## twenty-four to nil

Posted in Books, Kids, Statistics with tags , , , on September 16, 2022 by xi'an

Another puzzling question on X validated, where the expectation of a random sum of deterministic vectors is to be computed. (That is, the sum involves a random number of terms.) Without enough detail to understand why this proves a difficulty, given that each deterministic vector is to be invoked at most once. Nonetheless, my (straightforward) answer there

$Y_1\underbrace{\mathbb P(\tau\ge 1)}_{=1}+Y_2\mathbb P(\tau\ge 2)+\cdots+Y_N\underbrace{\mathbb P(\tau=N)}_{=0}$

proved much more popular (in terms of votes) that many of my much more involved answers there. Possibly because both question and answer are straightforward.

## Why do we draw parameters to draw from a marginal distribution that does not contain the parameters?

Posted in Statistics with tags , , , , , , , on November 3, 2019 by xi'an

A revealing question on X validated of a simulation concept students (and others) have trouble gripping with. Namely using auxiliary variates to simulate from a marginal distribution, since these auxiliary variables are later dismissed and hence appear to them (students) of no use at all. Even after being exposed to the accept-reject algorithm. Or to multiple importance sampling. In the sense that a realisation of a random variable can be associated with a whole series of densities in an importance weight, all of them being valid (but some more equal than others!).

## a probabilistic proof to a quasi-Monte Carlo lemma

Posted in Books, Statistics, Travel, University life with tags , , , , , on November 17, 2014 by xi'an

As I was reading in the Paris métro a new textbook on Quasi-Monte Carlo methods, Introduction to Quasi-Monte Carlo Integration and Applications, written by Gunther Leobacher and Friedrich Pillichshammer, I came upon the lemma that, given two sequences on (0,1) such that, for all i’s,

$|u_i-v_i|\le\delta\quad\text{then}\quad\left|\prod_{i=1}^s u_i-\prod_{i=1}^s v_i\right|\le 1-(1-\delta)^s$

and the geometric bound made me wonder if there was an easy probabilistic proof to this inequality. Rather than the algebraic proof contained in the book. Unsurprisingly, there is one based on associating with each pair (u,v) a pair of independent events (A,B) such that, for all i’s,

$A_i\subset B_i\,,\ u_i=\mathbb{P}(A_i)\,,\ v_i=\mathbb{P}(B_i)$

and representing

$\left|\prod_{i=1}^s u_i-\prod_{i=1}^s v_i\right| = \mathbb{P}(\cap_{i=1}^s A_i) - \mathbb{P}(\cap_{i=1}^s B_i)\,.$

Obviously, there is no visible consequence to this remark, but it was a good way to switch off the métro hassle for a while! (The book is under review and the review will hopefully be posted on the ‘Og as soon as it is completed.)