Invariance of results when scaling explanatory variables in logistic regression, is there a proof? Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar ManaraLogistic regression and scaling of featuresHow to do logistic regression subset selection?Is there a simple rule for interpretation of Interactions (and their directions) in binary logistic regression?Is there a binomial regression model that captures data with fat tails?Variance-Covariance Matrix for $l_1$ regularized binomial logistic regressionTesting for coefficients significance in Lasso logistic regressionIs it possible to simulate logistic regression without randomness?Logistic Regression is a Convex Problem but my results show otherwise?Logistic Regression - Covariance Matrix of “Response” Residual Vector“Return values” of univariate logistic regressionLogistic regression including instrumental variable (“ivprobit” in R) has coefficients with much reduced significance

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



Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar ManaraLogistic regression and scaling of featuresHow to do logistic regression subset selection?Is there a simple rule for interpretation of Interactions (and their directions) in binary logistic regression?Is there a binomial regression model that captures data with fat tails?Variance-Covariance Matrix for $l_1$ regularized binomial logistic regressionTesting for coefficients significance in Lasso logistic regressionIs it possible to simulate logistic regression without randomness?Logistic Regression is a Convex Problem but my results show otherwise?Logistic Regression - Covariance Matrix of “Response” Residual Vector“Return values” of univariate logistic regressionLogistic regression including instrumental variable (“ivprobit” in R) has coefficients with much reduced significance



.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;








6












$begingroup$


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



$$mathbfbeta=(mathbfX^T X)^-1mathbfX^T y$$



or



$(mathbfX^T X)mathbfbeta=mathbfX^T y tag2labeleq2$



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
$mathbfy=mathbfX beta tag3labeleq3$



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



Scaling the explanatory variables with a $(p+1) times (p+1) $ diagonal matrix $mathbfD$, whose entries are the scaling factors
$
mathbfX^s = mathbfXD tag4labeleq4$



$mathbfX^s$ and $mathbfbeta^s$ satisfy $eqrefeq2$:



$$(mathbfD^TX^T XD)mathbfbeta^s =mathbfD^TX^T y$$



so



$$mathbfX^T XDmathbfbeta^s =mathbfX^T y$$



$$Rightarrow mathbfD beta^s = (mathbfX^T X)^-1mathbfX^T y=mathbfbeta$$



$Rightarrow mathbfbeta^s=mathbfD^-1mathbfbeta tag5labeleq5$



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 $eqrefeq4,eqrefeq5,eqrefeq3$



$$mathbfy^s=mathbfX^s beta^s = mathbfX D D^-1beta=mathbfX beta=mathbfy$$
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



    $$mathbfbeta=(mathbfX^T X)^-1mathbfX^T y$$



    or



    $(mathbfX^T X)mathbfbeta=mathbfX^T y tag2labeleq2$



    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
    $mathbfy=mathbfX beta tag3labeleq3$



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



    Scaling the explanatory variables with a $(p+1) times (p+1) $ diagonal matrix $mathbfD$, whose entries are the scaling factors
    $
    mathbfX^s = mathbfXD tag4labeleq4$



    $mathbfX^s$ and $mathbfbeta^s$ satisfy $eqrefeq2$:



    $$(mathbfD^TX^T XD)mathbfbeta^s =mathbfD^TX^T y$$



    so



    $$mathbfX^T XDmathbfbeta^s =mathbfX^T y$$



    $$Rightarrow mathbfD beta^s = (mathbfX^T X)^-1mathbfX^T y=mathbfbeta$$



    $Rightarrow mathbfbeta^s=mathbfD^-1mathbfbeta tag5labeleq5$



    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 $eqrefeq4,eqrefeq5,eqrefeq3$



    $$mathbfy^s=mathbfX^s beta^s = mathbfX D D^-1beta=mathbfX beta=mathbfy$$
    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



      $$mathbfbeta=(mathbfX^T X)^-1mathbfX^T y$$



      or



      $(mathbfX^T X)mathbfbeta=mathbfX^T y tag2labeleq2$



      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
      $mathbfy=mathbfX beta tag3labeleq3$



