Let X , Y , Z be integrable random variables on the same probability space.Is it true that σ ( E [ X | Y ] , E [ X | Z ] ) ⊆ σ ( E [ X | Y , Z ] ) ?My intuition says yes: the approximation of X on σ ( Y , Z ) is richer in information than the one on σ ( Y ) and the one on σ ( Z ). How to prove this?If it is not true, what if we add σ ( Y ) ⊆ σ ( Z ) ? Does this make the above true?

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Let   X  ,  Y  ,  Z be integrable random variables on the same probability space.Is it true that   σ  (  E  [  X      |    Y  ]  ,  E  [  X      |    Z  ]  )  ⊆  σ  (  E  [  X      |    Y  ,  Z  ]  ) ?My intuition says yes: the approximation of   X on   σ  (  Y  ,  Z  ) is richer in information than the one on   σ  (  Y  ) and the one on   σ  (  Z  ). How to prove this?If it is not true, what if we add   σ  (  Y  )  ⊆  σ  (  Z  ) ? Does this make the above true?

Marlee Guerra · Accepted Answer

No, not even with the added condition. The issue is that conditioning on a smaller   σ-field may be able to &quot;split apart&quot; two events that   X does not otherwise distinguish.Consider a probability space with 4 outcomes,   Ω  =  {  a  ,  b  ,  c  ,  d  }, each having probability 1/4. We take       F    =      2    Ω   so all sets are measurable.Let   Y  =      1          {      a      ,      b      }       and let   Z be injective, so that   σ  (  Z  )  =      F   and in particular   σ  (  Y  )  ⊂  σ  (  Z  ). Define   X as follows:                    ω                    X        (        ω        )                            a                    0                            b                    2                            c                    2                            d                    8            We have   E  [  X  ∣  Z  ]  =  E  [  X  ∣  Y  ,  Z  ]  =  X. So   σ  (  E  [  X  ∣  Y  ,  Z  ]  )  =  σ  (  X  ) is generated by the events   {  a  }  ,  {  b  ,  c  }  ,  {  d  } and is not all of       F  . In particular   {  b  }  ∉  σ  (  E  [  X  ∣  Y  ,  Z  ]  ).However, you may check that   E  [  X  ∣  Y  ] is given by                    ω                    E        [        X        ∣        Y        ]        (        ω        )                            a                    1                            b                    1                            c                    5                            d                    5            Thus   {  a  ,  c  }  ∈  σ  (  E  [  X  ∣  Y  ]  ). So   σ  (  E  [  X  ∣  Y  ]  ,  E  [  X  ∣  Z  ]  ) contains both   {  a  ,  b  } and   {  b  ,  c  } so it is all of       F  . In particular, it contains   {  b  }  =  {  E  [  X  ∣  Z  ]  =  2  }  ∩  {  E  [  X  ∣  Y  ]  =  1  }.

Let X , Y , Z be integrable random variables on the same probability space. Is i

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