Minimizing the sum of absolute difference of each point from the average does it minimize the variance?












0












$begingroup$


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 :)










share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












$begingroup$


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 :)










share|cite|improve this question









$endgroup$












  • $begingroup$
    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














0












0








0





$begingroup$


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 :)










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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


















  • $begingroup$
    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










0






active

oldest

votes











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068512%2fminimizing-the-sum-of-absolute-difference-of-each-point-from-the-average-does-it%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























0






active

oldest

votes








0






active

oldest

votes









active

oldest

votes






active

oldest

votes
















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3068512%2fminimizing-the-sum-of-absolute-difference-of-each-point-from-the-average-does-it%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

SQL update select statement

WPF add header to Image with URL pettitions [duplicate]