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

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  • #1
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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,
 
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Answers and Replies

  • #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.)
 
  • #3
mathman
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I have E[X] = 0.

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

Best regards,
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.
 
  • #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.
 
  • #5
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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?
 
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  • #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.
 
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