Statistics Sum of Squares x*y Proof

[laz0r]

Homework Statement

Prove that

\sum[(x_{i} - \overline{x})(y_{i} - \overline{y})] = \sum[(x_{i} - \overline{x})y_{i}]

Homework Equations

None.

The Attempt at a Solution

I tried using the fact that the sum of the mean values is just the mean value, because the sum of a constant is simply a constant, and I expanded out the first sum, but I didn't end up anywhere.

\sum[x_{i}y_{i}] - \overline{y}\sum(x_{i}) - \overline{x}\sum(y_{i}) + \overline(x)\overline(y)

I have no idea where to go from here, so some inspiration would be much appreciated.

 

[Ray Vickson]

Homework Statement

Prove that

\sum[(x_{i} - \overline{x})(y_{i} - \overline{y})] = \sum[(x_{i} - \overline{x})y_{i}]

Homework Equations

None.

The Attempt at a Solution

I tried using the fact that the sum of the mean values is just the mean value, because the sum of a constant is simply a constant, and I expanded out the first sum, but I didn't end up anywhere.

\sum[x_{i}y_{i}] - \overline{y}\sum(x_{i}) - \overline{x}\sum(y_{i}) + \overline(x)\overline(y)

I have no idea where to go from here, so some inspiration would be much appreciated.

I assume you have n values $x_1, \ldots, x_n$ and $y_1, \ldots,y_n$. In that case your last term above, $\bar{x} \bar{y}$ is incorrect. Can You see why? You can also do other simplifications, but if I say more I will essentially be doing the question for you.

 

[LCKurtz]

Homework Statement

Prove that

\sum[(x_{i} - \overline{x})(y_{i} - \overline{y})] = \sum[(x_{i} - \overline{x})y_{i}]

Homework Equations

None.

The Attempt at a Solution

I tried using the fact that the sum of the mean values is just the mean value, because the sum of a constant is simply a constant, and I expanded out the first sum, but I didn't end up anywhere.

\sum[x_{i}y_{i}] - \overline{y}\sum(x_{i}) - \overline{x}\sum(y_{i}) + \overline(x)\overline(y)

I have no idea where to go from here, so some inspiration would be much appreciated.

Notice that your first and third terms give what is on the right side of the equation. So you are left with showing$$
-\sum x_i\bar y + \sum \bar y \bar x = 0$$You're pretty close...

 

[laz0r]

Ok, so I actually did end up getting somewhere and didn't realize it =)

EDIT:

I've done proof

\sum[(x_{i}-\overline{x})y_{i}] + \sum[\overline{y}(\overline{x} - x_{i})]

Playing around in excel has taught me that \sum[\overline{y}(\overline{x} - x_{i})] is actually just equal to zero, because your subtracting an array of x values from the mean and then just multiplying them by a constant, \overline{y}, then adding them, which nets you zero.

 

[Ray Vickson]

Ok, so I actually did end up getting somewhere and didn't realize it =)

I've done most of the proof and I've gotten up until this point

\sum[(x_{i}-\overline{x})y_{i}] + \sum[\overline{y}(\overline{x} - x_{i})]

So I take it that either \overline{y} or (\overline{x} - x_{i}) must equal zero, but I'm not entirely sure why.

Try writing out a few small examples, such as for n = 2 or n = 3. These are small enough that you can write down everything explicitly and see exactly what is going on.

 

[laz0r]

I've figured out why, but I'm not sure how to explain it symbolically, do you think I would need to elaborate more or is my edited explanation good enough in your view?

Thanks for the help!

 

[LCKurtz]

$\bar x$ and $\sum x_i$ are related to each other. How?

 

[laz0r]

$\bar x$ and $\sum x_i$ are related to each other. How?

I'm a little bit rusty on my operation of sums, so excuse me if this is incorrect, but can I do this?

<br /> <br /> [(\overline{y})/n][\sum x_{i}] - \overline{y}[\sum x_{i}]<br /> <br /> = [(\overline{y}^2)/n][\sum(x_{i} - x_{i})]<br /> <br />

Then what tends to zero appears to be obvious

 

[LCKurtz]

$\bar x$ and $\sum x_i$ are related to each other. How?

I'm a little bit rusty on my operation of sums, so excuse me if this is incorrect, but can I do this?

<br /> <br /> [(\overline{y})/n][\sum x_{i}] - \overline{y}[\sum x_{i}]<br /> <br /> = [(\overline{y}^2)/n][\sum(x_{i} - x_{i})]<br /> <br />

Then what tends to zero appears to be obvious

I don't follow that at all. Why don't you just answer my question at the top?

 

[laz0r]

I don't follow that at all. Why don't you just answer my question at the top?

They're related by

<br /> <br /> \overline{x} = [\sum x_{i}]/n<br /> <br />

 

[LCKurtz]

They're related by

<br /> <br /> \overline{x} = [\sum x_{i}]/n<br /> <br />

So you could use $\sum x_i = n\bar x$. Do you see how you might use that to get$$
\sum \bar y(\bar x - x_i) = 0\text{?}$$

 

[laz0r]

So you could use $\sum x_i = n\bar x$. Do you see how you might use that to get$$
\sum \bar y(\bar x - x_i) = 0\text{?}$$

Wow, it appears my brain took a nap today.

Thank you for your help lol, I appreciate it.

 

[LCKurtz]

So amuse me and show the finishing steps...

 

[laz0r]

Sorry, I'm very slow at typing this latex code..

<br /> \overline{y}*[\sum(\overline{x} - x_{i})] = 0<br />

<br /> [\sum(\overline{x} - x_{i})] = 0<br />

<br /> \sum(\overline{x}) - \sum(x_{i}) = 0<br />

<br /> n(\overline{x}) - n(\overline{x}) = 0<br />

LHS = RHS

used

<br /> \sum(c) = n*c<br />

Where n is the upper bound and 1 is the lower bound of the sum

 

[LCKurtz]

Good. Thanks for finishing it up.