Machine Learning: Difference between revisions

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Replaced content with "Machine Learning ==Loss functions== ===(Mean) Squared Error=== The squared error is:<br> <math>J(\theta) = \sum|h_{\theta}(x^{(i)}) - y^{(i)}|^2</math><br> If our model i..."
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Machine Learning
Machine Learning
Machine Learning


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so the hessian is positive semi-definite
so the hessian is positive semi-definite
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==Optimization==
===Gradient Descent===
Also known as direction of steepest gradient.<br>
To minimize a loss function, just take steps in the opposite direction of the gradient.
===Stochastic Gradient Descent===
Oftentimes with large amounts of data, you can't take the gradient of all your data at once.
In this case, we use batches of data.<br>
We can take batches by getting a random order of indices without replacement.
<pre>
for epoch=1 to n
  generate batches
  for i=1 to m
    take gradient w.r.t batch i
    update using above gradient
</pre>
;Batch Size
* [https://medium.com/mini-distill/effect-of-batch-size-on-training-dynamics-21c14f7a716e A medium post empirically evaluating the effect of batch_size]
===Coordinate Block Descent===
===Learning Rate===
==SVM==
[http://cs229.stanford.edu/notes/cs229-notes3.pdf Andrew Ng Notes]<br>
Support Vector Machine<br>
This is a linear classifier the same as a perceptron except the goal is not to just classify our data properly but to also maximize the margin.<br>
<math>h_{w,b}(x) = g(W*x+b)</math> where <math>g(x) = I[x>=0]-I[x<0]</math> is the sign function.<br>
===Margins===
The margin denoted by <math>\gamma</math> is the distance between our classifier and the closest point.<br>
;Functional Margin
The margin corresponding to one example is:<br>
<math>\hat{\gamma}^{(i)} = y^{(i)}(w^Tx^{(i)}+b)</math>
The margin for our entire sample is the smallest margin per example.
;Geometric Margin
The geometric margin is the actual distance.<br>
<math>\hat{\gamma}^{(i)} = y^{(i)}((\frac{w}{\Vert w \Vert})^Tx^{(i)}+\frac{b}{\Vert w \Vert})</math><br>
;Notes
* <math>\mathbf{w}</math> is the normal vector of our hyperplane so <math>\frac{w}{\Vert w \Vert})^Tx^{(i)}</math> is the length of the projection of x onto our normal vector.
: This is the distance to our hyperplane.
===Lagrangians===
The goal for svm is to maximize the margin:<br>
<math>
\begin{aligned}
\max_{\hat{\gamma}, w, b} &\frac{\hat{\gamma}}{\Vert w \Vert}\\
\text{s.t. }& y^{(i)}(w^Tx^{(i)} + b) \geq \hat{\gamma} \quad \forall i
\end{aligned}
</math><br>
which is equivalent to, by setting <math>\hat{\gamma}=1</math>,<br>
<math>
\begin{aligned}
\min_{\gamma, w, b} &\Vert w \Vert ^2\\
\text{s.t. }& y^{(i)}(w^Tx^{(i)} + b) \geq 1 \quad \forall i
\end{aligned}
</math><br><br>
In general, given an optimization in the (primal) form:<br>
<math>
\begin{aligned}
\min_w & f(w)\\
\text{s.t. }& h_i(w) \leq 0 \quad \forall i\\
& g_i(w) = 0
\end{aligned}
</math><br>
we can rewrite the optimization as <br>
<math>
\min_{w}\max_{\alpha, \beta \mid \alpha \geq 0} \mathcal{L}(w, \alpha, \beta)
</math><br>
where <math>\mathcal{L}(w, \alpha, \beta) = f(w) + \sum \alpha_i g_i(w) + \sum \beta_i h_i(w)</math> is called the lagrangian.<br>
Since <math>\max \min \leq \min \max</math>,<br>
we have:<br>
<math>
