Show lemma on probability of n independent setsLet be ( Ω , A , P ) a probability space and A 1 , … , A n ∈ A . For sake of convenience we define A 1 := A and A c := Ω ∖ A. Let be ( k 1 , … , k n ) ∈ { 1 , c } n . Show that A 1 , … , A n are stochastically independent ⟹ P ( ⋂ j = 1 n A j k j ) = ∏ j = 1 n P ( A j k j ) .

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Show lemma on probability of n independent setsLet be   (  Ω  ,      A    ,  P  ) a probability space and       A    1    ,  …  ,      A    n    ∈      A  . For sake of convenience we define       A    1    :=  A and       A    c    :=  Ω  ∖  A. Let be   (      k    1    ,  …  ,      k    n    )  ∈  {  1  ,  c      }    n  . Show that      A    1    ,  …  ,      A    n     are stochastically independent     ⟹    P      (          ⋂              j        =        1            n              A      j                        k          j                      )    =      ∏          j      =      1        n    P  (      A    j                  k        j              )  .

Hector Hamilton · Accepted Answer

Step 1The induction on m is sufficient: we show by induction on m that for each number   n  ⩾  m, stochastically independent sets       A    1    ,  …  ,      A    n    ∈      A   and   k  ∈  {  1  ,  c      }    n   containing m-many complement symbols the equality from the question holds.For   m  =  0 the claim is trivial. Fix   m  ⩾  1, assume the claim holds for   m  −  1 and take       A    1    ,  …  ,      A    n   and   k  ∈  {  1  ,  c      }    n   as in the statement. By symmetry we can assume that       k    n    =  c. We have that      ⋂          j      =      1              n      −      1            A    j                  k        j              =      (          ⋂              j        =        1                    n        −        1                    A      j                        k          j                      ∩          A      n        )    ∪      (          ⋂              j        =        1                    n        −        1                    A      j                        k          j                      ∩          A      n      c        )  and the sets on the right are disjoint, so  P      (          ⋂              j        =        1                    n        −        1                    A      j                        k          j                      )    =  P      (          ⋂              j        =        1                    n        −        1                    A      j                        k          j                      ∩          A      n        )    +  P      (          ⋂              j        =        1                    n        −        1                    A      j                        k          j                      ∩          A      n      c        )    .Step 2The first two probabilities can be computed using the induction hypothesis since among the &quot;exponents&quot; there are exactly   m  −  1-many complement signs:      ∏          j      =      1              n      −      1        P  (      A    j                  k        j              )  =      ∏          j      =      1              n      −      1        P  (      A    j                  k        j              )  ⋅  P  (      A    n    )  +  P      (          ⋂              j        =        1                    n        −        1                    A      j                        k          j                      ∩          A      n      c        )    .It follows that  P      (          ⋂              j        =        1            n              A      j                        k          j                      )    =  P      (          ⋂              j        =        1                    n        −        1                    A      j                        k          j                      ∩          A      n      c        )    =      ∏          j      =      1              n      −      1        P  (      A    j                  k        j              )  ⋅            (        1  −  P  (      A    n    )            )        =      ∏          j      =      1        n    P  (      A    j                  k        j              )  ,as desired

Let be (Omega,A,P) a probability space and A_1,…,A_n in A. For sake of convenience we define A^1:=A and A^c:=Omega∖A.

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