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‘0) 6 3% 0.: LOMVW Qﬁumﬁlﬁ \‘0 OD WWIm we.“ W‘w} gammy 2M fag” ﬂ 4w C «5“ " . I I 'h‘ \dw {iﬂl'ﬂ H$§YEQ£3 ’ijﬂ': ;,a tor QWN‘ M34, N 7/ W44 [Wiﬁéég I (b) .1, Proof. Let f : X x Y —> [0, 00] be M x Wmeasurable. From now on we’ll write, H for M X N, and gr for the
product measure (and if for the completions). Then there exists a sequence of ﬁmeasurable simple functions 0 5 (pl S 9:22 5 . . . S f that converge pointwise to f. For each n, we have: N(")
‘Pn = Z andAmi
i=1 where Am is ﬁmeasurable. Then (since we’ve taken completions), we know that there exists a Hrneasurable set
AL”; g Aw such that AW \AL, is a subset of a ﬁnull set, which (by the previous problemset) is in turn contained in a 1r~null set, and hence Am \ AA}, is inside a 7r—null set. Then consider: which is a H—measurable simple function, which (by finite additivity) differs from 9071 only possibly on a arnull set E”. By countable additivity, E = U” En is also r—null, and on the complement E6 of which we have (on = 90;, for all 4/Q) n. Now for each n, let: n := XEC max{(,o’1,... ,goil} which is H—meas rable, .5 f, and 0 5 1,01 5 1,02 5 ... g f. 0n E“ we have ”din = so; = (pm and hence convergence to f.
On E we have n = 0 for all n. Consequently, the limit 9 = f X Ec of the 1% is meeasurable, non—negative, and wae. equal to f. Since E is ir—null, we have by Fubini’s theorem for non—completions that XE(x, ) is N—measurable for all :r; E X , and 0=nyxE:LLXE($iy)dV(y)dI~L(x) and hence since the integrand is non—negative: furthermore that: may) dye) = 0
Y for uae. 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 vnull, 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 Mmeasurable by Fubini for non—completions) differs from F only on a Mnllu 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 noncompletions. Since 9 is Hmeasurable, and ﬁ 2 II, this equals: _ = f 9 d?
X X Y  . 6
and ﬁnally, since 9 = f 1ra.e. (and hence ﬁa.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;
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31¢} «a gm. mumM, 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“ $11” 2420/ Consequently, ax, cry contain preimages (under X and Y) of intervals of the form [(1, b), which (the intervals) form
an algebra. But unions / intersections of preimages are preimages of unions/ intersections, and hence the preimages
themselves form an algebra. That is, we've established independence on two algebras, and hence by (a), we have
independence on the induces cralgebras. But we know that the intervals [(1, b) generate the Borel sets, and hence we conclude that cm, 03/ are independent. E! ((1) Proof. By deﬁnition, it sufﬁces to show that the events { f (X ) < a} and (Y) < b} are independent, for all a, b E R.
Indeed, observe that f ‘1((—oo, a)) and g“1((—oo, 5)) are Borel (pre' e f Borel sets under Borel functions), and
by hence {f(X) < a} E X‘1(f‘1((~oo,a))) G ex and {g(Y) < b ‘1(g‘1((—oo,b))) E O'y. Thus events are independent, since egg, 0}» are independent (by (c)). III if . {Oil‘s/L“ X L.) I‘ \— 6—75 p)“: 3 p "Pu
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for, 4 Wﬂzsum qu 3P?“ P a .4, (f) Proof. Let X, Y be such that IE [X] ,lE [Y] < 00, Le. X, Y are integrable with respect to PX, Py. We claim XY is
integrable with respect to PXy. Indeed, observe that jXYl 2 O and is measurable, so by Fubini’s Theorem: teem dPXY($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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