We want to compare features of the model to features of the data:

Visual diagnostics

1. PDF plot
2. CDF of data vs. CDF of model
3. Quantile-Quantile plot
4. Calibration Plot

Summative Metrics

1. KL Divergence
2. Expected Calibration Error
3. Maximum Calibration Error

Marginalization Ignores Covariances

Notice on the figure on the right captures distribution much better, yet the marginal distributions don’t show this. This is because marginalizing over the datasets ignores the covariances. Hence, remember to keep dimensions and any projections hould capture covariances, etc.

An event is sub-subset of the sample space \(E \in S\). These are some subset to which you ascribe some meaning.

Have: \(m\) training data points \((\theta_{i}, \phi_{i})\) generated from the true/approximated function \(\phi_{i} = f\qty (\theta_{i})\) (which uses physical simulation/CV techniques). Training data here is *very expensive and lots of errors

Want: \(\hat{f}\qty(\theta) = f\qty(\theta)\)

Problem: as joints rotate (which is highly nonlinear), cloth verticies move in complex and non-linear ways which are difficult to handle with a standard neural network—there are highly non-linear rotations! which is not really easy to make with standard model functions using \(\hat{f}\).

Preamble

As notation differs between Alg4DM (which the presentation and notes use) and the paper, we provide a note here to standardize the notation of the PGA formulation to avoid confusion.

Recall that the non-linear program formulation of the naive PGA implementation gives:

\begin{align} \max_{\theta}\ &f(\theta) \\ \text{such that}\ &J\theta = \bold{1} \\ & \theta \geq \bold{0} \\ & h_{i}(\theta) \leq \epsilon_{i},\ \forall i \end{align}

for:

\begin{equation} f(\theta) = \beta^{\top} \bold{Z}^{-1} \bold{r}_{\theta} \end{equation}

A result so important it gets a page.

Every operator on a finite-dimensional, non-zero, complex vector space has an eigenvalue.

Proof:

Suppose \(V\) is a complex vector space with dimension \(n > 0\), and \(T \in \mathcal{L}(V)\). Choose \(v \in V, v\neq 0\) (possible as \(V\) is non-zero):

Construct a list of \(n+1\) vectors:

\begin{equation} v, Tv, \dots T^{n} v \end{equation}

because we managed to cram \(n+1\) vectors into a list for a vector space with dimension \(n\), that list is linearly dependent.