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All of Yaroslav_Bulatov's Comments + Replies

0 And 1 Are Not Probabilities
Yaroslav_Bulatov16y1

Disallowing a symbol for "all events" breaks the definition of a probability space. It's probably easier to allow extended reals and break some field axioms than figure out do rigorous probability without a sigma-algebra.

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Trust in Bayes
Yaroslav_Bulatov16y0

Gray Area -- good question, thanks for bring my attention to reservoir sampling. I found a compact description of it in Devroye's "Non-Uniform ..." and for sampling just 1 integer x it looks as follows

  1. At step 1, let x=1
  2. At step k, let x=k with probability 1/k

sum_i 1/i diverges, this means x will never stop growing

pdf23ds -- I think you can use reservoir sampling for sampling from infinite streams, an interesting question is when it works. For instance, consider an infinite stream of IID bits, 1-element reservoir sampling converges after 1 ste... (read more)

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Trust in Bayes
Yaroslav_Bulatov16y0

Interesting post, thanks for the Jaynes link. Related book which is a great read is Szekely's Paradoxes in Probability Theory and Statistics

I think the most intriguing paradoxes are the ones that experts can not agree how to resolve. For instance, take the two envelope paradox: you are presented with two envelopes, one has twice as much money as the other. You are told that first envelope contains x dollars, which envelope should you choose? From expected value calculations, the other envelope has $1.25x which is larger regardless of x. Turns out that the ... (read more)

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3CronoDAS13y
I've read a pretty good resolution to the two envelopes problem. You have to have some prior distribution over the possible amounts of money in the envelopes. The expected value of switching is equal to 1/2xP(I have the envelope with more money | I opened an envelope containing x dollars) + 2xP(I have the envelope with less money | I opened an envelope containing x dollars). This means that, once you know what x is, if you think that you have less than a 2/3 chance of having the envelope with more money, you should switch. According to what I read, if your prior is such that there is no finite X for which you would decide that you have less than a 2/3 chance of having the envelope with more money, then your expected value for x, before you learned what it was, was infinite - and if you were expecting an infinite amount of money, then of course any finite value is disappointing. (Note that having a proper prior doesn't keep you from expecting an infinite value for x). So it doesn't matter what you actually find in the envelope, once you find out what it is, you should switch - but there's no reason to switch before you open at least one envelope.