Invariance of results when scaling explanatory variables in logistic regression, is there a proof?












6












$begingroup$


There is a standard result for linear regression that the regression coefficients are given by



$$mathbf{beta}=(mathbf{X^T X})^{-1}mathbf{X^T y}$$



or



$(mathbf{X^T X})mathbf{beta}=mathbf{X^T y} tag{2}label{eq2}$



Scaling the explanatory variables does not affect the predictions. I have tried to show this algebraically as follows.



The response is related to the explanatory variables via the matrix equation
$mathbf{y}=mathbf{X beta} tag{3}label{eq3}$



$mathbf{X}$ is an $n times (p+1)$ matrix of n observations on p explanatory variables. The first column of $mathbf{X}$ is a column of ones.



Scaling the explanatory variables with a $(p+1) times (p+1) $ diagonal matrix $mathbf{D}$, whose entries are the scaling factors
$
mathbf{X^s} = mathbf{XD} tag{4}label{eq4}$



$mathbf{X^s}$ and $mathbf{beta^s}$ satisfy $eqref{eq2}$:



$$(mathbf{D^TX^T XD})mathbf{beta^s} =mathbf{D^TX^T y}$$



so



$$mathbf{X^T XD}mathbf{beta^s} =mathbf{X^T y}$$



$$Rightarrow mathbf{D beta^s} = (mathbf{X^T X)^{-1}}mathbf{X^T y}=mathbf{beta}$$



$Rightarrow mathbf{beta^s}=mathbf{D}^{-1}mathbf{beta} tag{5}label{eq5}$



This means if an explanatory variable is scaled by $d_i$ then the regression coefficient $beta_i$is scaled by $1/d_i$ and the effect of the scaling cancels out, i.e.
considering predictions based on scaled values, and using $eqref{eq4},eqref{eq5},eqref{eq3}$



$$mathbf{y^s}=mathbf{X^s beta^s} = mathbf{X D D^{-1}beta}=mathbf{X beta}=mathbf{y}$$
as expected.



Now to the question.



For logistic regression without any regularization, it is suggested, by doing regressions with and without scaling the same effect is seen




fit <- glm(vs ~ mpg, data=mtcars,family=binomial)

print(fit)

Coefficients:
(Intercept) mpg
-8.8331 0.4304


mtcars$mpg <- mtcars$mpg * 10

fit <- glm(vs ~ mpg, data=mtcars,family=binomial)

print(fit)

Coefficients:
(Intercept) mpg
-8.83307 0.04304


When the variable mpg is scaled up by 10, the corresponding coefficient is scaled down by 10.




  1. How could this scaling property be proved (or disproved ) algebraically for logistic regression?


I found a similar question relating to the effect on AUC when regularization is used.




  1. Is there any point to scaling explanatory variables in logistic regression, in the absence of regularization?


Thanks.










share|cite|improve this question









$endgroup$

















    6












    $begingroup$


    There is a standard result for linear regression that the regression coefficients are given by



    $$mathbf{beta}=(mathbf{X^T X})^{-1}mathbf{X^T y}$$



    or



    $(mathbf{X^T X})mathbf{beta}=mathbf{X^T y} tag{2}label{eq2}$



    Scaling the explanatory variables does not affect the predictions. I have tried to show this algebraically as follows.



    The response is related to the explanatory variables via the matrix equation
    $mathbf{y}=mathbf{X beta} tag{3}label{eq3}$



    $mathbf{X}$ is an $n times (p+1)$ matrix of n observations on p explanatory variables. The first column of $mathbf{X}$ is a column of ones.



    Scaling the explanatory variables with a $(p+1) times (p+1) $ diagonal matrix $mathbf{D}$, whose entries are the scaling factors
    $
    mathbf{X^s} = mathbf{XD} tag{4}label{eq4}$



    $mathbf{X^s}$ and $mathbf{beta^s}$ satisfy $eqref{eq2}$:



    $$(mathbf{D^TX^T XD})mathbf{beta^s} =mathbf{D^TX^T y}$$



    so



    $$mathbf{X^T XD}mathbf{beta^s} =mathbf{X^T y}$$



    $$Rightarrow mathbf{D beta^s} = (mathbf{X^T X)^{-1}}mathbf{X^T y}=mathbf{beta}$$



    $Rightarrow mathbf{beta^s}=mathbf{D}^{-1}mathbf{beta} tag{5}label{eq5}$



    This means if an explanatory variable is scaled by $d_i$ then the regression coefficient $beta_i$is scaled by $1/d_i$ and the effect of the scaling cancels out, i.e.
    considering predictions based on scaled values, and using $eqref{eq4},eqref{eq5},eqref{eq3}$



    $$mathbf{y^s}=mathbf{X^s beta^s} = mathbf{X D D^{-1}beta}=mathbf{X beta}=mathbf{y}$$
    as expected.



    Now to the question.



