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Deriving Expectations (i.e. mean)

  1. Mar 24, 2008 #1
    Deriving Expectations (i.e. means)

    I'm looking at my Introduction to Econometrics book and trying to figure out the derivations in the 2nd Chapter.

    First, [tex]E(Y^{2}) = \sigma^{2}_{Y}+\mu^{2}_{Y}[/tex]

    The derivation goes like this:

    [tex]E(Y^{2}) = E{[(Y - \mu_{Y})+ \mu_{Y}]^{2}} = E[(Y- \mu_{Y})^2] + 2\mu_{Y}E(Y-\mu_{Y})+ \mu^{2}_{Y} = \sigma^{2}_{Y} + \mu^{2}_{Y}[/tex] because [tex]E(Y - \mu_{Y}) = 0[/tex]

    If this last thing equals zero, then why doesn't everything but [tex]\mu^{2}_{Y}[/tex] drop out?
    Last edited: Mar 24, 2008
  2. jcsd
  3. Mar 24, 2008 #2
    Already the first "equality" is certainly not true.
    Last edited by a moderator: Dec 6, 2008
  4. Mar 24, 2008 #3
    Err, yeah. Sorry -- fixed.
  5. Mar 24, 2008 #4
    Because [itex]E(Y-\mu_Y)=0[/itex] does not imply that [itex]E\left[(Y-\mu_Y)^2\right]=0[/itex]. Otherwise, every random variable would be degenerate. Note that the square is *inside* the expectation.
    Last edited by a moderator: Dec 6, 2008
  6. Mar 29, 2008 #5
    Ah, yeah. [tex]E\left[(Y-\mu_Y)^2\right]=\sigma^{2}[/tex], right?

    So now I'm wondering how [tex]E(Y^{2}) = E{[(Y - \mu_{Y})+ \mu_{Y}]^{2}} [/tex]
  7. Mar 30, 2008 #6


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    Are you sure it is not E[((Y-\mu_Y)+ \mu_Y)^2][/itex]? That would then be [itex]E[(Y-\mu_Y)^2- 2\mu_Y(Y- \mu_Y)+ \mu_y^2][/itex] which gives the rest.
  8. Mar 30, 2008 #7
    Err, sorry. That's what it is. But I still don't see how you go from [tex]E(Y^2)[/tex] to [tex]E[((Y-\mu_Y)+\mu_Y)^2][/tex]. How'd they come up with the latter definition? Did they just add and subtract [tex]\mu_{Y}[/tex] and then add it, then group them using the associativity property?

    If I didn't add and subtract [tex]\mu_{Y}\[/tex], it seems to me that I would get [tex]E(Y^2) = \mu_Y^2[/tex]...

    How do you do the proof of [tex]E(Y) = \mu_Y[/tex]?

    Nevermind, I see. You have to add and subtract [tex]\mu_Y[/tex] (why is this formatting strangely?) to even get it. The proof of [tex]E(Y) = \mu_Y[/tex] is, I guess

    [tex]E(Y) = E[(Y -\mu_Y) + \mu_Y] = E(Y - \mu_Y) + E(\mu_Y)[/tex]

    Was what I did above legal? Can I distribute the E, and would [tex]E(\mu_Y) = \mu_Y[/tex]
    Last edited: Mar 30, 2008
  9. Mar 30, 2008 #8
    No, because

    E(Y^2)\neq \left[E(Y)\right]^2 = \mu_Y^2

    That's a definition.

    It was "legal", but to see that the first term vanishes you use [itex]E(Y-\mu_Y)=0[/itex] which is equivalent to [itex]E(Y)=\mu_Y[/itex] which is what you want to show. [itex]\mu_Y[/itex] is just a short notation for [itex]E(Y)[/itex]
    Last edited by a moderator: Dec 6, 2008
  10. Mar 31, 2008 #9
    The notation could be a source of confusion. Some people interpret the square in the second term to act on Y rather than on the expectation as a whole. (To the Original poster): perhaps its better if you write

    [tex]E^{2}(Y) = (E(Y))^2 = (\mu_Y)^{2} = \mu_{Y}^2[/tex]

    (PS--I wrote this because some books define variance(X) as [itex]E(X-\mu_{X})^2[/itex]. They actually mean [itex]E[(X-\mu_{X})^2][/itex], that is, expectation of the square of the difference between X and its mean.)
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