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



      Scaling the explanatory variables with a $(p+1) times (p+1) $ diagonal matrix $mathbfD$, whose entries are the scaling factors
      $
      mathbfX^s = mathbfXD tag4labeleq4$



      $mathbfX^s$ and $mathbfbeta^s$ satisfy $eqrefeq2$:



      $$(mathbfD^TX^T XD)mathbfbeta^s =mathbfD^TX^T y$$



      so



      $$mathbfX^T XDmathbfbeta^s =mathbfX^T y$$



      $$Rightarrow mathbfD beta^s = (mathbfX^T X)^-1mathbfX^T y=mathbfbeta$$



      $Rightarrow mathbfbeta^s=mathbfD^-1mathbfbeta tag5labeleq5$



      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 $eqrefeq4,eqrefeq5,eqrefeq3$



      $$mathbfy^s=mathbfX^s beta^s = mathbfX D D^-1beta=mathbfX beta=mathbfy$$
      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



      $$mathbfbeta=(mathbfX^T X)^-1mathbfX^T y$$



      or



      $(mathbfX^T X)mathbfbeta=mathbfX^T y tag2labeleq2$



      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
      $mathbfy=mathbfX beta tag3labeleq3$



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



      Scaling the explanatory variables with a $(p+1) times (p+1) $ diagonal matrix $mathbfD$, whose entries are the scaling factors
      $
      mathbfX^s = mathbfXD tag4labeleq4$



      $mathbfX^s$ and $mathbfbeta^s$ satisfy $eqrefeq2$:



      $$(mathbfD^TX^T XD)mathbfbeta^s =mathbfD^TX^T y$$



      so



      $$mathbfX^T XDmathbfbeta^s =mathbfX^T y$$



      $$Rightarrow mathbfD beta^s = (mathbfX^T X)^-1mathbfX^T y=mathbfbeta$$



      $Rightarrow mathbfbeta^s=mathbfD^-1mathbfbeta tag5labeleq5$



      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 $eqrefeq4,eqrefeq5,eqrefeq3$



      $$mathbfy^s=mathbfX^s beta^s = mathbfX D D^-1beta=mathbfX beta=mathbfy$$
      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 Mar 25 at 14:36









      PM.PM.

      2161211




      2161211




















          2 Answers
          2






          active

          oldest

          votes


















          9












          $begingroup$

          Here is a heuristic idea:



          The likelihood for a logistic regression model is
          $$ ell(beta|y) propto prod_ileft(fracexp(x_i'beta)1+exp(x_i'beta)right)^y_ileft(frac11+exp(x_i'beta)right)^1-y_i
          $$

          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.
            Mar 25 at 15:36






          • 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
            Mar 25 at 15:40


















          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 $(XD) (D^-1beta)$", 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 $y_i | x_i stackrelind.sim mathrmbernoullileft[ mathrmlogit^-1 (x_i^T beta) right]$ be independent draws according to the logistic regression model, where $x_i in mathbbR^p+1$ 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 $barx_i = D x_i$, where $D$ is diagonal matrix with all diagonal entries nonzero. By definition of maximum likelihood estimation, we know that maximum likelihood estimators $hatbeta$ of the data $ x_i$ satisfy that $$sum_i=1^n l(y_i, x_i^T beta) leq sum_i=1^n l(y_i, x_i^T hatbeta) tag1$$ for all coefficients $beta in mathbbR^p$, and that maximum likelihood estimators for the data $y_i $ satisfy that $$sum_i=1^n l(y_i, barx_i^T alpha) leq sum_i=1^n l(y_i, barx_i^T hatalpha) tag2$$ for all coefficients $alpha in mathbbR^p$.



