Original Question

Having used XGBoost a fair bit, clearly changing the learning rate dramatically affects the algorithm's performance. That said, I really can't understand the theoretical justification for it. It makes sense in "vanilla" gradient boosting, when you don't make use of second derivatives. Analogously, one does not need a learning rate if using Newton-Raphson to perform optimisation by finding the zeros of the derivative of the cost function.

I thought it might have to do with ensuring that the updates one makes at every step are small, and thus the gradient expansion to second order is valid, but it seems to me like one could achieve the same thing more effectively by regularising?

Also, the XGBoost docs have a theoretical introduction to XGBoost and don't mention a learning rate anywhere (https://xgboost.readthedocs.io/en/latest/tutorials/model.html)

Is it as simple as "it is experimentally observed to improve performance" and if so, is it possible to rationalise post-fact ?

Update: Almost a year on, I thought I'd update my thinking on this and somewhat refine my question

While it might be the case that the need for a learning rate was ascertained experimentally, it seems likely to me that the reason it is necessary, is to do with the fact that XGBOOST assumes the that the total loss $L$ of a classifier consisting of an existing classifier $F_{t}(x)$ plus a new classifier $f_{t+1}(x)$, can be written as a Taylor Expansion of $L$ about $F_{t}(x)$, which requires $f_{t+1}(x)$ to represent a "small enough" correction to $F_{t}(x)$, that we don't need to expand to too high an order.

My suspicion for a while has been that using lots of regularisation should take care of this, hence why use a learning rate at all? Another approach, could be to say that the tree $f_{t+1}(x)$, which splits the space into a number of distinct regions (terminal nodes) $\{R_{j}\}$, outputs a constant $\epsilon \cdot w_{j}$ in the $j^{th}$ region. By choosing a sufficiently small $\epsilon$, we can ensure that $\epsilon \cdot w_{j}$ will be sufficiently small for any partitioning and any j.

It turns out however, that if you follow the derivation in the XGBOOST docs but take this approach, and use no regularisation, the weight $w_{j}^{*}$ you should assign to the region $R_{j}$ is given by

$w_{j}^{*} = - \frac{\sum_{i \in R_{j}}\frac{\partial \ell}{\partial \hat{y}_{i}}\bigg|_{F_{t}(x_{i})}}{\epsilon \sum_{i \in R_{j}}\frac{\partial ^{2}\ell}{\partial \hat{y}_{i}^{2}}}$

in which $L[F_{t}(x)+f_{t+1}(x)] = \sum_{i=1}^{N}\ell (y_{i}, \hat{y}_{i})=\sum_{i=1}^{N}\ell (y_{i}, F_{t}(x_{i}) + f_{t+1}(x_{i}))$

In other words, if you state that the output of each tree at each leaf will be a constant $w_{j}$ multiplied by a very small number $\epsilon$, small enough to ensure that the product is always small, $w_{j}^{*}$ will simply compensate, so the smaller you make $\epsilon$, the larger you make $w_{j}^{*}$, and the product remains unchanged. Crucially, the product will not necessarily be "small enough" for the Taylor series to quickly converge and justify the second order expansion. However, if a little bit of regularisation is used, enough to stop $w_{j}$ becoming infinite and thus ensure the product is always small, then you're good.

In essence, you have two approaches:

  1. Set $\lambda$ to be "very large", this will ensure that $w_{j}^{*}$ is small, and thus the expansion is valid
  2. Use a learning rate parameter $\epsilon$, and have some small amount of regularisation, to ensure that $w_{j}^{*}$ cannot be arbitrarily large

They sound the same, at a phenomenological level, but let's investigate the $w_{j}^{*}$ they imply. Using approach 1, not having a learning rate, we obtain (as in the xgboost docs, linked above)

