contrast in probabilities when using lrm

[sdaza]

Apparently, I cannot use a function such as 1/(1 + exp(-x)) in the contrast command.

age <- rnorm(200,40,12)
sex <- factor(sample(c('female','male'),200,TRUE))
logit <- (sex=='male') + (age-40)/5
y <- ifelse(runif(200) <= plogis(logit), 1, 0)
f <- lrm(y ~ pol(age,2)*sex)
getProb <- function (x)  { 1/(1 + exp(-x)) }
contrast(f, list(sex='female', age=30), list(sex='male', age=40), fun = getProb)

Any ideas on how to get a contrast in probabilities?

[harrelfe]

fun is an argument to print.contrast.rms. So do

k <- contrast(...)

print(k, fun=plogis) # or fun=getProb (same thing)

---

Frank E Harrell Jr Professor and Chairman School of Medicine

Department of Biostatistics Vanderbilt University

On Thu, Jun 30, 2016 at 11:48 AM, Sebastian Daza notifications@github.com
wrote:

Apparently, I cannot use a function such as 1/(1 + exp(-x)) in the
contrast command.

age <- rnorm(200,40,12)
sex <- factor(sample(c('female','male'),200,TRUE))
logit <- (sex=='male') + (age-40)/5
y <- ifelse(runif(200) <= plogis(logit), 1, 0)
f <- lrm(y ~ pol(age,2)*sex)
getProb <- function (x) { 1/(1 + exp(-x)) }
contrast(f, list(sex='female', age=30), list(sex='male', age=40), fun = getProb)

Any ideas on how to get a contrast in probabilities?

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[sdaza]

Thank you, I got it. But this is not what I want. I would like to contrast predicted probabilities: if a logit coefficient is negative, the contrast in predicted probabilities should be also negative, this is not what I get. So, the transformation should be applied, I guess, at the prediction procedure. Do you think that is possible? Do you have any suggestion on how to do it?

Thank you again,
Sebastian

[harrelfe]

You would have to program complex standard error calculations to be able to
do that. Probability differences are highly specific to the adjustment
covariate settings so I do not recommend that approach. If you really want
to do it, bootstrap the entire process to get confidence intervals.

---

Frank E Harrell Jr Professor and Chairman School of Medicine

Department of Biostatistics Vanderbilt University

On Fri, Jul 1, 2016 at 8:12 AM, Sebastian Daza notifications@github.com
wrote:

Thank you, I got. But this is not what I want. I would like to contrast
predicted probabilities: if a logit coefficient is negative, the contrast
in predicted probabilities should be also negative, this is not what I get.
So, the transformation needs to be done, I guess, at the prediction
procedure. Do you think that is possible? Do you have any suggestion on how
to do it?

Thank you again,
Sebastian

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[ruddjm]

Hi Frank,

I'm puzzled by why using the transformation in print.contrast.rms
doesn't do what he wants. Wouldn't doing print(k, fun=plogis) as you
suggest return as contrast in probabilities? How is that result
different from doing the complex standard error calculations (assuming
he did it correctly)?

I copied his original question and your reply below for reference.

Thanks,

JoAnn

Your reply:
fun is an argument to print.contrast.rms. So do

k <- contrast(...)

print(k, fun=plogis) # or fun=getProb (same thing)

Original question:

Apparently, I cannot use a function such as |1/(1 + exp(-x))| in the
contrast command.

|age <- rnorm(200,40,12)
sex <- factor(sample(c('female','male'),200,TRUE))
logit <- (sex=='male') + (age-40)/5
y <- ifelse(runif(200) <= plogis(logit), 1, 0)
f <- lrm(y ~ pol(age,2)*sex)
getProb <- function (x) { 1/(1 + exp(-x)) }
contrast(f, list(sex='female', age=30), list(sex='male', age=40), fun = getProb)
|

Any ideas on how to get a contrast in probabilities?

JoAnn Rudd Alvarez
http://biostat.mc.vanderbilt.edu/JoAnnAlvarez

On 07/01/2016 08:53 AM, Frank Harrell wrote:

You would have to program complex standard error calculations to be
able to
do that. Probability differences are highly specific to the adjustment
covariate settings so I do not recommend that approach. If you really want
to do it, bootstrap the entire process to get confidence intervals.

---

Frank E Harrell Jr Professor and Chairman School of Medicine

Department of Biostatistics Vanderbilt University

On Fri, Jul 1, 2016 at 8:12 AM, Sebastian Daza notifications@github.com
wrote:

Thank you, I got. But this is not what I want. I would like to contrast
predicted probabilities: if a logit coefficient is negative, the
contrast
in predicted probabilities should be also negative, this is not what
I get.
So, the transformation needs to be done, I guess, at the prediction
procedure. Do you think that is possible? Do you have any suggestion
on how
to do it?