          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 $hatbeta$ be a maximum likelihood estimator of the data $ x_i$. Now, writing $beta = D alpha$, we can use equation (1) to show that $$sum_i=1^n l(y_i, barx_i^T alpha) = sum_i=1^n lleft(y_i, (x_i^T D) (D^-1 beta)right) leq sum_i=1^n l(y_i, x_i^T hatbeta) = sum_i=1^n l(y_i, barx_i^T D^-1 hatbeta).$$ That is, $D^-1 hatbeta$ satisfies equation (2) and is therefore a maximum likelihood estimator with respect to the data $y_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











          $endgroup$













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            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
            $$ ell(beta|y) propto prod_ileft(fracexp(x_i'beta)1+exp(x_i'beta)right)^y_ileft(frac11+exp(x_i'beta)right)^1-y_i
            $$

            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.
              Mar 25 at 15:36






            • 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
              Mar 25 at 15:40















            9












            $begingroup$

            Here is a heuristic idea:



            The likelihood for a logistic regression model is
            $$ ell(beta|y) propto prod_ileft(fracexp(x_i'beta)1+exp(x_i'beta)right)^y_ileft(frac11+exp(x_i'beta)right)^1-y_i
            $$

            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.
              Mar 25 at 15:36






            • 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
              Mar 25 at 15:40













            9












            9








            9





            $begingroup$

            Here is a heuristic idea:



            The likelihood for a logistic regression model is
            $$ ell(beta|y) propto prod_ileft(fracexp(x_i'beta)1+exp(x_i'beta)right)^y_ileft(frac11+exp(x_i'beta)right)^1-y_i
            $$

            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
            $$ ell(beta|y) propto prod_ileft(fracexp(x_i'beta)1+exp(x_i'beta)right)^y_ileft(frac11+exp(x_i'beta)right)^1-y_i
            $$

            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 Mar 25 at 15:21









            Christoph HanckChristoph Hanck

            18k34275




            18k34275











            • $begingroup$
              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.
              Mar 25 at 15:36






            • 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
              Mar 25 at 15:40
















            • $begingroup$
              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.
              Mar 25 at 15:36






            • 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
              Mar 25 at 15:40















            $begingroup$
            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.
            Mar 25 at 15:36




            $begingroup$
            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.
            Mar 25 at 15:36




            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
            Mar 25 at 15:40




            $begingroup$
            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
            Mar 25 at 15:40













            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 $(XD) (D^-1beta)$", 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 $y_i | x_i stackrelind.sim mathrmbernoullileft[ mathrmlogit^-1 (x_i^T beta) right]$ be independent draws according to the logistic regression model, where $x_i in mathbbR^p+1$ 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 $barx_i = D x_i$, where $D$ is diagonal matrix with all diagonal entries nonzero. By definition of maximum likelihood estimation, we know that maximum likelihood estimators $hatbeta$ of the data $ x_i$ satisfy that $$sum_i=1^n l(y_i, x_i^T beta) leq sum_i=1^n l(y_i, x_i^T hatbeta) tag1$$ for all coefficients $beta in mathbbR^p$, and that maximum likelihood estimators for the data $y_i $ satisfy that $$sum_i=1^n l(y_i, barx_i^T alpha) leq sum_i=1^n l(y_i, barx_i^T hatalpha) tag2$$ for all coefficients $alpha in mathbbR^p$.



            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 $hatbeta$ be a maximum likelihood estimator of the data $ x_i$. Now, writing $beta = D alpha$, we can use equation (1) to show that $$sum_i=1^n l(y_i, barx_i^T alpha) = sum_i=1^n lleft(y_i, (x_i^T D) (D^-1 beta)right) leq sum_i=1^n l(y_i, x_i^T hatbeta) = sum_i=1^n l(y_i, barx_i^T D^-1 hatbeta).$$ That is, $D^-1 hatbeta$ satisfies equation (2) and is therefore a maximum likelihood estimator with respect to the data $y_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











            $endgroup$

















              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 $(XD) (D^-1beta)$", 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 $y_i | x_i stackrelind.sim mathrmbernoullileft[ mathrmlogit^-1 (x_i^T beta) right]$ be independent draws according to the logistic regression model, where $x_i in mathbbR^p+1$ 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 $barx_i = D x_i$, where $D$ is diagonal matrix with all diagonal entries nonzero. By definition of maximum likelihood estimation, we know that maximum likelihood estimators $hatbeta$ of the data $ x_i$ satisfy that $$sum_i=1^n l(y_i, x_i^T beta) leq sum_i=1^n l(y_i, x_i^T hatbeta) tag1$$ for all coefficients $beta in mathbbR^p$, and that maximum likelihood estimators for the data $y_i $ satisfy that $$sum_i=1^n l(y_i, barx_i^T alpha) leq sum_i=1^n l(y_i, barx_i^T hatalpha) tag2$$ for all coefficients $alpha in mathbbR^p$.