$w_{j}^{*}= - \frac{\sum_{i \in R_{j}}\frac{\partial \ell}{\partial \hat{y}_{i}}\bigg|_{F_{t}(x_{i})}}{\lambda + \sum_{i \in R_{j}}\frac{\partial^{2} \ell}{\partial \hat{y}_{i}^{2}}\bigg|_{F_{t}(x_{i})}}$

whereas if we use a learning rate as well, we obtain

$w_{j}^{*}= - \frac{\sum_{i \in R_{j}}\frac{\partial \ell}{\partial \hat{y}_{i}}\bigg|_{F_{t}(x_{i})}}{\frac{\lambda}{\epsilon} + \epsilon \cdot \sum_{i \in R_{j}}\frac{\partial^{2} \ell}{\partial \hat{y}_{i}^{2}}\bigg|_{F_{t}(x_{i})}}$

They look very similar, and in both cases, as you up the amount of regularisation by increasing $\lambda$, the curvature term becomes less relevant. In the case when you have a learning rate, you can get this effect either by increasing $\lambda$ or decreasing $\epsilon$.

No matter which way I think about the problem, either approach seems conceptually the same, but they do give slightly different solutions. Furthermore, in practice, the learning rate is perhaps the most important hyperparamter to tune in XGBOOST, although I haven't seen anybody explore whether similarly good results could be obtained by tuning the regularisation parameter more. In particular, am I missing something jumping out at me from these two equations?

Another Update: Another Year On

Thanks to Andreas for his answer below which got me onto this.

Because the loss function is assumed to be approximated by a function quadratic in $w_{j}$, which is valid if $w_{j}$ is small, it will only have one minimum (assuming we're doing loss minimisation). Thus the loss evaluated at $\epsilon \cdot w^{*}_{j}$ will be greater than the loss evaluated at $w^{*}_{j}$, but less than the loss evaluated at $w_{j}=0$, in other words, by updating your prediction by $\epsilon \cdot w^{*}_{j}$, you are guaranteed to decrease your training loss. If $\epsilon$ is very small, this process happens very slowly but if $\epsilon$ is too large, then the Taylor series might not be valid. The key point here is that it's not about finding the optimal $w_{j}$, it's about finding a $w_{j}$ that guarantees the training loss decreases at every iteration.

I think the logic must go something like this, but it can't quite be this. While I agree that if we know $w^{*}_{j}$, then $\epsilon w^{*}_{j}$ will also decrease training loss, but this logic seems circular to me. If we actually knew $w^{*}_{j}$, then while we could multiply by $\epsilon$, why would we?

Conversely, if we want to find the optimal $w_{j}$ subject to the assumption that $w_{j}$ is sufficiently small, it doesn't seem correct to find the optimal $w_{j}$ assuming that $w_{j}$ is small, finding that it isn't small, and then multiplying it by a small number to make it small.

  • $\begingroup$ I guess that the answer still goes alonog these lines: stats.stackexchange.com/questions/282544/… (even though you are bringin up the second derivative): setting a LR to a value < 1 helps the algorithm evite spiraling around the minimum. You are right: If we use the second derivative we should get a better feeling of how much we should go into the direction given by the gradient, so yes, for the regression case, the second derivative vanishes, so I am puzzled by your argument as well, however... $\endgroup$ – Fabian Werner Jul 5 '18 at 9:52
  • $\begingroup$ in the classification case, I think that there are more than 2 derivatives and xgboost only uses the first two out of an infinite selection (third derivative helps not to overinterpret the results from the second and so forth) $\endgroup$ – Fabian Werner Jul 5 '18 at 9:54
  • $\begingroup$ My interpretation of using only two derivatives, is that one can use regularisation to ensure that the correction will always be "relatively small", and it is then justified to assume that the second order expansion is a good approximation to how the loss will change when this correction is added. By the way, I don't think the second derivative disappears in "the regression case" (specifically, I think you mean least squares loss), it's a constant, is it not? $\endgroup$ – gazza89 Jul 5 '18 at 14:23
  • $\begingroup$ 1) Yes, I meant that it does not give you any insightful information that you could possibly use in order to get a feeling of how long you should follow the gradient. 2) Thought about it a little longer. Say you want to minimize the function $f(x) = |x|$. You start at $x=0.5$ and you always follow the gradient, i.e. you "spiral" between $x=+0.5$ and $x=-0.5$. How exactly does the second derivative help you in that case to judge how much weight you should put on the gradient? I think this is where you are ... '''wrong''': the second derivative also only describes the change in $\endgroup$ – Fabian Werner Jul 5 '18 at 14:26
  • $\begingroup$ the derivative in a region that might be reeeeeally small (around the point you are right now). So in principle, it does not help you to know how reliable the gradient is... In terms of Taylor series, it helps you to 'know' the function a bit better than with only one derivative but in these simple cases like $x^2$ or $|x|$ you know the function already somehow... $\endgroup$ – Fabian Werner Jul 5 '18 at 14:27