Thank you again,
Sebastian

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[harrelfe]

JoAnn,

Why would a nonlinear transformation work in this way? contrast will
correctly get you confidence intervals for transformations of contrasts on
the linear predictor scale. It has no way to know about differences in
nonlinear transformations of the linear predictor. fun= in
print.contrast.rms refers to a last-second transformation of the contrast

contrast.rms handles bootstrapped estimates and it may be possible to
extend the function to use bootstrap reps for confidence intervals on the
right scale.

BUT note that differences in probabilities is not a great scale anyway.

Frank

---

Frank E Harrell Jr Professor and Chairman School of Medicine

Department of Biostatistics Vanderbilt University

On Fri, Jul 1, 2016 at 11:39 AM, JoAnn Rudd Alvarez <
notifications@github.com> wrote:

Hi Frank,

I'm puzzled by why using the transformation in print.contrast.rms
doesn't do what he wants. Wouldn't doing print(k, fun=plogis) as you
suggest return as contrast in probabilities? How is that result
different from doing the complex standard error calculations (assuming
he did it correctly)?

I copied his original question and your reply below for reference.

Thanks,

JoAnn

Your reply:
fun is an argument to print.contrast.rms. So do

k <- contrast(...)

print(k, fun=plogis) # or fun=getProb (same thing)

Original question:

Apparently, I cannot use a function such as |1/(1 + exp(-x))| in the
contrast command.

|age <- rnorm(200,40,12)
sex <- factor(sample(c('female','male'),200,TRUE))
logit <- (sex=='male') + (age-40)/5
y <- ifelse(runif(200) <= plogis(logit), 1, 0)
f <- lrm(y ~ pol(age,2)*sex)
getProb <- function (x) { 1/(1 + exp(-x)) }
contrast(f, list(sex='female', age=30), list(sex='male', age=40), fun =
getProb)
|

Any ideas on how to get a contrast in probabilities?

JoAnn Rudd Alvarez
http://biostat.mc.vanderbilt.edu/JoAnnAlvarez

On 07/01/2016 08:53 AM, Frank Harrell wrote:

You would have to program complex standard error calculations to be
able to
do that. Probability differences are highly specific to the adjustment
covariate settings so I do not recommend that approach. If you really
want
to do it, bootstrap the entire process to get confidence intervals.

---

Frank E Harrell Jr Professor and Chairman School of Medicine

Department of Biostatistics Vanderbilt University

On Fri, Jul 1, 2016 at 8:12 AM, Sebastian Daza <notifications@github.com

wrote:

Thank you, I got. But this is not what I want. I would like to contrast
predicted probabilities: if a logit coefficient is negative, the
contrast
in predicted probabilities should be also negative, this is not what
I get.
So, the transformation needs to be done, I guess, at the prediction
procedure. Do you think that is possible? Do you have any suggestion
on how
to do it?

Thank you again,
Sebastian

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[ruddjm]

JoAnn Rudd Alvarez
http://biostat.mc.vanderbilt.edu/JoAnnAlvarez

On 07/01/2016 12:52 PM, Frank Harrell wrote:

JoAnn,

Why would a nonlinear transformation work in this way? contrast will
correctly get you confidence intervals for transformations of contrasts on
the linear predictor scale. It has no way to know about differences in
nonlinear transformations of the linear predictor. fun= in
print.contrast.rms refers to a last-second transformation of the contrast
Yes, so it will just apply plogis to the betahat and to each of the
limits. But what is the difference between this last-second
transformation and what Sebastian wants?

contrast.rms handles bootstrapped estimates and it may be possible to
extend the function to use bootstrap reps for confidence intervals on the
right scale.

BUT note that differences in probabilities is not a great scale anyway.
I thought differences on a probability scale are nice because they're
more understandable than odds ratios.

By the way, thanks for answering my questions.

Frank

---

Frank E Harrell Jr Professor and Chairman School of Medicine

Department of Biostatistics Vanderbilt University

On Fri, Jul 1, 2016 at 11:39 AM, JoAnn Rudd Alvarez <
notifications@github.com> wrote:

Hi Frank,

I'm puzzled by why using the transformation in print.contrast.rms
doesn't do what he wants. Wouldn't doing print(k, fun=plogis) as you
suggest return as contrast in probabilities? How is that result
different from doing the complex standard error calculations (assuming
he did it correctly)?

I copied his original question and your reply below for reference.

