- #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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- Thread starter areslagae
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- #1

- 11

- 0

I have E[X] = 0.

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

Best regards,

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- #2

statdad

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- #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.I have E[X] = 0.

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

Best regards,

- #4

statdad

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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?

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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