I have been trying to understand gradient boosting reading various blogs, websites and trying to find my answer by looking through for example the XGBoost source code. However, I cannot seem to find an understandable explanation of how gradient boosting algorithms produce probability estimates. So, how do they calculate the probabilities?
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3$\begingroup$ This essentially asks and answers the same question, in case a different explanation would be useful to you: stats.stackexchange.com/questions/204154/… $\endgroup$– Matthew DruryJul 8, 2018 at 21:33
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TL;DR: The log-odds for a sample is the sum of the weights of its terminal leafs. The probability of the sample belonging to class 1 is the inverse-logit transformation of the sum.
Analogously to logistic regression, the logistic function computes probabilities that are linear on the logit scale:
$$ z = Xw \\ \mathbb{P}(y=1|X) = \frac{1}{1 + \exp(-z)} $$
Unlike logistic regression, the "features" in $X$ are constructed as the terminal nodes of an ensemble of decision trees using the boosting procedure. Each row of $X$ collects the terminal leafs for each sample; the row is a $T$-hot binary vector, for $T$ the number of trees. (Each XGBoost tree is generated according to a particular algorithm, but that's not relevant here.)
There are $n$ columns in $X$, one column for each terminal node. There is no expression for the total number of terminal nodes, because the number of nodes can vary between trees (and usually does, in my experience).
Each leaf in the tree has an associated "weight." That weight is recorded in $w$. To be conformable with $X$, there are $n$ elements in $w$. The weights themselves are derived from the gradient boosting procedure; see: In XGboost are weights estimated for each sample and then averaged
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$\begingroup$ This is very helpful, thanks. How many elements would the beta-vector contain? Would it be equal to the number of total leaf nodes across all trees? (And there would be equal number of columns in the X matrix, correct?) $\endgroup$– VishalJun 6, 2018 at 20:00
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$\begingroup$ Thank you for the updated answer. Does this mean that there's a unique
Xmatrix as well as a unique set of betas for each sample/observation (i)? In other words, for every sample/observation for which you want to calculate the probability of belonging to class 1, you'd need to determine the unique values of theXmatrix and beta vector? $\endgroup$– VishalJul 3, 2018 at 15:50 -
1$\begingroup$ Each row of $X$ stores the terminal leafs for a sample. $\endgroup$– Sycorax ♦Jul 3, 2018 at 15:54
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$\begingroup$ @Sycorax when you say each row of X collects terminal leafs of each sample, can you give an example? $\endgroup$– Maths12Sep 2, 2020 at 14:43
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1$\begingroup$ Suppose you have 3 trees and each tree has 8 leaves (24 in total), which we can index 1,2,3...8. A sample might end up in leaves [1,2,5] of each tree respectively, so a row of $X$ is all zeros except columns 1, 10 and 21, which contain a 1. $\endgroup$– Sycorax ♦Sep 2, 2020 at 14:48