\max_{\alpha, \beta \mid \alpha \geq 0}\min_{w} \mathcal{L}(w, \alpha, \beta) \leq \min_{w}\max_{\alpha, \beta \mid \alpha \geq 0} \mathcal{L}(w, \alpha, \beta)
</math><br>
The left term is called the dual problem.<br>
If the solution to the dual problem satisfy some conditions called the KKT conditions, then it is also the solution to the original problem.
===Kernel Trick===
Oftentimes, using linear classifiers such as perceptron and SVM may fail to classify data for which the true decision boundary is non-linear.<br>
In this case, one way to get around this is to perform a non-linear preprocessing of the data <math>\phi(x)</math>.<br>
For example <math>\phi(x) = \begin{bmatrix}x \\ x^2 \\ x^3\end{bmatrix}</math>
If our original model and training only used <math>\langle x, z\rangle</math> then we only need <math>\phi(x)^T\phi(z)</math><br>
A kernel <math>K(x,z)</math> is a function that can be expressed as <math>K(x,z)=\phi(x)^T\phi(z)</math> for some function <math>\phi</math><br>
====Identifying if a function is a kernel====
Basic check:
Since the kernel is an inner-product between <math>\phi(x), \phi(z)</math>, it should satisfy the axioms of inner products, namely <math>K(x,z)=K(z,x)</math>, otherwise it is not a kernel.<br>
====Mercer's Theorem====
Let our kernel function be <math>K(z,x)</math>.
Then for any sample S, the corresponding matrix <math>\mathbf{K}</math> where <math>K_{ij} = K(x^{(i)},x^{(j)})</math> is symmetric positive definite.
{{hidden | Proof |
Symmetry: <math>K_{ij} = K(\mathbf{x}^{(i)},\mathbf{x}^{(j)} = \phi(\mathbf{x}^{(i)})^T\phi(\mathbf{x}^{(j)}) = \phi(\mathbf{x}^{(j)})^T\phi(\mathbf{x}^{(i)}) = K_{ji}</math><br>
Positive Definite:<br>
Let <math>\mathbf{v} \in \mathbb{R}^n</math>.<br>
Then<br>
<math>
\begin{aligned}
\mathbf{v}^T \mathbf{K} \mathbf{v}
&= v^T [\sum_j K_{ij}v_j]\\
&= \sum_i \sum_j v_{i}K_{ij}v_{j}\\
&= \sum_i \sum_j v_{i}\phi(\mathbf{x}^{(i)})^T\phi(\mathbf{x}^{(j)})v_{j}\\
&= \sum_i \sum_j v_{i} \sum_k \phi_k(\mathbf{x}^{(i)}) \phi_k(\mathbf{x}^{(j)})v_{j}\\
&= \sum_k \sum_i \sum_j v_{i} \phi_k(\mathbf{x}^{(i)}) \phi_k(\mathbf{x}^{(j)})v_{j}\\
&= \sum_k \sum_i  v_{i} \phi_k(\mathbf{x}^{(i)}) \sum_j \phi_k(\mathbf{x}^{(j)})v_{j}\\
&= \sum_k (\sum_i  v_{i} \phi_k(\mathbf{x}^{(i)}))^{2}\\
&\geq 0 \quad
\end{aligned}
</math>
}}
==Learning Theory==
===PAC Learning===
Probably Approximately Correct (PAC)<br>
A hypothesis class <math>H</math> is PAC learnable if given <math>0 < \epsilon, \delta < 1</math>, there is some function <math>m(\epsilon, \delta)</math> polynomial in <math>1/\epsilon, 1/\delta</math> such that if we have a sample size <math>\geq m(\epsilon, \delta)</math> then with probability <math>1-\delta</math> the hypothesis we will learn will have an average error <math>\leq \epsilon</math>.
===Uniform Convergence===
If for all hypothesis <math>h</math>, <math>|L_S(h)-L_D(h)| \leq \epsilon</math>, then the training set <math>S</math> is called <math>\epsilon</math>-representative.<br>
Then
<math>
L_D(h_s)
\leq L_S(h_S) + \epsilon / 2
\leq L_S(h_D) + \epsilon / 2
\leq L_D(h_D) + \epsilon
</math>.<br>
A hypothesis class <math>H</math> has uniform convergence if there exists <math>m^{UC}(\epsilon, \delta)</math> such that for every <math>\epsilon, \delta</math>, if we draw a sample <math>S</math> then with probability <math>1-\delta</math>, <math>S</math> is <math>\epsilon</math>-representative.
===VC dimension===
[https://en.wikipedia.org/wiki/Vapnik%E2%80%93Chervonenkis_dimension Wikipedia Page]<br>
[https://www.stat.berkeley.edu/~bartlett/courses/2013spring-stat210b/notes/10notes.pdf Some slides]
