Answers.Body: <p>In short, there are two major reasons:</p>
  <ol>
  <li><p>The optimization landscape in parameter space is non-convex even with convex loss function (e.g., MSE). Therefore, you need to do small update steps (i.e., the gradient scaled by the learning rate) to find a suitable local minimum and avoid divergence.</p>
  </li>
  <li><p>The gradient is estimated on a batch of samples, which does not represent the full let's say &quot;population&quot; of data. Even by using batch gradient descent, which uses the full dataset each step, the resulting gradient does not point in the direction of the true gradient. So you need to introduce a step size (i.e., the learning rate.) Moreover, at least in principle, it is possible to correct the gradient direction by including second order information (e.g., the Hessian of the loss w.r.t. parameters) although it is usually infeasible to compute.</p>
  </li>
  </ol>
  
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Answers.Text: Point 1 doesn't really have anything to do with non-convexity, GD can diverge even on a simple 1D convex function like $x \mapsto x^2$ if the learning rate is too high. In fact, if anything, non-convexity is a reason to use _larger_ (and/or stochastic) steps, at least at first, to avoid converging on a poor local minimum near the starting point.
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Answers.Text: @leftaroundabout My point was to explain that with a step size of 1, the optimization is really unstable unless the problem is trivial, because the updates are just too large. Indeed, in DL, to avoid premature convergence and ensure stable training you need at least a well-designed architecture with suitable weights initialization, a learning rate schedule, gradient clipping, and some regularization as well. But all of these are outside the scope of the question.
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Answers.Text: @LucaAnzalone he is saying that even in trivial cases you need a stepsize, the only case where you don't need it, is if the conditioning of the problem is 1, but your point is valid, just maybe too specific
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Answers.Body: <p>Indeed your intuition of <a href="https://en.wikipedia.org/wiki/Learning_rate" rel="nofollow noreferrer">Learning rate</a> is right regarding its additional control in training the model, and the precise terminology for your somewhat vague 'additional control' intuition even in a convex loss landscape is called hyperparameter or tuning parameter in ML which specifies details of the learning process in contrast to parameters which determine the model itself, such as learning rate, momentum (Adam, Nesterov), batch size, regularization strength, etc. Specifically the linked reference clarifies the necessity of your concerned learning rate hyperparameter as follows.</p>
  <blockquote>
  <p>Since it influences to what extent newly acquired information overrides old information, it metaphorically represents the <strong>speed</strong> at which a machine learning model &quot;learns&quot;... In setting a learning rate, there is a trade-off between the rate of convergence and overshooting. While the descent direction is usually determined from the gradient of the loss function, the learning rate determines how big a step is taken in that direction... In order to achieve faster convergence, prevent oscillations and getting stuck in undesirable local minima the learning rate is often varied during training either in accordance to a learning rate schedule or by using an adaptive learning rate.</p>
  </blockquote>
  <p>Usually a fixed learning rate is set very small to ensure convergence, however, in practice a learning rate schedule is often used by decay and momentum which help faster convergence and prevent oscillations as discussed in my answer to another <a href="https://ai.stackexchange.com/questions/35648/at-which-point-does-the-momentum-based-gd-helps-really-in-this-figure/46171#46171">post</a>.</p>
  