In particular, am I missing something jumping out at me from these two equations?

From what I've looked at in Friedman's paper, the 'learning rate' $\epsilon$ (there, called 'shrinkage' and denoted by $\nu$) is applied after choosing those weights $w_j^*$ which minimise the cost function. That is, we determine the boost's optimal weights, $w_j^*$ first, and only then do we consider multiplying by $\epsilon$.

What would this mean?

This would mean that neither of the equations in the question which feature both $\epsilon$ and $w_j^*$, are used in the XGBoost algorithm.

Also, that $\lambda$ is still necessary in order to guarantee the Taylor expansion validity, and has a non-uniform effect on the $w_j$, its effect depending on the partial derivatives of $\ell$ as you wrote before: \begin{align*} w_{j}^{*}= - \frac{\sum_{i \in R_{j}}\frac{\partial \ell}{\partial \hat{y}_{i}}\bigg|_{F_{t}(x_{i})}}{\lambda + \sum_{i \in R_{j}}\frac{\partial^{2} \ell}{\partial \hat{y}_{i}^{2}}\bigg|_{F_{t}(x_{i})}} \end{align*}

The learning rate doesn't come in until after this point, when, having determined the optimal weights of the new tree $\lbrace w_j^* \rbrace_{j=1}^T$, we decide that, actually, we don't want to add what we've just deemed to be the 'optimal boost' straight-up, but instead, update our additive predictor $F_t$ by adding a scaled version of $f_{t+1}$: scaling each weight $w_j^*$ uniformly by $\epsilon$, and thus scaling the contribution of the whole of $f_{t+1}$ by $\epsilon$, too.

From where I'm sitting, there is some (weak-ish) analogy with the learning rate in gradient descent optimization: gently aggregating the predictors in order to iterate towards what we believe a general and descriptive predictor to be, but maintaining control over how fast we get there. In contrast, a high learning rate will mean that we use up all of our predictive power relatively quickly. If we do so too quickly with too few trees then any subsequent boost might need to make large corrections, causing the loss to remain at a relatively high plateau, after a few steps of which the algorithm terminates.

Keeping a lower learning rate, would aid generalisability because we are relying less upon the predictions of the new boosting tree, and instead permitting subsequent boosts to have more predictive power. It will mean that we need more boosts, and that training will take longer to terminate - in line with the empirical results shown in @Sycorax's answer.

In summary:

My understanding is that:

  1. $\lambda$ is used when regularising the weights $\lbrace w_j\rbrace$ and to justify the 2nd order truncation of the loss function's Taylor expansion, enabling us to find the 'optimal' weights $\lbrace w_j^*\rbrace$. This has a non-uniform effect on each of the weights $w_j$.

  2. $\epsilon$ is used only after determination of the optimal weights $w_j^*$ and applied by scaling all of the weights uniformly to give $\epsilon\, w_j^*$.

