Gradient Descent

Suppose we have a cost function $J$ and want to minimize it

  • say it takes 2 parameters $\theta_0$ and $\theta_1$
  • So we have $J(\theta_0, \theta_1)$ and want to find $\min_{\theta_0, \theta_1} J(\theta_0, \theta_1)$


Idea:

  • start with some $(\theta_0, \theta_1)$ (say, $(0,0)$)
  • keep changing $(\theta_0, \theta_1)$ to reduce $J(\theta_0, \theta_1)$ until we end up in minimum


In the pseudo code

  • repeat until converges
    • for $j = 0$ and $j = 1$
    • $\theta_j = \theta_j - \alpha \cfrac{\partial}{\partial \theta_i} J(\theta_0, \theta_1)$

$\alpha$ is the learning rate, value that specifies how small are steps we take


Simultaneous Update

Note that the update for $(\theta_0, \theta_1)$ has to be simultaneous. That is

  • $\tau_0 = \theta_0 - \alpha \cfrac{\partial}{\partial \theta_0} J(\theta_0, \theta_1)$
  • $\tau_1 = \theta_1 - \alpha \cfrac{\partial}{\partial \theta_1} J(\theta_0, \theta_1)$
  • $\theta_0 = \tau_0$
  • $\theta_1 = \tau_1$

As you see, $\theta_0$ is used to calculate new value for $\theta_1$, so we cannot update it before we calculate new value for $\theta_0$.

This is called simultaneous update


Intuition

Let's see how it works

  • assuming that there's only one variable $\theta_1$ and $\theta_1 \in \mathbb{R}$

$\theta_1 = \theta_1 - \alpha \cfrac{d}{d \theta_1} J(\theta_1)$


let's have a look at the partial derivative:

  • $\beta = \alpha \cfrac{d}{d \theta_1} J(\theta_1)$

if the derivative is positive

  • we're moving left:
$\theta_1 = \theta_1 - \beta$

if the derivative is negative

  • we're moving right:
$\theta_1 = \theta_1 + \beta$


For two variables the cost function would look like that:


Learning Rate

  • when $\alpha$ is too small - we're taking very small steps - too slow
  • when $\alpha$ is too large - we're taking too big steps and may miss the minimum
in this case not only may it fail to converge, but even diverge!

Approaching the minimum

  • As we're approaching the local minimum, it takes smaller and smaller steps
  • If $\theta_1$ is at the local minimum, then $\beta = 0$ and $\theta_1$ won't change


Convex Function

The cost function $J$ has to be convex if we don't want to end up in a local minimum.


Univarivate Linear Regression

  • Given input data set $\{(x^{(i)}, y^{(i)}\}$ of size $m$
  • We have our hypothesis $h(x) = \theta_0 + \theta_1 x$
  • how to choose $\theta_0$ and $\theta_1$ so $h(x)$ is closest to the set of input data

We need to minimize the cost function:

  • $J(\theta_1, \theta_2) = \cfrac{1}{2 m} \sum_{i = 1}{m} (h_0 x^{(i)} - y^{(i)} )^2$
  • This is the squared error cost function

Let's simplify our expression:

$\cfrac{\partial}{\partial \theta_j} J(\theta_0, \theta_1) = \cfrac{\partial}{\partial \theta_j} \cfrac{1}{2m} \sum (h_{\theta}(x^{i}) - y^{(i)} )^2 = \cfrac{\partial}{\partial \theta_j} \cfrac{1}{2m} \sum (\theta_0 + \theta_1 x^{i} - y^{(i)} )^2 $


Now we calculate the derivatives and have:

  • for $\theta_0$:
$\cfrac{\partial}{\partial \theta_0} J(\theta_0, \theta_1) = \cfrac{1}{m} \sum (h_{\theta} (x^{(i)}) - y^{(i)})$
  • for $\theta_1$:
$\cfrac{\partial}{\partial \theta_1} J(\theta_0, \theta_1) = \cfrac{1}{m} \sum (h_{\theta} (x^{(i)}) - y^{(i)}) \cdot x^{(i)}$


So for the regression the algorithm is

  • repeat until converges
    • $\theta_0 = \theta_0 - \alpha \cfrac{\partial}{\partial \theta_0} J(\theta_0, \theta_1) = \cfrac{1}{m} \sum (h_{\theta} (x^{(i)}) - y^{(i)})$
    • $\theta_1 = \theta_1 - \alpha \cfrac{\partial}{\partial \theta_1} J(\theta_0, \theta_1) = \cfrac{1}{m} \sum (h_{\theta} (x^{(i)}) - y^{(i)}) \cdot x^{(i)}$
  • (update simultaneously)


The square error cost function is convex, so we always converge to the global minimum.


Gradient Descent for Multivariate Linear Regression

For Multivariate Linear Regression we have $x^{(i)} \in \mathbb{R}^{n + 1} $and $\theta = \in \mathbb{R}^{n+1}$, where

  • $n$ - is number of features
  • $m$ - number of training examples
  • and $x_0^{(i)} = 1$ for all $i$ (the slope)

So out cost function takes the following form:

$J(\theta) = J(\theta_0, ... , \theta_n) = \cfrac{1}{2m} \sum_{i = 1}^{m} (h_{\theta} (x^{(i)}) - y^{(i)} )^2$


The algorithm:

  • repeat
    • simultaneously for all $i$
    • $\theta_j = \theta_j - \alpha \cfrac{\partial}{\partial \theta_j} J(\theta)$

or, having calculated the derivatives:

  • repeat
    • simultaneously for all $i$
    • $\theta_j = \theta_j - \alpha \cfrac{1}{m} \sum (h_{\theta}) (x^{(i)}) - y^{(i)} ) \cdot x_j^{(i)}$


Gradient Descent in Practice

Feature Scaling

Use Feature Scaling to help GD converge faster

Learning Rate

  • How to choose $\alpha$?
  • If GD works properly, cost function should decrease after each iteration
  • if $J$ decreases by less than $\epsilon = 10^{-3}$ in one iteration - declare convergence
  • If $J$ is increasing instead - need to make $\alpha$ smaller

Choosing $\alpha$:

  • But don't choose $\alpha$ too small - it'll take too long to converge
  • To choose $\alpha$ try
$..., 0.001, 0.01, 0.1$, ... or increase it 3-fold
  • And see what is acceptable


Applications

Apart from Linear Regression, Gradient Descent may also be used for


Sources

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