Why does adding a redundant predictor to randomForest improve prediction?
$begingroup$
I write in hopes of understanding an odd behavior of the randomForest package. I am trying to predict a factor y with 9 levels using 8 binary factors X1-X8. I get good accuracy (0.8959), and the following confusion matrix:
y
A B C D E F G H I
A 75 0 0 0 0 0 0 0 0
B 0 121 0 5 0 0 0 0 0
C 0 0 156 1 0 0 0 1 0
D 0 0 0 0 0 0 0 0 0
E 1 6 0 73 172 3 0 1 1
F 0 0 0 0 0 90 0 0 0
G 0 0 0 1 0 0 31 0 0
H 0 0 0 1 0 0 0 84 0
I 0 0 0 3 0 0 0 0 106
Notice that RF makes no predictions for row D of the confusion matrix. Now I perform the following experiment: I make a copy of the first column of the predictor matrix, call it "junk", and append it to the predictor matrix. Now randomForest gives improved accuracy (0.9657) and the
following confusion matrix:
y
A B C D E F G H I
A 73 0 0 0 0 0 0 0 0
B 2 119 0 5 0 0 0 0 0
C 0 2 156 1 0 0 0 1 0
D 1 6 0 73 4 3 0 1 1
E 0 0 0 0 168 0 0 0 0
F 0 0 0 0 0 90 0 0 0
G 0 0 0 1 0 0 31 0 0
H 0 0 0 1 0 0 0 84 0
I 0 0 0 3 0 0 0 0 106
Note that randomForest now makes good predictions for row D of the confusion matrix.
In summary, appending a redundant copy of one of the predictor variables to the predictor matrix improves accuracy of randomForest. Further, it doesn't make much difference which predictor you append. They all give roughly the same accuracy and roughly the same confusion matrix.
I append code and data below. Can someone explain what is happening?
Code:
### Save the file and change the location
setwd("C:\tmp")
rm(list=ls())
library(randomForest)
# input compressed data and restore the
# number observed for each row
compressed <- read.csv("compressed.csv")
num <- compressed$NUM
newnum <- rep(1:length(num),num)
dat <- compressed[newnum,2:10]
y <- dat$y
x <- dat[,2:9]
# original data produces bad results
# for row D of confusion matrix
set.seed(323)
badrf=randomForest(y=y,x=x)
badpred=predict(badrf,newdata=x)
badtable <- table(badpred, y)
badtable
badaccuracy=sum(diag(badtable))/sum(badtable)
badaccuracy
# duplicate, say, x-matrix column 1
ndx <- 1
junk <- x[,ndx]
newx <- cbind(x,junk)
# re-analysis with superfluous new variable
# gives good results
set.seed(323)
goodrf=randomForest(y=y,x=newx)
goodpred=predict(goodrf,newdata=newx)
goodtable <- table(goodpred, y)
goodtable
goodaccuracy=sum(diag(goodtable))/sum(goodtable)
goodaccuracy
Data:
"NUM","y","X1","X2","X3","X4","X5","X6","X7","X8"
1,"A","NO","NO","NO","NO","NO","NO","NO","NO"
69,"A","NO","NO","YES","NO","NO","NO","NO","NO"
2,"A","NO","NO","YES","NO","NO","NO","NO","YES"
4,"A","NO","YES","YES","NO","NO","NO","NO","NO"
6,"B","NO","NO","NO","NO","NO","NO","NO","NO"
119,"B","NO","NO","NO","NO","NO","NO","NO","YES"
2,"B","YES","NO","NO","NO","NO","NO","NO","YES"
155,"C","YES","NO","NO","NO","NO","NO","NO","NO"
1,"C","YES","YES","NO","NO","NO","NO","NO","NO"
73,"D","NO","NO","NO","NO","NO","NO","NO","NO"
5,"D","NO","NO","NO","NO","NO","NO","NO","YES"
1,"D","NO","NO","NO","NO","NO","NO","YES","NO"
1,"D","NO","NO","NO","NO","YES","NO","NO","NO"
3,"D","NO","YES","NO","NO","NO","NO","NO","NO"
1,"D","YES","NO","NO","NO","NO","NO","NO","NO"
4,"E","NO","NO","NO","NO","NO","NO","NO","NO"
158,"E","NO","NO","NO","NO","NO","YES","NO","NO"
10,"E","YES","NO","NO","NO","NO","YES","NO","NO"
3,"F","NO","NO","NO","NO","NO","NO","NO","NO"
90,"F","NO","NO","NO","YES","NO","NO","NO","NO"
31,"G","NO","NO","NO","NO","NO","NO","YES","NO"
1,"H","NO","NO","NO","NO","NO","NO","NO","NO"
83,"H","NO","NO","NO","NO","YES","NO","NO","NO"
1,"H","NO","YES","NO","NO","YES","NO","NO","NO"
1,"H","YES","NO","NO","NO","YES","NO","NO","NO"
1,"I","NO","NO","NO","NO","NO","NO","NO","NO"
102,"I","NO","YES","NO","NO","NO","NO","NO","NO"
3,"I","NO","YES","NO","NO","NO","NO","NO","YES"
1,"I","NO","YES","NO","NO","NO","NO","YES","NO"
r machine-learning random-forest prediction
$endgroup$
migrated from stackoverflow.com Nov 24 '18 at 14:10
This question came from our site for professional and enthusiast programmers.