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  • $\begingroup$ I think you are right on most points, except that, from my understanding, it is the learning rate πœ–, not πœ†, that controls for validity of the Taylor expansion(TE). This is because πœ– scales the final step size taken towards the TE-minimum and for small πœ– TE clearly becomes a better approximation. Moreover, since the Hessian is diagonal in XGB, we are still guaranteed to monotonically shrink costs when walking towards the minimum, even if it's not the full step (πœ– = 1) that is taken. The experiments I've made so far with XGB are absolutely consistent with this interpretation. $\endgroup$ – Andreas Steimer Jul 6 at 19:32
  • $\begingroup$ It is certainly the case that using a regulariser $\lambda$ does not make the Taylor Expansion valid. You can see this by looking at the equation for $w^{*}_{j}$, which will not be a small value for all value of $\lambda$. XGboost allows tiny values of $\lambda$, but it only allows values $\in ]0,1[$ for $\epsilon$ and suggests values $\in ]0, 0.3[$ $\endgroup$ – gazza89 Jul 8 at 15:17

Parameters for Tree Booster eta [default=0.3, alias: learning_rate] step size shrinkage used in update to prevents overfitting. After each boosting step, we can directly get the weights of new features. and eta actually shrinks the feature weights to make the boosting process more conservative. range: [0,1]

From: manual

According to this source: math, learning_rate affects the value of the function of gradient calculation that incorporates both first and second order derivatives. I just looked into code, but I am not good at Py, so my answer is really a guide for you to explore more.

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  • 1
    $\begingroup$ thanks for the link. My c++ isn't great, but it looks to me like it's doing exactly what I thought, it's finding the best solution as shown in the XGBoost docs, and then multiplying by a learning rate. Unfortunately, justification is pretty thin on the ground, just that it "helps combat overfitting", but I'm still struggling to understand why, other than "it has been experimentally observed". I'm still curious as to whether the same effect could be achieved by regularising more heavily. $\endgroup$ – gazza89 Jul 5 '18 at 14:29

Adding to montols answer:

I think he is right on most points, except that, from my understanding, it is the learning rate πœ–, not πœ†, that controls for validity of the Taylor expansion(TE). This is because πœ– scales the final step size taken towards the TE-minimum and for small πœ– TE clearly becomes a better approximation. Moreover, since the Hessian is diagonal in XGB, we are still guaranteed to monotonically shrink costs when walking towards the minimum, even if it's not the full step (πœ– = 1) that is taken.

So far, the experiments I've made with XGB are absolutely consistent with this interpretation.

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  • $\begingroup$ I disagree/don't understand. It seems nuts to me, to calculate an optimal value for $w^{*}_{j}$ which is only accurate if the Taylor Expansion is valid, and thus only valid for small $w_{j}$, but does not guarantee $w_{j}$ to be small, and to then multiply that value by $\epsilon$ post-hoc to make it be small. $\endgroup$ – gazza89 Jul 8 at 15:20
  • $\begingroup$ What exactly is it that "seems" you to be nuts? The crucial point is that, since the Hessian is diagonal, we are still guaranteed to diminish, albeit not necessarily minimize, the cost. For an arbitrary cost function it is hard to estimate the region of validity for TE, thus adding πœ– as a hyperparameter provides a simple workaround $\endgroup$ – Andreas Steimer Jul 9 at 19:53
  • $\begingroup$ I expanded upon this by editing my original post. I take your point about the Hessian, and thus that if the optimal $w_{j}$ (i.e. $w^{*}_{j}$) were known, then $\epsilon \cdot w^{*}_{j}$ will guarantee that the loss improves. The problem is that I don't see how $w^{*}_{j}$ is known, it's calculated assuming $w_{j}$ to be small, but without any guarantees it is small. In fact, it's possible to find that $w^{*}_{j}$ isn't even small, as far as I can tell. $\endgroup$ – gazza89 Jul 11 at 12:10

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