Unsupervised Learning

From David's Wiki
Jump to navigation Jump to search
\( \newcommand{\P}[]{\unicode{xB6}} \newcommand{\AA}[]{\unicode{x212B}} \newcommand{\empty}[]{\emptyset} \newcommand{\O}[]{\emptyset} \newcommand{\Alpha}[]{Α} \newcommand{\Beta}[]{Β} \newcommand{\Epsilon}[]{Ε} \newcommand{\Iota}[]{Ι} \newcommand{\Kappa}[]{Κ} \newcommand{\Rho}[]{Ρ} \newcommand{\Tau}[]{Τ} \newcommand{\Zeta}[]{Ζ} \newcommand{\Mu}[]{\unicode{x039C}} \newcommand{\Chi}[]{Χ} \newcommand{\Eta}[]{\unicode{x0397}} \newcommand{\Nu}[]{\unicode{x039D}} \newcommand{\Omicron}[]{\unicode{x039F}} \DeclareMathOperator{\sgn}{sgn} \def\oiint{\mathop{\vcenter{\mathchoice{\huge\unicode{x222F}\,}{\unicode{x222F}}{\unicode{x222F}}{\unicode{x222F}}}\,}\nolimits} \def\oiiint{\mathop{\vcenter{\mathchoice{\huge\unicode{x2230}\,}{\unicode{x2230}}{\unicode{x2230}}{\unicode{x2230}}}\,}\nolimits} \)

Basics of Unsupervised Learning

Clustering

Given points \(\displaystyle \{x^{(1)},...,x^{(m)}\}\) we wish to group them into clusters 1,...,k.
Usually we are given the number of clusters, k.
The problem is to assign labels \(\displaystyle \{z^{(1)},...,z^{(m)}\}\) where \(\displaystyle z^{(i)} \in \{1,...,k\}\)

K-means

Here we wish to cluster them to minimize the distance to the mean of their cluster.
In our formulation we have k clusters.
The mean of each cluster \(\displaystyle \mu_i\) is called the centroid.

Optimization

Let \(\displaystyle \mathbf{\mu}\) denote the centroids and let \(\displaystyle \mathbf{z}\) denote the cluster labels for our data.
Our loss function is \(\displaystyle L(\mu, \mathbf{z}) = \sum_{i} \Vert x^{(i)} - \mu_{z^{(i)}} \Vert ^2\).

Using coordinate-block descent

\(\displaystyle L(\mu, \mathbf{z}) = \sum_{i} \Vert x^{(i)} - \mu_{z^{(i)}} \Vert ^2\).
If we fix \(\displaystyle \mu\) then we have only need to minimize \(\displaystyle z^{(i)}\).
Since each term of the sum is independent, we simply choose the closest centroid for each point
If we fix \(\displaystyle \mathbf{z}\) then we need to minimize \(\displaystyle L(\mu, \mathbf{z})\) wrt \(\displaystyle \mu\).
Taking the gradient and setting it to 0 we get:
\[ \begin{aligned} \nabla_{\mu} L(\mu, \mathbf{z}) &= \nabla_{\mu} \sum_{i} \Vert x^{(i)} - \mu_{z^{(i)}} \Vert ^2\\ &= \nabla_{\mu} \sum_{j=1}^{k} \sum_{i\mid z(i)=j} \Vert x^{(i)} - \mu_{z^{(i)}} \Vert ^2\\ &= \nabla_{\mu} \sum_{j=1}^{k} \sum_{i\mid z(i)=j} \Vert x^{(i)} - \mu_{j} \Vert ^2\\ &= \sum_{j=1}^{k} \sum_{i\mid z(i)=j} \nabla_{\mu} \Vert x^{(i)} - \mu_{j} \Vert ^2\\ &= -\sum_{j=1}^{k} \sum_{i\mid z(i)=j} 2(x^{(i)} - \mu_{j}) = 0\\ \implies \mu_{j} &= (\sum_{i\mid z(i)=j} x^{(i)})/(\sum_{i\mid z(i)=j} 1) \quad \forall j \end{aligned} \]

Notes

This procedure will yield us a sequence of parameters and losses.
\(\displaystyle \mu^{0}, z^{0}, \mu^1, z^1, ...\)
\(\displaystyle L(0) \geq L(1) \geq L(2) \geq ...\)

  • Since the loss is monotone decreasing and bounded below, it will converge by the monotone convergence theorem.
  • However, this does not imply that the parameters \(\displaystyle \mathbf{\mu}\) and \(\displaystyle \mathbf{z}\) will converge.