    For logistic regression without any regularization, it is suggested, by doing regressions with and without scaling the same effect is seen




    fit <- glm(vs ~ mpg, data=mtcars,family=binomial)

    print(fit)

    Coefficients:
    (Intercept) mpg
    -8.8331 0.4304


    mtcars$mpg <- mtcars$mpg * 10

    fit <- glm(vs ~ mpg, data=mtcars,family=binomial)

    print(fit)

    Coefficients:
    (Intercept) mpg
    -8.83307 0.04304


    When the variable mpg is scaled up by 10, the corresponding coefficient is scaled down by 10.




    1. How could this scaling property be proved (or disproved ) algebraically for logistic regression?


    I found a similar question relating to the effect on AUC when regularization is used.




    1. Is there any point to scaling explanatory variables in logistic regression, in the absence of regularization?


    Thanks.










    share|cite|improve this question









    $endgroup$















      6












      6








      6





      $begingroup$


      There is a standard result for linear regression that the regression coefficients are given by



      $$mathbf{beta}=(mathbf{X^T X})^{-1}mathbf{X^T y}$$



      or



      $(mathbf{X^T X})mathbf{beta}=mathbf{X^T y} tag{2}label{eq2}$



      Scaling the explanatory variables does not affect the predictions. I have tried to show this algebraically as follows.



      The response is related to the explanatory variables via the matrix equation
      $mathbf{y}=mathbf{X beta} tag{3}label{eq3}$



      $mathbf{X}$ is an $n times (p+1)$ matrix of n observations on p explanatory variables. The first column of $mathbf{X}$ is a column of ones.



      Scaling the explanatory variables with a $(p+1) times (p+1) $ diagonal matrix $mathbf{D}$, whose entries are the scaling factors
      $
      mathbf{X^s} = mathbf{XD} tag{4}label{eq4}$



      $mathbf{X^s}$ and $mathbf{beta^s}$ satisfy $eqref{eq2}$:



      $$(mathbf{D^TX^T XD})mathbf{beta^s} =mathbf{D^TX^T y}$$



      so



      $$mathbf{X^T XD}mathbf{beta^s} =mathbf{X^T y}$$



      $$Rightarrow mathbf{D beta^s} = (mathbf{X^T X)^{-1}}mathbf{X^T y}=mathbf{beta}$$



      $Rightarrow mathbf{beta^s}=mathbf{D}^{-1}mathbf{beta} tag{5}label{eq5}$



      This means if an explanatory variable is scaled by $d_i$ then the regression coefficient $beta_i$is scaled by $1/d_i$ and the effect of the scaling cancels out, i.e.
      considering predictions based on scaled values, and using $eqref{eq4},eqref{eq5},eqref{eq3}$



      $$mathbf{y^s}=mathbf{X^s beta^s} = mathbf{X D D^{-1}beta}=mathbf{X beta}=mathbf{y}$$
      as expected.



      Now to the question.



      For logistic regression without any regularization, it is suggested, by doing regressions with and without scaling the same effect is seen




      fit <- glm(vs ~ mpg, data=mtcars,family=binomial)

      print(fit)

      Coefficients:
      (Intercept) mpg
      -8.8331 0.4304


      mtcars$mpg <- mtcars$mpg * 10

      fit <- glm(vs ~ mpg, data=mtcars,family=binomial)

      print(fit)

      Coefficients:
      (Intercept) mpg
      -8.83307 0.04304


      When the variable mpg is scaled up by 10, the corresponding coefficient is scaled down by 10.




      1. How could this scaling property be proved (or disproved ) algebraically for logistic regression?


      I found a similar question relating to the effect on AUC when regularization is used.




      1. Is there any point to scaling explanatory variables in logistic regression, in the absence of regularization?


      Thanks.










      share|cite|improve this question









      $endgroup$




      There is a standard result for linear regression that the regression coefficients are given by



      $$mathbf{beta}=(mathbf{X^T X})^{-1}mathbf{X^T y}$$



      or



      $(mathbf{X^T X})mathbf{beta}=mathbf{X^T y} tag{2}label{eq2}$



      Scaling the explanatory variables does not affect the predictions. I have tried to show this algebraically as follows.



      The response is related to the explanatory variables via the matrix equation
      $mathbf{y}=mathbf{X beta} tag{3}label{eq3}$



      $mathbf{X}$ is an $n times (p+1)$ matrix of n observations on p explanatory variables. The first column of $mathbf{X}$ is a column of ones.



      Scaling the explanatory variables with a $(p+1) times (p+1) $ diagonal matrix $mathbf{D}$, whose entries are the scaling factors
      $
      mathbf{X^s} = mathbf{XD} tag{4}label{eq4}$



      $mathbf{X^s}$ and $mathbf{beta^s}$ satisfy $eqref{eq2}$:



      $$(mathbf{D^TX^T XD})mathbf{beta^s} =mathbf{D^TX^T y}$$



      so



      $$mathbf{X^T XD}mathbf{beta^s} =mathbf{X^T y}$$



      $$Rightarrow mathbf{D beta^s} = (mathbf{X^T X)^{-1}}mathbf{X^T y}=mathbf{beta}$$



      $Rightarrow mathbf{beta^s}=mathbf{D}^{-1}mathbf{beta} tag{5}label{eq5}$



      This means if an explanatory variable is scaled by $d_i$ then the regression coefficient $beta_i$is scaled by $1/d_i$ and the effect of the scaling cancels out, i.e.
      considering predictions based on scaled values, and using $eqref{eq4},eqref{eq5},eqref{eq3}$



      $$mathbf{y^s}=mathbf{X^s beta^s} = mathbf{X D D^{-1}beta}=mathbf{X beta}=mathbf{y}$$
      as expected.



