Why does X ∈ L p ( Ω , F , P ) , p ∈ [ 1 , ∞ ), implies that E [ X | A ] ∈ L p ( Ω , A , P ), where A ⊂ F is a sub- σ-algebra?The defintion of the conditional expectation says that E [ X 1 A ] = E [ E [ X | A ] 1 A ] for all A ∈ A but why can we conclude from this that E [ | X | p 1 A ] = E [ E [ | X | p | A ] 1 A ] for all A ∈ A ?

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Why does   X  ∈            L        p    (  Ω  ,      F    ,      P    )  ,  p  ∈  [  1  ,  ∞  ), implies that       E    [  X        |        A    ]  ∈            L        p    (  Ω  ,      A    ,      P    ), where       A    ⊂      F   is a sub-  σ-algebra?The defintion of the conditional expectation says that       E    [  X            1        A    ]  =      E    [      E    [  X      |        A    ]            1        A    ] for all   A  ∈      A   but why can we conclude from this that       E    [      |    X            |        p              1        A    ]  =      E    [      E    [      |    X            |        p        |        A    ]            1        A    ] for all   A  ∈      A  ?

Elias Flores · Accepted Answer

We will use Jensen&#039;s inequality for conditional expectation: If   ϕ is a convex function, then  ϕ            (            E    [  X  ∣      A    ]            )        ≤      E              [        ϕ  (  X  )  ∣      A              ]      Now   p  ∈  [  1  ,  ∞  ), so the map   t  ↦      |    t            |        p   is convex.Assume   X  ∈      L    p    (  Ω  ,      F    ,      P    ); that is       E    [      |    X            |        p    ]  &amp;lt;  ∞. By Jensen,            |            E    [  X  ∣      A    ]                    |              p    ≤      E              [            |    X            |        p    ∣      A              ]      so      E              [                  |            E    [  X  ∣      A    ]                    |              p              ]        ≤      E              [            E              [            |    X            |        p    ∣      A              ]                  ]        =      E              [            |    X            |        p              ]        &amp;lt;  ∞  .Thus   E  [  X  ∣      A    ]  ∈      L    p    (  Ω  ,      A    ,      P    ).

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