Algorithm

  1. Randomly initialize labels \(\displaystyle \mathbf{z}\).
  2. Then calculate the centroids \(\displaystyle \mathbf{\mu}\).
  3. Then update the labels for each example to the closest centroid.
  4. Update the centroids by taking the mean of every point in the cluster.
  5. Repeat steps 3 and 4

Soft K-means

We will develop a model for how our data is generated.
We will then find probabilities for each element being from a cluster (ala Bayesian paradigm).
Given \(\displaystyle k\) clusters, the probability of a point being from cluster k is \(\displaystyle \phi_k = P(z^{(i)} = k)\)
We will assume each cluster is from a normal distribution \(\displaystyle N(\mu_j, \sigma_j)\)

EM Algorithm

Expectation Maximization
The key idea is to introduce intermediate coefficients to apply Jensen's inequality.

Maximum Likelihood

Let \(\displaystyle \theta_j = [\phi_j, \mu_j, \sigma_j]\) denote our parameters.
The likelihood is \(\displaystyle L(\theta) = \Pi_{i=1}^{m} P(x^{(i)}; \theta)\) where \(\displaystyle P\) is our pdf/pmf.
Then the log-likelihood is \(\displaystyle l(\theta) = \sum_{i=1}^{m} \log P(x^{(i)}; \theta) = \sum_{i=1}^{m} \log \sum_{j=1}^{k}P(x^{(i)}, z^{(i)}=j; \theta)\)
By introducing an extra variable \(\displaystyle Q^{(i)}_{(j)}\) we'll be able to apply Jensen's inequality to the concave log function.
Assume \(\displaystyle Q^{(i)}{(j)}\) is a probability mass function.
\[ \begin{align*} l(\theta) &= \sum_{i=1}^{m} \log \sum_{j=1}^{k}P(x^{(i)}, z^{(i)}=j; \theta)\\ &=\sum_{i=1}^{m} \log \sum_{j=1}^{k} \frac{Q^{(i)}_{(j)}}{Q^{(i)}_{(j)}} P(x^{(i)}, z^{(i)}=j; \theta)\\ &\geq \sum_{i=1}^{m} \sum_{j=1}^{k} Q^{(i)}_{(j)} \log(\frac{P(x^{(i)}, z^{(i)}=j; \theta) }{Q^{(i)}_{(j)}})\\ \implies \log\left[E_{Q}\left(\frac{P(x^{(i)}, q^{(i)}; \theta)}{Q^{(i)}_{(j)}}\right)\right] &\geq E_{Q} \left[ \log \left(\frac{P(x^{(i)}, q^{(i)}; \theta)}{Q^{(i)}(j)} \right)\right] \end{align*} \]
Let \(\displaystyle J(\theta, Q) = \sum_{i=1}^{m} \sum_{j=1}^{k} Q^{(i)}_{(j)} \log \left( \frac{P(x^{(i)}, z^{(i)}=j;\theta)}{Q^{(i)}(j)} \right)\)
The EM algorithm is an iterative algorithm which alternates between an E-Step and an M-Step.

E-Step

We will fix \(\displaystyle \theta\) and maximize J wrt \(\displaystyle Q\).
Jensen's inequality holds with equality iff either the function is linear or if the random variable is degenerate.
We will assume \(\displaystyle \frac{P(x^{(i)}, z^{(i)}=j; \theta) }{Q^{(i)}_{(j)}}\) is a constant.
This implies \(\displaystyle Q^{(i)}(j) = c * P(x^{(i)}, z^{(i)} = j ; \theta)\).
Since Q is a pmf, we have \(\displaystyle Q^{(i)}(j) = \frac{1}{P(x^{(i)})} * P(x^{(i)}, z^{(i)} = j ; \theta) = P(z^{(i)} ; x^{(i)}, \theta)\)
Q is updated with \(\displaystyle Q^{(i)}(j) = \frac{P(z^{(i)}=j)P(x^{(i)}|z^{(i)}=j)}{P(x^{(i)})}\)

