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

                    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

                    Bunad

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