Conditional Expectation Property GeometricI need to calculate E ( X | Z ) where X, Y are Geometric r.v. and Z = X + Y. The conditional distribution of E ( X | Z ) is 1 n − 1 So now, E ( X | Z ) = ∑ k = 1 n − 1 k n − 1 = n 2 . However, E ( X ) = 1 p . And by property of conditional Expectation E ( E ( X | Z ) ) = E ( X ). So I get n 2 = 1 p . What am I doing wrong? How do I get E ( E ( X | Z ) ) = 1 p

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Conditional Expectation Property GeometricI need to calculate       E    (  X      |    Z  ) where X, Y are Geometric r.v. and   Z  =  X  +  Y. The conditional distribution of       E    (  X      |    Z  ) is       1          n      −      1       So now,       E    (  X      |    Z  )  =      ∑          k      =      1              n      −      1            k          n      −      1        =      n    2  . However,       E    (  X  )  =      1    p  . And by property of conditional Expectation       E    (      E    (  X      |    Z  )  )  =      E    (  X  ). So I get       n    2    =      1    p  . What am I doing wrong? How do I get       E    (      E    (  X      |    Z  )  )  =      1    p

Demarion Thornton · Accepted Answer

Step 1Your sin are sloppy notations. In fact, you (correctly) calculate       E    (  X      |    Z  =  n  )  =      ∑          k      =      1              n      −      1            k          n      −      1        =      n    2    .Step 2You could write that as       E    (  X      |    Z  )  =      1    2    Z, too. Then,       E    (      E    (  X      |    Z  )  )  =      E        (          1      2        Z    )    =      1    2    (      E    X  +      E    Y  )  =      1    p    .

I need to calculate E(X|Z) where X, Y are Geometric r.v. and Z=X+Y. The conditional distribution of E(X|Z) is 1/(n-1)

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