add a comment |
$begingroup$
I write in hopes of understanding an odd behavior of the randomForest package. I am trying to predict a factor y with 9 levels using 8 binary factors X1-X8. I get good accuracy (0.8959), and the following confusion matrix:
y
A B C D E F G H I
A 75 0 0 0 0 0 0 0 0
B 0 121 0 5 0 0 0 0 0
C 0 0 156 1 0 0 0 1 0
D 0 0 0 0 0 0 0 0 0
E 1 6 0 73 172 3 0 1 1
F 0 0 0 0 0 90 0 0 0
G 0 0 0 1 0 0 31 0 0
H 0 0 0 1 0 0 0 84 0
I 0 0 0 3 0 0 0 0 106
Notice that RF makes no predictions for row D of the confusion matrix. Now I perform the following experiment: I make a copy of the first column of the predictor matrix, call it "junk", and append it to the predictor matrix. Now randomForest gives improved accuracy (0.9657) and the
following confusion matrix:
y
A B C D E F G H I
A 73 0 0 0 0 0 0 0 0
B 2 119 0 5 0 0 0 0 0
C 0 2 156 1 0 0 0 1 0
D 1 6 0 73 4 3 0 1 1
E 0 0 0 0 168 0 0 0 0
F 0 0 0 0 0 90 0 0 0
G 0 0 0 1 0 0 31 0 0
H 0 0 0 1 0 0 0 84 0
I 0 0 0 3 0 0 0 0 106
Note that randomForest now makes good predictions for row D of the confusion matrix.
In summary, appending a redundant copy of one of the predictor variables to the predictor matrix improves accuracy of randomForest. Further, it doesn't make much difference which predictor you append. They all give roughly the same accuracy and roughly the same confusion matrix.
I append code and data below. Can someone explain what is happening?
Code:
### Save the file and change the location
setwd("C:\tmp")
rm(list=ls())
library(randomForest)
# input compressed data and restore the
# number observed for each row
compressed <- read.csv("compressed.csv")
num <- compressed$NUM
newnum <- rep(1:length(num),num)
dat <- compressed[newnum,2:10]
y <- dat$y
x <- dat[,2:9]
# original data produces bad results
# for row D of confusion matrix
set.seed(323)
badrf=randomForest(y=y,x=x)
badpred=predict(badrf,newdata=x)
badtable <- table(badpred, y)
badtable
badaccuracy=sum(diag(badtable))/sum(badtable)
badaccuracy
# duplicate, say, x-matrix column 1
ndx <- 1
junk <- x[,ndx]
newx <- cbind(x,junk)
# re-analysis with superfluous new variable
# gives good results
set.seed(323)
goodrf=randomForest(y=y,x=newx)
goodpred=predict(goodrf,newdata=newx)
goodtable <- table(goodpred, y)
goodtable
goodaccuracy=sum(diag(goodtable))/sum(goodtable)
goodaccuracy
Data:
"NUM","y","X1","X2","X3","X4","X5","X6","X7","X8"
1,"A","NO","NO","NO","NO","NO","NO","NO","NO"
69,"A","NO","NO","YES","NO","NO","NO","NO","NO"
2,"A","NO","NO","YES","NO","NO","NO","NO","YES"
4,"A","NO","YES","YES","NO","NO","NO","NO","NO"
6,"B","NO","NO","NO","NO","NO","NO","NO","NO"
119,"B","NO","NO","NO","NO","NO","NO","NO","YES"
2,"B","YES","NO","NO","NO","NO","NO","NO","YES"
155,"C","YES","NO","NO","NO","NO","NO","NO","NO"