Maximization w.r.t Q

\(\displaystyle J(\theta, Q) = \sum_{i=1}^{m} \sum_{j=1}^{k} Q^{(i)}_{(j)} \log \left( \frac{P(x^{(i)}, z^{(i)}=j;\theta)}{Q^{(i)}(j)} \right)\)
We are assuming that Q is a pmf so \(\displaystyle \sum_j Q = 1 \)
Our lagrangian is:
\(\displaystyle \max_{Q} \min_{\beta} \left[ \sum_{i=1}^{m} \sum_{j=1}^{k} Q^{(i)}_{(j)} \log \left( \frac{P(x^{(i)}, z^{(i)}=j;\theta)}{Q^{(i)}(j)} \right) +\beta (\sum_j Q^{(i)}_{(j)} - 1)\right] \) We can maximize each \(\displaystyle Q^{(i)}\) independently.
Taking the derivative wrt Q we get:
\[ \begin{aligned} \frac{\partial}{\partial Q^{(i)}_{(j)}} \sum_{j=1}^{k} Q^{(i)}_{(j)} \log \left( \frac{P(x^{(i)}, z^{(i)}=j;\theta)}{Q^{(i)}(j)} \right) +\beta (\sum_j Q^{(i)}_{(j)} - 1) &= \log(\frac{P(x^{(i)}, z^{(i)}=j;\theta)}{Q}) - Q \frac{Q}{P(x^{(i)}, z^{(i)}=j;\theta)} (P(x^{(i)}, z^{(i)}=j;\theta))(Q^{-2}) + \beta\\ &= \log(\frac{P(x^{(i)}, z^{(i)}=j;\theta)}{Q}) - 1 + \beta \\ &= 0\\ \implies Q^{(i)}_{(j)} &= (\frac{1}{exp(1-\beta)})P(x^{(i)}, z^{(i)}=j;\theta) \end{aligned} \]
Since Q is a pmf, we know it sums to 1 so we get the same result replacing \(\displaystyle (\frac{1}{exp(1-\beta)})\) with \(\displaystyle P(x^{(i)}\)

M-Step

We will fix \(\displaystyle Q\) and maximize J wrt \(\displaystyle \theta\) \(\displaystyle J(\theta, Q) = \sum_{i=1}^{m} \sum_{j=1}^{m} Q^{(i)}_{(j)} \log \left( \frac{P(x^{(i)}, z^{(i)}=j;\theta)}{Q^{(i)}(j)} \right)\)
Assume \(\displaystyle \Sigma_j = I\) for simplicity.
Then: \[ \begin{aligned} J(\theta, Q) &= \sum_{i=1}^{m} \sum_{j=1}^{m} Q^{(i)}_{(j)} \log \left( \frac{P(x^{(i)}, z^{(i)}=j;\theta)}{Q^{(i)}(j)} \right)\\ &= \sum_{i=1}^{m} \sum_{j=1}^{m} Q^{(i)}_{(j)} \log ( P(x^{(i)}, z^{(i)}=j;\theta)) + C_1 \\ &= \sum_{i=1}^{m} \sum_{j=1}^{m} Q^{(i)}_{(j)} \log ( P(x^{(i)} \mid z^{(i)}=j) P(z^{(i)}=j)) + C_1\\ &= \sum_{i=1}^{m} \sum_{j=1}^{m} Q^{(i)}_{(j)} \log ( P(x^{(i)} \mid z^{(i)}=j)) - Q^{(i)}_{(j)} \log( P(z^{(i)}=j)) + C_1\\ &= \sum_{i=1}^{m}\left[ \sum_{j=1}^{m} Q^{(i)}_{(j)} \log ( (2\pi)^{-n/2}exp(-\Vert x^{(i)} + \mu_j \Vert^2 / 2)) - Q^{(i)}_{(j)} \log( \phi_j) \right]+ C_1 \\ &= \sum_{i=1}^{m}\left[ \sum_{j=1}^{m} Q^{(i)}_{(j)} -\Vert x^{(i)} - \mu_j \Vert^2 / 2) + Q^{(i)}_{(j)} \log( \phi_j) \right]+ C_2 \end{aligned} \]
Maximizing wrt \(\displaystyle \mu\), we get \(\displaystyle \mu_j^* = (\sum_{i} Q^{(i)}_{(i)}x^{(i)}) / (\sum_{i}Q^{(i)}_{(j)})\).
Maximizing wrt \(\displaystyle \phi\) we get \(\displaystyle \phi_j^* = \frac{1}{m} \sum_{i=1}^{m}Q^{(i)}_{(j)}\)

