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






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      asked 12 hours ago









      PM.PM.

      2021210




      2021210






















          2 Answers
          2






          active

          oldest

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          8












          $begingroup$

          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.
            11 hours ago






          • 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
            11 hours ago



















          4












          $begingroup$

          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.






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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            8












            $begingroup$

            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.
              11 hours ago






            • 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
              11 hours ago
















            8












            $begingroup$

            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.
              11 hours ago






            • 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
              11 hours ago














            8












            8








            8





            $begingroup$

            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 11 hours ago









            Christoph HanckChristoph Hanck

            17.8k34274




            17.8k34274












            • $begingroup$
              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.
              11 hours ago






            • 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
              11 hours ago


















            • $begingroup$
              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.
              11 hours ago






            • 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
              11 hours ago
















            $begingroup$
            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.
            11 hours ago




            $begingroup$
            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.
            11 hours ago




            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
            11 hours ago




            $begingroup$
            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
            11 hours ago













            4












            $begingroup$

            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.






            $endgroup$


















              4












              $begingroup$

              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?)






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                4












                4








                4





                $begingroup$

                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










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                $endgroup$



                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










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                answered 7 hours ago









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