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What happens if there is no difference between the mean and the weighted mean?
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I was wondering. What happens if there is no difference between the arithmetic mean and the weighted mean? Why would that happen?
means
asked Oct 3 '17 at 22:07
Nathan FleischmanNathan Fleischman
32
$endgroup$
add a comment |
$begingroup$
I was wondering. What happens if there is no difference between the arithmetic mean and the weighted mean? Why would that happen?
means
asked Oct 3 '17 at 22:07
Nathan FleischmanNathan Fleischman
32
$endgroup$
$begingroup$
Weighted means can be different things in different contexts. Your question doesn't provide enough details for us to give any kind of answer.
$endgroup$
– Jam
Oct 3 '17 at 22:10
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@Jam I was thinking I could until I read your comment. however I was thinking the arithmetic mean won't work for my idea if it's a geometric mean or a harmonic mean I bet.
$endgroup$
– user451844
Oct 3 '17 at 22:12
$begingroup$
Are you wondering about this then $n^{-1}(x_1+dotso+x_n)-n^{-1}(w_1 x_1 +dotso +w_n x_n)=0$? Why not explore?
$endgroup$
– LoveTooNap29
Oct 3 '17 at 22:13
add a comment |
$begingroup$
I was wondering. What happens if there is no difference between the arithmetic mean and the weighted mean? Why would that happen?
means
asked Oct 3 '17 at 22:07
Nathan FleischmanNathan Fleischman
32
$endgroup$
I was wondering. What happens if there is no difference between the arithmetic mean and the weighted mean? Why would that happen?
means
means
asked Oct 3 '17 at 22:07
Nathan FleischmanNathan Fleischman
32
asked Oct 3 '17 at 22:07
Nathan FleischmanNathan Fleischman
32
edited Jan 7 at 2:06
Nathan Fleischman
asked Oct 3 '17 at 22:07
Nathan FleischmanNathan Fleischman
32
asked Oct 3 '17 at 22:07
Nathan FleischmanNathan Fleischman
32
asked Oct 3 '17 at 22:07
Nathan FleischmanNathan Fleischman
32
32
$begingroup$
Weighted means can be different things in different contexts. Your question doesn't provide enough details for us to give any kind of answer.
$endgroup$
– Jam
Oct 3 '17 at 22:10
$begingroup$
@Jam I was thinking I could until I read your comment. however I was thinking the arithmetic mean won't work for my idea if it's a geometric mean or a harmonic mean I bet.
$endgroup$
– user451844
Oct 3 '17 at 22:12
$begingroup$
Are you wondering about this then $n^{-1}(x_1+dotso+x_n)-n^{-1}(w_1 x_1 +dotso +w_n x_n)=0$? Why not explore?
$endgroup$
– LoveTooNap29
Oct 3 '17 at 22:13
add a comment |
$begingroup$
Weighted means can be different things in different contexts. Your question doesn't provide enough details for us to give any kind of answer.
$endgroup$
– Jam
Oct 3 '17 at 22:10
$begingroup$
@Jam I was thinking I could until I read your comment. however I was thinking the arithmetic mean won't work for my idea if it's a geometric mean or a harmonic mean I bet.
$endgroup$
– user451844
Oct 3 '17 at 22:12
$begingroup$
Are you wondering about this then $n^{-1}(x_1+dotso+x_n)-n^{-1}(w_1 x_1 +dotso +w_n x_n)=0$? Why not explore?
$endgroup$
– LoveTooNap29
Oct 3 '17 at 22:13
$begingroup$
Weighted means can be different things in different contexts. Your question doesn't provide enough details for us to give any kind of answer.
$endgroup$
– Jam
Oct 3 '17 at 22:10
$begingroup$
Weighted means can be different things in different contexts. Your question doesn't provide enough details for us to give any kind of answer.
