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

1. Dec 20, 2009

### 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,

Last edited: Dec 21, 2009
2. Dec 20, 2009

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. Dec 20, 2009

### mathman

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. Dec 20, 2009

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.

5. Dec 21, 2009

### 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?

Last edited: Dec 21, 2009
6. Dec 21, 2009

### 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.

Last edited: Dec 21, 2009