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# hw6 - :g A j“{M S 52{man‘s 6"{‘57 V Fiﬁ(MACWAU...

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Unformatted text preview: :g A j“ {M S 52' {man‘s} 6"}? {‘57 V Fiﬁ . \ (MACWAU 3 SECWHVQ (”pit“) “w m N J 7 mi 00 Q %% W A?! 3.6 \kf) '-* 2 9 ‘31“ «xx-«x M M‘“ w-wﬁizs “W“ a @ P \ m ‘0) 6 3% 0.: LOMVW Qﬁumﬁlﬁ \‘0 OD WWI-m we.“ W‘w} gammy 2M fag” ﬂ 4w C «5“ " . I I 'h‘ \dw {iﬂl'ﬂ H\$§YEQ£3 ’ijﬂ': ;,a tor QWN‘ M34, N 7/ W44 [Wiﬁéég I (b) .1,- Proof. Let f : X x Y —> [0, 00] be M x W-measurable. From now on we’ll write, H for M X N, and gr for the product measure (and if for the completions). Then there exists a sequence of ﬁ-measurable simple functions 0 5 (pl S 9:22 5 . . . S f that converge pointwise to f. For each n, we have: N(") ‘Pn = Z and-Ami i=1 where Am- is ﬁ-measurable. Then (since we’ve taken completions), we know that there exists a H-rneasurable set AL”; g Aw- such that AW- \AL, is a subset of a ﬁ-null set, which (by the previous problemset) is in turn contained in a 1r~null set, and hence Am \ AA}, is inside a 7r—null set. Then consider: which is a H—measurable simple function, which (by finite additivity) differs from 9071 only possibly on a ar-null set E”. By countable additivity, E = U” En is also r—null, and on the complement E6 of which we have (on = 90;, for all 4/Q) n. Now for each n, let: n := XEC max{(,o’1,... ,goil} which is H—meas rable, .5 f, and 0 5 1,01 5 1,02 5 ... g f. 0n E“ we have ”din = so; = (pm and hence convergence to f. On E we have n = 0 for all n. Consequently, the limit 9 = f X Ec of the 1% is meeasurable, non—negative, and w-ae. equal to f. Since E is ir—null, we have by Fubini’s theorem for non—completions that XE(x, -) is N—measurable for all :r; E X , and 0=nyxE:LLXE(\$iy)dV(y)d-I~L(x) and hence since the integrand is non—negative: furthermore that: may) dye) = 0 Y for u-ae. a; E X. That is, for ate. a: E X, the section 33:13 is u—null, and hence since f(:c, -) = g(x, -) except on 33E, which is v-null, we conclude that f (:c, -) is N—measurable, for ate. a: E X (since g(:c, -) is measurable, as a consequence of Fubini’s theorem for non—completions). Therefore we may deﬁne: F( ) fY ﬁx, y) clv(y) E fY g(:c, y) dv(y) for a.e. cc 6 X (same set as above) CC I: 0 otherwise Notice that the function: Gtm) =: fy gins, y) duty) (which we know to be M-measurable by Fubini for non—completions) differs from F only on a M-nllu set, and hence F is M—measurable, and: 7 “fx Lanadvejduo)“: LF(m)du(m)= Lemme): [X Lg(\$iy)dl’(y)ilﬂl(\$)= fxxygd'” t where the last equality follows from Fubini for non-completions. Since 9 is H-measurable, and ﬁ 2 II, this equals: _ = f 9 d? X X Y - . 6 and ﬁnally, since 9 = f 1r-a.e. (and hence ﬁ-a.e.): ' 5 =nyde Of course we may switch the order of integration, and apply the same argument (verbatim, with a: and y, as well as M and N switched) to show that the iterated integrals are equal. III 5/8 [an m, UciAQMc'w -_.m if“, (Liam l’ihw o; :f-ﬂ ( :23) éMx *- Co: ’4 JAR : 750 \ V’L f" [011 A [(333 ’35? {3 4T1'V1'au RCOI‘MS + ’. L" 0 r‘ {Inc-“"1“, a“) ‘Fbmln‘jg T "r {A ._ g5” RM, ' ‘- #0 5.2:» ,MH cm X 0 cl (k) "LL - 33-2 ”-5-? O O to "Ln .. e 3 (pf: Mammy??? U .— . J”; "M3 4 (Lu/xix?) 494:0 : a! .3. 1J C 3 (JACK) ,. SLNt/n, a” 91‘ “DU: measum .‘i g A mam-«m m we. '4. A “a. , I. E 3‘ gm 43M 0\ £2120?! EE \$36091 U 1 Reg/L E , \ vwng‘xzvkmwﬁ : “M PKBOQAA We szx < m : ““1, P Les w ' I” 3‘ I Cottﬂﬁww 9;»? CtUx posm‘bel W' (P3?—a You] An MW itn We» WW5“ ”owi' Xrﬂ,—§ 24113 XLM Lard) ; +\- ‘XC H043» Cox-(L) -‘»k —| , 0% A: W] ; ’“iﬂ L A" H”: 2'6 MM)- 1‘: : §F4MJ1AQ gm)“; {1N0 '— (mw ﬁrm—Jr Ohmic Corela'a (WWW ZS x4 (A) - H ‘ Mm cm «’n M '““' 793m) - .. .. MW (curd Au; amps“. a xdf'o‘) 3 (d ”rm (WA); Luruuui-'\w A M «Wot, Mach H A #5 ,rtmw. “ V 7. 