Show the two forms of the sample variance are equivalent

[TeenieBopper]

Homework Statement

Showthe two forms of the sample variance are equivalent:
$\frac{1}{n-1}$$\sum_{i=1}^\n (Y_i-Ybar)²$ = $\frac{1}{n(n-1)}$$\sum_{i=1}^\n \sum_{j>i}\n (Y_i-Y_j)²$

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

$\sum (Y_i²-2Y_iYbar + Ybar²$ = $\frac{1}{n}$$\sum \sum (Y_i²-2Y_iY_j+Y_j²)$

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:

$\sum Y_i²$ -2$\sum Y_iYbar$ + $\sum Ybar²$ = $\frac{1}{n}$$\sum Y_i² - \frac{2}{n}$$\sum \sum Y_iY_j$+$\sum \sum Y_j²$

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.

 

[Mark44]

Homework Statement

Showthe two forms of the sample variance are equivalent:
$\frac{1}{n-1}$$\sum_{i=1}^\n (Y_i-Ybar)²$ = $\frac{1}{n(n-1)}$$\sum_{i=1}^\n \sum_{j>i}\n (Y_i-Y_j)²$

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

$\sum (Y_i²-2Y_iYbar + Ybar²$ = $\frac{1}{n}$$\sum \sum (Y_i²-2Y_iY_j+Y_j²)$

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:

$\sum Y_i²$ -2$\sum Y_iYbar$ + $\sum Ybar²$ = $\frac{1}{n}$$\sum Y_i² - \frac{2}{n}$$\sum \sum Y_iY_j$+$\sum \sum Y_j²$

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.

What you wrote is very hard to read. The problem you're having with formatting comes from mixing HTML tags (e.g., <sub>and <sup>) with LaTeX script. Use one or the other, but not both.

Here's your first equation, cleaned up:
$\frac{1}{n-1} \sum_{i=1}^n (Y_i - Ybar)^2 = \frac{1}{n(n-1)}\sum_{i=1}^n \sum_{j>i}^n (Y_i-Y_j)^2$</sup></sub>

 

[Ray Vickson]

What you wrote is very hard to read. The problem you're having with formatting comes from mixing HTML tags (e.g., <sub>and <sup>) with LaTeX script. Use one or the other, but not both.

Here's your first equation, cleaned up:
$\frac{1}{n-1} \sum_{i=1}^n (Y_i - Ybar)^2 = \frac{1}{n(n-1)}\sum_{i=1}^n \sum_{j>i}^n (Y_i-Y_j)^2$</sup></sub>

_{^{Even better:
$$\frac{1}{n-1} \sum_{i=1}^n (Y_i - \bar{Y})^2 = \frac{1}{n(n-1)}\sum_{i=1}^{n-1} \sum_{j=i+1}^n (Y_i-Y_j)^2$$
Note that in the second form, if we have j>i with j going up to n, then i can only go up to (n-1).}}

 

[TeenieBopper]

Ah, I see. I was just using the buttons at the top and side. Hopefully I'll get this right this time.

The original equation.

$$\frac{1}{n-1} \sum_{i=1}^n (Y_i - \bar{Y})^2 = \frac{1}{n(n-1)}\sum_{i=1}^{n-1} \sum_{j=i+1}^n (Y_i-Y_j)^2$$

Like I said, the (n-1) from both sides cancel, and then I expand both sides

$$\sum_{i=1}^n(Y_i^2 -2Y_i \bar{Y} + \bar{Y}^2 = \frac{1}{n}\sum_{i=1}^{n-1} \sum_{j=i+1}^n(Y_i^2 - 2Y_iY_j + 2Y_j^2)$$

I then distribute the summations to all the terms

$\sum_{i=1}^nY_i^2-2 \sum_{i=1}^nY_i\bar{Y} + \sum_{i=1}^n\bar{Y}^2 = \frac{1}{n}\sum_{i=1}^{n-1} \sum_{j=i+1}^nY_i^2 - \frac{2}{n}\sum_{i=1}^{n-1} \sum_{j=i+1}^n Y_iY_j + \frac{1}{n}\sum_{i=1}^{n-1} \sum_{j=i+1}^nY_j^2$

I don't really know where to go from here. I know that $\frac{1}{n}\sum_{i=1}^nY_i=\bar{Y}$, but I'm not sure what that does from me. I'll still have the j summations to worry about on the right hand side, as well as the Y_j terms.