      Now to the question.



      For logistic regression without any regularization, it is suggested, by doing regressions with and without scaling the same effect is seen




      fit <- glm(vs ~ mpg, data=mtcars,family=binomial)

      print(fit)

      Coefficients:
      (Intercept) mpg
      -8.8331 0.4304


      mtcars$mpg <- mtcars$mpg * 10

      fit <- glm(vs ~ mpg, data=mtcars,family=binomial)

      print(fit)

      Coefficients:
      (Intercept) mpg
      -8.83307 0.04304


      When the variable mpg is scaled up by 10, the corresponding coefficient is scaled down by 10.




      1. How could this scaling property be proved (or disproved ) algebraically for logistic regression?


      I found a similar question relating to the effect on AUC when regularization is used.




      1. Is there any point to scaling explanatory variables in logistic regression, in the absence of regularization?


      Thanks.







      regression logistic regression-coefficients






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked yesterday









      PM.PM.

      2021211




      2021211






















          2 Answers
          2






          active

          oldest

          votes


















          9












          $begingroup$

          Here is a heuristic idea:



          The likelihood for a logistic regression model is
          $$ ell(beta|y) propto prod_ileft(frac{exp(x_i'beta)}{1+exp(x_i'beta)}right)^{y_i}left(frac{1}{1+exp(x_i'beta)}right)^{1-y_i}
          $$

          and the MLE is the arg max of that likelihood. When you scale a regressor, you also need to accordingly scale the coefficients to achieve the original maximal likelihood.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            That's useful +1 thank-you. Anything to add regarding the second point about the reasons for doing (or not doing) the scaling?
            $endgroup$
            – PM.
            yesterday






          • 2




            $begingroup$
            Not sure. Given that in practice, MLEs are computed via numerical (e.g. Newton-Raphson) algorithms, they might be more susceptible to stability issues when regressors live on extremely different scales. Also, of course, different users may prefer different units of measurement (say, km vs miles).
            $endgroup$
            – Christoph Hanck
            yesterday



















          6












          $begingroup$

          Christoph has a great answer (+1). Just writing this because I can't comment there.



          The crucial point here is that the likelihood only depends on the coefficients $beta$ through the linear term $X beta$. This makes the likelihood unable to distinguish between "$X beta$" and $(XD) (D^{-1}beta)$", causing the invariance you've noticed.



          To be specific about this, we need to introduce some notation (which we can do since we're writing an answer!). Let $y_i | x_i stackrel{ind.}{sim} mathrm{bernoulli}left[ mathrm{logit}^{-1} (x_i^T beta) right]$ be independent draws according to the logistic regression model, where $x_i in mathbb{R}^{p+1}$ is the measured covariates. Write the likelihood of the $i^{th}$ observation as $l(y_i, x_i^T beta)$.



          To introduce the change of coordinates, write $bar{x}_i = D x_i$, where $D$ is diagonal matrix with all diagonal entries nonzero. By definition of maximum likelihood estimation, we know that maximum likelihood estimators $hat{beta}$ of the data ${y_i | x_i}$ satisfy that $$sum_{i=1}^n l(y_i, x_i^T beta) leq sum_{i=1}^n l(y_i, x_i^T hatbeta) tag{1}$$ for all coefficients $beta in mathbb{R}^p$, and that maximum likelihood estimators for the data ${y_i | bar{x}_i}$ satisfy that $$sum_{i=1}^n l(y_i, bar{x}_i^T alpha) leq sum_{i=1}^n l(y_i, bar{x}_i^T hatalpha) tag{2}$$ for all coefficients $alpha in mathbb{R}^p$.



          In your argument, you used a closed form of the maximum likelihood estimator to derive the result. It turns out, though, (as Cristoph suggested above), all you need to do is work with the likelihood. Let $hat{beta}$ be a maximum likelihood estimator of the data ${y_i | x_i}$. Now, writing $beta = D alpha$, we can use equation (1) to show that $$sum_{i=1}^n l(y_i, bar{x}_i^T alpha) = sum_{i=1}^n lleft(y_i, (x_i^T D) (D^{-1} beta)right) leq sum_{i=1}^n l(y_i, x_i^T hatbeta) = sum_{i=1}^n l(y_i, bar{x}_i^T D^{-1} hat{beta}).$$ That is, $D^{-1} hat{beta}$ satisfies equation (2) and is therefore a maximum likelihood estimator with respect to the data ${y_i | bar{x}_i}$. This is the invariance property you noticed.