1,"C","YES","YES","NO","NO","NO","NO","NO","NO"
73,"D","NO","NO","NO","NO","NO","NO","NO","NO"
5,"D","NO","NO","NO","NO","NO","NO","NO","YES"
1,"D","NO","NO","NO","NO","NO","NO","YES","NO"
1,"D","NO","NO","NO","NO","YES","NO","NO","NO"
3,"D","NO","YES","NO","NO","NO","NO","NO","NO"
1,"D","YES","NO","NO","NO","NO","NO","NO","NO"
4,"E","NO","NO","NO","NO","NO","NO","NO","NO"
158,"E","NO","NO","NO","NO","NO","YES","NO","NO"
10,"E","YES","NO","NO","NO","NO","YES","NO","NO"
3,"F","NO","NO","NO","NO","NO","NO","NO","NO"
90,"F","NO","NO","NO","YES","NO","NO","NO","NO"
31,"G","NO","NO","NO","NO","NO","NO","YES","NO"
1,"H","NO","NO","NO","NO","NO","NO","NO","NO"
83,"H","NO","NO","NO","NO","YES","NO","NO","NO"
1,"H","NO","YES","NO","NO","YES","NO","NO","NO"
1,"H","YES","NO","NO","NO","YES","NO","NO","NO"
1,"I","NO","NO","NO","NO","NO","NO","NO","NO"
102,"I","NO","YES","NO","NO","NO","NO","NO","NO"
3,"I","NO","YES","NO","NO","NO","NO","NO","YES"
1,"I","NO","YES","NO","NO","NO","NO","YES","NO"
r machine-learning random-forest prediction
$endgroup$
migrated from stackoverflow.com Nov 24 '18 at 14:10
This question came from our site for professional and enthusiast programmers.
$begingroup$
Tune the rf model and it will output similar predictions for the two cases.
$endgroup$
– missuse
Nov 15 '18 at 13:03
add a comment |
$begingroup$
I write in hopes of understanding an odd behavior of the randomForest package. I am trying to predict a factor y with 9 levels using 8 binary factors X1-X8. I get good accuracy (0.8959), and the following confusion matrix:
y
A B C D E F G H I
A 75 0 0 0 0 0 0 0 0
B 0 121 0 5 0 0 0 0 0
C 0 0 156 1 0 0 0 1 0
D 0 0 0 0 0 0 0 0 0
E 1 6 0 73 172 3 0 1 1
F 0 0 0 0 0 90 0 0 0
G 0 0 0 1 0 0 31 0 0
H 0 0 0 1 0 0 0 84 0
I 0 0 0 3 0 0 0 0 106
Notice that RF makes no predictions for row D of the confusion matrix. Now I perform the following experiment: I make a copy of the first column of the predictor matrix, call it "junk", and append it to the predictor matrix. Now randomForest gives improved accuracy (0.9657) and the
following confusion matrix:
y
A B C D E F G H I
A 73 0 0 0 0 0 0 0 0
B 2 119 0 5 0 0 0 0 0
C 0 2 156 1 0 0 0 1 0
D 1 6 0 73 4 3 0 1 1
E 0 0 0 0 168 0 0 0 0
F 0 0 0 0 0 90 0 0 0
G 0 0 0 1 0 0 31 0 0
H 0 0 0 1 0 0 0 84 0
I 0 0 0 3 0 0 0 0 106
Note that randomForest now makes good predictions for row D of the confusion matrix.
In summary, appending a redundant copy of one of the predictor variables to the predictor matrix improves accuracy of randomForest. Further, it doesn't make much difference which predictor you append. They all give roughly the same accuracy and roughly the same confusion matrix.
I append code and data below. Can someone explain what is happening?
