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

1. Oct 29, 2012

### 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

2. Oct 29, 2012

### 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.

3. Oct 30, 2012

### 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?

4. Oct 31, 2012

### dipole

Yes except the last term should be positive.

5. Nov 1, 2012

### Staff: Mentor

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

6. Nov 1, 2012

### mitch_1211

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?

7. Nov 2, 2012

### Staff: Mentor

I think that equation has an error.

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