          (For what it's worth, there's a lot of room for generalizing this argument beyond logistic regression: did we need independent observations? did we need the matrix $D$ to be diagonal? did we need a binary response? did we need the use logit? What notation would you change for this argument to work in different scenarios?)






          share|cite|improve this answer










          New contributor




          user551504 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.






          $endgroup$













            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: "65"
            };
            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: false,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: null,
            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
            },
            onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });














            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f399318%2finvariance-of-results-when-scaling-explanatory-variables-in-logistic-regression%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









            9












            $begingroup$

            Here is a heuristic idea:



            The likelihood for a logistic regression model is
            $$ ell(beta|y) propto prod_ileft(frac{exp(x_i'beta)}{1+exp(x_i'beta)}right)^{y_i}left(frac{1}{1+exp(x_i'beta)}right)^{1-y_i}
            $$

            and the MLE is the arg max of that likelihood. When you scale a regressor, you also need to accordingly scale the coefficients to achieve the original maximal likelihood.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              That's useful +1 thank-you. Anything to add regarding the second point about the reasons for doing (or not doing) the scaling?
              $endgroup$
              – PM.
              yesterday






            • 2




              $begingroup$
              Not sure. Given that in practice, MLEs are computed via numerical (e.g. Newton-Raphson) algorithms, they might be more susceptible to stability issues when regressors live on extremely different scales. Also, of course, different users may prefer different units of measurement (say, km vs miles).
              $endgroup$
              – Christoph Hanck
              yesterday
















            9












            $begingroup$

            Here is a heuristic idea:



            The likelihood for a logistic regression model is
            $$ ell(beta|y) propto prod_ileft(frac{exp(x_i'beta)}{1+exp(x_i'beta)}right)^{y_i}left(frac{1}{1+exp(x_i'beta)}right)^{1-y_i}
            $$

            and the MLE is the arg max of that likelihood. When you scale a regressor, you also need to accordingly scale the coefficients to achieve the original maximal likelihood.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              That's useful +1 thank-you. Anything to add regarding the second point about the reasons for doing (or not doing) the scaling?
              $endgroup$
              – PM.
              yesterday






            • 2




              $begingroup$
              Not sure. Given that in practice, MLEs are computed via numerical (e.g. Newton-Raphson) algorithms, they might be more susceptible to stability issues when regressors live on extremely different scales. Also, of course, different users may prefer different units of measurement (say, km vs miles).
              $endgroup$
              – Christoph Hanck
              yesterday














            9












            9








            9





            $begingroup$

            Here is a heuristic idea:



            The likelihood for a logistic regression model is
            $$ ell(beta|y) propto prod_ileft(frac{exp(x_i'beta)}{1+exp(x_i'beta)}right)^{y_i}left(frac{1}{1+exp(x_i'beta)}right)^{1-y_i}
            $$

            and the MLE is the arg max of that likelihood. When you scale a regressor, you also need to accordingly scale the coefficients to achieve the original maximal likelihood.






            share|cite|improve this answer









            $endgroup$



            Here is a heuristic idea:



            The likelihood for a logistic regression model is
            $$ ell(beta|y) propto prod_ileft(frac{exp(x_i'beta)}{1+exp(x_i'beta)}right)^{y_i}left(frac{1}{1+exp(x_i'beta)}right)^{1-y_i}
            $$

            and the MLE is the arg max of that likelihood. When you scale a regressor, you also need to accordingly scale the coefficients to achieve the original maximal likelihood.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered yesterday









            Christoph HanckChristoph Hanck

            17.9k34274




            17.9k34274












            • $begingroup$
              That's useful +1 thank-you. Anything to add regarding the second point about the reasons for doing (or not doing) the scaling?
              $endgroup$
              – PM.
              yesterday






            • 2




              $begingroup$
              Not sure. Given that in practice, MLEs are computed via numerical (e.g. Newton-Raphson) algorithms, they might be more susceptible to stability issues when regressors live on extremely different scales. Also, of course, different users may prefer different units of measurement (say, km vs miles).
              $endgroup$
              – Christoph Hanck
              yesterday


















            • $begingroup$
              That's useful +1 thank-you. Anything to add regarding the second point about the reasons for doing (or not doing) the scaling?
              $endgroup$
              – PM.
              yesterday






            • 2




              $begingroup$
              Not sure. Given that in practice, MLEs are computed via numerical (e.g. Newton-Raphson) algorithms, they might be more susceptible to stability issues when regressors live on extremely different scales. Also, of course, different users may prefer different units of measurement (say, km vs miles).
              $endgroup$
              – Christoph Hanck
              yesterday
















            $begingroup$
            That's useful +1 thank-you. Anything to add regarding the second point about the reasons for doing (or not doing) the scaling?
            $endgroup$
            – PM.
            yesterday




            $begingroup$
            That's useful +1 thank-you. Anything to add regarding the second point about the reasons for doing (or not doing) the scaling?
            $endgroup$
            – PM.
            yesterday




            2




            2




            $begingroup$
            Not sure. Given that in practice, MLEs are computed via numerical (e.g. Newton-Raphson) algorithms, they might be more susceptible to stability issues when regressors live on extremely different scales. Also, of course, different users may prefer different units of measurement (say, km vs miles).
            $endgroup$
            – Christoph Hanck
            yesterday




            $begingroup$
            Not sure. Given that in practice, MLEs are computed via numerical (e.g. Newton-Raphson) algorithms, they might be more susceptible to stability issues when regressors live on extremely different scales. Also, of course, different users may prefer different units of measurement (say, km vs miles).
            $endgroup$
            – Christoph Hanck
            yesterday













            6












            $begingroup$

            Christoph has a great answer (+1). Just writing this because I can't comment there.