              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 $hatbeta$ be a maximum likelihood estimator of the data $ x_i$. Now, writing $beta = D alpha$, we can use equation (1) to show that $$sum_i=1^n l(y_i, barx_i^T alpha) = sum_i=1^n lleft(y_i, (x_i^T D) (D^-1 beta)right) leq sum_i=1^n l(y_i, x_i^T hatbeta) = sum_i=1^n l(y_i, barx_i^T D^-1 hatbeta).$$ That is, $D^-1 hatbeta$ satisfies equation (2) and is therefore a maximum likelihood estimator with respect to the data $y_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











              $endgroup$















                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 $(XD) (D^-1beta)$", 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 $y_i | x_i stackrelind.sim mathrmbernoullileft[ mathrmlogit^-1 (x_i^T beta) right]$ be independent draws according to the logistic regression model, where $x_i in mathbbR^p+1$ 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 $barx_i = D x_i$, where $D$ is diagonal matrix with all diagonal entries nonzero. By definition of maximum likelihood estimation, we know that maximum likelihood estimators $hatbeta$ of the data $ x_i$ satisfy that $$sum_i=1^n l(y_i, x_i^T beta) leq sum_i=1^n l(y_i, x_i^T hatbeta) tag1$$ for all coefficients $beta in mathbbR^p$, and that maximum likelihood estimators for the data $y_i $ satisfy that $$sum_i=1^n l(y_i, barx_i^T alpha) leq sum_i=1^n l(y_i, barx_i^T hatalpha) tag2$$ for all coefficients $alpha in mathbbR^p$.



                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 $hatbeta$ be a maximum likelihood estimator of the data $ x_i$. Now, writing $beta = D alpha$, we can use equation (1) to show that $$sum_i=1^n l(y_i, barx_i^T alpha) = sum_i=1^n lleft(y_i, (x_i^T D) (D^-1 beta)right) leq sum_i=1^n l(y_i, x_i^T hatbeta) = sum_i=1^n l(y_i, barx_i^T D^-1 hatbeta).$$ That is, $D^-1 hatbeta$ satisfies equation (2) and is therefore a maximum likelihood estimator with respect to the data $y_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











                $endgroup$



                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 $(XD) (D^-1beta)$", 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 $y_i | x_i stackrelind.sim mathrmbernoullileft[ mathrmlogit^-1 (x_i^T beta) right]$ be independent draws according to the logistic regression model, where $x_i in mathbbR^p+1$ 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 $barx_i = D x_i$, where $D$ is diagonal matrix with all diagonal entries nonzero. By definition of maximum likelihood estimation, we know that maximum likelihood estimators $hatbeta$ of the data $ x_i$ satisfy that $$sum_i=1^n l(y_i, x_i^T beta) leq sum_i=1^n l(y_i, x_i^T hatbeta) tag1$$ for all coefficients $beta in mathbbR^p$, and that maximum likelihood estimators for the data $y_i $ satisfy that $$sum_i=1^n l(y_i, barx_i^T alpha) leq sum_i=1^n l(y_i, barx_i^T hatalpha) tag2$$ for all coefficients $alpha in mathbbR^p$.



                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 $hatbeta$ be a maximum likelihood estimator of the data $ x_i$. Now, writing $beta = D alpha$, we can use equation (1) to show that $$sum_i=1^n l(y_i, barx_i^T alpha) = sum_i=1^n lleft(y_i, (x_i^T D) (D^-1 beta)right) leq sum_i=1^n l(y_i, x_i^T hatbeta) = sum_i=1^n l(y_i, barx_i^T D^-1 hatbeta).$$ That is, $D^-1 hatbeta$ satisfies equation (2) and is therefore a maximum likelihood estimator with respect to the data $y_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














                share|cite|improve this answer



                share|cite|improve this answer








                edited Mar 25 at 19:27

























                answered Mar 25 at 19:13









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