Code:
### Save the file and change the location
setwd("C:\tmp")
rm(list=ls())
library(randomForest)
# input compressed data and restore the
# number observed for each row
compressed <- read.csv("compressed.csv")
num <- compressed$NUM
newnum <- rep(1:length(num),num)
dat <- compressed[newnum,2:10]
y <- dat$y
x <- dat[,2:9]
# original data produces bad results
# for row D of confusion matrix
set.seed(323)
badrf=randomForest(y=y,x=x)
badpred=predict(badrf,newdata=x)
badtable <- table(badpred, y)
badtable
badaccuracy=sum(diag(badtable))/sum(badtable)
badaccuracy
# duplicate, say, x-matrix column 1
ndx <- 1
junk <- x[,ndx]
newx <- cbind(x,junk)
# re-analysis with superfluous new variable
# gives good results
set.seed(323)
goodrf=randomForest(y=y,x=newx)
goodpred=predict(goodrf,newdata=newx)
goodtable <- table(goodpred, y)
goodtable
goodaccuracy=sum(diag(goodtable))/sum(goodtable)
goodaccuracy
Data:
"NUM","y","X1","X2","X3","X4","X5","X6","X7","X8"
1,"A","NO","NO","NO","NO","NO","NO","NO","NO"
69,"A","NO","NO","YES","NO","NO","NO","NO","NO"
2,"A","NO","NO","YES","NO","NO","NO","NO","YES"
4,"A","NO","YES","YES","NO","NO","NO","NO","NO"
6,"B","NO","NO","NO","NO","NO","NO","NO","NO"
119,"B","NO","NO","NO","NO","NO","NO","NO","YES"
2,"B","YES","NO","NO","NO","NO","NO","NO","YES"
155,"C","YES","NO","NO","NO","NO","NO","NO","NO"
1,"C","YES","YES","NO","NO","NO","NO","NO","NO"
73,"D","NO","NO","NO","NO","NO","NO","NO","NO"
5,"D","NO","NO","NO","NO","NO","NO","NO","YES"
1,"D","NO","NO","NO","NO","NO","NO","YES","NO"
1,"D","NO","NO","NO","NO","YES","NO","NO","NO"
3,"D","NO","YES","NO","NO","NO","NO","NO","NO"
1,"D","YES","NO","NO","NO","NO","NO","NO","NO"
4,"E","NO","NO","NO","NO","NO","NO","NO","NO"
158,"E","NO","NO","NO","NO","NO","YES","NO","NO"
10,"E","YES","NO","NO","NO","NO","YES","NO","NO"
3,"F","NO","NO","NO","NO","NO","NO","NO","NO"
90,"F","NO","NO","NO","YES","NO","NO","NO","NO"
31,"G","NO","NO","NO","NO","NO","NO","YES","NO"
1,"H","NO","NO","NO","NO","NO","NO","NO","NO"
83,"H","NO","NO","NO","NO","YES","NO","NO","NO"
1,"H","NO","YES","NO","NO","YES","NO","NO","NO"
1,"H","YES","NO","NO","NO","YES","NO","NO","NO"
1,"I","NO","NO","NO","NO","NO","NO","NO","NO"
102,"I","NO","YES","NO","NO","NO","NO","NO","NO"
3,"I","NO","YES","NO","NO","NO","NO","NO","YES"
1,"I","NO","YES","NO","NO","NO","NO","YES","NO"
r machine-learning random-forest prediction
$endgroup$
I write in hopes of understanding an odd behavior of the randomForest package. I am trying to predict a factor y with 9 levels using 8 binary factors X1-X8. I get good accuracy (0.8959), and the following confusion matrix:
y
A B C D E F G H I
A 75 0 0 0 0 0 0 0 0
B 0 121 0 5 0 0 0 0 0
C 0 0 156 1 0 0 0 1 0
D 0 0 0 0 0 0 0 0 0
E 1 6 0 73 172 3 0 1 1
F 0 0 0 0 0 90 0 0 0
G 0 0 0 1 0 0 31 0 0
H 0 0 0 1 0 0 0 84 0
I 0 0 0 3 0 0 0 0 106
Notice that RF makes no predictions for row D of the confusion matrix. Now I perform the following experiment: I make a copy of the first column of the predictor matrix, call it "junk", and append it to the predictor matrix. Now randomForest gives improved accuracy (0.9657) and the
following confusion matrix:
y
A B C D E F G H I
A 73 0 0 0 0 0 0 0 0
B 2 119 0 5 0 0 0 0 0
C 0 2 156 1 0 0 0 1 0
D 1 6 0 73 4 3 0 1 1
E 0 0 0 0 168 0 0 0 0
F 0 0 0 0 0 90 0 0 0
G 0 0 0 1 0 0 31 0 0
H 0 0 0 1 0 0 0 84 0
I 0 0 0 3 0 0 0 0 106
Note that randomForest now makes good predictions for row D of the confusion matrix.