====Shattering====
A model <math>f</math> parameterized by <math>\theta</math> is said to shatter a set of points <math>\{x_1, ..., x_n\}</math> if for every possible set of binary labellings <math>\{y_1,...,y_n\}</math> there exists <math>\theta</math> such that <math>f</math> makes no errors.
====Definition====
Intuitively, the VC dimension of a hypothesis set is how complex of a model it is.<br>
Stolen from wikipedia:<br>
The VC dimension of a model <math>f</math> is the maximum number of points that can be arranged so that <math>f</math> shatters them.
More formally, it is the maximum cardinal <math>D</math> such that some data point set of cardinality <math>D</math> can be shattered by <math>f</math>.
;Notes
* To show VCdim is at least n, find a training set of size n that can be shattered by our hypothesis class.
* To show VCdim is leq n, prove no training set of size n+1 can be shattered by our hypothesis class.
* Number of parameters do not necessarily correspond to VC dimension.
: <math>H=\{h(x)=\sin(\theta x)\}</math> has infinite VC dimension with one parameter
====Theory====
[https://nowak.ece.wisc.edu/SLT09/lecture8.pdf Reference]<br>
In the case where the Hypothesis class <math>\mathcal{H}</math> is finite, we have
* <math>|L_D(h) - L_S(h)| < \sqrt{
\frac{\log|\mathcal{H}| + \log(1/\delta)}{2m}}
</math>
: where <math>m</math> is the size of the sample.
For all h in H,
* <math>|L_D(h) - L_S(h)| < K_1 \sqrt{
\frac{VCdim + K_2 \log(2/\delta)}{2m}}
</math>
: for some constants <math>K_1, K_2</math>
===Bias-Variance Tradeoff===
* Let <math>L_D(h)</math> be the true loss of hypothesis h
: and <math>L_S(h)</math> be the true loss of hypothesis h
* <math>L_D(h_s^*) = L_D(h_D^*) + [L_D(h_s^*) - L_D(h_D^*)]</math>
* The term <math>L_D(h_D^*)</math> is called the bias
* The term <math>[L_D(h_s^*) - L_D(h_D^*)]</math> is called variance.
* Larger hypothesis class will get smaller bias but larger variance.
* Overfitting vs. underfitting
===Rademacher Complexity===
====Definition====
The Rademacher complexity, like the VC dimension, measures how "rich" the hypothesis space is.<br>
In this case, we see how well we can fit random noise.<br>
Given a set <math>A \subset \mathbb{R}^m</math> the Rademacher complexity is:<br>
<math>R(A) = \frac{1}{m}E_{\sigma} [\sup_{a \in A} \sum_{i=1}^{m} \sigma_i a_i]</math><br>
where each <math>\sigma_i</math> are from a discrete uniform distribution <math>\{-1, 1\}</math><br>
Given a sample <math>S=\{z_1,...,z_n\}</math> and a function class <math>F</math>, the empirical rademacher complexity is:<br>
<math>R(F \circ S)</math><br>
where <math>F \circ S = \{(f(z_1),...,f(z_n)) \mid f \in F\}</math><br>
;Notes
===Concentration Bounds===
====Hoeffding's inequality====
Let <math>X_1,...,X_n</math> be bounded in (a,b)<br>
Then <math>P(|\bar{X}-E[\bar{X}]| \geq t) \leq 2\exp(-\frac{2nt^2}{(b-a)^2})</math>

Revision as of 16:12, 10 November 2019

Machine Learning

Loss functions

(Mean) Squared Error

The squared error is:
\(\displaystyle J(\theta) = \sum|h_{\theta}(x^{(i)}) - y^{(i)}|^2\)
If our model is linear regression \(\displaystyle h(x)=w^tx\) then this is convex.

Proof

\(\displaystyle \begin{aligned} \nabla_{w} J(w) &= \nabla_{w} \sum (w^tx^{(i)} - y^{(i)})^2\\ &= 2\sum (w^t x^{(i)} - y^{(i)})x \\ \implies \nabla_{w}^2 J(w) &= \nabla 2\sum (w^T x^{(i)} - y^{(i)})x^{(i)}\\ &= 2 \sum x^{(i)}(x^{(i)})^T \end{aligned} \)
so the hessian is positive semi-definite