Let ( Ω , A , P ) be a probability space and F i ⊆ A be a σ-algebra on Ω.If ( F 1 , F 2 , F 3 ) is independent, does it follow that F 1 ∨ F 2 := σ ( F 1 ∪ F 2 ) is independent of F 3 ?In the special case, where F i = σ ( X i ) for some random variable X i taking values in a measurable space ( E i , E i ), we've got (1) F 1 ∨ F 2 = ( X 1 , X 2 ) − 1 ( E 1 ⊗ E 2 ) = ( X 1 , X 2 ) − 1 ( σ ( G 1 × G 2 ) , where G i ⊆ E i is arbitrary with σ ( G i ) = E i ). From (1) the desired claim immediately follows.So, I wondered whether the same holds in general or why it breaks down.

Question

Let   (  Ω  ,      A    ,  P  ) be a probability space and             F        i    ⊆      A   be a   σ-algebra on   Ω.If   (            F        1    ,            F        2    ,            F        3    ) is independent, does it follow that             F        1    ∨            F        2    :=  σ  (            F        1    ∪            F        2    ) is independent of             F        3  ?In the special case, where             F        i    =  σ  (      X    i    ) for some random variable       X    i   taking values in a measurable space   (      E    i    ,            E        i    ), we&#039;ve got                    (1)                                          F                    1                ∨                              F                    2                =        (                  X          1                ,                  X          2                          )                      −            1                          (                              E                    1                ⊗                              E                    2                )        =        (                  X          1                ,                  X          2                          )                      −            1                          (        σ        (                              G                    1                ×                              G                    2                )        ,            where             G        i    ⊆            E        i   is arbitrary with   σ  (            G        i    )  =            E        i    ). From (1) the desired claim immediately follows.So, I wondered whether the same holds in general or why it breaks down.

Cordell Crosby · Accepted Answer

Sets of the form   A  ∩  B with   A  ∈            F        1    ,  B  ∈            F        2   form a   π system which generates             F        1    ∨            F        2  . The equation   P  (  C  ∩  D  )  =  P  (  C  )  P  (  D  ) holds if   C is above type and   D  ∈            F        3  . Apply Dymkin&#039;s   π  −  λ Theorem (with   D  ∈            F        3   fixed) to finish.

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