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Calculate weighted mean based on simple mean
The Next CEO of Stack OverflowDifferent approaches to calculate weighted meanshow to estimate weight for weighted product?Is there a way to generally transform a weighted arithmetic mean to a weighted geometric mean?What is the “best” way to calculate the average of individual averages?Average of fractions: Weighted sum of numerators over weighted sum of denominators?Exclude worst value from weighted arithmetic meanCorrect way to calculate mean values for timed dataUnderstanding the formalization of weighted means for matchingCalculate the winner of a match based on weighted parametersQuestion about an inequality about geometric mean
$begingroup$
I need to calculate the mobile weighted mean of a quantity, $a_i$, based on $b_i$.
I have a tool that allows me to calculate the mobile mean of a given series, so we can say I have a function:
$F(a_i) = frac1NSigma_i a_i $
Is it right to say that $ fracSigma_i a_i*b_iSigma_i b_i = fracF(a_i*b_i)F(b_i) (1)$ ?
Is there any condition on which this equality holds that I'm assuming implicitly? I can only think of $N ne 0$, is there anything else I'm missing?
EDIT:
To clarify the situation, I have a source of dynamic data from which I get $a_i$ and $b_i$ and I have a module which I can give these data to and that will calculate the mobile mean, taking care of excluding older samples. The function $F(a_i)$ is meant to represent the module.
Since I only get the simple mean from the module, and not the sum, my idea was to calculate the simple mean of $a_i*b_i$ and of $b_i$ and divide the two means to obtain the weighted mean, but before doing that I wanted to be sure about the conditions at which equation (1) holds.
average means
asked Mar 18 at 11:07
bracco23bracco23
1013
$endgroup$
add a comment |
$begingroup$
I need to calculate the mobile weighted mean of a quantity, $a_i$, based on $b_i$.
I have a tool that allows me to calculate the mobile mean of a given series, so we can say I have a function:
$F(a_i) = frac1NSigma_i a_i $
Is it right to say that $ fracSigma_i a_i*b_iSigma_i b_i = fracF(a_i*b_i)F(b_i) (1)$ ?
Is there any condition on which this equality holds that I'm assuming implicitly? I can only think of $N ne 0$, is there anything else I'm missing?
EDIT:
To clarify the situation, I have a source of dynamic data from which I get $a_i$ and $b_i$ and I have a module which I can give these data to and that will calculate the mobile mean, taking care of excluding older samples. The function $F(a_i)$ is meant to represent the module.
Since I only get the simple mean from the module, and not the sum, my idea was to calculate the simple mean of $a_i*b_i$ and of $b_i$ and divide the two means to obtain the weighted mean, but before doing that I wanted to be sure about the conditions at which equation (1) holds.
average means
asked Mar 18 at 11:07
bracco23bracco23
1013
$endgroup$
$begingroup$
I don't really follow what you mean by the equation that you wrote. If $F$ is an average operator, the number of entries just divides out and your equation is true. But what is this $b_i$? Are those the weights? If you have the average of $a_i$'s and you want to to calculate the weighted average (average of $a_i times b_i$), I don't think there's a way to do it if you don't have all the values of $a_i$'s ...
$endgroup$
– Matti P.
Mar 18 at 11:23
$begingroup$
Edited the question to better specify the situation, thanks for the comment @MattiP.
$endgroup$
– bracco23
Mar 18 at 11:32
add a comment |
$begingroup$
I need to calculate the mobile weighted mean of a quantity, $a_i$, based on $b_i$.
I have a tool that allows me to calculate the mobile mean of a given series, so we can say I have a function:
$F(a_i) = frac1NSigma_i a_i $
Is it right to say that $ fracSigma_i a_i*b_iSigma_i b_i = fracF(a_i*b_i)F(b_i) (1)$ ?
Is there any condition on which this equality holds that I'm assuming implicitly? I can only think of $N ne 0$, is there anything else I'm missing?
EDIT:
To clarify the situation, I have a source of dynamic data from which I get $a_i$ and $b_i$ and I have a module which I can give these data to and that will calculate the mobile mean, taking care of excluding older samples. The function $F(a_i)$ is meant to represent the module.
Since I only get the simple mean from the module, and not the sum, my idea was to calculate the simple mean of $a_i*b_i$ and of $b_i$ and divide the two means to obtain the weighted mean, but before doing that I wanted to be sure about the conditions at which equation (1) holds.
average means
asked Mar 18 at 11:07
bracco23bracco23
1013
$endgroup$
I need to calculate the mobile weighted mean of a quantity, $a_i$, based on $b_i$.
