how to prove chi-square statistics conforms to chi-square distribution with contingency table?
chi-square test(principle used in C4.5's CVP Pruning),
also called chi-square statistics,
also called chi-square goodness-of fit
How to prove
$sum_{i=1}^{i=r}sum_{j=1}^{j=c}frac{(x_{ij}-E_{ij} )^2}{E_{ij}} = chi^2_{(r-1)(c-1)}$
where
$E_{ij}=frac{N_i·N_j}{N}$,
$N$ is the total counts of the whole datasets.
$N_i$ are the counts of the sub-datasets of the same-value of feature
$N_j$ are the counts of the sub-datasets of the same-class
please help,thanks～！
here is contingency table
/------------------------------------------------
here are some references which are not clear:
https://arxiv.org/pdf/1808.09171.pdf (not mention why $k-1$ is used in formula(5))
https://www.math.utah.edu/~davar/ps-pdf-files/Chisquared.pdf (Not mention why $Theta<1$
from (9)->(10))
https://arxiv.org/pdf/1808.09171 (page 4th not mention what is X*with a line on it)
http://personal.psu.edu/drh20/asymp/fall2006/lectures/ANGELchpt07.pdf
(Page 109th,Not mention why $Cov(X_{ij},X_{il}=-p_ip_l)$)
probability probability-theory statistics normal-distribution chi-squared
asked Nov 26 '18 at 7:43
appleyuchi
113
add a comment |
chi-square test(principle used in C4.5's CVP Pruning),
also called chi-square statistics,
also called chi-square goodness-of fit
How to prove
$sum_{i=1}^{i=r}sum_{j=1}^{j=c}frac{(x_{ij}-E_{ij} )^2}{E_{ij}} = chi^2_{(r-1)(c-1)}$
where
$E_{ij}=frac{N_i·N_j}{N}$,
$N$ is the total counts of the whole datasets.
$N_i$ are the counts of the sub-datasets of the same-value of feature
$N_j$ are the counts of the sub-datasets of the same-class
please help,thanks～！
here is contingency table
/------------------------------------------------
here are some references which are not clear:
https://arxiv.org/pdf/1808.09171.pdf (not mention why $k-1$ is used in formula(5))
https://www.math.utah.edu/~davar/ps-pdf-files/Chisquared.pdf (Not mention why $Theta<1$
from (9)->(10))
https://arxiv.org/pdf/1808.09171 (page 4th not mention what is X*with a line on it)
http://personal.psu.edu/drh20/asymp/fall2006/lectures/ANGELchpt07.pdf
(Page 109th,Not mention why $Cov(X_{ij},X_{il}=-p_ip_l)$)
probability probability-theory statistics normal-distribution chi-squared
asked Nov 26 '18 at 7:43
appleyuchi
113
add a comment |
chi-square test(principle used in C4.5's CVP Pruning),
also called chi-square statistics,
also called chi-square goodness-of fit
How to prove
$sum_{i=1}^{i=r}sum_{j=1}^{j=c}frac{(x_{ij}-E_{ij} )^2}{E_{ij}} = chi^2_{(r-1)(c-1)}$
where
$E_{ij}=frac{N_i·N_j}{N}$,
$N$ is the total counts of the whole datasets.
$N_i$ are the counts of the sub-datasets of the same-value of feature
$N_j$ are the counts of the sub-datasets of the same-class
please help,thanks～！
here is contingency table
/------------------------------------------------
here are some references which are not clear:
https://arxiv.org/pdf/1808.09171.pdf (not mention why $k-1$ is used in formula(5))
https://www.math.utah.edu/~davar/ps-pdf-files/Chisquared.pdf (Not mention why $Theta<1$
from (9)->(10))
https://arxiv.org/pdf/1808.09171 (page 4th not mention what is X*with a line on it)
http://personal.psu.edu/drh20/asymp/fall2006/lectures/ANGELchpt07.pdf
(Page 109th,Not mention why $Cov(X_{ij},X_{il}=-p_ip_l)$)
probability probability-theory statistics normal-distribution chi-squared
asked Nov 26 '18 at 7:43
appleyuchi
113
chi-square test(principle used in C4.5's CVP Pruning),
also called chi-square statistics,
also called chi-square goodness-of fit
How to prove
$sum_{i=1}^{i=r}sum_{j=1}^{j=c}frac{(x_{ij}-E_{ij} )^2}{E_{ij}} = chi^2_{(r-1)(c-1)}$
where
$E_{ij}=frac{N_i·N_j}{N}$,
$N$ is the total counts of the whole datasets.
