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

[areslagae]

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,

 

[statdad]

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.)

 

[mathman]

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.

 

[statdad]

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

Does E[Y/X] mean

$$E[\frac Y X]$$

(Y divided by X)

or is it

$$E[Y \mid X]$$

(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.

 

[areslagae]

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?

 

[mathman]

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.