In summary, appending a redundant copy of one of the predictor variables to the predictor matrix improves accuracy of randomForest. Further, it doesn't make much difference which predictor you append. They all give roughly the same accuracy and roughly the same confusion matrix.
I append code and data below. Can someone explain what is happening?
Code:
### Save the file and change the location
setwd("C:\tmp")
rm(list=ls())
library(randomForest)
# input compressed data and restore the
# number observed for each row
compressed <- read.csv("compressed.csv")
num <- compressed$NUM
newnum <- rep(1:length(num),num)
dat <- compressed[newnum,2:10]
y <- dat$y
x <- dat[,2:9]
# original data produces bad results
# for row D of confusion matrix
set.seed(323)
badrf=randomForest(y=y,x=x)
badpred=predict(badrf,newdata=x)
badtable <- table(badpred, y)
badtable
badaccuracy=sum(diag(badtable))/sum(badtable)
badaccuracy
# duplicate, say, x-matrix column 1
ndx <- 1
junk <- x[,ndx]
newx <- cbind(x,junk)
# re-analysis with superfluous new variable
# gives good results
set.seed(323)
goodrf=randomForest(y=y,x=newx)
goodpred=predict(goodrf,newdata=newx)
goodtable <- table(goodpred, y)
goodtable
goodaccuracy=sum(diag(goodtable))/sum(goodtable)
goodaccuracy
Data:
"NUM","y","X1","X2","X3","X4","X5","X6","X7","X8"
1,"A","NO","NO","NO","NO","NO","NO","NO","NO"
69,"A","NO","NO","YES","NO","NO","NO","NO","NO"
2,"A","NO","NO","YES","NO","NO","NO","NO","YES"
4,"A","NO","YES","YES","NO","NO","NO","NO","NO"
6,"B","NO","NO","NO","NO","NO","NO","NO","NO"
119,"B","NO","NO","NO","NO","NO","NO","NO","YES"
2,"B","YES","NO","NO","NO","NO","NO","NO","YES"
155,"C","YES","NO","NO","NO","NO","NO","NO","NO"
1,"C","YES","YES","NO","NO","NO","NO","NO","NO"
73,"D","NO","NO","NO","NO","NO","NO","NO","NO"
5,"D","NO","NO","NO","NO","NO","NO","NO","YES"
1,"D","NO","NO","NO","NO","NO","NO","YES","NO"
1,"D","NO","NO","NO","NO","YES","NO","NO","NO"
3,"D","NO","YES","NO","NO","NO","NO","NO","NO"
1,"D","YES","NO","NO","NO","NO","NO","NO","NO"
4,"E","NO","NO","NO","NO","NO","NO","NO","NO"
158,"E","NO","NO","NO","NO","NO","YES","NO","NO"
10,"E","YES","NO","NO","NO","NO","YES","NO","NO"
3,"F","NO","NO","NO","NO","NO","NO","NO","NO"
90,"F","NO","NO","NO","YES","NO","NO","NO","NO"
31,"G","NO","NO","NO","NO","NO","NO","YES","NO"
1,"H","NO","NO","NO","NO","NO","NO","NO","NO"
83,"H","NO","NO","NO","NO","YES","NO","NO","NO"
1,"H","NO","YES","NO","NO","YES","NO","NO","NO"
1,"H","YES","NO","NO","NO","YES","NO","NO","NO"
1,"I","NO","NO","NO","NO","NO","NO","NO","NO"
102,"I","NO","YES","NO","NO","NO","NO","NO","NO"
3,"I","NO","YES","NO","NO","NO","NO","NO","YES"
1,"I","NO","YES","NO","NO","NO","NO","YES","NO"
r machine-learning random-forest prediction
r machine-learning random-forest prediction
asked Nov 15 '18 at 10:58
Neal Oden
migrated from stackoverflow.com Nov 24 '18 at 14:10
This question came from our site for professional and enthusiast programmers.
migrated from stackoverflow.com Nov 24 '18 at 14:10
This question came from our site for professional and enthusiast programmers.