$N_i$ are the counts of the sub-datasets of the same-value of feature
$N_j$ are the counts of the sub-datasets of the same-class
please help,thanks～！
here is contingency table
/------------------------------------------------
here are some references which are not clear:
https://arxiv.org/pdf/1808.09171.pdf (not mention why $k-1$ is used in formula(5))
https://www.math.utah.edu/~davar/ps-pdf-files/Chisquared.pdf (Not mention why $Theta<1$
from (9)->(10))
https://arxiv.org/pdf/1808.09171 (page 4th not mention what is X*with a line on it)
http://personal.psu.edu/drh20/asymp/fall2006/lectures/ANGELchpt07.pdf
(Page 109th,Not mention why $Cov(X_{ij},X_{il}=-p_ip_l)$)
probability probability-theory statistics normal-distribution chi-squared
probability probability-theory statistics normal-distribution chi-squared
asked Nov 26 '18 at 7:43
appleyuchi
113
asked Nov 26 '18 at 7:43
appleyuchi
113
edited Nov 26 '18 at 11:44
asked Nov 26 '18 at 7:43
appleyuchi
113
asked Nov 26 '18 at 7:43
appleyuchi
113
asked Nov 26 '18 at 7:43
appleyuchi
113
113
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
The proof uses $x_{ij}approxoperatorname{Poisson}(E_{ij})approx N(E_{ij},,E_{ij})$. The reason for $k-1$ is that $sum_i N_i=N$ removes a degree of freedom. The reason for $Thetale 1$ is because the $theta_i$ are probabilities.
answered Nov 26 '18 at 7:52
J.G.
23k22137
thanks for your replies,could you please give a proof with details about the xij≈Poisson(Eij)≈N(Eij,Eij)?THANKS.
– appleyuchi
Nov 26 '18 at 8:05
what's the meaning of xij≈Poisson(Eij)≈N(Eij,Eij)？
– appleyuchi
Nov 26 '18 at 8:25
For the Poisson distribution the mean is equal to the variance.
– Karl
Nov 26 '18 at 12:32
thanks for your replies,but what's the meaning of "xij≈Poisson(Eij)"?
– appleyuchi
Nov 27 '18 at 6:52
@appleyuchi That $x_{ij}$ is approximately Poisson-distributed.
– J.G.
Nov 27 '18 at 6:53
|
show 3 more comments
https://blog.csdn.net/appleyuchi/article/details/84567158
I try to prove it from multi-nominal distribution.
The above link is my record,NOT very rigorous,
If there are something wrong ,please let me know,thanks.
If there are other proof which is much easier to understand ,please let me know,thanks.
Many thanks for all your help~!
answered Nov 27 '18 at 7:51
appleyuchi
113
add a comment |
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%2f3014004%2fhow-to-prove-chi-square-statistics-conforms-to-chi-square-distribution-with-cont%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
The proof uses $x_{ij}approxoperatorname{Poisson}(E_{ij})approx N(E_{ij},,E_{ij})$. The reason for $k-1$ is that $sum_i N_i=N$ removes a degree of freedom. The reason for $Thetale 1$ is because the $theta_i$ are probabilities.
answered Nov 26 '18 at 7:52
J.G.
23k22137
thanks for your replies,could you please give a proof with details about the xij≈Poisson(Eij)≈N(Eij,Eij)?THANKS.
– appleyuchi
Nov 26 '18 at 8:05
what's the meaning of xij≈Poisson(Eij)≈N(Eij,Eij)？
– appleyuchi
Nov 26 '18 at 8:25
For the Poisson distribution the mean is equal to the variance.
– Karl
Nov 26 '18 at 12:32
thanks for your replies,but what's the meaning of "xij≈Poisson(Eij)"?
– appleyuchi
Nov 27 '18 at 6:52
@appleyuchi That $x_{ij}$ is approximately Poisson-distributed.
– J.G.
Nov 27 '18 at 6:53
|
show 3 more comments
The proof uses $x_{ij}approxoperatorname{Poisson}(E_{ij})approx N(E_{ij},,E_{ij})$. The reason for $k-1$ is that $sum_i N_i=N$ removes a degree of freedom. The reason for $Thetale 1$ is because the $theta_i$ are probabilities.
answered Nov 26 '18 at 7:52
J.G.
23k22137
thanks for your replies,could you please give a proof with details about the xij≈Poisson(Eij)≈N(Eij,Eij)?THANKS.
– appleyuchi
Nov 26 '18 at 8:05
what's the meaning of xij≈Poisson(Eij)≈N(Eij,Eij)？
– appleyuchi
Nov 26 '18 at 8:25
For the Poisson distribution the mean is equal to the variance.
– Karl
Nov 26 '18 at 12:32
thanks for your replies,but what's the meaning of "xij≈Poisson(Eij)"?
– appleyuchi
Nov 27 '18 at 6:52
@appleyuchi That $x_{ij}$ is approximately Poisson-distributed.
– J.G.
Nov 27 '18 at 6:53
|
show 3 more comments
The proof uses $x_{ij}approxoperatorname{Poisson}(E_{ij})approx N(E_{ij},,E_{ij})$. The reason for $k-1$ is that $sum_i N_i=N$ removes a degree of freedom. The reason for $Thetale 1$ is because the $theta_i$ are probabilities.
answered Nov 26 '18 at 7:52
J.G.
23k22137
The proof uses $x_{ij}approxoperatorname{Poisson}(E_{ij})approx N(E_{ij},,E_{ij})$. The reason for $k-1$ is that $sum_i N_i=N$ removes a degree of freedom. The reason for $Thetale 1$ is because the $theta_i$ are probabilities.
answered Nov 26 '18 at 7:52
J.G.
