Please correct my thinking about Ridge Regression
If ridge regression biases ALL beta coefficients of a regression model towards zero, wouldn't the model massively mispredict the y-variable?
I know my logic must be wrong here, but I'd appreciate if someone could point out the flaw. I am just starting to get into this topic, so would appreciate as simple an answer as possible.
For example: Say you have the coefficients of a linear regression output (and for simplicity, all coefficients are positive, and say the x-variables can only take on positive values). Then, if you biased each beta coefficient towards zero, wouldn't the model consistently underpredict the y-variable?
Would appreciate any intuition on how ridge regression shrinks all the coefficients but still maintains predictive ability. Thank you!
statistics regression mathematical-modeling linear-regression regression-analysis
add a comment |
If ridge regression biases ALL beta coefficients of a regression model towards zero, wouldn't the model massively mispredict the y-variable?
I know my logic must be wrong here, but I'd appreciate if someone could point out the flaw. I am just starting to get into this topic, so would appreciate as simple an answer as possible.
For example: Say you have the coefficients of a linear regression output (and for simplicity, all coefficients are positive, and say the x-variables can only take on positive values). Then, if you biased each beta coefficient towards zero, wouldn't the model consistently underpredict the y-variable?
Would appreciate any intuition on how ridge regression shrinks all the coefficients but still maintains predictive ability. Thank you!
statistics regression mathematical-modeling linear-regression regression-analysis
add a comment |
If ridge regression biases ALL beta coefficients of a regression model towards zero, wouldn't the model massively mispredict the y-variable?
I know my logic must be wrong here, but I'd appreciate if someone could point out the flaw. I am just starting to get into this topic, so would appreciate as simple an answer as possible.
For example: Say you have the coefficients of a linear regression output (and for simplicity, all coefficients are positive, and say the x-variables can only take on positive values). Then, if you biased each beta coefficient towards zero, wouldn't the model consistently underpredict the y-variable?
Would appreciate any intuition on how ridge regression shrinks all the coefficients but still maintains predictive ability. Thank you!
statistics regression mathematical-modeling linear-regression regression-analysis
If ridge regression biases ALL beta coefficients of a regression model towards zero, wouldn't the model massively mispredict the y-variable?
I know my logic must be wrong here, but I'd appreciate if someone could point out the flaw. I am just starting to get into this topic, so would appreciate as simple an answer as possible.
For example: Say you have the coefficients of a linear regression output (and for simplicity, all coefficients are positive, and say the x-variables can only take on positive values). Then, if you biased each beta coefficient towards zero, wouldn't the model consistently underpredict the y-variable?
Would appreciate any intuition on how ridge regression shrinks all the coefficients but still maintains predictive ability. Thank you!
statistics regression mathematical-modeling linear-regression regression-analysis
statistics regression mathematical-modeling linear-regression regression-analysis
asked Dec 2 at 17:18
bob
1089
1089
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
The objective function in ridge regression has two terms. You are correct that the penalty term $lambda |beta|_2^2$ encourages the coefficients to be small. However, you forget that the main term $|y - X beta|^2$ encourages the choice of coefficients to fit $y$ well. So there is a tension between these two competing goals of fitting $y$ well and keeping the coefficients small.
Regarding "constantly underpredict," you are correct that ridge regression is biased. But this is not the only criteria for evaluating the performance of an estimator (bias-variance tradeoff).
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%2f3022920%2fplease-correct-my-thinking-about-ridge-regression%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
The objective function in ridge regression has two terms. You are correct that the penalty term $lambda |beta|_2^2$ encourages the coefficients to be small. However, you forget that the main term $|y - X beta|^2$ encourages the choice of coefficients to fit $y$ well. So there is a tension between these two competing goals of fitting $y$ well and keeping the coefficients small.
Regarding "constantly underpredict," you are correct that ridge regression is biased. But this is not the only criteria for evaluating the performance of an estimator (bias-variance tradeoff).
add a comment |
The objective function in ridge regression has two terms. You are correct that the penalty term $lambda |beta|_2^2$ encourages the coefficients to be small. However, you forget that the main term $|y - X beta|^2$ encourages the choice of coefficients to fit $y$ well. So there is a tension between these two competing goals of fitting $y$ well and keeping the coefficients small.
Regarding "constantly underpredict," you are correct that ridge regression is biased. But this is not the only criteria for evaluating the performance of an estimator (bias-variance tradeoff).
add a comment |
The objective function in ridge regression has two terms. You are correct that the penalty term $lambda |beta|_2^2$ encourages the coefficients to be small. However, you forget that the main term $|y - X beta|^2$ encourages the choice of coefficients to fit $y$ well. So there is a tension between these two competing goals of fitting $y$ well and keeping the coefficients small.
Regarding "constantly underpredict," you are correct that ridge regression is biased. But this is not the only criteria for evaluating the performance of an estimator (bias-variance tradeoff).
The objective function in ridge regression has two terms. You are correct that the penalty term $lambda |beta|_2^2$ encourages the coefficients to be small. However, you forget that the main term $|y - X beta|^2$ encourages the choice of coefficients to fit $y$ well. So there is a tension between these two competing goals of fitting $y$ well and keeping the coefficients small.
Regarding "constantly underpredict," you are correct that ridge regression is biased. But this is not the only criteria for evaluating the performance of an estimator (bias-variance tradeoff).
answered Dec 2 at 17:43
angryavian
38.9k23180
38.9k23180
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%2f3022920%2fplease-correct-my-thinking-about-ridge-regression%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