10 “t . ”M - 31¢} «a gm. mum-M, 1.0m {mag , 0W3 C 1 {{n “HM armoim vowiabb- dum’gvj Tim OWW- « 41"” ﬁat/me- « U, W Cart“- Angﬁrmm, (c) 5 Proof. (4:) Let us assume that ex, 0y are independent. Then for any ,b the events {X < a} and {Y < b} belong to 0X , O'y, as preimages of Borel sets, and hence they are independent. (=>) Let us assume that {X < a} and {Y < b} are independent for all (1,!) E R. Let us ﬁrst show that the events {a g X < b} and {Y < c} are independent, for all e < b, and 0. Indeed: lP[{aS X < b}ﬂ{Y <c}] =IP’[({X < b}ﬂ{Y <c})\({X < a}ﬂ{Y<c})] :IP[{XI< b}n{Y <c}]—IF’[{X < a}ﬂ{Y<c}] And by the independence of X , Y this equals: . é { =IPV[X < b]]P’[Y g c] —lP{X < a]]P{Y <c] =IP[e_<_X < b]lP’[Y < c] and hence the events are independent. By using the same method, i.e. writing {c_ < Y < at}: {Y < saw \ {Y < c}, it follows that {a g X < b} and {c g Y < d} are independent. ﬁwb“ \$11” 2420/ Consequently, ax, cry contain preimages (under X and Y) of intervals of the form [(1, b), which (the intervals) form an algebra. But unions / intersections of preimages are preimages of unions/ intersections, and hence the preimages themselves form an algebra. That is, we've established independence on two algebras, and hence by (a), we have independence on the induces cr-algebras. But we know that the intervals [(1, b) generate the Borel sets, and hence we conclude that cm, 03/ are independent. E! ((1) Proof. By deﬁnition, it sufﬁces to show that the events { f (X ) < a} and (Y) < b} are independent, for all a, b E R. Indeed, observe that f ‘1((—oo, a)) and g“1((—oo, 5)) are Borel (pre' e f Borel sets under Borel functions), and by hence {f(X) < a} E X‘1(f‘1((~oo,a))) G ex and {g(Y) < b ‘1(g‘1((—oo,b))) E O'y. Thus even-ts are independent, since egg, 0}» are independent (by (c)). III if . {Oil‘s/L“ X L.) I‘ \— 6—75 p)“: 3 p "Pu 8 é€¥a '(ﬁ‘v‘l‘ d‘h‘l—(‘lgui‘rons PX :I‘ P°%,II(S\ {or . c X [3) ‘ 11— a we ”mm mm , 900020 n *5“th : P {TAKES} ,‘ tire}; at 151 _ 9. PK A31 sew, (Ass) L “some, t; ' ' =P (Aﬁplo) Sim ii a: 11. Wt— or: 1 =P(X" m PCs—113D I .30 Kath—i ‘ Uﬁ by are in IE ;4+ v * C‘: ‘ j :33 5'” 5' oF'L- M _yC\$/\i‘ “-3) (or-i“ (535 Y U 1‘41484d19/HL1 1. 1%” ”rake. A E e (3‘ TIM Pi Xe = P (21" (11m) 19 we: 1’ P (Vim “hf!" Aid a: txlmnei above. ‘PCX‘YAW3PCLTJP‘ng 'Sa'ngﬂ X U ‘ ‘III’M - 414113 P to) TRUE P g _ ,. -,_ A - _ B(} 33‘ ﬁtﬁ wl‘tL P NYE : Pg (4 i1 {83 :33 (um UN :53 6‘6 for, 4 Wﬂzsum qu 3P?“ P a .4, (f) Proof. Let X, Y be such that IE [|X|] ,lE [|Y|] < 00, Le. X, Y are integrable with respect to PX, Py. We claim XY is integrable with respect to PXy. Indeed, observe that jXYl 2 O and is measurable, so by Fubini’s Theorem: teem dPXY(\$sy) = f list 01M) f m dPys) = IE [1me [1Y1 < oo R2 R [R so it is indeed integrable. Therefore, we may apply Fubini’s theorem to X Y now to get: my diam/(93’s) = j m 01me / y dis/(y) = IE [VXJIE [Y] R2 R R - Because (as, y) x—r my is continuous, hence Borel, so by the same argument used for 3(1)) in the previ§u%blemset, wehave: spa/1: f9 X(w)Y(w)le’[w]m fwwy dPXy(w,y)=lE[X]]E[Y] and the first part follows. Finally, we have: 7 (X+Y—E[X]ms[y])2 = (X—IE[X])2+(Y-—IE[Y])2+2XY+21E[X]IE[Y} —2]E[X]Y—2IE[Y]X So taking expectations and using the linearity of expectation (Le. linearity of integration): Var [X + Y] : Var [X] + liar [Y] + 21E [XY} + 21E [JE [X]]E{Y]] — 2E {E [X] Y} — 2E[IE.[Y] X] Recalling that E [X] ,]E [Y] are just constants, and X , Y are indepedent, (e) gives:' Var [X + Y] = Var [X] + Var [Y] + 2IE [X] ]E [Y] + 21E [X] E {Y} — 21s [X] ]E [Y] -— 2113 [X] IE [Y] = Var [X] + Var {Y} W ...
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hw6 - :g A j“{M S 52{man‘s 6"{‘57 V Fiﬁ(MACWAU...

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