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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;
$begingroup$
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.
- 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.
- Is there any point to scaling explanatory variables in logistic regression, in the absence of regularization?
Thanks.
regression logistic regression-coefficients
$endgroup$
add a comment |
$begingroup$
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.
- 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.
- Is there any point to scaling explanatory variables in logistic regression, in the absence of regularization?
Thanks.
regression logistic regression-coefficients
$endgroup$
add a comment |
$begingroup$
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.
- 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.
- Is there any point to scaling explanatory variables in logistic regression, in the absence of regularization?
Thanks.
regression logistic regression-coefficients
$endgroup$
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.
- 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.
- Is there any point to scaling explanatory variables in logistic regression, in the absence of regularization?
Thanks.
regression logistic regression-coefficients
regression logistic regression-coefficients
asked Mar 25 at 14:36
PM.PM.
2161211
2161211
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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?)
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
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?)
$endgroup$
add a comment |
$begingroup$
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?)
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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?)
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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?)
edited Mar 25 at 19:27
answered Mar 25 at 19:13
user551504user551504
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