Perhaps the worst (published) straw-man argument against cosmological fine tuning as an apologetic

In the following article, @david.heddle critiques a paper that was intended to debunk an aspect of the cosmological fine-tuning argument:

@david.heddle says:

With some of the worst analysis I have ever seen published. To paraphrase (read it if you think I am being uncharitable) they argue that if a constant k can have any real value (which, by the way, nobody claims, but OK) , then any finite range of possible values (even big ranges, let alone the small ranges demanded by fine-tuning) has effectively a zero probability. Therefore, they conclude, improbable habitable universes are, in fact, impossible.

I’m not kidding, that’s their argument.

3 Likes

“It is known that there are an infinite number of worlds, simply because there is an infinite amount of space for them to be in. However, not every one of them is inhabited. Therefore, there must be a finite number of inhabited worlds. Any finite number divided by infinity is as near to nothing as makes no odds, so the average population of all the planets in the Universe can be said to be zero. From this it follows that the population of the whole Universe is also zero, and that any people you may meet from time to time are merely the products of a deranged imagination.”

― Douglas Adams, The Restaurant at the End of the Universe

:slight_smile:

2 Likes

If we REALLY want to quibble, then I’d like to point out that probability is defined on the rational numbers, not the real numbers. :stuck_out_tongue_closed_eyes:

1 Like

Really? That’s not rational.

2 Likes

How could that be the case? Define θ to be an angle between 0 and 2π, with uniform probability density over that interval. What is the probability that θ < 1 radian?

1 Like

If we REALLY REALLY want to quibble, the probability that a series of coin-tosses mapped onto a binary expansion 0.??? using heads=1,tails=0 will produce a number less than a specified binary expansion x is itself x. This is true for any value of x, even irrational values such as pi/4 or 1/sqrt(2).
:mage:

1 Like

Probability is defined over countably infinite sets, making it a discontinuous function everywhere. Most of the time this doesn’t matter, because it approximates a continuous function arbitrarily well. But in the deep-end of probability theory, where you might want to make formal proofs rather than approximations, you need to use Lebesque integrals instead of Reinmann integrals to do these calculations.

Thus ends today’s lesson in statistical trivia. :wink: /fnord

1 Like

It is far far TOO rational … :laughing:

1 Like

I have the same question as @glipsnort. How can it be that probabilities are not defined on the reals?

Hmm, I’m not convinced. Of course you’re the statistician and I’ll defer to your expertise, but the sigma-algebra in a probability space can be countable or uncountable, as long as the sets that it consists are measurable.

2 Likes

It’s been a looooong time since I last looked at that stuff, but my recollection is the sigma-algebras should be countably infinite, or “closed under countably infinite unions” if I am allowed to Google. :slight_smile:

1 Like

There’s plenty of material out there about probability on uncountably infinite spaces. I gather it gets rather weird (as in non-intuitive), but it does exist as far as I know.

3 Likes

Hmm, I don’t think that’s true. For example: from Wikipedia’s page on Probability Space (Probability space - Wikipedia), under General Case and Non-Atomic Case, they talked about what happens when \Omega is uncountable. Since the sigma algebra has to contain \Omega, this means that the sigma-algebra can admit uncountable sets.

Again, I defer to your expertise as a statistician, it could definitely be the case that I am misunderstanding something.

1 Like

I could be wrong - I’m not a probability theorist, and seldom venture into that deep-end without a flotation device. :wink: Let me do a bit of digging …

2 Likes

I think it is fairly trivial to create examples of probability distributions on uncountable sets, and to demonstrate that they do not break the rules of probability.

It seems that, necessarily, an uncountably large subset of the domain has to have a probability distribution of zero in order to integrate to 1. That might be the only consequence of using an uncountably large domain.

In the case of real numbers on a line (continuous distribution), we can prove that the discretizing the distribution can approximate the distribution to arbitrary accuracy by decreasing the step size. This gives a way of defining the probability distribution such that it doesn’t not matter, at limit, whether we are talking about reals or just irrationals. It is the same result, at the limit, either way.

2 Likes

Anything beyond what’s been mentioned here is over my head, for sure.

2 Likes