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Answers.Body: <p>Your intuition is correct and the short answer is: The learning rate is necessary in order to control the <em>size of the update</em>!</p>
  <p>You probably already know that gradient descent (GD) is derived from the [Newton-Raphson algorithm] (<a href="https://en.wikipedia.org/wiki/Newton%27s_method" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Newton%27s_method</a> &quot;or Newton's method&quot;).
  In Machine Learning applications, the learning rate in GD plays the same role as the second order derivative which encodes the <em>curvature information</em>. To be more explicit, let's consider two convex functions (which, if you know your math, have positive second derivatives), <span class="math-container">$f_{1}(x) = 2x^{2}$</span> and <span class="math-container">$f_{2}(x) = \frac{x^{2}}{2}$</span>.
  We can see that their gradient and second-order derivative are: <span class="math-container">$(f'_{1}(x) = 4x, f''_{1}(x) = 4)$</span> and <span class="math-container">$(f'_{2}(x) = x, f''_{2}(x) = 1)$</span>.</p>
  <p>Now, if we consider the Newton-Raphson algorithm, then we see that:</p>
  <p><span class="math-container">$x_{k+1} = x_{k} - \frac{f'_{1}(x_{k})}{f''_{1}(x_{k})} = x_{k} - \frac{4x_{k}}{4} = x_{k} - \frac{f'_{2}(x_{k})}{f''_{2}(x_{k})} = x_{k} - \frac{x_{k}}{1} = 0, \;\forall x_{k}\in\mathbb{R}$</span>.</p>
  <p>We can see that the second order derivative rescales the gradient so that <span class="math-container">$\forall x_{k}\in\mathbb{R}$</span>, the update always takes you directly to the minimum (this works in one step because the functions are quadratic. When the function is not precisely quadratic, the algorithm will take more iterations, but <em>will</em> converge similarly to the minimizing solution).</p>
  <p>Hence, in the case of <span class="math-container">$f_{1}$</span>, the scaling factor <span class="math-container">$\frac{1}{f''_{1}(x_{k})} = \frac{1}{4} = \alpha^{*}_{1}$</span> and for <span class="math-container">$f_{2}$</span>, the scaling factor <span class="math-container">$\frac{1}{f''_{2}(x_{k})} = 1 = \alpha^{*}_{2}$</span>. These represent the optimal learning rate values <span class="math-container">$\alpha^{*}_{1}$</span> and <span class="math-container">$\alpha^{*}_{2}$</span> for each function.</p>
  <p>But, what if instead of these values we used values that are smaller or larger? Well, let's see:</p>
  <ul>
  <li>Let <span class="math-container">$\alpha_{1} = \frac{1}{6} &lt; \alpha^{*}_{1}$</span>, then:
  <span class="math-container">$x_{k+1} = x_{k} - \frac{4x_{k}}{6} = x_{k} - \frac{2x_{k}}{3} = \frac{x_{k}}{3} \neq 0$</span>. However, since this will be the case for every time step <span class="math-container">$k$</span>, the update rule will give you a geometric progression (e.g. if we start with <span class="math-container">$x_{0} = 1$</span>, then <span class="math-container">$x_{1} = \frac{1}{3}$</span>, <span class="math-container">$x_{2} = \frac{1}{9}$</span>, <span class="math-container">$x_{3} = \frac{1}{27}$</span>, <span class="math-container">$\cdots$</span>, <span class="math-container">$x_{k} = (\frac{1}{3})^k$</span>. If <span class="math-container">$k$</span> is large enough, then the final solution will be very close to 0, i.e. <span class="math-container">$k\to\infty \implies x_{k}\to 0$</span>). Hence, quite intuitively, <strong>a smaller learning rate leads to a slow convergence</strong>.</li>
  <li>Let <span class="math-container">$\alpha_{1} = \frac{1}{2} &gt; \alpha^{*}_{1}$</span>, then:
  <span class="math-container">$x_{k+1} = x_{k} - \frac{4x_{k}}{2} = x_{k} - 2x_{k} = -x_{k} \neq 0$</span>. As can be seen, this does not converge at all to the optimal solution, 0! In fact it oscillates back and forth between <span class="math-container">$x_{0}$</span> and <span class="math-container">$-x_{0}$</span>. This is a divergent progression! In fact, if we chose <span class="math-container">$\alpha_{1} = \alpha^{*}_{2} = 1 &gt; \alpha^{*}_{1}$</span>, then: <span class="math-container">$x_{k+1} = x_{k} - 4x_{k} = x_{k} - 4x_{k} = -3x_{k} \neq 0$</span>. which implies a geometric progression with ratio <span class="math-container">$r=-3$</span> which is divergent. Hence, <strong>a larger learning rate leads to divergence</strong>.</li>
  </ul>
  <p>In summary, the learning rate scales the gradient in order to control the amount of change in the optimized parameters at every step and it plays the same role as the second order derivative, i.e. it is an estimate of the function's curvature information or shape. If properly estimated, then the algorithm will converge quickly. If underestimated, the convergence will be slow (but guaranteed to converge after a potentially large number of updates). If overestimated, the algorithm will overshoot and diverge, never settling down to a specific value.</p>
  <p>P.S.: Chris Cundy mentions in his comment the Lipschitz coefficient and in general, you want your learning rate to satisfy <span class="math-container">$0 &lt; \alpha &lt; \frac{1}{L}$</span> where <span class="math-container">$L$</span> is the Lipschitz coefficient. This is also related to the second order derivative information and is valid for functions Lipschitz continuous gradients (this connects to L-smoothness). Indeed, if <span class="math-container">$f(x)$</span> is twice differentiable and if there exists <span class="math-container">$L &lt; \infty$</span> such that its Hessian matrix (second order derivative) has a bounded spectral norm, i.e. <span class="math-container">$||f''(x)|| \leq L$</span>, then <span class="math-container">$f(x)$</span> has a Lipschitz continuous gradient with Lipschitz constant <span class="math-container">$L$</span>.</p>
  <p>I recommend checking the theorem 3.4. in [1] <a href="https://gowerrobert.github.io/pdf/M2_statistique_optimisation/grad_conv.pdf" rel="nofollow noreferrer">https://gowerrobert.github.io/pdf/M2_statistique_optimisation/grad_conv.pdf</a></p>
  
