Summation Operator

[Mathcatchup]

Not sure how we get there.

 

[mmm4444bot]

Not sure how we get there.

Where is "there"?

It seems like maybe you've only posted the second half of your question. That is, I'm thinking you have some sort of mathematical statement, and you were told that it can be rewritten. The image you posted looks like a justification for why you're allowed to rewrite the statement.

Are you asking why this justification validates the specific rewrite, or are you simply asking why the posted equation is true? If it's the former, we need to see the statement that is being rewritten.

From where does this question come? Are you in a math class? :cool:

 

[Dr.Peterson]

View attachment 9842
Not sure how we get there.

It would be helpful if you showed the context, but I think you are just asking how to get from the left side to the right, which isn't hard.

Are you aware that $\displaystyle \sum (x_i-\bar{x})=0$?

This is, in a sense, the definition of the mean, and also is how I check my mean when I am working out a standard deviation by hand.

Try applying this to the summation you are asking about; the distributive property will be handy, along with the fact that $\displaystyle \bar{y}$ is a constant.

 

[Mathcatchup]

It’s from an econometrics class. My math is quite rusty...

My question is how we get to the right side of the equation.

So are you saying that the equation holds just because both sides are equal to zero anyway? I am trying to expand the left side and get to the right but I can’t get there.

 

[Dr.Peterson]

It’s from an econometrics class. My math is quite rusty...

My question is how we get to the right side of the equation.

So are you saying that the equation holds just because both sides are equal to zero anyway? I am trying to expand the left side and get to the right but I can’t get there.

No, both sides are not equal to zero. That wouldn't be very useful.

Break up the summand by distributing, but only half-way: keep the first factor together, but multiply it by each term in the second. Since y-bar is just a number, you can pull it outside the sum (that is, factor it out).

You'll find the sum that is equal to zero in there, and replacing it with zero will give the desired result.

 

[JeffM]

I suspect a formal proof would be a bit lengthy and might require weak mathematical induction. But it is easy to show the basic idea when n = 2.

$\displaystyle \bar x = \dfrac{x_1 + x_2}{2}\implies 2 \bar x = (x_1 + x_2).$

$\displaystyle \displaystyle \left ( \sum_{j=1}^2(x_j - \bar x)(y_j - \bar y) \right ) =\left ( \sum_{j=1}^2 x_jy_j - x_j \bar y - \bar x y_j + \bar x \bar y \right ) =$

$\displaystyle (x_1y_1 - x_1 \bar y - \bar x y_1 + \bar x \bar y) + (x_2y_2 - x_2 \bar y - \bar x y_2 + \bar x \bar y) =$

$\displaystyle \{(x_1y_1 - \bar x y_1) + (x_2y_2 - \bar x y_2)\} - \bar y (x_1 + x_2) + 2 \bar x \bar y =$

$\displaystyle \displaystyle \left ( \sum_{j=1}^2 x_jy_j - \bar x y_j \right) - \bar y (2 \bar x) + 2 \bar x \bar y =$

$\displaystyle \displaystyle \left ( \sum_{j=1}^2 (x_j - \bar x)y_j \right ) - 2 \bar x \bar y + 2 \bar x \bar y = \left ( \sum_{j=1}^2(x_j - \bar x)y_j \right ).$

 

[Dr.Peterson]

I suspect a formal proof would be a bit lengthy and might require weak mathematical induction. But it is easy to show the basic idea when n = 2.

