orthonormal comments on Open Thread: March 2010 - Less Wrong

5 Post author: AdeleneDawner 01 March 2010 09:25AM

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

Article Navigation

Comments (658)
Tags: open_thread

Comments (658)

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

Comment author: [deleted] 06 March 2010 03:24:51PM *  5 points [-]

Pick some reasonable priors and use them to answer the following question.

On week 1, Grandma calls on Thursday to say she is coming over, and then comes over on Friday. On week 2, Grandma once again calls on Thursday to say she is coming over, and then comes over on Friday. On week 3, Grandma does not call on Thursday to say she is coming over. What is the probability that she will come over on Friday?

ETA: This is a problem, not a puzzle. Disclose your reasoning, and your chosen priors, and don't use ROT13.

Comment author: orthonormal 08 March 2010 10:11:34PM *  2 points [-]

Let

  • AN = "Grandma calls on Thursday of week N",
  • BN = "Grandma comes on Friday of week N".

A toy version of my prior could be reasonably close to the following:

P(AN)=p, P(AN,BN)=pq, P(~AN,BN)=(1-p)r

where

  • the distribution of p is uniform on [0,1]
  • the distribution of q is concentrated near 1 (distribution proportional to f(x)=x on [0,1], let's say)
  • the distribution of r is concentrated near 0 (distribution proportional to f(x)=1-x on [0,1], let's say)

Thus, the joint probability distribution of (p,q,r) is given by 4q(1-r) once we normalize. Now, how does the evidence affect this? The likelihood ratio for (A1,B1,A2,B2) is proportional to (pq)^2, so after multiplying and renormalizing, we get a joint probability distribution of 24p^2q^3(1-r). Thus P(~A3|A1,B1,A2,B2)=1/4 and P(~A3,B3|A1,B1,A2,B2)=1/12, so I wind up with a 1 in 3 chance that Grandma will come on Friday, if I've done all my math correctly.

Of course, this is all just a toy model, as I shouldn't assume things like "different weeks are independent", but to first order, this looks like the right behavior.

Comment author: orthonormal 09 March 2010 08:42:33AM 1 point [-]

I should have realized this sooner: P(B3|~A3) is just the updated value of r, which isn't affected at all by (A1,B1,A2,B2). So of course the answer according to this model should be 1/3, as it's the expected value of r in the prior distribution.

Still, it was a good exercise to actually work out a Bayesian update on a continuous prior. I suggest everyone try it for themselves at least once!

Powered by Reddit Powered by Reddit