            The crucial point here is that the likelihood only depends on the coefficients $beta$ through the linear term $X beta$. This makes the likelihood unable to distinguish between "$X beta$" and $(XD) (D^{-1}beta)$", causing the invariance you've noticed.



            To be specific about this, we need to introduce some notation (which we can do since we're writing an answer!). Let $y_i | x_i stackrel{ind.}{sim} mathrm{bernoulli}left[ mathrm{logit}^{-1} (x_i^T beta) right]$ be independent draws according to the logistic regression model, where $x_i in mathbb{R}^{p+1}$ is the measured covariates. Write the likelihood of the $i^{th}$ observation as $l(y_i, x_i^T beta)$.



            To introduce the change of coordinates, write $bar{x}_i = D x_i$, where $D$ is diagonal matrix with all diagonal entries nonzero. By definition of maximum likelihood estimation, we know that maximum likelihood estimators $hat{beta}$ of the data ${y_i | x_i}$ satisfy that $$sum_{i=1}^n l(y_i, x_i^T beta) leq sum_{i=1}^n l(y_i, x_i^T hatbeta) tag{1}$$ for all coefficients $beta in mathbb{R}^p$, and that maximum likelihood estimators for the data ${y_i | bar{x}_i}$ satisfy that $$sum_{i=1}^n l(y_i, bar{x}_i^T alpha) leq sum_{i=1}^n l(y_i, bar{x}_i^T hatalpha) tag{2}$$ for all coefficients $alpha in mathbb{R}^p$.



            In your argument, you used a closed form of the maximum likelihood estimator to derive the result. It turns out, though, (as Cristoph suggested above), all you need to do is work with the likelihood. Let $hat{beta}$ be a maximum likelihood estimator of the data ${y_i | x_i}$. Now, writing $beta = D alpha$, we can use equation (1) to show that $$sum_{i=1}^n l(y_i, bar{x}_i^T alpha) = sum_{i=1}^n lleft(y_i, (x_i^T D) (D^{-1} beta)right) leq sum_{i=1}^n l(y_i, x_i^T hatbeta) = sum_{i=1}^n l(y_i, bar{x}_i^T D^{-1} hat{beta}).$$ That is, $D^{-1} hat{beta}$ satisfies equation (2) and is therefore a maximum likelihood estimator with respect to the data ${y_i | bar{x}_i}$. This is the invariance property you noticed.



            (For what it's worth, there's a lot of room for generalizing this argument beyond logistic regression: did we need independent observations? did we need the matrix $D$ to be diagonal? did we need a binary response? did we need the use logit? What notation would you change for this argument to work in different scenarios?)






            share|cite|improve this answer










            New contributor




            user551504 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
            Check out our Code of Conduct.






            $endgroup$


















              6












              $begingroup$

              Christoph has a great answer (+1). Just writing this because I can't comment there.



              The crucial point here is that the likelihood only depends on the coefficients $beta$ through the linear term $X beta$. This makes the likelihood unable to distinguish between "$X beta$" and $(XD) (D^{-1}beta)$", causing the invariance you've noticed.



              To be specific about this, we need to introduce some notation (which we can do since we're writing an answer!). Let $y_i | x_i stackrel{ind.}{sim} mathrm{bernoulli}left[ mathrm{logit}^{-1} (x_i^T beta) right]$ be independent draws according to the logistic regression model, where $x_i in mathbb{R}^{p+1}$ is the measured covariates. Write the likelihood of the $i^{th}$ observation as $l(y_i, x_i^T beta)$.



              To introduce the change of coordinates, write $bar{x}_i = D x_i$, where $D$ is diagonal matrix with all diagonal entries nonzero. By definition of maximum likelihood estimation, we know that maximum likelihood estimators $hat{beta}$ of the data ${y_i | x_i}$ satisfy that $$sum_{i=1}^n l(y_i, x_i^T beta) leq sum_{i=1}^n l(y_i, x_i^T hatbeta) tag{1}$$ for all coefficients $beta in mathbb{R}^p$, and that maximum likelihood estimators for the data ${y_i | bar{x}_i}$ satisfy that $$sum_{i=1}^n l(y_i, bar{x}_i^T alpha) leq sum_{i=1}^n l(y_i, bar{x}_i^T hatalpha) tag{2}$$ for all coefficients $alpha in mathbb{R}^p$.