Maximization wrt phi

We want to maximize \(\displaystyle Q^{(i)}_{(j)} \log( \phi_j)\) subject to the condition \(\displaystyle \sum \phi_j = 1\) since \(\displaystyle \phi_j = P(z^{(i)}=j)\) is a pmf.
The lagrangian for this is \(\displaystyle \max_{\phi_1,...,\phi_k} \min_{\beta} \left[ \sum_{i=1}^{m} \sum_{j=1}^{k}Q^{(i)}_{(j)} \log( \phi_j) + \beta(\sum_{j=1}^{k}\phi_j - 1) \right]\)
This can be rewritten as \(\displaystyle \max_{\phi_1,...,\phi_k} \min_{\beta} \sum_{j=1}^{k}\left[\log( \phi_j) \sum_{i=1}^{m} Q^{(i)}_{(j)} + \beta(\phi_j - 1/k) \right]\)
The dual of this problem is \(\displaystyle \min_{\beta} \max_{\phi_1,...,\phi_k} \sum_{j=1}^{k} \left[\log( \phi_j) \sum_{i=1}^{m} Q^{(i)}_{(j)} + \beta(\phi_j - 1/k) \right]\)
Taking the gradient w.r.t \(\displaystyle \phi\), we get \(\displaystyle \frac{1}{\phi_j}\sum_{i}Q^{(i)}_{(j)} + \beta = 0 \implies \phi_j = \frac{-1}{\beta}(\sum_{i}Q^{(i)}_{(j)})\)
Plugging this into our dual problem we get:
\(\displaystyle \min_{\beta} \sum_{j=1}^{k} \left[\log(\frac{-1}{\beta}(\sum_{i}Q^{(i)}_{(j)})) \sum_{i=1}^{m} Q^{(i)}_{(j)} + \beta(\frac{-1}{\beta}(\sum_{i}Q^{(i)}_{(j)}) - 1/k) \right]\)
\(\displaystyle = \min_{\beta} \sum_{j=1}^{k} \left[\log(\frac{-1}{\beta}(\sum_{i}Q^{(i)}_{(j)})) \sum_{i=1}^{m} Q^{(i)}_{(j)} -(\sum_{i}Q^{(i)}_{(j)}) - (\beta/k) \right]\)
Taking the derivative with respect to \(\displaystyle \beta\), we get:
\[ \begin{aligned} \sum_{j=1}^{k} [(\frac{1}{(-1/\beta)(\sum Q)})(-\sum Q)(-\beta^{-2})(\sum Q) - \frac{1}{k}] &= \sum_{j=1}^{k} [ (\beta)(-\beta^{-2})(\sum Q) - \frac{1}{k}] \\ &= \sum_{j=1}^{k} [\frac{-1}{\beta}(\sum_{i=1}^{m} Q) - \frac{1}{k}]\\ &= [\sum_{i=1}^{m} \frac{-1}{\beta} \sum_{j=1}^{k}P(z^{(i)} = j | x^{(i)}) - \sum_{j=1}^{k}\frac{1}{k}]\\ &= [\frac{-1}{\beta}\sum_{i=1}^{m}1 - 1]\\ &= \frac{-m}{\beta} - 1 = 0\\ \implies \beta &= -m \end{aligned} \]
Plugging in \(\displaystyle \beta = -m\) into our equation for \(\displaystyle \phi_j\) we get \(\displaystyle \phi_j = \frac{1}{m}\sum_{i=1}^{m}Q^{(i)}_{(j)}\)