$\displaystyle \bar x = \dfrac{x_1 + x_2}{2}\implies 2 \bar x = (x_1 + x_2).$

$\displaystyle \displaystyle \left ( \sum_{j=1}^2(x_j - \bar x)(y_j - \bar y) \right ) =\left ( \sum_{j=1}^2 x_jy_j - x_j \bar y - \bar x y_j + \bar x \bar y \right ) =$

$\displaystyle (x_1y_1 - x_1 \bar y - \bar x y_1 + \bar x \bar y) + (x_2y_2 - x_2 \bar y - \bar x y_2 + \bar x \bar y) =$

$\displaystyle \{(x_1y_1 - \bar x y_1) + (x_2y_2 - \bar x y_2)\} - \bar y (x_1 + x_2) + 2 \bar x \bar y =$

$\displaystyle \displaystyle \left ( \sum_{j=1}^2 x_jy_j - \bar x y_j \right) - \bar y (2 \bar x) + 2 \bar x \bar y =$

$\displaystyle \displaystyle \left ( \sum_{j=1}^2 (x_j - \bar x)y_j \right ) - 2 \bar x \bar y + 2 \bar x \bar y = \left ( \sum_{j=1}^2(x_j - \bar x)y_j \right ).$

It's a lot easier than that if you have theorems for manipulating summations.

$\displaystyle \displaystyle \sum(x_i - \bar x)(y_i - \bar y) = \sum\left( (x_i - \bar x)y_i + (x_i - \bar x)\bar y \right) = \sum (x_i - \bar x)y_i + \sum (x_i - \bar x)\bar y =$
$\displaystyle \displaystyle \sum (x_i - \bar x)y_i + \bar y \sum(x_i - \bar x) = \sum (x_i - \bar x)y_i + 0 = \sum(x_i - \bar x)y_i$

 

[JeffM]

It's a lot easier than that if you have theorems for manipulating summations.

$\displaystyle \displaystyle \sum(x_i - \bar x)(y_i - \bar y) = \sum\left( (x_i - \bar x)y_i + (x_i - \bar x)\bar y \right) = \sum (x_i - \bar x)y_i + \sum (x_i - \bar x)\bar y =$
$\displaystyle \displaystyle \sum (x_i - \bar x)y_i + \bar y \sum(x_i - \bar x) = \sum (x_i - \bar x)y_i + 0 = \sum(x_i - \bar x)y_i$

Yes. I thought about it some more and came back to edit and found your response. I cannot imagine now why I thought induction would be necessary because, as shown below, my approach generalizes. Of course, your approach is more elegant and avoids induction as well.

$\displaystyle \bar x = \dfrac{\displaystyle \sum_{j=1}^n x_j}{n} \implies n \bar x = \displaystyle \sum_{j=1}^n x_j.$

$\displaystyle \displaystyle \therefore \left ( \sum_{j=1}^n (x_j - \bar x)(y_j - \bar y) \right ) =$

$\displaystyle \displaystyle \left ( \sum_{j=1}^n x_jy_j - x_j \bar y - \bar x y_j + \bar x \bar y) \right ) = \left ( \sum_{j=1}^n x_jy_j - \bar x y_j - x_j \bar y + \bar x \bar y) \right ) =$

$\displaystyle \displaystyle \left ( \sum_{j=1}^n x_jy_j - \bar x y_j \right ) - \left ( \sum_{j=1}^n x_ j \bar y \right ) + \left ( \sum_{j=1}^n \bar x \bar y) \right ) =$

$\displaystyle \displaystyle \left ( \sum_{j=1}^n x_jy_j - \bar x y_j \right ) - \left \{ \bar y * \left ( \sum_{j=1}^n x_ j \right ) \right \} + n \bar x \bar y =$

$\displaystyle \displaystyle \left ( \sum_{j=1}^n x_jy_j - \bar x y_j \right ) - \{ \bar y * (n \bar x)\} + n \bar x \bar y =$

$\displaystyle \displaystyle \left ( \sum_{j=1}^n x_jy_j - \bar x y_j \right ) - n \bar x \bar y + n \bar x \bar y =$

$\displaystyle \displaystyle \left ( \sum_{j=1}^n x_jy_j - \bar x y_j \right ) =$

$\displaystyle \displaystyle \left ( \sum_{j=1}^n (x_j - \bar x )y_j \right ).$

No induction needed. The n = 2 case generalizes directly, and the only theorem on sums needed is that relating to the sum of a constant.

 

[Mathcatchup]

Thanks everyone, I get it now.