# Deriving Expectations (i.e. mean)

1. Mar 24, 2008

### kurvmax

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, $$E(Y^{2}) = \sigma^{2}_{Y}+\mu^{2}_{Y}$$

The derivation goes like this:

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

If this last thing equals zero, then why doesn't everything but $$\mu^{2}_{Y}$$ drop out?

Last edited: Mar 24, 2008
2. Mar 24, 2008

### Pere Callahan

Already the first "equality" is certainly not true.

Last edited by a moderator: Dec 6, 2008
3. Mar 24, 2008

### kurvmax

Err, yeah. Sorry -- fixed.

4. Mar 24, 2008

Because $E(Y-\mu_Y)=0$ does not imply that $E\left[(Y-\mu_Y)^2\right]=0$. Otherwise, every random variable would be degenerate. Note that the square is *inside* the expectation.

Last edited by a moderator: Dec 6, 2008
5. Mar 29, 2008

### kurvmax

Ah, yeah. $$E\left[(Y-\mu_Y)^2\right]=\sigma^{2}$$, right?

So now I'm wondering how $$E(Y^{2}) = E{[(Y - \mu_{Y})+ \mu_{Y}]^{2}}$$

6. Mar 30, 2008

### HallsofIvy

Staff Emeritus
Are you sure it is not E[((Y-\mu_Y)+ \mu_Y)^2][/itex]? That would then be $E[(Y-\mu_Y)^2- 2\mu_Y(Y- \mu_Y)+ \mu_y^2]$ which gives the rest.

7. Mar 30, 2008

### kurvmax

Err, sorry. That's what it is. But I still don't see how you go from $$E(Y^2)$$ to $$E[((Y-\mu_Y)+\mu_Y)^2]$$. How'd they come up with the latter definition? Did they just add and subtract $$\mu_{Y}$$ and then add it, then group them using the associativity property?

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

How do you do the proof of $$E(Y) = \mu_Y$$?

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

$$E(Y) = E[(Y -\mu_Y) + \mu_Y] = E(Y - \mu_Y) + E(\mu_Y)$$

Was what I did above legal? Can I distribute the E, and would $$E(\mu_Y) = \mu_Y$$

Last edited: Mar 30, 2008
8. Mar 30, 2008

### Pere Callahan

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 $E(Y-\mu_Y)=0$ which is equivalent to $E(Y)=\mu_Y$ which is what you want to show. $\mu_Y$ is just a short notation for $E(Y)$

Last edited by a moderator: Dec 6, 2008
9. Mar 31, 2008

### maverick280857

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

$$E^{2}(Y) = (E(Y))^2 = (\mu_Y)^{2} = \mu_{Y}^2$$

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