I have a tool that allows me to calculate the mobile mean of a given series, so we can say I have a function:
$F(a_i) = frac1NSigma_i a_i $
Is it right to say that $ fracSigma_i a_i*b_iSigma_i b_i = fracF(a_i*b_i)F(b_i) (1)$ ?
Is there any condition on which this equality holds that I'm assuming implicitly? I can only think of $N ne 0$, is there anything else I'm missing?
EDIT:
To clarify the situation, I have a source of dynamic data from which I get $a_i$ and $b_i$ and I have a module which I can give these data to and that will calculate the mobile mean, taking care of excluding older samples. The function $F(a_i)$ is meant to represent the module.
Since I only get the simple mean from the module, and not the sum, my idea was to calculate the simple mean of $a_i*b_i$ and of $b_i$ and divide the two means to obtain the weighted mean, but before doing that I wanted to be sure about the conditions at which equation (1) holds.
average means
average means
asked Mar 18 at 11:07
bracco23bracco23
1013
asked Mar 18 at 11:07
bracco23bracco23
1013
edited Mar 18 at 11:32
bracco23
asked Mar 18 at 11:07
bracco23bracco23
1013
asked Mar 18 at 11:07
bracco23bracco23
1013
asked Mar 18 at 11:07
bracco23bracco23
1013
1013
$begingroup$
I don't really follow what you mean by the equation that you wrote. If $F$ is an average operator, the number of entries just divides out and your equation is true. But what is this $b_i$? Are those the weights? If you have the average of $a_i$'s and you want to to calculate the weighted average (average of $a_i times b_i$), I don't think there's a way to do it if you don't have all the values of $a_i$'s ...
$endgroup$
– Matti P.
Mar 18 at 11:23
$begingroup$
Edited the question to better specify the situation, thanks for the comment @MattiP.
$endgroup$
– bracco23
Mar 18 at 11:32
add a comment |
$begingroup$
I don't really follow what you mean by the equation that you wrote. If $F$ is an average operator, the number of entries just divides out and your equation is true. But what is this $b_i$? Are those the weights? If you have the average of $a_i$'s and you want to to calculate the weighted average (average of $a_i times b_i$), I don't think there's a way to do it if you don't have all the values of $a_i$'s ...
$endgroup$
– Matti P.
Mar 18 at 11:23
$begingroup$
Edited the question to better specify the situation, thanks for the comment @MattiP.
$endgroup$
– bracco23
Mar 18 at 11:32
$begingroup$
I don't really follow what you mean by the equation that you wrote. If $F$ is an average operator, the number of entries just divides out and your equation is true. But what is this $b_i$? Are those the weights? If you have the average of $a_i$'s and you want to to calculate the weighted average (average of $a_i times b_i$), I don't think there's a way to do it if you don't have all the values of $a_i$'s ...
$endgroup$
– Matti P.
Mar 18 at 11:23
$begingroup$
I don't really follow what you mean by the equation that you wrote. If $F$ is an average operator, the number of entries just divides out and your equation is true. But what is this $b_i$? Are those the weights? If you have the average of $a_i$'s and you want to to calculate the weighted average (average of $a_i times b_i$), I don't think there's a way to do it if you don't have all the values of $a_i$'s ...
$endgroup$
– Matti P.
Mar 18 at 11:23
$begingroup$
Edited the question to better specify the situation, thanks for the comment @MattiP.
$endgroup$
– bracco23
Mar 18 at 11:32
$begingroup$
Edited the question to better specify the situation, thanks for the comment @MattiP.
$endgroup$
– bracco23
Mar 18 at 11:32
add a comment |
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$begingroup$
I don't really follow what you mean by the equation that you wrote. If $F$ is an average operator, the number of entries just divides out and your equation is true. But what is this $b_i$? Are those the weights? If you have the average of $a_i$'s and you want to to calculate the weighted average (average of $a_i times b_i$), I don't think there's a way to do it if you don't have all the values of $a_i$'s ...
$endgroup$
– Matti P.
Mar 18 at 11:23
$begingroup$
Edited the question to better specify the situation, thanks for the comment @MattiP.
$endgroup$
– bracco23
Mar 18 at 11:32