[RESOLVED] Weighted Average

[oswaler]

OK, its been a long time since I had to do a weighted average and I want to make sure that I understand it conceptually.

If I have the following numbers:
1 2 3 4 5

And I want to assign them the following weights:

.1 .3 .2 .1 .3

Since the weights add to 1 I would just multiply each number by its weight and sum the products: 1(.1)+2(.3)...

I am doing this because when you find the weighted average you sum the products of each number and its weight and then divide by the sum of the weights, but in this case the weights add to 1 so you're just dividing by 1. I believe this is called a convex combination.


Now if I take the numbers 1 through 5 again, but assign more arbitrary weights like:

.3 .5 .7 .9 5

Then the average would be [1(.3)+2(.5)+3(.7)+4(.9)+5(5)]/(.3+.5+.7+.9+5)=4.32
so in this case taking [1(.3)+2(.5)+3(.7)+4(.9)+5(5)]/5 where the 5 in the divisor just represents the fact that there are 5 numbers in the original list that are being multiplied is mathematical garbage (which I can tell right away because this would give the average of 1 through 5 as being 6.4)


My understanding then is that every average is a weighted average, except that normally all numbers are treated equally so that their weights are all 1 and then you divide by the sum of the weights which is 1 for each number so the divisor ends up being the count of the numbers being averaged.


Could somebody tell me if anything I just said is wrong?

Thanks - Eric

[jemidiah]

OK, its been a long time since I had to do a weighted average and I want to make sure that I understand it conceptually.

If I have the following numbers:
1 2 3 4 5

And I want to assign them the following weights:

.1 .3 .2 .1 .3

Since the weights add to 1 I would just multiply each number by its weight and sum the products: 1(.1)+2(.3)...

Yes.

I am doing this because when you find the weighted average you sum the products of each number and its weight and then divide by the sum of the weights, but in this case the weights add to 1 so you're just dividing by 1. I believe this is called a convex combination.

Yes.

Now if I take the numbers 1 through 5 again, but assign more arbitrary weights like:

.3 .5 .7 .9 5

Then the average would be [1(.3)+2(.5)+3(.7)+4(.9)+5(5)]/(.3+.5+.7+.9+5)=4.32

Yes.

so in this case taking [1(.3)+2(.5)+3(.7)+4(.9)+5(5)]/5 where the 5 in the divisor just represents the fact that there are 5 numbers in the original list that are being multiplied is mathematical garbage (which I can tell right away because this would give the average of 1 through 5 as being 6.4)

Yes. Dividing by 5 is a leftover from using uniformly weighted averages and is, as you said, garbage.

My understanding then is that every average is a weighted average, except that normally all numbers are treated equally so that their weights are all 1 and then you divide by the sum of the weights which is 1 for each number so the divisor ends up being the count of the numbers being averaged.

Yes, this is one of several equivalent ways to think about it.

Could somebody tell me if anything I just said is wrong?

Yes.

[oswaler]

Thanks!