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

  • Thread starter areslagae
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In summary, the conversation discusses the implication of E[X] = 0 on the expected value of Y given X, E[Y/X]. It is mentioned that if X and Y are independent, then E[Y/X] = E[Y]E[1/X], and it remains to be shown that E[1/X] = 0 given E[X] = 0 and X is symmetric. The potential issue of 0 being a possible value for X is also addressed.
  • #1
areslagae
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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?

 
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  • #2
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
areslagae said:
I have E[X] = 0.

Does this imply that E[Y/X] = 0, for Y independent of X?
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
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
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
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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What does the equation "E[x] = 0" mean?

The equation "E[x] = 0" represents the expected value of a random variable x, which is the average value of x over a large number of trials.

What does the notation "e[y/x]" mean?

The notation "e[y/x]" indicates the conditional expectation of y given x, which is the expected value of y when x is known or fixed.

What is the relationship between "E[x] = 0" and "e[y/x] = 0"?

If the expected value of x is equal to 0, then the conditional expectation of y given x is also equal to 0. In other words, the average value of x is 0 and the average value of y when x is known is also 0.

What does it mean if "E[x] = 0 => e[y/x] = 0" is true?

If the statement "E[x] = 0 => e[y/x] = 0" is true, it means that whenever the expected value of x is 0, the conditional expectation of y given x is also 0. This can be interpreted as a relationship between the averages of two random variables.

Can "E[x] = 0 => e[y/x] = 0" be reversed to "e[y/x] = 0 => E[x] = 0"?

No, the statement "E[x] = 0 => e[y/x] = 0" cannot be reversed to "e[y/x] = 0 => E[x] = 0". While the first statement is true, the second statement is not necessarily true. This is because the expected value of x being 0 does not always imply that the conditional expectation of y given x is also 0.

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