              In your argument, you used a closed form of the maximum likelihood estimator to derive the result. It turns out, though, (as Cristoph suggested above), all you need to do is work with the likelihood. Let $hat{beta}$ be a maximum likelihood estimator of the data ${y_i | x_i}$. Now, writing $beta = D alpha$, we can use equation (1) to show that $$sum_{i=1}^n l(y_i, bar{x}_i^T alpha) = sum_{i=1}^n lleft(y_i, (x_i^T D) (D^{-1} beta)right) leq sum_{i=1}^n l(y_i, x_i^T hatbeta) = sum_{i=1}^n l(y_i, bar{x}_i^T D^{-1} hat{beta}).$$ That is, $D^{-1} hat{beta}$ satisfies equation (2) and is therefore a maximum likelihood estimator with respect to the data ${y_i | bar{x}_i}$. This is the invariance property you noticed.



              (For what it's worth, there's a lot of room for generalizing this argument beyond logistic regression: did we need independent observations? did we need the matrix $D$ to be diagonal? did we need a binary response? did we need the use logit? What notation would you change for this argument to work in different scenarios?)






              share|cite|improve this answer










              New contributor




              user551504 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.






              $endgroup$
















                6












                6








                6





                $begingroup$

                Christoph has a great answer (+1). Just writing this because I can't comment there.



                The crucial point here is that the likelihood only depends on the coefficients $beta$ through the linear term $X beta$. This makes the likelihood unable to distinguish between "$X beta$" and $(XD) (D^{-1}beta)$", causing the invariance you've noticed.



                To be specific about this, we need to introduce some notation (which we can do since we're writing an answer!). Let $y_i | x_i stackrel{ind.}{sim} mathrm{bernoulli}left[ mathrm{logit}^{-1} (x_i^T beta) right]$ be independent draws according to the logistic regression model, where $x_i in mathbb{R}^{p+1}$ is the measured covariates. Write the likelihood of the $i^{th}$ observation as $l(y_i, x_i^T beta)$.



                To introduce the change of coordinates, write $bar{x}_i = D x_i$, where $D$ is diagonal matrix with all diagonal entries nonzero. By definition of maximum likelihood estimation, we know that maximum likelihood estimators $hat{beta}$ of the data ${y_i | x_i}$ satisfy that $$sum_{i=1}^n l(y_i, x_i^T beta) leq sum_{i=1}^n l(y_i, x_i^T hatbeta) tag{1}$$ for all coefficients $beta in mathbb{R}^p$, and that maximum likelihood estimators for the data ${y_i | bar{x}_i}$ satisfy that $$sum_{i=1}^n l(y_i, bar{x}_i^T alpha) leq sum_{i=1}^n l(y_i, bar{x}_i^T hatalpha) tag{2}$$ for all coefficients $alpha in mathbb{R}^p$.



                In your argument, you used a closed form of the maximum likelihood estimator to derive the result. It turns out, though, (as Cristoph suggested above), all you need to do is work with the likelihood. Let $hat{beta}$ be a maximum likelihood estimator of the data ${y_i | x_i}$. Now, writing $beta = D alpha$, we can use equation (1) to show that $$sum_{i=1}^n l(y_i, bar{x}_i^T alpha) = sum_{i=1}^n lleft(y_i, (x_i^T D) (D^{-1} beta)right) leq sum_{i=1}^n l(y_i, x_i^T hatbeta) = sum_{i=1}^n l(y_i, bar{x}_i^T D^{-1} hat{beta}).$$ That is, $D^{-1} hat{beta}$ satisfies equation (2) and is therefore a maximum likelihood estimator with respect to the data ${y_i | bar{x}_i}$. This is the invariance property you noticed.



                (For what it's worth, there's a lot of room for generalizing this argument beyond logistic regression: did we need independent observations? did we need the matrix $D$ to be diagonal? did we need a binary response? did we need the use logit? What notation would you change for this argument to work in different scenarios?)






                share|cite|improve this answer










                New contributor




                user551504 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                $endgroup$



                Christoph has a great answer (+1). Just writing this because I can't comment there.



                The crucial point here is that the likelihood only depends on the coefficients $beta$ through the linear term $X beta$. This makes the likelihood unable to distinguish between "$X beta$" and $(XD) (D^{-1}beta)$", causing the invariance you've noticed.



                To be specific about this, we need to introduce some notation (which we can do since we're writing an answer!). Let $y_i | x_i stackrel{ind.}{sim} mathrm{bernoulli}left[ mathrm{logit}^{-1} (x_i^T beta) right]$ be independent draws according to the logistic regression model, where $x_i in mathbb{R}^{p+1}$ is the measured covariates. Write the likelihood of the $i^{th}$ observation as $l(y_i, x_i^T beta)$.



