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E[x] = 0 => e[y/x] = 0 ?

  1. Dec 20, 2009 #1
    E[x] = 0 => e[y/x] = 0 ? (division, not cond prob)

    I have E[X] = 0.

    Does this imply that E[Y/X] = 0, for Y independent of X?

    Best regards,
     
    Last edited: Dec 21, 2009
  2. jcsd
  3. Dec 20, 2009 #2

    statdad

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    Write out the joint distribution, find the required marginal distribution, and check. (No, you don't need to know the specific forms of the distributions.)
     
  4. Dec 20, 2009 #3

    mathman

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    If Y is independent of X, then E(Y/X) = E(Y)E(1/X), so you really don't know what the result is unless you have more information about the distributions of X and Y.
     
  5. Dec 20, 2009 #4

    statdad

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    Okay, my old eyes may be guilty of misreading the OP.

    Does E[Y/X] mean

    [tex]
    E[\frac Y X]
    [/tex]

    (Y divided by X)

    or is it

    [tex]
    E[Y \mid X]
    [/tex]

    (expected value of Y GIVEN X)?

    My response was made under the assumption that a conditional expectation was desired. If that is not the case, I'm not sure any other answer could possibly be more incomplete or incorrect than mine.
     
  6. Dec 21, 2009 #5
    I actually mean division.

    So, if E[X] = 0, does E[Y/X] = 0 follow, for Y independent of X?

    Additionally, I have that X is symmetric.

    Intuitively, it seems that this does hold.

    E[Y/X] = E[Y] E[1/X] (X and Y are independent)

    So, it remains to be shown that, if E[X]=0, and X is symmetric, then E[1/X]=0.

    Is this trivial? Since the PDF of X is symmetric around 0, the PDF of 1/X will be symmetric around zero. Is this argument correct?
     
    Last edited: Dec 21, 2009
  7. Dec 21, 2009 #6

    mathman

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    If you exclude an interval around 0 for possible values of X, then the symmetry will work. However if 0 is a possible value for X, E(1/X) may not exist.
     
    Last edited: Dec 21, 2009
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