$begingroup$
Tune the rf model and it will output similar predictions for the two cases.
$endgroup$
– missuse
Nov 15 '18 at 13:03
add a comment |
$begingroup$
Tune the rf model and it will output similar predictions for the two cases.
$endgroup$
– missuse
Nov 15 '18 at 13:03
$begingroup$
Tune the rf model and it will output similar predictions for the two cases.
$endgroup$
– missuse
Nov 15 '18 at 13:03
$begingroup$
Tune the rf model and it will output similar predictions for the two cases.
$endgroup$
– missuse
Nov 15 '18 at 13:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For each node in a tree, the random forest algorithm does not pick the best dimension to split on but the best dimension among a (small) sample of them in order to add variance in the prediction.
Here, the column you add must be important and by adding it you make it appear more often in the sample of dimensions among which the random forest chooses. You make it more important in your prediction. And by doing so, you do not decrease too much the variability of your trees.
This behavior could explain your findings
PS: Indeed, ideally, it should be moved to stats.stackexchange
answered Nov 15 '18 at 12:17
PopPop
953619
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "65"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f378548%2fwhy-does-adding-a-redundant-predictor-to-randomforest-improve-prediction%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For each node in a tree, the random forest algorithm does not pick the best dimension to split on but the best dimension among a (small) sample of them in order to add variance in the prediction.
Here, the column you add must be important and by adding it you make it appear more often in the sample of dimensions among which the random forest chooses. You make it more important in your prediction. And by doing so, you do not decrease too much the variability of your trees.
This behavior could explain your findings
PS: Indeed, ideally, it should be moved to stats.stackexchange
answered Nov 15 '18 at 12:17
PopPop
953619
$endgroup$
add a comment |
$begingroup$
For each node in a tree, the random forest algorithm does not pick the best dimension to split on but the best dimension among a (small) sample of them in order to add variance in the prediction.
Here, the column you add must be important and by adding it you make it appear more often in the sample of dimensions among which the random forest chooses. You make it more important in your prediction. And by doing so, you do not decrease too much the variability of your trees.
This behavior could explain your findings
PS: Indeed, ideally, it should be moved to stats.stackexchange
answered Nov 15 '18 at 12:17
PopPop
953619
$endgroup$
add a comment |
$begingroup$
For each node in a tree, the random forest algorithm does not pick the best dimension to split on but the best dimension among a (small) sample of them in order to add variance in the prediction.
Here, the column you add must be important and by adding it you make it appear more often in the sample of dimensions among which the random forest chooses. You make it more important in your prediction. And by doing so, you do not decrease too much the variability of your trees.
This behavior could explain your findings
PS: Indeed, ideally, it should be moved to stats.stackexchange
answered Nov 15 '18 at 12:17
PopPop
953619
$endgroup$
For each node in a tree, the random forest algorithm does not pick the best dimension to split on but the best dimension among a (small) sample of them in order to add variance in the prediction.
Here, the column you add must be important and by adding it you make it appear more often in the sample of dimensions among which the random forest chooses. You make it more important in your prediction. And by doing so, you do not decrease too much the variability of your trees.
This behavior could explain your findings
PS: Indeed, ideally, it should be moved to stats.stackexchange
answered Nov 15 '18 at 12:17
PopPop
953619
answered Nov 15 '18 at 12:17
PopPop
953619
answered Nov 15 '18 at 12:17
PopPop
953619
answered Nov 15 '18 at 12:17
PopPop
953619
953619
add a comment |
add a comment |
Thanks for contributing an answer to Cross Validated!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fstats.stackexchange.com%2fquestions%2f378548%2fwhy-does-adding-a-redundant-predictor-to-randomforest-improve-prediction%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
Tune the rf model and it will output similar predictions for the two cases.
$endgroup$
– missuse
Nov 15 '18 at 13:03