Thanks,

JoAnn

Your reply:
fun is an argument to print.contrast.rms. So do

k <- contrast(...)

print(k, fun=plogis) # or fun=getProb (same thing)

Original question:

Apparently, I cannot use a function such as |1/(1 + exp(-x))| in the
contrast command.

|age <- rnorm(200,40,12)
sex <- factor(sample(c('female','male'),200,TRUE))
logit <- (sex=='male') + (age-40)/5
y <- ifelse(runif(200) <= plogis(logit), 1, 0)
f <- lrm(y ~ pol(age,2)*sex)
getProb <- function (x) { 1/(1 + exp(-x)) }
contrast(f, list(sex='female', age=30), list(sex='male', age=40), fun =
getProb)
|

Any ideas on how to get a contrast in probabilities?

JoAnn Rudd Alvarez
http://biostat.mc.vanderbilt.edu/JoAnnAlvarez

On 07/01/2016 08:53 AM, Frank Harrell wrote:

You would have to program complex standard error calculations to be
able to
do that. Probability differences are highly specific to the adjustment
covariate settings so I do not recommend that approach. If you really
want
to do it, bootstrap the entire process to get confidence intervals.

---

Frank E Harrell Jr Professor and Chairman School of Medicine

Department of Biostatistics Vanderbilt University

On Fri, Jul 1, 2016 at 8:12 AM, Sebastian Daza
<notifications@github.com

wrote:

Thank you, I got. But this is not what I want. I would like to
contrast
predicted probabilities: if a logit coefficient is negative, the
contrast
in predicted probabilities should be also negative, this is not what
I get.
So, the transformation needs to be done, I guess, at the prediction
procedure. Do you think that is possible? Do you have any suggestion
on how
to do it?

Thank you again,
Sebastian

—
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#26 (comment),
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[sdaza]

The function is applying plogis to log odds ratios (as @harrelfe pointed out). If the difference is zero, you get a value of 0.5. The difference between predicted probabilities should be also 0.

I agree about the use of odd ratios and logit coefficients. I think we shouldn't use them to communicate our results and models. Look at this.

You can use the package Zelig to simulate predicted values.

[harrelfe]

It applies plogis to the log odds RATIO which doesnt make sense.
On Jul 1, 2016 11:00 AM, "JoAnn Rudd Alvarez" notifications@github.com
wrote:

JoAnn Rudd Alvarez
http://biostat.mc.vanderbilt.edu/JoAnnAlvarez

On 07/01/2016 12:52 PM, Frank Harrell wrote:

JoAnn,

Why would a nonlinear transformation work in this way? contrast will
correctly get you confidence intervals for transformations of contrasts
on
the linear predictor scale. It has no way to know about differences in
nonlinear transformations of the linear predictor. fun= in
print.contrast.rms refers to a last-second transformation of the contrast
Yes, so it will just apply plogis to the betahat and to each of the
limits. But what is the difference between this last-second
transformation and what Sebastian wants?

contrast.rms handles bootstrapped estimates and it may be possible to
extend the function to use bootstrap reps for confidence intervals on the
right scale.

BUT note that differences in probabilities is not a great scale anyway.
I thought differences on a probability scale are nice because they're
more understandable than odds ratios.

By the way, thanks for answering my questions.

Frank

---

Frank E Harrell Jr Professor and Chairman School of Medicine

Department of Biostatistics Vanderbilt University

On Fri, Jul 1, 2016 at 11:39 AM, JoAnn Rudd Alvarez <
notifications@github.com> wrote:

Hi Frank,

I'm puzzled by why using the transformation in print.contrast.rms
doesn't do what he wants. Wouldn't doing print(k, fun=plogis) as you
suggest return as contrast in probabilities? How is that result
different from doing the complex standard error calculations (assuming
he did it correctly)?

I copied his original question and your reply below for reference.

Thanks,

JoAnn

Your reply:
fun is an argument to print.contrast.rms. So do

k <- contrast(...)

print(k, fun=plogis) # or fun=getProb (same thing)

Original question:

Apparently, I cannot use a function such as |1/(1 + exp(-x))| in the
contrast command.

|age <- rnorm(200,40,12)
sex <- factor(sample(c('female','male'),200,TRUE))
logit <- (sex=='male') + (age-40)/5
y <- ifelse(runif(200) <= plogis(logit), 1, 0)
f <- lrm(y ~ pol(age,2)*sex)
getProb <- function (x) { 1/(1 + exp(-x)) }
contrast(f, list(sex='female', age=30), list(sex='male', age=40), fun =
getProb)
|

Any ideas on how to get a contrast in probabilities?