$endgroup$
– Jam
Oct 3 '17 at 22:10
$begingroup$
@Jam I was thinking I could until I read your comment. however I was thinking the arithmetic mean won't work for my idea if it's a geometric mean or a harmonic mean I bet.
$endgroup$
– user451844
Oct 3 '17 at 22:12
$begingroup$
@Jam I was thinking I could until I read your comment. however I was thinking the arithmetic mean won't work for my idea if it's a geometric mean or a harmonic mean I bet.
$endgroup$
– user451844
Oct 3 '17 at 22:12
$begingroup$
Are you wondering about this then $n^{-1}(x_1+dotso+x_n)-n^{-1}(w_1 x_1 +dotso +w_n x_n)=0$? Why not explore?
$endgroup$
– LoveTooNap29
Oct 3 '17 at 22:13
$begingroup$
Are you wondering about this then $n^{-1}(x_1+dotso+x_n)-n^{-1}(w_1 x_1 +dotso +w_n x_n)=0$? Why not explore?
$endgroup$
– LoveTooNap29
Oct 3 '17 at 22:13
add a comment |
1 Answer
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$begingroup$
Let's look at a simple case, where you only have 2 values $x_1$ and $x_2$ with weights $w_1$ and $w_2$. So the unweighted mean is $frac{1}{2}(x_1 + x_2)$, and the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2)$.
So why would these be equal? Well, there are a lot of values involved, so it could just be a lucky coincidence - in fact, if you fix any three of the values, there is (usually) a value for the fourth that will make everything equal.
However, there are a few special cases that you can see:
The weights are equal. If $w_1 = w_2$ then the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2) = frac{1}{w_1 + w_1}(w_1 x_1 + w_1 x_2) = frac{1}{2w_1} times w_1 (x_1 + x_2) = frac{1}{2}(x_1 + x_2)$, which is the unweighted mean.
The sample values are equal. If $x_1 = x_2$ then the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2) = frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_1) = frac{w_1 + w_2}{w_1 + w_2} x_1 = x_1$, and it's pretty easy to show that the unweighted mean has the same value.
You can show that these two cases are true for any number of values, but I'll leave the working to you.
answered Oct 3 '17 at 22:48
ConManConMan
7,7021324
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1 Answer
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1 Answer
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$begingroup$
Let's look at a simple case, where you only have 2 values $x_1$ and $x_2$ with weights $w_1$ and $w_2$. So the unweighted mean is $frac{1}{2}(x_1 + x_2)$, and the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2)$.
So why would these be equal? Well, there are a lot of values involved, so it could just be a lucky coincidence - in fact, if you fix any three of the values, there is (usually) a value for the fourth that will make everything equal.
However, there are a few special cases that you can see:
The weights are equal. If $w_1 = w_2$ then the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2) = frac{1}{w_1 + w_1}(w_1 x_1 + w_1 x_2) = frac{1}{2w_1} times w_1 (x_1 + x_2) = frac{1}{2}(x_1 + x_2)$, which is the unweighted mean.
The sample values are equal. If $x_1 = x_2$ then the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2) = frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_1) = frac{w_1 + w_2}{w_1 + w_2} x_1 = x_1$, and it's pretty easy to show that the unweighted mean has the same value.
You can show that these two cases are true for any number of values, but I'll leave the working to you.
answered Oct 3 '17 at 22:48
ConManConMan
7,7021324
$endgroup$
add a comment |
$begingroup$
Let's look at a simple case, where you only have 2 values $x_1$ and $x_2$ with weights $w_1$ and $w_2$. So the unweighted mean is $frac{1}{2}(x_1 + x_2)$, and the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2)$.
So why would these be equal? Well, there are a lot of values involved, so it could just be a lucky coincidence - in fact, if you fix any three of the values, there is (usually) a value for the fourth that will make everything equal.