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Body: <p>I am currently following course 1 of Andrew Ng's Machine Learning Specialization. I understood we need a convex cost function to reach the global minimum of the loss. But that means the gradient will reduce and approach 0 by itself as long as we choose a convex function.</p>
  <p>Then what is the need of the learning rate?</p>
  <p>My intuition tells me its an additional measure to give us more control in training the model, but I want a proper answer and evidence to solidify my understanding.</p>
  
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Comments.Text: Just pointing out a misconception in the question: It is not true that gradient descent with a fixed learning rate will converge, even for a convex function. Consider $f(x) = x^2$ with learning rate 10 and starting point $2$. The sequence of iterates is $-38, 722, \ldots$. Generally you need a condition like $\eta < 2/L$, where $L$ is the Lipschitz coefficient
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Comments.Text: @ChrisCundy by η do you mean the learning rate? I was accustomed to alpha so trying to confirm.
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Comments.Text: Yes,  is another variable typically used for the learning rate
Comments.UserDisplayName: Chris Cundy
Tags: <machine-learning><gradient-descent><learning-rate>
Title: Why exactly do we need the learning rate in gradient descent?
UserDisplayName: Namirah Rasul
ThreadText: # Why exactly do we need the learning rate in gradient descent?
  
  (Asked by: username_0 on 2024-07-25)
  
  I am currently following course 1 of Andrew Ng's Machine Learning Specialization. I understood we need a convex cost function to reach the global minimum of the loss. But that means the gradient will reduce and approach 0 by itself as long as we choose a convex function.
  
  Then what is the need of the learning rate?
  
  My intuition tells me its an additional measure to give us more control in training the model, but I want a proper answer and evidence to solidify my understanding.
  
  **Tags:** machine-learning, gradient-descent, learning-rate
  
  > **Comments:**
  > **username_1**: Just pointing out a misconception in the question: It is not true that gradient descent with a fixed learning rate will converge, even for a convex function. Consider $f(x) = x^2$ with learning rate 10 and starting point $2$. The sequence of iterates is $-38, 722, \ldots$. Generally you need a condition like $\eta < 2/L$, where $L$ is the Lipschitz coefficient  
  > **username_0**: @username_1 by η do you mean the learning rate? I was accustomed to alpha so trying to confirm.  
  > **username_1**: Yes,  is another variable typically used for the learning rate  
  
  ---
  
  ## 3 answers
  
  ### Answer by username_2 on 2024-07-25. Score: 5
  
  In short, there are two major reasons:
  
  1. The optimization landscape in parameter space is non-convex even with convex loss function (e.g., MSE). Therefore, you need to do small update steps (i.e., the gradient scaled by the learning rate) to find a suitable local minimum and avoid divergence.
  2. The gradient is estimated on a batch of samples, which does not represent the full let's say "population" of data. Even by using batch gradient descent, which uses the full dataset each step, the resulting gradient does not point in the direction of the true gradient. So you need to introduce a step size (i.e., the learning rate.) Moreover, at least in principle, it is possible to correct the gradient direction by including second order information (e.g., the Hessian of the loss w.r.t. parameters) although it is usually infeasible to compute.
  