Mean-Shift

DBSCAN

Generative Models

Goal: Generate realistic but fake samples. Applications: Denoising, impainting

VAEs

Variational Auto-Encoders
Tutorial

KL Divergence

Notes
  • KL is always >= 0
  • KL is not symmetric
  • Jensen-Shannon Divergence
    • \(\displaystyle JSD(P \Vert Q) = \frac{1}{2}KL(P \Vert Q) + \frac{1}{2}KL(Q \Vert P)\)
    • This is symmetric

Model

Our model for how the data is generated is as follows:

  • Generate latent variables \(\displaystyle z^{(1)},...,z^{(m)} \in \mathbb{R}^r\) iid where dimension r is less than n.
    • We assume \(\displaystyle Z^{(i)} \sim N(\mathbf{0},\mathbf{I})\)
  • Generate \(\displaystyle x^{(i)}\) where \(\displaystyle X^{(i)} \vert Z^{(i)} \sim N(g_{\theta}(z), \sigma^2 \mathbf{I})\)
    • For some function \(\displaystyle g_{\theta_1}\) parameterized by \(\displaystyle \theta_1\)

Variational Bound

The variational bound is:

  • \(\displaystyle \log P(x^{(i)}) \geq E_{Z \sim Q_i}[\log P(X^{(i)} \vert Z)] - KL(Q_i(z) \Vert P(z))\)
Derivation

We know from Baye's rule that \(\displaystyle P(z|X) = \frac{P(X|z)P(z)}{P(X)}\).
Plugging this into the equation for \(\displaystyle KL(Q_i(z) \Vert P(z|X))\) yields our inequality.
\[ \begin{aligned} KL(Q_i(z) \Vert P(z|X)) &= E_{Q} \left[ \log(\frac{Q_i(z)}{P(z|X)}) \right]\\ &=E_Q(\log(\frac{Q_i(z) P(X^{(i)})}{P(X_z)P(z)})\\ &=E_Q(\log(\frac{Q_i(z)}{P(z)})) + \log(P(x^{(i)})) - E_Q(\log(P(X|z))\\ &=KL(Q_i(z) \Vert P(z)) + \log(P(x^{(i)}) - E_Q(\log(P(X|z)) \end{aligned} \] Rearranging terms we get:
\(\displaystyle \log P(x^{(i)}) - KL(Q_i(z) \Vert P(z|X)) = E_Q(\log(P(X|z)) - KL(Q_i(z) \Vert P(z))\)
Since the KL divergence is greater than or equal to 0, our variational bound follows.

GANs

Wasserstein GAN

Paper
The main idea is to ensure the that discriminator is lipschitz continuous and to limit the lipschitz constant (i.e. the derivative) of the discriminator.
If the correct answer is 1.0 and the generator produces 1.0001, we don't want the discriminator to give us a very high loss.

Earth mover's distance

The minimum cost of converting one pile of dirt to another.
Where cost is the cost of moving (amount * distance)
Given a set \(\displaystyle P\) with m clusters and a set \(\displaystyle Q\) with n clusters:
... \(\displaystyle EMD(P, Q) = \frac{\sum_{i=1}^{m}\sum_{j=1}^{n}f_{i,j}d_{i,j}}{\sum_{i=1}^{m}\sum_{j=1}^{n}f_{i,j}}\)

Notes
  • Also known as Wasserstein metric

Dimension Reduction

Goal: Reduce the dimension of a dataset.
If each example \(\displaystyle x \in \mathbb{R}^n\), we want to reduce each example to be in \(\displaystyle \mathbb{R}^k\) where \(\displaystyle k \lt n\)

PCA

Principal Component Analysis
Preprocessing: Subtract the sample mean from each example so that the new sample mean is 0.
Goal: Find a vector \(\displaystyle v_1\) such that the projection \(\displaystyle v_1 \cdot x\) has maximum variance.
Result: These principal components are the eigenvectors of \(\displaystyle X^TX\).
Idea: Maximize the variance of the projections.
\(\displaystyle \max \frac{1}{m}\sum (v_1 \cdot x^{(i)})^2\)
Note that \(\displaystyle \sum (v_1 \cdot x^{(i)})^2 = \sum v_1^T x^{(i)} (x^{(i)})^T v_1 = v_1^T (\sum x^{(i)} (x^{(i)})^T) v_1\).

Kernel PCA

Autoencoder

You have a encoder and a decoder which are both neural networks.

Resources

Clustering