23k22137
answered Nov 26 '18 at 7:52
J.G.
23k22137
answered Nov 26 '18 at 7:52
J.G.
23k22137
answered Nov 26 '18 at 7:52
J.G.
23k22137
23k22137
thanks for your replies,could you please give a proof with details about the xij≈Poisson(Eij)≈N(Eij,Eij)?THANKS.
– appleyuchi
Nov 26 '18 at 8:05
what's the meaning of xij≈Poisson(Eij)≈N(Eij,Eij)？
– appleyuchi
Nov 26 '18 at 8:25
For the Poisson distribution the mean is equal to the variance.
– Karl
Nov 26 '18 at 12:32
thanks for your replies,but what's the meaning of "xij≈Poisson(Eij)"?
– appleyuchi
Nov 27 '18 at 6:52
@appleyuchi That $x_{ij}$ is approximately Poisson-distributed.
– J.G.
Nov 27 '18 at 6:53
|
show 3 more comments
thanks for your replies,could you please give a proof with details about the xij≈Poisson(Eij)≈N(Eij,Eij)?THANKS.
– appleyuchi
Nov 26 '18 at 8:05
what's the meaning of xij≈Poisson(Eij)≈N(Eij,Eij)？
– appleyuchi
Nov 26 '18 at 8:25
For the Poisson distribution the mean is equal to the variance.
– Karl
Nov 26 '18 at 12:32
thanks for your replies,but what's the meaning of "xij≈Poisson(Eij)"?
– appleyuchi
Nov 27 '18 at 6:52
@appleyuchi That $x_{ij}$ is approximately Poisson-distributed.
– J.G.
Nov 27 '18 at 6:53
thanks for your replies,could you please give a proof with details about the xij≈Poisson(Eij)≈N(Eij,Eij)?THANKS.
– appleyuchi
Nov 26 '18 at 8:05
thanks for your replies,could you please give a proof with details about the xij≈Poisson(Eij)≈N(Eij,Eij)?THANKS.
– appleyuchi
Nov 26 '18 at 8:05
what's the meaning of xij≈Poisson(Eij)≈N(Eij,Eij)？
– appleyuchi
Nov 26 '18 at 8:25
what's the meaning of xij≈Poisson(Eij)≈N(Eij,Eij)？
– appleyuchi
Nov 26 '18 at 8:25
For the Poisson distribution the mean is equal to the variance.
– Karl
Nov 26 '18 at 12:32
For the Poisson distribution the mean is equal to the variance.
– Karl
Nov 26 '18 at 12:32
thanks for your replies,but what's the meaning of "xij≈Poisson(Eij)"?
– appleyuchi
Nov 27 '18 at 6:52
thanks for your replies,but what's the meaning of "xij≈Poisson(Eij)"?
– appleyuchi
Nov 27 '18 at 6:52
@appleyuchi That $x_{ij}$ is approximately Poisson-distributed.
– J.G.
Nov 27 '18 at 6:53
@appleyuchi That $x_{ij}$ is approximately Poisson-distributed.
– J.G.
Nov 27 '18 at 6:53
|
show 3 more comments
https://blog.csdn.net/appleyuchi/article/details/84567158
I try to prove it from multi-nominal distribution.
The above link is my record,NOT very rigorous,
If there are something wrong ,please let me know,thanks.
If there are other proof which is much easier to understand ,please let me know,thanks.
Many thanks for all your help~!
answered Nov 27 '18 at 7:51
appleyuchi
113
add a comment |
https://blog.csdn.net/appleyuchi/article/details/84567158
I try to prove it from multi-nominal distribution.
The above link is my record,NOT very rigorous,
If there are something wrong ,please let me know,thanks.
If there are other proof which is much easier to understand ,please let me know,thanks.
Many thanks for all your help~!
answered Nov 27 '18 at 7:51
appleyuchi
113
add a comment |
https://blog.csdn.net/appleyuchi/article/details/84567158
I try to prove it from multi-nominal distribution.
The above link is my record,NOT very rigorous,
If there are something wrong ,please let me know,thanks.
If there are other proof which is much easier to understand ,please let me know,thanks.
Many thanks for all your help~!
answered Nov 27 '18 at 7:51
appleyuchi
113
https://blog.csdn.net/appleyuchi/article/details/84567158
I try to prove it from multi-nominal distribution.
The above link is my record,NOT very rigorous,
If there are something wrong ,please let me know,thanks.
If there are other proof which is much easier to understand ,please let me know,thanks.
Many thanks for all your help~!
answered Nov 27 '18 at 7:51
appleyuchi
113
edited Nov 27 '18 at 11:12
answered Nov 27 '18 at 7:51
appleyuchi
113
answered Nov 27 '18 at 7:51
appleyuchi
113
answered Nov 27 '18 at 7:51
appleyuchi
113
113
add a comment |
add a comment |
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.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- 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.
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%2f3014004%2fhow-to-prove-chi-square-statistics-conforms-to-chi-square-distribution-with-cont%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
