IlyaShpitser comments on A Fervent Defense of Frequentist Statistics - Less Wrong

43 Post author: jsteinhardt 18 February 2014 08:08PM

You are viewing a comment permalink. View the original post to see all comments and the full post content.

Comments (125)

You are viewing a single comment's thread. Show more comments above.

Comment author: IlyaShpitser 12 February 2014 08:56:01AM *  1 point [-]

Oh, saying A,B,C,D are in [0,1] restricts quite a bit. It eliminates distributions with support over all the reals

???

There are easy to compute bijections from R to [0,1], etc.

The Bayesian approach to this problem uses an hierarchical distribution with two levels: one specifying the distribution p[A,B,C,D | X] in terms of some parameter vector X, and the other specifying the distribution p[X]

Yes, parametric Bayes does this. I am giving you a problem where you can't write down p(A,B,C,D | X) explicitly and then asking you to solve something frequentists are quite happy solving. Yes I am aware I can do a prior for this in the discrete case. I am sure a paper will come of it eventually.

Latent variables are still a pain, though.

The whole point of things like the beautiful distribution is you don't have to deal with latent variables. By the way the reason to think about H1 is that it represents all independences over A,B,C,D in this latent variable DAG:

A <- u1 -> B <- u2 -> C <- u3 -> D <- u4 -> A

where we marginalize out the ui variables.


which can indeed be written quite elegantly as an exponential family distribution with features for each clique in the graph

I think you might be confusing undirected and bidirected graph models. The former form linear exponential families and can be parameterized via cliques, the latter form curved exponential families, and can be parameterized via connected sets.

Comment author: labachevskij 19 February 2014 12:44:49PM *  1 point [-]

There are easy to compute bijections from R to [0,1], etc.

This is not true, there are bijections between R and (0,1), but not the closed interval.

Anyway there are more striking examples, for example if you know that A, B, C, D are in a discrete finite set, it restricts yout choices quite a lot.

Comment author: nshepperd 19 February 2014 05:04:50PM 3 points [-]

Did you mean to say continuous bijections? Obviously adding two points wouldn't change the cardinality of an infinite set, but "easy to compute" might change.

Comment author: labachevskij 21 February 2014 09:52:50AM 1 point [-]

You're right, I meant continuous bijections, as the context was a transformation of a probability distribution.

Comment author: V_V 19 February 2014 02:19:03PM *  3 points [-]

This is not true, there are bijections between R and (0,1), but not the closed interval.

No.

Comment author: IlyaShpitser 19 February 2014 12:49:16PM 0 points [-]

This is not true, there are bijections between R and (0,1), but not the closed interval.

You are right, apologies.