# variance problem (i think)

• Jan 20th 2010, 07:47 AM
chemekydie
variance problem (i think)
Sorry, I was having difficulty with the equation editor on here. I've attached my problem as a picture file.

I need to show that the two equations are equal, and I'm not sure where to begin.

Thanks!
• Jan 20th 2010, 08:01 AM
Plato
Assuming that $\displaystyle \overline{z}$ is the conjugate of the complex number $\displaystyle z$, that statement is false.
$\displaystyle \overline{({z-\overline{z}})^2~}=\overline{z~}^2 -2\overline{z}z+z^2$.

Note that $\displaystyle \overline{\overline{z}}=z$ and $\displaystyle \overline{z^2}=\overline{z}^2$
• Jan 20th 2010, 09:37 AM
CaptainBlack
Quote:

Originally Posted by chemekydie
Sorry, I was having difficulty with the equation editor on here. I've attached my problem as a picture file.

I need to show that the two equations are equal, and I'm not sure where to begin.

Thanks!

$\displaystyle \overline{(x-\overline{x})^2}=\overline{x^2-2x\overline{x}+\overline{x}^2}$ $\displaystyle =\overline{x^2}-2\overline{x}\overline{x}+\overline{x}^2$

etc.

CB
• Jan 20th 2010, 12:15 PM
chemekydie
@Plato - x-bar is not a conjugate

@CaptainBlack -

In your process, what happens to the x double bar? How does that get factored out?
• Jan 20th 2010, 12:33 PM
Plato
Quote:

Originally Posted by chemekydie
@Plato - x-bar is not a conjugate

@CaptainBlack -

In your process, what happens to the x double bar? How does that get factored out?

OK, then what is $\displaystyle \overline{x}?$
• Jan 20th 2010, 01:14 PM
CaptainBlack
Quote:

Originally Posted by Plato
OK, then what is $\displaystyle \overline{x}?$

Expectation of $\displaystyle x$ (the clue is in the thread title)

CB
• Jan 20th 2010, 01:16 PM
CaptainBlack
Quote:

Originally Posted by chemekydie
@Plato - x-bar is not a conjugate

@CaptainBlack -

In your process, what happens to the x double bar? How does that get factored out?

$\displaystyle \overline{x}$ is just a number its expected value is itself :

$\displaystyle \overline{(\overline{x})}=\overline{x}$

CB
• Jan 20th 2010, 01:38 PM
chemekydie
x-bar is the mean.

I was under the assumption that x double bar is the average of the means. I'm not aware that x double bar can be transformed back to regular x bar.

Am I missing something?
• Jan 20th 2010, 01:44 PM
CaptainBlack
Quote:

Originally Posted by chemekydie
x-bar is the mean.

I was under the assumption that x double bar is the average of the means. I'm not aware that x double bar can be transformed back to regular x bar.

Am I missing something?

x double bar as you call it is the mean of the mean, but the mean is just a number, its mean is itself.

To see this just look at your definition of a mean.

CB
• Jan 21st 2010, 06:02 AM
chemekydie
Quote:

Originally Posted by CaptainBlack
$\displaystyle \overline{(x-\overline{x})^2}=\overline{x^2-2x\overline{x}+\overline{x}^2}$ $\displaystyle =\overline{x^2}-2\overline{x}\overline{x}+\overline{x}^2$

etc.

CB

I'm getting a $\displaystyle +\overline{x}^2$ instead of negative. And I'm not sure how to factor out the 2.
• Jan 21st 2010, 07:16 AM
Plato
Quote:

Originally Posted by chemekydie
I'm getting a $\displaystyle +\overline{x}^2$ instead of negative. And I'm not sure how to factor out the 2.

$\displaystyle \overline{-2x\overline{x}}=-2\overline{x}~\overline{\overline{x}}=-2\overline{x}~\overline{x}=-2\overline{x}^2$