Deriving Expectations (i.e. mean)

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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?
 
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[tex]E(Y^{2}) = E[(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]
Already the first "equality" is certainly not true.
 
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Err, yeah. Sorry -- fixed.
 
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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?
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.
 
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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]
 
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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.
 
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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]
 
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it seems to me that I would get [tex]E(Y^2) = \mu_Y^2[/tex]
No, because

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


How do you do the proof of [tex]E(Y) = \mu_Y[/tex]?
That's a definition.

Nevermind, I see. You have to add and subtract [tex]\mu_Y[/tex] 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]
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]
 
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No, because

[tex]
E(Y^2)\neq E(Y)^2 = \mu_Y^2
[/tex]
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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