Minimizing the sum of absolute difference of each point from the average does it minimize the variance?
$begingroup$
I have a problem in which there is a set of $N$ points. Each point has a set of possible weight (lets call them $W_i$ where $i in [1,N])$. My goal is to select the correct weight $W_i$ for each point $i$ so that the variance of the weights is minimized.
If I find the selection of weights $W_i$ so that the function $f=sum_{i=1}^{N} mid W_i - mu mid$, where $mu$ is the mean value of the $W_i$, is minimized, does that $ Rightarrow $ that the variation of the $W_i$ is also minimized?
Is there a way to formally prove it?
Sorry for my bad terminology and use of English.
Any help will be greatly appreciated :)
statistics variance
asked Jan 10 at 11:14
Dimitris StathisDimitris Stathis
1
$endgroup$
add a comment |
$begingroup$
I have a problem in which there is a set of $N$ points. Each point has a set of possible weight (lets call them $W_i$ where $i in [1,N])$. My goal is to select the correct weight $W_i$ for each point $i$ so that the variance of the weights is minimized.
If I find the selection of weights $W_i$ so that the function $f=sum_{i=1}^{N} mid W_i - mu mid$, where $mu$ is the mean value of the $W_i$, is minimized, does that $ Rightarrow $ that the variation of the $W_i$ is also minimized?
Is there a way to formally prove it?
Sorry for my bad terminology and use of English.
Any help will be greatly appreciated :)
statistics variance
asked Jan 10 at 11:14
Dimitris StathisDimitris Stathis
1
$endgroup$
$begingroup$
I'm not sure what's preventing you from simply setting $W_i=1, forall i$. This gives variance $0$ and your $f$ is also $0$. But generally, the variance (which uses the square of differences) is not minimized at the same time as your function (which uses just the absolute value).
$endgroup$
– Ingix
Jan 10 at 11:40
$begingroup$
The set of values that each $W_i$ can take is pre- decided and can not change. Also each $W_i$ has a different set of possible values. My simple thinking was that since the variance is the average of the square then if you minimize the sum of the absolute values then that means that you minimize the sum of the squares as well => to minimize the mean. I am not an expert in statics though and I am afraid I might be missing something.
$endgroup$
– Dimitris Stathis
Jan 11 at 16:10
add a comment |
$begingroup$
I have a problem in which there is a set of $N$ points. Each point has a set of possible weight (lets call them $W_i$ where $i in [1,N])$. My goal is to select the correct weight $W_i$ for each point $i$ so that the variance of the weights is minimized.
If I find the selection of weights $W_i$ so that the function $f=sum_{i=1}^{N} mid W_i - mu mid$, where $mu$ is the mean value of the $W_i$, is minimized, does that $ Rightarrow $ that the variation of the $W_i$ is also minimized?
Is there a way to formally prove it?
Sorry for my bad terminology and use of English.
Any help will be greatly appreciated :)
statistics variance
asked Jan 10 at 11:14
Dimitris StathisDimitris Stathis
1
$endgroup$
I have a problem in which there is a set of $N$ points. Each point has a set of possible weight (lets call them $W_i$ where $i in [1,N])$. My goal is to select the correct weight $W_i$ for each point $i$ so that the variance of the weights is minimized.
If I find the selection of weights $W_i$ so that the function $f=sum_{i=1}^{N} mid W_i - mu mid$, where $mu$ is the mean value of the $W_i$, is minimized, does that $ Rightarrow $ that the variation of the $W_i$ is also minimized?
Is there a way to formally prove it?
Sorry for my bad terminology and use of English.
Any help will be greatly appreciated :)
statistics variance
statistics variance
asked Jan 10 at 11:14
Dimitris StathisDimitris Stathis
1
asked Jan 10 at 11:14
Dimitris StathisDimitris Stathis
1
asked Jan 10 at 11:14
Dimitris StathisDimitris Stathis
1
asked Jan 10 at 11:14
Dimitris StathisDimitris Stathis
1
asked Jan 10 at 11:14
Dimitris StathisDimitris Stathis
1
1
$begingroup$
I'm not sure what's preventing you from simply setting $W_i=1, forall i$. This gives variance $0$ and your $f$ is also $0$. But generally, the variance (which uses the square of differences) is not minimized at the same time as your function (which uses just the absolute value).
$endgroup$
– Ingix
Jan 10 at 11:40
$begingroup$
The set of values that each $W_i$ can take is pre- decided and can not change. Also each $W_i$ has a different set of possible values. My simple thinking was that since the variance is the average of the square then if you minimize the sum of the absolute values then that means that you minimize the sum of the squares as well => to minimize the mean. I am not an expert in statics though and I am afraid I might be missing something.
$endgroup$
– Dimitris Stathis
Jan 11 at 16:10
add a comment |
$begingroup$
I'm not sure what's preventing you from simply setting $W_i=1, forall i$. This gives variance $0$ and your $f$ is also $0$. But generally, the variance (which uses the square of differences) is not minimized at the same time as your function (which uses just the absolute value).
$endgroup$
– Ingix
Jan 10 at 11:40
$begingroup$
The set of values that each $W_i$ can take is pre- decided and can not change. Also each $W_i$ has a different set of possible values. My simple thinking was that since the variance is the average of the square then if you minimize the sum of the absolute values then that means that you minimize the sum of the squares as well => to minimize the mean. I am not an expert in statics though and I am afraid I might be missing something.
$endgroup$
– Dimitris Stathis
Jan 11 at 16:10
$begingroup$
I'm not sure what's preventing you from simply setting $W_i=1, forall i$. This gives variance $0$ and your $f$ is also $0$. But generally, the variance (which uses the square of differences) is not minimized at the same time as your function (which uses just the absolute value).
$endgroup$
– Ingix
Jan 10 at 11:40
$begingroup$
I'm not sure what's preventing you from simply setting $W_i=1, forall i$. This gives variance $0$ and your $f$ is also $0$. But generally, the variance (which uses the square of differences) is not minimized at the same time as your function (which uses just the absolute value).
$endgroup$
– Ingix
Jan 10 at 11:40
$begingroup$
The set of values that each $W_i$ can take is pre- decided and can not change. Also each $W_i$ has a different set of possible values. My simple thinking was that since the variance is the average of the square then if you minimize the sum of the absolute values then that means that you minimize the sum of the squares as well => to minimize the mean. I am not an expert in statics though and I am afraid I might be missing something.
$endgroup$
– Dimitris Stathis
Jan 11 at 16:10
$begingroup$
The set of values that each $W_i$ can take is pre- decided and can not change. Also each $W_i$ has a different set of possible values. My simple thinking was that since the variance is the average of the square then if you minimize the sum of the absolute values then that means that you minimize the sum of the squares as well => to minimize the mean. I am not an expert in statics though and I am afraid I might be missing something.
$endgroup$
– Dimitris Stathis
Jan 11 at 16:10
add a comment |
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$begingroup$
I'm not sure what's preventing you from simply setting $W_i=1, forall i$. This gives variance $0$ and your $f$ is also $0$. But generally, the variance (which uses the square of differences) is not minimized at the same time as your function (which uses just the absolute value).
$endgroup$
– Ingix
Jan 10 at 11:40
$begingroup$
The set of values that each $W_i$ can take is pre- decided and can not change. Also each $W_i$ has a different set of possible values. My simple thinking was that since the variance is the average of the square then if you minimize the sum of the absolute values then that means that you minimize the sum of the squares as well => to minimize the mean. I am not an expert in statics though and I am afraid I might be missing something.
$endgroup$
– Dimitris Stathis
Jan 11 at 16:10