JoAnn Rudd Alvarez
http://biostat.mc.vanderbilt.edu/JoAnnAlvarez

On 07/01/2016 08:53 AM, Frank Harrell wrote:

You would have to program complex standard error calculations to be
able to
do that. Probability differences are highly specific to the
adjustment
covariate settings so I do not recommend that approach. If you really
want
to do it, bootstrap the entire process to get confidence intervals.

---

Frank E Harrell Jr Professor and Chairman School of Medicine

Department of Biostatistics Vanderbilt University

On Fri, Jul 1, 2016 at 8:12 AM, Sebastian Daza
<notifications@github.com

wrote:

Thank you, I got. But this is not what I want. I would like to
contrast
predicted probabilities: if a logit coefficient is negative, the
contrast
in predicted probabilities should be also negative, this is not
what
I get.
So, the transformation needs to be done, I guess, at the prediction
procedure. Do you think that is possible? Do you have any
suggestion
on how
to do it?

Thank you again,
Sebastian

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[ruddjm]

Oh, ok, thanks.

JoAnn Rudd Alvarez
http://biostat.mc.vanderbilt.edu/JoAnnAlvarez

On 07/01/2016 03:16 PM, Frank Harrell wrote:

It applies plogis to the log odds RATIO which doesnt make sense.
On Jul 1, 2016 11:00 AM, "JoAnn Rudd Alvarez" notifications@github.com
wrote:

JoAnn Rudd Alvarez
http://biostat.mc.vanderbilt.edu/JoAnnAlvarez

On 07/01/2016 12:52 PM, Frank Harrell wrote:

JoAnn,

Why would a nonlinear transformation work in this way? contrast will
correctly get you confidence intervals for transformations of
contrasts
on
the linear predictor scale. It has no way to know about differences in
nonlinear transformations of the linear predictor. fun= in
print.contrast.rms refers to a last-second transformation of the
contrast
Yes, so it will just apply plogis to the betahat and to each of the
limits. But what is the difference between this last-second
transformation and what Sebastian wants?

contrast.rms handles bootstrapped estimates and it may be possible to
extend the function to use bootstrap reps for confidence intervals
on the
right scale.

BUT note that differences in probabilities is not a great scale
anyway.
I thought differences on a probability scale are nice because they're
more understandable than odds ratios.

By the way, thanks for answering my questions.

Frank

---

Frank E Harrell Jr Professor and Chairman School of Medicine

Department of Biostatistics Vanderbilt University

On Fri, Jul 1, 2016 at 11:39 AM, JoAnn Rudd Alvarez <
notifications@github.com> wrote:

Hi Frank,

I'm puzzled by why using the transformation in print.contrast.rms
doesn't do what he wants. Wouldn't doing print(k, fun=plogis) as you
suggest return as contrast in probabilities? How is that result
different from doing the complex standard error calculations
(assuming
he did it correctly)?

I copied his original question and your reply below for reference.

Thanks,

JoAnn

Your reply:
fun is an argument to print.contrast.rms. So do

k <- contrast(...)

print(k, fun=plogis) # or fun=getProb (same thing)

Original question:

Apparently, I cannot use a function such as |1/(1 + exp(-x))| in the
contrast command.

|age <- rnorm(200,40,12)
sex <- factor(sample(c('female','male'),200,TRUE))
logit <- (sex=='male') + (age-40)/5
y <- ifelse(runif(200) <= plogis(logit), 1, 0)
f <- lrm(y ~ pol(age,2)*sex)
getProb <- function (x) { 1/(1 + exp(-x)) }
contrast(f, list(sex='female', age=30), list(sex='male',
age=40), fun =
getProb)
|

Any ideas on how to get a contrast in probabilities?

JoAnn Rudd Alvarez
http://biostat.mc.vanderbilt.edu/JoAnnAlvarez

On 07/01/2016 08:53 AM, Frank Harrell wrote:

You would have to program complex standard error calculations
to be
able to
do that. Probability differences are highly specific to the
adjustment
covariate settings so I do not recommend that approach. If you
really
want
to do it, bootstrap the entire process to get confidence
intervals.

---

Frank E Harrell Jr Professor and Chairman School of Medicine

Department of Biostatistics Vanderbilt University

On Fri, Jul 1, 2016 at 8:12 AM, Sebastian Daza
<notifications@github.com

wrote:

Thank you, I got. But this is not what I want. I would like to
contrast
predicted probabilities: if a logit coefficient is negative, the
contrast
in predicted probabilities should be also negative, this is not
what
I get.
So, the transformation needs to be done, I guess, at the
prediction
procedure. Do you think that is possible? Do you have any
suggestion
on how
to do it?

Thank you again,
Sebastian

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