways, using progressively less and less ma- chinery. Exercise 10. (i) Deduce the DCT directly from Fatou’s Lemma, us- ing the fact thatR xn!R x()lim inf nR xn=R x= lim supnR xn. (ii) Deduce the DCT from the MCT, using xn!x=)supn¸mjxn¡ xj #0 as m! 1 . (iii) Deduce the DCT “from scratch”, usingR jxn¡xj ·PK k=1jxn(k)¡x(k)j+ 2 sup`>Kd(`) for each K. ¦ Exercise 11. (a) Let Ube a standard uniform random variable and letWa standard exponential random variable. Show that for t¸0 E(e¡tU) =1¡e¡t t; E(e¡tW) =1 1 +t; E(W2e¡tW) =2 (1 +t)3; use the convention that (1 ¡e¡0)=0 = 1. (b) Let XandYbe two (possibly dependent) nonnegative real-valued random variables. Set L(s; t) =E(e¡sX¡tY) fors¸0 and t¸0;Lis called the joint Laplace transform of XandY. Show that Ls(0+; t) := lim s#0L(s; t)¡L(0; t) s=¡E(Xe¡tY); (25) Z1 0¡ ¡Ls(0+; t)¢ dt=E³X Y´ ; (26) use the convention that x=y= 0 if x= 0 = y. (c) Let VandW be two independent standard exponential random variables. Show that E¡ (V2+W2)=(V+W)¢ = 4=3:[Hint: use part (b), but don’t completely evaluate L(s; t).] ¦ 10 – 18 Exercise 12. LetXbe an integrable random variable with an arbi- trary distribution function F. For real numbers x, set c(x) :=E¡ jx¡Xj¢ : (a) Show that lim h#0c(x+h)¡c(x) h=E(IfX·xg¡Ifx<Xg) = 2 F(x)¡1; the limit on the left is taken as htends down to 0 from above. [Hint: There are at least two ways to solve this problem; one uses the fact that for real numbers aandb, one has¯¯jbj ¡ jaj¯¯· jb¡aj.] (b) What is the necessary and sufficient condition on Ffor the function cto be differentiable at x? Explain briefly. ¦ The following exercise deals with a topic in Markov chains. The only facts you need to know about MC’s are set out below. Let X0,X1,: : :be an irreducible aperiodic Markov chain with countable state space I(which you may take to be N) and transition probability matrixP= (Pij)i2I;j2I; thus Pij=P[Xn=jjXn¡1=i] for each n. It follows from the Markov property that P[Xn=jjX0=i] = (Pn)ij wherePndenotes the nthpower ofP. Using renewal theory, one can show that ¼j:= lim n!1P[Xn=jjX0=i] (27) exists for each j2Iand does not depend on the initial state i. 10 – 19Exercise 13. Let the Xn’s,I,P, and ¼be as above. (a) Use Fatou’s Lemma to show that X j2I¼j·1: (28) (b) Use Fatou’s Lemma and Fubini’s theorem to show that the ¼k’s satisfy the equations ¼k=X j2I¼jPjk; k2I: (29) [Hint: Start by letting n! 1 in the equation P(n) ik=P jP(n¡1) ijPjk.] (c) Let º= (ºj)j2Ibe a vector of nonnegative numbers such that c:=P j2Jºj<1andºk=P j2JºjPjkfor each k2J. Use Fubini and the DCT to show that º=c¼. [Hint: Start by showing º=ºPn for each n.] ¦ 10 – 20