Matrix Factorisation with Linear Filters (and discussion)

I submitted a preprint on matrix factorisations and linear filters. I managed to derive some factorisation algorithms as linear filtering algorithms. In the paper, I left a discussion to here, estimating parameters via maximising marginal likelihood. So here it is.


Online Matrix Factorization via Broyden Updates

I arXived a new preprint titled Online Matrix Factorization via Broyden Updates.

Around this April, I was reading quasi-Newton methods (from this very nice paper of Hennig and Kiefel) and upon seeing the derivation of the Broyden update, I immediately realized that this idea may be used for computing matrix factorizations. Furthermore, it will lead to an online scheme, more preferable!

The idea is to solve the following optimization problem at each iteration $k$:\begin{align*} \min_{x_k,C_k} \big\| y_k - C_k x_k \big\|_2^2 + \lambda \big\|C_k - C_{k-1}\big\|_F^2.\end{align*}
Although the basic idea was explicit, I tried to set a few goals. First of all, I wanted to develop a method that one can sample any column of the dataset and use it immediately. So I modified the notation a bit, as you can see from Eq. (2) in the manuscript. Secondly, I wanted that one must be able to use mini-batches as well, a group of columns at each time. Thirdly, it was obvious that a modern matrix factorization method must handle the missing data, so I extended the algorithm to handle the missing data. Consequently, I have sorted out all of this except a rule for missing data with mini-batches which turned out to be harder, so I left that out from this work.


Tinkering around logistic map

I was tinkering around logistic map $x_{n+1} = a x_n (1 - x_n)$ today and I wondered what happens if I plot the histogram of the generated sequence $(x_n)_{n\geq 0}$. Can it possess some statistical properties?


Monte Carlo as Intuition

Suppose we have a continuous random variable $X \sim p(x)$ and we would like to estimate its tail probability, i.e. the probability of the event $\{X \geq t\}$ for some $t \in \mathbb{R}$. What is the most intuitive way to do this?