  > **Comments:**
  > **username_3**: Point 1 doesn't really have anything to do with non-convexity, GD can diverge even on a simple 1D convex function like $x \mapsto x^2$ if the learning rate is too high. In fact, if anything, non-convexity is a reason to use _larger_ (and/or stochastic) steps, at least at first, to avoid converging on a poor local minimum near the starting point.  
  > **username_2**: @username_3 My point was to explain that with a step size of 1, the optimization is really unstable unless the problem is trivial, because the updates are just too large. Indeed, in DL, to avoid premature convergence and ensure stable training you need at least a well-designed architecture with suitable weights initialization, a learning rate schedule, gradient clipping, and some regularization as well. But all of these are outside the scope of the question.  
  > **username_4**: @username_2 he is saying that even in trivial cases you need a stepsize, the only case where you don't need it, is if the conditioning of the problem is 1, but your point is valid, just maybe too specific  
  
  ---
  
  ### Answer by username_5 on 2024-07-25. Score: 3
  
  Indeed your intuition of [Learning rate](https://en.wikipedia.org/wiki/Learning_rate) is right regarding its additional control in training the model, and the precise terminology for your somewhat vague 'additional control' intuition even in a convex loss landscape is called hyperparameter or tuning parameter in ML which specifies details of the learning process in contrast to parameters which determine the model itself, such as learning rate, momentum (Adam, Nesterov), batch size, regularization strength, etc. Specifically the linked reference clarifies the necessity of your concerned learning rate hyperparameter as follows.
  
  > Since it influences to what extent newly acquired information overrides old information, it metaphorically represents the **speed** at which a machine learning model "learns"... In setting a learning rate, there is a trade-off between the rate of convergence and overshooting. While the descent direction is usually determined from the gradient of the loss function, the learning rate determines how big a step is taken in that direction... In order to achieve faster convergence, prevent oscillations and getting stuck in undesirable local minima the learning rate is often varied during training either in accordance to a learning rate schedule or by using an adaptive learning rate.
  
  Usually a fixed learning rate is set very small to ensure convergence, however, in practice a learning rate schedule is often used by decay and momentum which help faster convergence and prevent oscillations as discussed in my answer to another [post](https://ai.stackexchange.com/questions/35648/at-which-point-does-the-momentum-based-gd-helps-really-in-this-figure/46171#46171).
  
  
  ---
  
  ### Answer by username_6 on 2024-07-26. Score: 3
  
  Your intuition is correct and the short answer is: The learning rate is necessary in order to control the *size of the update*!
  
  You probably already know that gradient descent (GD) is derived from the [Newton-Raphson algorithm] (<https://en.wikipedia.org/wiki/Newton%27s_method> "or Newton's method").
  In Machine Learning applications, the learning rate in GD plays the same role as the second order derivative which encodes the *curvature information*. To be more explicit, let's consider two convex functions (which, if you know your math, have positive second derivatives), $f\_{1}(x) = 2x^{2}$ and $f\_{2}(x) = \frac{x^{2}}{2}$.
  We can see that their gradient and second-order derivative are: $(f'\_{1}(x) = 4x, f''\_{1}(x) = 4)$ and $(f'\_{2}(x) = x, f''\_{2}(x) = 1)$.
  