                To introduce the change of coordinates, write $bar{x}_i = D x_i$, where $D$ is diagonal matrix with all diagonal entries nonzero. By definition of maximum likelihood estimation, we know that maximum likelihood estimators $hat{beta}$ of the data ${y_i | x_i}$ satisfy that $$sum_{i=1}^n l(y_i, x_i^T beta) leq sum_{i=1}^n l(y_i, x_i^T hatbeta) tag{1}$$ for all coefficients $beta in mathbb{R}^p$, and that maximum likelihood estimators for the data ${y_i | bar{x}_i}$ satisfy that $$sum_{i=1}^n l(y_i, bar{x}_i^T alpha) leq sum_{i=1}^n l(y_i, bar{x}_i^T hatalpha) tag{2}$$ for all coefficients $alpha in mathbb{R}^p$.



                In your argument, you used a closed form of the maximum likelihood estimator to derive the result. It turns out, though, (as Cristoph suggested above), all you need to do is work with the likelihood. Let $hat{beta}$ be a maximum likelihood estimator of the data ${y_i | x_i}$. Now, writing $beta = D alpha$, we can use equation (1) to show that $$sum_{i=1}^n l(y_i, bar{x}_i^T alpha) = sum_{i=1}^n lleft(y_i, (x_i^T D) (D^{-1} beta)right) leq sum_{i=1}^n l(y_i, x_i^T hatbeta) = sum_{i=1}^n l(y_i, bar{x}_i^T D^{-1} hat{beta}).$$ That is, $D^{-1} hat{beta}$ satisfies equation (2) and is therefore a maximum likelihood estimator with respect to the data ${y_i | bar{x}_i}$. This is the invariance property you noticed.



                (For what it's worth, there's a lot of room for generalizing this argument beyond logistic regression: did we need independent observations? did we need the matrix $D$ to be diagonal? did we need a binary response? did we need the use logit? What notation would you change for this argument to work in different scenarios?)







                share|cite|improve this answer










                New contributor




                user551504 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                share|cite|improve this answer



                share|cite|improve this answer








                edited 22 hours ago





















                New contributor




                user551504 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.









                answered 23 hours ago









                user551504user551504

                612




                612




                New contributor




                user551504 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.





                New contributor





                user551504 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






                user551504 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.






























                    draft saved

                    draft discarded




















































                    Thanks for contributing an answer to Cross Validated!


                    • 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%2fstats.stackexchange.com%2fquestions%2f399318%2finvariance-of-results-when-scaling-explanatory-variables-in-logistic-regression%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

                    Færeyskur hestur Heimild | Tengill | Tilvísanir | LeiðsagnarvalRossið - síða um færeyska hrossið á færeyskuGott ár hjá færeyska hestinum

                    He _____ here since 1970 . Answer needed [closed]What does “since he was so high” mean?Meaning of “catch birds for”?How do I ensure “since” takes the meaning I want?“Who cares here” meaningWhat does “right round toward” mean?the time tense (had now been detected)What does the phrase “ring around the roses” mean here?Correct usage of “visited upon”Meaning of “foiled rail sabotage bid”It was the third time I had gone to Rome or It is the third time I had been to Rome

