variance problem (i think)

**chemekydie** · January 20th 2010, 07:47 AM

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!

**Plato** · January 20th 2010, 08:01 AM

Assuming that $\overline{z}$ is the conjugate of the complex number $z$ , that statement is false.
$\overline{({z-\overline{z}})^2~}=\overline{z~}^2 -2\overline{z}z+z^2$ .

Note that $\overline{\overline{z}}=z$ and $\overline{z^2}=\overline{z}^2$

**CaptainBlack** · January 20th 2010, 09:37 AM

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!

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

etc.

CB

**chemekydie** · January 20th 2010, 12:15 PM

@Plato - x-bar is not a conjugate

@CaptainBlack -

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

**Plato** · January 20th 2010, 12:33 PM

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 $\overline{x}?$

**CaptainBlack** · January 20th 2010, 01:14 PM

Originally Posted by Plato

OK, then what is $\overline{x}?$

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

CB

**CaptainBlack** · January 20th 2010, 01:16 PM

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?

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

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

CB

**chemekydie** · January 20th 2010, 01:38 PM

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?

**CaptainBlack** · January 20th 2010, 01:44 PM

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

**chemekydie** · January 21st 2010, 06:02 AM

Originally Posted by CaptainBlack

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

etc.

CB

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

**Plato** · January 21st 2010, 07:16 AM

Originally Posted by chemekydie

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

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

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