Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

  1. Dec 20, 2009 #1

    areslagae

    User Avatar

    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
    areslagae, Dec 20, 2009
  2. jcsd
    Phys.org - latest science and technology news stories on Phys.org
  3. Dec 20, 2009 #2

    statdad

    User Avatar
    Homework Helper

    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.)
     
    statdad, Dec 20, 2009
  4. Dec 20, 2009 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    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.
     
    mathman, Dec 20, 2009
  5. Dec 20, 2009 #4

    statdad

    User Avatar
    Homework Helper

    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.
     
    statdad, Dec 20, 2009
  6. Dec 21, 2009 #5

    areslagae

    User Avatar

    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
    areslagae, Dec 21, 2009
  7. Dec 21, 2009 #6

    mathman

    User Avatar
    Science Advisor
    Gold Member

    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
    mathman, Dec 21, 2009
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: E[x] = 0 => e[y/x] = 0 ?

  1. Statistics: E(X) = Integral(0 to infinity) of (1-F(x))dx (Replies: 12)

  2. X=0 almost surely => E(X)=0 (Replies: 13)

  3. If X ≤ Y, then E(X) ≤ E(Y) (Replies: 6)

  4. P(X>x)=e^-ax, x less/equal 0 (Replies: 1)

  5. Why E[x^2] = autocorrelation of x evaluated in 0 ? (Replies: 1)

Loading...