However, there are a few special cases that you can see:
The weights are equal. If $w_1 = w_2$ then the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2) = frac{1}{w_1 + w_1}(w_1 x_1 + w_1 x_2) = frac{1}{2w_1} times w_1 (x_1 + x_2) = frac{1}{2}(x_1 + x_2)$, which is the unweighted mean.
The sample values are equal. If $x_1 = x_2$ then the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2) = frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_1) = frac{w_1 + w_2}{w_1 + w_2} x_1 = x_1$, and it's pretty easy to show that the unweighted mean has the same value.
You can show that these two cases are true for any number of values, but I'll leave the working to you.
answered Oct 3 '17 at 22:48
ConManConMan
7,7021324
$endgroup$
add a comment |
$begingroup$
Let's look at a simple case, where you only have 2 values $x_1$ and $x_2$ with weights $w_1$ and $w_2$. So the unweighted mean is $frac{1}{2}(x_1 + x_2)$, and the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2)$.
So why would these be equal? Well, there are a lot of values involved, so it could just be a lucky coincidence - in fact, if you fix any three of the values, there is (usually) a value for the fourth that will make everything equal.
However, there are a few special cases that you can see:
The weights are equal. If $w_1 = w_2$ then the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2) = frac{1}{w_1 + w_1}(w_1 x_1 + w_1 x_2) = frac{1}{2w_1} times w_1 (x_1 + x_2) = frac{1}{2}(x_1 + x_2)$, which is the unweighted mean.
The sample values are equal. If $x_1 = x_2$ then the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2) = frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_1) = frac{w_1 + w_2}{w_1 + w_2} x_1 = x_1$, and it's pretty easy to show that the unweighted mean has the same value.
You can show that these two cases are true for any number of values, but I'll leave the working to you.
answered Oct 3 '17 at 22:48
ConManConMan
7,7021324
$endgroup$
Let's look at a simple case, where you only have 2 values $x_1$ and $x_2$ with weights $w_1$ and $w_2$. So the unweighted mean is $frac{1}{2}(x_1 + x_2)$, and the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2)$.
So why would these be equal? Well, there are a lot of values involved, so it could just be a lucky coincidence - in fact, if you fix any three of the values, there is (usually) a value for the fourth that will make everything equal.
However, there are a few special cases that you can see:
The weights are equal. If $w_1 = w_2$ then the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2) = frac{1}{w_1 + w_1}(w_1 x_1 + w_1 x_2) = frac{1}{2w_1} times w_1 (x_1 + x_2) = frac{1}{2}(x_1 + x_2)$, which is the unweighted mean.
The sample values are equal. If $x_1 = x_2$ then the weighted mean is $frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_2) = frac{1}{w_1 + w_2}(w_1 x_1 + w_2 x_1) = frac{w_1 + w_2}{w_1 + w_2} x_1 = x_1$, and it's pretty easy to show that the unweighted mean has the same value.
You can show that these two cases are true for any number of values, but I'll leave the working to you.
answered Oct 3 '17 at 22:48
ConManConMan
7,7021324
answered Oct 3 '17 at 22:48
ConManConMan
7,7021324
answered Oct 3 '17 at 22:48
ConManConMan
7,7021324
answered Oct 3 '17 at 22:48
ConManConMan
7,7021324
7,7021324
add a comment |
add a comment |
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$begingroup$
Weighted means can be different things in different contexts. Your question doesn't provide enough details for us to give any kind of answer.
$endgroup$
– Jam
Oct 3 '17 at 22:10
$begingroup$
@Jam I was thinking I could until I read your comment. however I was thinking the arithmetic mean won't work for my idea if it's a geometric mean or a harmonic mean I bet.
$endgroup$
– user451844
Oct 3 '17 at 22:12
$begingroup$
Are you wondering about this then $n^{-1}(x_1+dotso+x_n)-n^{-1}(w_1 x_1 +dotso +w_n x_n)=0$? Why not explore?
$endgroup$
– LoveTooNap29
Oct 3 '17 at 22:13