Random forest - Out-of-bag estimates
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I am reading the chapter on random forests by Leo Breiman (found here: https://www.stat.berkeley.edu/~breiman/randomforest2001.pdf).
In section 3.1 Using out-of-bag estimates to monitor error, strength, and correlation (page 11), it says:
In each bootstrap training set, about one-third of the instances are
left out. Therefore, the out-of-bag estimates are based on combining
only about one-third as many classifiers as in the ongoing main
combination.
I am not sure I understand how the first sentence (that about one-third of cases are left out of each bootstrap sample) implies the second (that each case is OOB in about one-third of the trees)?
random-forest bootstrap
asked Jan 2 at 0:14
user51462user51462
1234
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add a comment |
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I am reading the chapter on random forests by Leo Breiman (found here: https://www.stat.berkeley.edu/~breiman/randomforest2001.pdf).
In section 3.1 Using out-of-bag estimates to monitor error, strength, and correlation (page 11), it says:
In each bootstrap training set, about one-third of the instances are
left out. Therefore, the out-of-bag estimates are based on combining
only about one-third as many classifiers as in the ongoing main
combination.
I am not sure I understand how the first sentence (that about one-third of cases are left out of each bootstrap sample) implies the second (that each case is OOB in about one-third of the trees)?
random-forest bootstrap
asked Jan 2 at 0:14
user51462user51462
1234
$endgroup$
add a comment |
$begingroup$
I am reading the chapter on random forests by Leo Breiman (found here: https://www.stat.berkeley.edu/~breiman/randomforest2001.pdf).
In section 3.1 Using out-of-bag estimates to monitor error, strength, and correlation (page 11), it says:
In each bootstrap training set, about one-third of the instances are
left out. Therefore, the out-of-bag estimates are based on combining
only about one-third as many classifiers as in the ongoing main
combination.
I am not sure I understand how the first sentence (that about one-third of cases are left out of each bootstrap sample) implies the second (that each case is OOB in about one-third of the trees)?
random-forest bootstrap
asked Jan 2 at 0:14
user51462user51462
1234
$endgroup$
I am reading the chapter on random forests by Leo Breiman (found here: https://www.stat.berkeley.edu/~breiman/randomforest2001.pdf).
In section 3.1 Using out-of-bag estimates to monitor error, strength, and correlation (page 11), it says:
In each bootstrap training set, about one-third of the instances are
left out. Therefore, the out-of-bag estimates are based on combining
only about one-third as many classifiers as in the ongoing main
combination.
I am not sure I understand how the first sentence (that about one-third of cases are left out of each bootstrap sample) implies the second (that each case is OOB in about one-third of the trees)?
random-forest bootstrap
random-forest bootstrap
asked Jan 2 at 0:14
user51462user51462
1234
asked Jan 2 at 0:14
user51462user51462
1234
edited Jan 3 at 4:07
user51462
asked Jan 2 at 0:14
user51462user51462
1234
asked Jan 2 at 0:14
user51462user51462
1234
asked Jan 2 at 0:14
user51462user51462
1234
1234
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1 Answer
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In a bootstrap sample we replace. The probability of a subject x being sampled with replacement is at about 2/3. If we build for example 1000 trees with different bootstrap samples, we therefore expect that x will be in (2/3)*1000 of these samples. One third of the trees were therefore not build with subject x. The OOB only calculates the error for the trees that have not been build with x, which is (1/3)*1000 in that case.
answered Jan 2 at 0:50
peteRpeteR
11317
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+1 Nice explanation. I think that you could make your first sentence more clear if you write "Bootstrapping samples from the original data set with replacement." Also, stats.SE supports math typesetting. More information: math.meta.stackexchange.com/questions/5020/…
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– Sycorax
Jan 2 at 0:53
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1 Answer
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1 Answer
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In a bootstrap sample we replace. The probability of a subject x being sampled with replacement is at about 2/3. If we build for example 1000 trees with different bootstrap samples, we therefore expect that x will be in (2/3)*1000 of these samples. One third of the trees were therefore not build with subject x. The OOB only calculates the error for the trees that have not been build with x, which is (1/3)*1000 in that case.
answered Jan 2 at 0:50
peteRpeteR
11317
$endgroup$
$begingroup$
+1 Nice explanation. I think that you could make your first sentence more clear if you write "Bootstrapping samples from the original data set with replacement." Also, stats.SE supports math typesetting. More information: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sycorax
Jan 2 at 0:53
add a comment |
$begingroup$
In a bootstrap sample we replace. The probability of a subject x being sampled with replacement is at about 2/3. If we build for example 1000 trees with different bootstrap samples, we therefore expect that x will be in (2/3)*1000 of these samples. One third of the trees were therefore not build with subject x. The OOB only calculates the error for the trees that have not been build with x, which is (1/3)*1000 in that case.
answered Jan 2 at 0:50
peteRpeteR
11317
$endgroup$
$begingroup$
+1 Nice explanation. I think that you could make your first sentence more clear if you write "Bootstrapping samples from the original data set with replacement." Also, stats.SE supports math typesetting. More information: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sycorax
Jan 2 at 0:53
add a comment |
$begingroup$
In a bootstrap sample we replace. The probability of a subject x being sampled with replacement is at about 2/3. If we build for example 1000 trees with different bootstrap samples, we therefore expect that x will be in (2/3)*1000 of these samples. One third of the trees were therefore not build with subject x. The OOB only calculates the error for the trees that have not been build with x, which is (1/3)*1000 in that case.
answered Jan 2 at 0:50
peteRpeteR
11317
$endgroup$
In a bootstrap sample we replace. The probability of a subject x being sampled with replacement is at about 2/3. If we build for example 1000 trees with different bootstrap samples, we therefore expect that x will be in (2/3)*1000 of these samples. One third of the trees were therefore not build with subject x. The OOB only calculates the error for the trees that have not been build with x, which is (1/3)*1000 in that case.
answered Jan 2 at 0:50
peteRpeteR
11317
answered Jan 2 at 0:50
peteRpeteR
11317
answered Jan 2 at 0:50
peteRpeteR
11317
answered Jan 2 at 0:50
peteRpeteR
11317
11317
$begingroup$
+1 Nice explanation. I think that you could make your first sentence more clear if you write "Bootstrapping samples from the original data set with replacement." Also, stats.SE supports math typesetting. More information: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sycorax
Jan 2 at 0:53
add a comment |
$begingroup$
+1 Nice explanation. I think that you could make your first sentence more clear if you write "Bootstrapping samples from the original data set with replacement." Also, stats.SE supports math typesetting. More information: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sycorax
Jan 2 at 0:53
$begingroup$
+1 Nice explanation. I think that you could make your first sentence more clear if you write "Bootstrapping samples from the original data set with replacement." Also, stats.SE supports math typesetting. More information: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sycorax
Jan 2 at 0:53
$begingroup$
+1 Nice explanation. I think that you could make your first sentence more clear if you write "Bootstrapping samples from the original data set with replacement." Also, stats.SE supports math typesetting. More information: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– Sycorax
Jan 2 at 0:53
add a comment |
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