                    Slayer Innehåll Historia | Stil, komposition och lyrik | Bandets betydelse och framgångar | Sidoprojekt och samarbeten | Kontroverser | Medlemmar | Utmärkelser och nomineringar | Turnéer och festivaler | Diskografi | Referenser | Externa länkar | Navigeringsmenywww.slayer.net”Metal Massacre vol. 1””Metal Massacre vol. 3””Metal Massacre Volume III””Show No Mercy””Haunting the Chapel””Live Undead””Hell Awaits””Reign in Blood””Reign in Blood””Gold & Platinum – Reign in Blood””Golden Gods Awards Winners”originalet”Kerrang! Hall Of Fame””Slayer Looks Back On 37-Year Career In New Video Series: Part Two””South of Heaven””Gold & Platinum – South of Heaven””Seasons in the Abyss””Gold & Platinum - Seasons in the Abyss””Divine Intervention””Divine Intervention - Release group by Slayer””Gold & Platinum - Divine Intervention””Live Intrusion””Undisputed Attitude””Abolish Government/Superficial Love””Release “Slatanic Slaughter: A Tribute to Slayer” by Various Artists””Diabolus in Musica””Soundtrack to the Apocalypse””God Hates Us All””Systematic - Relationships””War at the Warfield””Gold & Platinum - War at the Warfield””Soundtrack to the Apocalypse””Gold & Platinum - Still Reigning””Metallica, Slayer, Iron Mauden Among Winners At Metal Hammer Awards””Eternal Pyre””Eternal Pyre - Slayer release group””Eternal Pyre””Metal Storm Awards 2006””Kerrang! Hall Of Fame””Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Bullet-For My Valentine booed at Metal Hammer Golden Gods Awards””Unholy Aliance””The End Of Slayer?””Slayer: We Could Thrash Out Two More Albums If We're Fast Enough...””'The Unholy Alliance: Chapter III' UK Dates Added”originalet”Megadeth And Slayer To Co-Headline 'Canadian Carnage' Trek”originalet”World Painted Blood””Release “World Painted Blood” by Slayer””Metallica Heading To Cinemas””Slayer, Megadeth To Join Forces For 'European Carnage' Tour - Dec. 18, 2010”originalet”Slayer's Hanneman Contracts Acute Infection; Band To Bring In Guest Guitarist””Cannibal Corpse's Pat O'Brien Will Step In As Slayer's Guest Guitarist”originalet”Slayer’s Jeff Hanneman Dead at 49””Dave Lombardo Says He Made Only $67,000 In 2011 While Touring With Slayer””Slayer: We Do Not Agree With Dave Lombardo's Substance Or Timeline Of Events””Slayer Welcomes Drummer Paul Bostaph Back To The Fold””Slayer Hope to Unveil Never-Before-Heard Jeff Hanneman Material on Next Album””Slayer Debut New Song 'Implode' During Surprise Golden Gods Appearance””Release group Repentless by Slayer””Repentless - Slayer - Credits””Slayer””Metal Storm Awards 2015””Slayer - to release comic book "Repentless #1"””Slayer To Release 'Repentless' 6.66" Vinyl Box Set””BREAKING NEWS: Slayer Announce Farewell Tour””Slayer Recruit Lamb of God, Anthrax, Behemoth + Testament for Final Tour””Slayer lägger ner efter 37 år””Slayer Announces Second North American Leg Of 'Final' Tour””Final World Tour””Slayer Announces Final European Tour With Lamb of God, Anthrax And Obituary””Slayer To Tour Europe With Lamb of God, Anthrax And Obituary””Slayer To Play 'Last French Show Ever' At Next Year's Hellfst””Slayer's Final World Tour Will Extend Into 2019””Death Angel's Rob Cavestany On Slayer's 'Farewell' Tour: 'Some Of Us Could See This Coming'””Testament Has No Plans To Retire Anytime Soon, Says Chuck Billy””Anthrax's Scott Ian On Slayer's 'Farewell' Tour Plans: 'I Was Surprised And I Wasn't Surprised'””Slayer””Slayer's Morbid Schlock””Review/Rock; For Slayer, the Mania Is the Message””Slayer - Biography””Slayer - Reign In Blood”originalet”Dave Lombardo””An exclusive oral history of Slayer”originalet”Exclusive! Interview With Slayer Guitarist Jeff Hanneman”originalet”Thinking Out Loud: Slayer's Kerry King on hair metal, Satan and being polite””Slayer Lyrics””Slayer - Biography””Most influential artists for extreme metal music””Slayer - Reign in Blood””Slayer guitarist Jeff Hanneman dies aged 49””Slatanic Slaughter: A Tribute to Slayer””Gateway to Hell: A Tribute to Slayer””Covered In Blood””Slayer: The Origins of Thrash in San Francisco, CA.””Why They Rule - #6 Slayer”originalet”Guitar World's 100 Greatest Heavy Metal Guitarists Of All Time”originalet”The fans have spoken: Slayer comes out on top in readers' polls”originalet”Tribute to Jeff Hanneman (1964-2013)””Lamb Of God Frontman: We Sound Like A Slayer Rip-Off””BEHEMOTH Frontman Pays Tribute To SLAYER's JEFF HANNEMAN””Slayer, Hatebreed Doing Double Duty On This Year's Ozzfest””System of a Down””Lacuna Coil’s Andrea Ferro Talks Influences, Skateboarding, Band Origins + More””Slayer - Reign in Blood””Into The Lungs of Hell””Slayer rules - en utställning om fans””Slayer and Their Fans Slashed Through a No-Holds-Barred Night at Gas Monkey””Home””Slayer””Gold & Platinum - The Big 4 Live from Sofia, Bulgaria””Exclusive! Interview With Slayer Guitarist Kerry King””2008-02-23: Wiltern, Los Angeles, CA, USA””Slayer's Kerry King To Perform With Megadeth Tonight! - Oct. 21, 2010”originalet”Dave Lombardo - Biography”Slayer Case DismissedArkiveradUltimate Classic Rock: Slayer guitarist Jeff Hanneman dead at 49.”Slayer: "We could never do any thing like Some Kind Of Monster..."””Cannibal Corpse'S Pat O'Brien Will Step In As Slayer'S Guest Guitarist | The Official Slayer Site”originalet”Slayer Wins 'Best Metal' Grammy Award””Slayer Guitarist Jeff Hanneman Dies””Kerrang! Awards 2006 Blog: Kerrang! Hall Of Fame””Kerrang! Awards 2013: Kerrang! Legend”originalet”Metallica, Slayer, Iron Maien Among Winners At Metal Hammer Awards””Metal Hammer Golden Gods Awards””Bullet For My Valentine Booed At Metal Hammer Golden Gods Awards””Metal Storm Awards 2006””Metal Storm Awards 2015””Slayer's Concert History””Slayer - Relationships””Slayer - Releases”Slayers officiella webbplatsSlayer på MusicBrainzOfficiell webbplatsSlayerSlayerr1373445760000 0001 1540 47353068615-5086262726cb13906545x(data)6033143kn20030215029