27.2. CONDITIONAL PROBABILITY 737

Definition 27.2.2 Let {X1, · · · ,Xn} be random vectors defined on a probabilityspace having values in Rp1 , · · · ,Rpn respectively. The random vectors are independent iffor every E a Borel set in Rp1 ×·· ·×Rpn ,∫

Rp1×···×RpnXEdλ (X1,··· ,Xn)

=∫Rp1· · ·∫Rpn

XEdλXndλXn−1 · · ·dλX2dλX1 (27.2)

and the iterated integration may be taken in any order. If A is any set of random vectorsdefined on a probability space, A is independent if any finite set of random vectors fromA is independent.

Thus, the random vectors are independent exactly when the dependence on the givensin 27.1 can be dropped.

Does this amount to the same thing as discussed earlier? Suppose you have three ran-dom variables X,Y,Z. Let A =X−1 (E), B = Y −1 (F) ,C = Z−1 (G) where E,F,G areBorel sets. Thus these inverse images are typical sets in

σ (X) ,σ (Y ) ,σ (Z)

respectively. First suppose that the random variables are independent in the earlier sense.Then

P(A∩B∩C) = P(A)P(B)P(C)

=∫Rp1

XE (x)dλX

∫Rp2

XF (y)dλY

∫Rp3

XG (z)dλZ

=∫Rp1

∫Rp2

∫Rp3

XE (x)XF (y)XG (z)dλZdλY dλX

AlsoP(A∩B∩C) =

∫Rp1×Rp2×Rp3

XE (x)XF (y)XG (z)dλ (X,Y,Z)

=∫Rp1

∫Rp2

∫Rp3

XE (x)XF (y)XG (z)dλZ|xydλY |xdλX

Thus ∫Rp1

∫Rp2

∫Rp3

XE (x)XF (y)XG (z)dλZdλY dλX

=∫Rp1

∫Rp2

∫Rp3

XE (x)XF (y)XG (z)dλZ|xydλY |xdλX

Now letting G = Rp3 , it follows that∫Rp1

∫Rp2

XE (x)XF (y)dλY dλX =∫Rp1

∫Rp2

XE (x)XF (y)dλY |xdλX

By uniqueness of the slicing measures or an application of the Besikovitch differentiationtheorem, it follows that for λX a.e. x,

λY = λY |x

27.2. CONDITIONAL PROBABILITY 737Definition 27.2.2 Le {X1,:-+,Xn} be random vectors defined on a probabilityspace having values in R?!,--- ,R?" respectively. The random vectors are independent iffor every E a Borel set in R?! x --- x R’,[ REM (x, ...Xy)RP1 x... RPn=| [| Lpdhx,dax,,---dax,dax, (27.2)RI RPaand the iterated integration may be taken in any order. If & is any set of random vectorsdefined on a probability space, & is independent if any finite set of random vectors fromA is independent.Thus, the random vectors are independent exactly when the dependence on the givensin 27.1 can be dropped.Does this amount to the same thing as discussed earlier? Suppose you have three ran-dom variables X,Y,Z. Let A = X~'(E), B= Y~'(F),C = Z| (G) where E,F,G areBorel sets. Thus these inverse images are typical sets ino(X),o(Y),o(Z)respectively. First suppose that the random variables are independent in the earlier sense.ThenP(ANBNC) = P(A) P(B)P(C)[ WA (x)dax | WA (y)ary | XG (z)dazR’1 R?2 R?3[ / Xe (a) Xe (y) Xo (z)ddgddydhxR?1 JR?P2 JRP3AlsoPiansne)= [| Xe (x) Lr (y) 26 (2) dA(x,y,z)R?1 xR?2 xR?3— / | 2g (@) Rr (y) Ro (Zz) da zlaydy ed xR?1 JR?2 JR?Thus| [ | Xe (w) Xe (y) Xo (z)dagdaydaxR?1 JR?2 JR?3- / | | Le (w) Re (y) Roz) AA gay dhy jad xR?1 JR?2 JR?3Now letting G = R?3, it follows that| |, %@ %wdavdax= [Fe (w) Fr (y)ddypdhxR?1 JIR?2R?1 JRP2By uniqueness of the slicing measures or an application of the Besikovitch differentiationtheorem, it follows that for A x ae. x,Ay = Ay |e