How to expand the average <(N - <N>)^2>

[mitch_1211]

Basically I would like to know how to expand:

$\left\langle$(N - $\left\langle$ N $\right\rangle$)²$\right\rangle$

<(N - <N>)^2>

Where < and > represent $\left\rangle\right\langle$ and denote the average of the quantity the enclose.

So this is pretty much the average of [N - N(average)] ²

thank you

 

[dipole]

Well first expand the square, and the average of three terms you'll get is the sum of their averages.

And note that the average of a constant is the same constant.

 

[mitch_1211]

So by expanding i get:

<N²-2N<N> - <N>²>

and you're saying this is equivalent to:

<N²> - <2N<N>> - <<N>²>

is that right?

 

[dipole]

So by expanding i get:

<N²-2N<N> - <N>²>

and you're saying this is equivalent to:

<N²> - <2N<N>> - <<N>²>

is that right?

Yes except the last term should be positive.

 

[mfb]

You can simplify <2N<N>> and <<N>²> and combine them to get a nice, short result.

 

[mitch_1211]

You can simplify <2N<N>> and <<N>²> and combine them to get a nice, short result.

Yea I figured that, I simplified

<N²> - 2N<N> + <N>²

into

<N²> - 2<N>² + <N>²

so I get

<N²> - <N>²

Also I know that <N>² = <N> + <N>²

Is there a simple way of showing that without resulting to a probability with sums of exponentials etc?

 

[mfb]

Also I know that <N>² = <N> + <N>²

I think that equation has an error.

Your expression is simply the variance: <N²> - <N>² = Var(N)