  Now, if we consider the Newton-Raphson algorithm, then we see that:
  
  $x\_{k+1} = x\_{k} - \frac{f'\_{1}(x\_{k})}{f''\_{1}(x\_{k})} = x\_{k} - \frac{4x\_{k}}{4} = x\_{k} - \frac{f'\_{2}(x\_{k})}{f''\_{2}(x\_{k})} = x\_{k} - \frac{x\_{k}}{1} = 0, \;\forall x\_{k}\in\mathbb{R}$.
  
  We can see that the second order derivative rescales the gradient so that $\forall x\_{k}\in\mathbb{R}$, the update always takes you directly to the minimum (this works in one step because the functions are quadratic. When the function is not precisely quadratic, the algorithm will take more iterations, but *will* converge similarly to the minimizing solution).
  
  Hence, in the case of $f\_{1}$, the scaling factor $\frac{1}{f''\_{1}(x\_{k})} = \frac{1}{4} = \alpha^{\*}\_{1}$ and for $f\_{2}$, the scaling factor $\frac{1}{f''\_{2}(x\_{k})} = 1 = \alpha^{\*}\_{2}$. These represent the optimal learning rate values $\alpha^{\*}\_{1}$ and $\alpha^{\*}\_{2}$ for each function.
  
  But, what if instead of these values we used values that are smaller or larger? Well, let's see:
  
  * Let $\alpha\_{1} = \frac{1}{6} < \alpha^{\*}\_{1}$, then:
    $x\_{k+1} = x\_{k} - \frac{4x\_{k}}{6} = x\_{k} - \frac{2x\_{k}}{3} = \frac{x\_{k}}{3} \neq 0$. However, since this will be the case for every time step $k$, the update rule will give you a geometric progression (e.g. if we start with $x\_{0} = 1$, then $x\_{1} = \frac{1}{3}$, $x\_{2} = \frac{1}{9}$, $x\_{3} = \frac{1}{27}$, $\cdots$, $x\_{k} = (\frac{1}{3})^k$. If $k$ is large enough, then the final solution will be very close to 0, i.e. $k\to\infty \implies x\_{k}\to 0$). Hence, quite intuitively, **a smaller learning rate leads to a slow convergence**.
  * Let $\alpha\_{1} = \frac{1}{2} > \alpha^{\*}\_{1}$, then:
    $x\_{k+1} = x\_{k} - \frac{4x\_{k}}{2} = x\_{k} - 2x\_{k} = -x\_{k} \neq 0$. As can be seen, this does not converge at all to the optimal solution, 0! In fact it oscillates back and forth between $x\_{0}$ and $-x\_{0}$. This is a divergent progression! In fact, if we chose $\alpha\_{1} = \alpha^{\*}\_{2} = 1 > \alpha^{\*}\_{1}$, then: $x\_{k+1} = x\_{k} - 4x\_{k} = x\_{k} - 4x\_{k} = -3x\_{k} \neq 0$. which implies a geometric progression with ratio $r=-3$ which is divergent. Hence, **a larger learning rate leads to divergence**.
  
  In summary, the learning rate scales the gradient in order to control the amount of change in the optimized parameters at every step and it plays the same role as the second order derivative, i.e. it is an estimate of the function's curvature information or shape. If properly estimated, then the algorithm will converge quickly. If underestimated, the convergence will be slow (but guaranteed to converge after a potentially large number of updates). If overestimated, the algorithm will overshoot and diverge, never settling down to a specific value.
  
  P.S.: username_1 username_1 mentions in his comment the Lipschitz coefficient and in general, you want your learning rate to satisfy $0 < \alpha < \frac{1}{L}$ where $L$ is the Lipschitz coefficient. This is also related to the second order derivative information and is valid for functions Lipschitz continuous gradients (this connects to L-smoothness). Indeed, if $f(x)$ is twice differentiable and if there exists $L < \infty$ such that its Hessian matrix (second order derivative) has a bounded spectral norm, i.e. $||f''(x)|| \leq L$, then $f(x)$ has a Lipschitz continuous gradient with Lipschitz constant $L$.
  
  I recommend checking the theorem 3.4. in [1] <https://gowerrobert.github.io/pdf/M2_statistique_optimisation/grad_conv.pdf>
  
  
  ---
  
  