- #1

TeenieBopper

- 29

- 0

## Homework Statement

Showthe two forms of the sample variance are equivalent:

[itex]\frac{1}{n-1}[/itex][itex]\sum_{i=1}^\n (Y

_{i}-Ybar)

^{2}[/itex] = [itex]\frac{1}{n(n-1)}[/itex][itex]\sum_{i=1}^\n \sum_{j>i}\n (Y

_{i}-Y

_{j})

^{2}[/itex]

The first summation is from i=1 to n, the second is i=1 to n and the third is j>i to n. Sorry, I don't know how to format those.

## Homework Equations

## The Attempt at a Solution

I don't really know where to begin, so I tried just expanding and cancelling where I could. I know the (n-1) in the denominator on both sides cancel, and then after expanding I get

[itex]\sum (Y

_{i}

^{2}-2Y

_{i}Ybar + Ybar

^{2}[/itex] = [itex]\frac{1}{n}[/itex][itex]\sum \sum (Y

_{i}

^{2}-2Y

_{i}Y

_{j}+Y

_{j}

^{2})[/itex]

Then, I can distribute the 1/n and the summations on the right side. If I do that, and I have a term that does not have j (such as Y

_{i}), I can essentially drop the j summation from that term, correct? After I do that, I have the following:

[itex]\sum Y

_{i}

^{2}[/itex] -2[itex]\sum Y

_{i}Ybar[/itex] + [itex]\sum Ybar

^{2}[/itex] = [itex]\frac{1}{n}[/itex][itex]\sum Y

_{i}

^{2}- \frac{2}{n}[/itex][itex]\sum \sum Y

_{i}Y

_{j}[/itex]+[itex]\sum \sum Y

_{j}

^{2}[/itex]

And here's where I'm stuck (assuming I even did everything right to get here, which I doubt). I don't know how to deal with the Y

_{j}, among other things. Any help would be greatly appreciated.

edit: I'm sorry about the terrible formatting. I tried using the LaTex tag buttons and I've looked at the FAQ; not sure what I'm doing wrong.

Last edited: