Mixing
k-means cost functions comparison
$begingroup$
$J_{avg^2} = sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2$ and $J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$, where $m_j = frac{1}{C_j}sum_{xin C_j}x$. Now I want to show that $J_{IC} = 2J_{avg^2}$.
I have so far,
$J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$
$sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}sum_{i = 1}^{d}(x_i - y_i)^2 = $
$sum_{j=1}^{k}sum_{xin C_j}frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - 2x_iy_i + y_i^2 = $
$sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - frac{2}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_iy_i + frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{i = 1}^{d}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}sum_{i = 1}^{d}(x_isum_{y in c_j}y_i) + frac{1}{|C_j|}sum_{i = 1}^{d}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{1}{|C_j|}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}x_isum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{|C_j|}{|C_j|}x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg]$
Then I have,
$2J_{avg^2} = 2sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2 = 2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}(x_i - frac{1}{|C_j|}sum_{yin C_j}y_i)^2 = $
$2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}x_i^2 - frac{2x_i}{|C_j|}sum_{yin C_j}y_i + frac{1}{|C_j|^2}big(sum_{yin C_j}y_ibig)^2$
I am lost right now on how to proceed and would be very thankful if someone could help point out any mistakes or provide any guidance on what approach to take so that $J_{IC} = 2J_{avg^2}$.
summation machine-learning clustering
asked Jan 21 at 21:08
KayandaKayanda
12
$endgroup$
add a comment |
$begingroup$
$J_{avg^2} = sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2$ and $J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$, where $m_j = frac{1}{C_j}sum_{xin C_j}x$. Now I want to show that $J_{IC} = 2J_{avg^2}$.
I have so far,
$J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$
$sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}sum_{i = 1}^{d}(x_i - y_i)^2 = $
$sum_{j=1}^{k}sum_{xin C_j}frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - 2x_iy_i + y_i^2 = $
$sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - frac{2}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_iy_i + frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{i = 1}^{d}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}sum_{i = 1}^{d}(x_isum_{y in c_j}y_i) + frac{1}{|C_j|}sum_{i = 1}^{d}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{1}{|C_j|}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}x_isum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{|C_j|}{|C_j|}x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg]$
Then I have,
$2J_{avg^2} = 2sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2 = 2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}(x_i - frac{1}{|C_j|}sum_{yin C_j}y_i)^2 = $
$2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}x_i^2 - frac{2x_i}{|C_j|}sum_{yin C_j}y_i + frac{1}{|C_j|^2}big(sum_{yin C_j}y_ibig)^2$
I am lost right now on how to proceed and would be very thankful if someone could help point out any mistakes or provide any guidance on what approach to take so that $J_{IC} = 2J_{avg^2}$.
summation machine-learning clustering
asked Jan 21 at 21:08
KayandaKayanda
12
$endgroup$
add a comment |
$begingroup$
$J_{avg^2} = sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2$ and $J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$, where $m_j = frac{1}{C_j}sum_{xin C_j}x$. Now I want to show that $J_{IC} = 2J_{avg^2}$.
I have so far,
$J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$
$sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}sum_{i = 1}^{d}(x_i - y_i)^2 = $
$sum_{j=1}^{k}sum_{xin C_j}frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - 2x_iy_i + y_i^2 = $
$sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - frac{2}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_iy_i + frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{i = 1}^{d}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}sum_{i = 1}^{d}(x_isum_{y in c_j}y_i) + frac{1}{|C_j|}sum_{i = 1}^{d}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{1}{|C_j|}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}x_isum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{|C_j|}{|C_j|}x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg]$
Then I have,
$2J_{avg^2} = 2sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2 = 2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}(x_i - frac{1}{|C_j|}sum_{yin C_j}y_i)^2 = $
$2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}x_i^2 - frac{2x_i}{|C_j|}sum_{yin C_j}y_i + frac{1}{|C_j|^2}big(sum_{yin C_j}y_ibig)^2$
I am lost right now on how to proceed and would be very thankful if someone could help point out any mistakes or provide any guidance on what approach to take so that $J_{IC} = 2J_{avg^2}$.
summation machine-learning clustering
asked Jan 21 at 21:08
KayandaKayanda
12
$endgroup$
$J_{avg^2} = sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2$ and $J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$, where $m_j = frac{1}{C_j}sum_{xin C_j}x$. Now I want to show that $J_{IC} = 2J_{avg^2}$.
I have so far,
$J_{IC} = sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}d(x,y)^2$
$sum_{j=1}^{k}frac{1}{|C_j|}sum_{xin C_j}sum_{y in c_j}sum_{i = 1}^{d}(x_i - y_i)^2 = $
$sum_{j=1}^{k}sum_{xin C_j}frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - 2x_iy_i + y_i^2 = $
$sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_i^2 - frac{2}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}x_iy_i + frac{1}{|C_j|}sum_{y in c_j}sum_{i = 1}^{d}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}bigg[frac{1}{|C_j|}sum_{i = 1}^{d}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}sum_{i = 1}^{d}(x_isum_{y in c_j}y_i) + frac{1}{|C_j|}sum_{i = 1}^{d}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{1}{|C_j|}x_i^2sum_{y in c_j}1 - frac{2}{|C_j|}x_isum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[frac{|C_j|}{|C_j|}x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg] = $
$sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}bigg[x_i^2 - frac{2x_i}{|C_j|}sum_{y in c_j}y_i + frac{1}{|C_j|}sum_{y in c_j}y_i^2 bigg]$
Then I have,
$2J_{avg^2} = 2sum_{j=1}^{k}sum_{xin C_j}d(x,m_j)^2 = 2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}(x_i - frac{1}{|C_j|}sum_{yin C_j}y_i)^2 = $
$2sum_{j=1}^{k}sum_{xin C_j}sum_{i = 1}^{d}x_i^2 - frac{2x_i}{|C_j|}sum_{yin C_j}y_i + frac{1}{|C_j|^2}big(sum_{yin C_j}y_ibig)^2$
I am lost right now on how to proceed and would be very thankful if someone could help point out any mistakes or provide any guidance on what approach to take so that $J_{IC} = 2J_{avg^2}$.
summation machine-learning clustering
summation machine-learning clustering
asked Jan 21 at 21:08
KayandaKayanda
12
asked Jan 21 at 21:08
KayandaKayanda
12
edited Jan 21 at 22:10
Kayanda
asked Jan 21 at 21:08
KayandaKayanda
12
asked Jan 21 at 21:08
KayandaKayanda
12
asked Jan 21 at 21:08
KayandaKayanda
12
12
add a comment |
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3082417%2fk-means-cost-functions-comparison%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
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3082417%2fk-means-cost-functions-comparison%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
