Some days are spent in my home office, scribbling on paper or typing TeX on the computer. Some days are spent on campus, attending meetings, seminars, or colloquia, meeting students in my office, or teaching classes. Some days are spent traveling to conferences, where I meet and socialize with other probabilists, attend talks, give talks, and even sometimes scribble on paper with other people. Most of my collaborations are with researchers from other institutions. The bulk of that work is done over email. When we do meet, we typically spend the time in coffee shops, brainstorming and again, scribbling on paper.

Here is a quote I saw from Lubos:

In probability, "independent" has a very specific mathematical meaning. In objective Bayesianism, which is the interpretation espoused by Jaynes, it is interpreted as a kind of logical independence. Generally speaking, independence is a bad thing in objective Bayesianism. We cannot learn from independent evidence. A simple example is in the Monty Hall problem. The door that Monty chooses to open is (probabilistically) independent of what is behind our chosen door. As a result, we learn nothing about what is behind our door from Monty's choice. There was a 1/3 chance that the prize was behind our door before Monty did anything, and there is still a 1/3 chance of it after Monty opens a door. In symbols, if B is independent of A, then P(A | B) = P(A). We learn nothing about A by observing B.

Also, when Bayesian updating is done properly, it is not possible to "double count" anything. This is because P(A | B) = P(A | B and B). Trying to "double count" evidence in Bayesian calculations is like trying to "reuse" a premise in logic. In a logical argument, I can state a particular premise only once or I can restate it 732 times. It makes no difference at all. The logic, and the result, is ultimately the same.

Just before you posted that link, I mentioned that physicists make me cringe the least. Perhaps I should have knocked on wood.

What university do you work for? I understand if this is too personal.

I think this will have to be the first question in this thread that I do not answer. Sorry.

I thought what he was saying was pretty obvious. There could be 2 papers on 2 very different looking systems that both say AdS/CFT works. It seems very unlikely that a theory that was not motivated by either of these systems somehow gets correct answers on both of them by chance while still being wrong. His point is that there may be deep mathematical reasons, which nobody yet understands, as to why these two seemingly different tests are really the exact same thing, so passing both is no different from passing 1. Just using everyday meanings of the words, his point seems valid.

So on the basis of what we do know and understand, what is the probability that there are deep mathematical reasons for these two seemingly different tests to really be the exact same thing? In other words, the conditional correlation of these two tests, given AdS/CFT, needs to be part of the prior probabilities. Once it is specified, probability theory will weight them appropriately. They need not be independent, and they need not be the exact same thing. There is a whole continuum of possibilities in between.

Incidentally, it is more than a little strange to wait until this deep into the analysis to raise the issue of objectivity. By his own admission, Berenstein's most primitive probability assignments are subjective in nature. If Lubos wants to argue that there is no objective way to decide something, he could have argued that from the beginning, in a much simpler setting. On the other hand, if Lubos grants Berenstein the right to assign subjective probabilities for his analysis, then Berenstein may also assign a probability describing the correlation between these two tests.

This is valid. I really see no way to evaluate this distribution though. Maybe the situation is more like this. Lets say somebody wrote down the quadratic formula but couldn't derive it and for the next ten years quite a few people worked out examples and nobody ever found one were it failed. So when estimating the chances that the formula is true, the key question isn't whether our tests are independent but whether or not they sort of span the space of all possible equations. These questions are related but I think the latter is closer to the actual situation. Personally all I would be willing to say about this without an actual proof is that it is above 99% but the general uncertainty of there being these situations that nobody has thought of is to high to start talking about decimal places. Of course even if we all missed some situations and it is not always correct, it is still worth studying if it is almost correct for a variety of situations.


Yeah, good point. If you are ok with what David did from the start then he could add conditional probabilities for pretty much anything that could arrise.

Any thoughts on this thread `shuffled numbers' in the probability forum?

http://forumserver.twoplustwo.com/25...umbers-492839/

My first reaction is that this is not a well-posed problem, since a probability distribution on permutations of naturals was not specified. If we were shuffling a finite set {1,...,N}, I would consider the problem well-posed, with the implied distribution being uniform. Given that, I think the best answer is the obvious one, given in the main thread: 1/e, which is the limit of the answer to the finite question as N → ∞. I am skeptical about the possibility of creating a distribution directly on permutations of the naturals which captures this idea.

A natural extension of uniform permutations of {1,...,N} to the infinite case is a random ordering (which is not a permutation). Let X₁, X₂, ... be iid random variables with continuous distribution. Then {X₁, ..., X_N} induces a uniform permutation σ on {1, ..., N} via X_σ(1) < ... < X_σ(N). In the limit, the entire sequence X₁, X₂, ... induces a random linear ordering on the naturals given by i < j iff X_i < X_j. Of course, this does not help much with the original question, since we lose the concept of fixed points. But it is a pretty natural probabilistic construction, so it may be worth considering.

Hi. I have a problem I'm stumped with:

Consider the Branching Process {Xn, n = 0,1,2,3...} where Xn is the population size at the nth generation. Assume P(Xo = 1) = 1 and that the pgf of the common offspring distribution N is A(z) = 1/(3-2z).

Express A(z) as a power series and hence find P(N = 6).

If qn = P(Xn = 0) for n = 0,1,..., write down an equation relating q(n+1) and qn. Hence or otherwise, evaluate qn for n = 0,1,2.

Find the extinction probability q = lim(n->inf) qn.

What is it exactly that is stumping you?

Pretty much the entire subject. I guess I should just go try and learn that ****...

I assume this problem is for a course. If so, I recommend you discuss it with your instructor. For now, though, here are a few comments that might help.

What you have here is a Galton-Watson process. It might help if you look at the mathematical definition and the extinction criterion.

Also, consider the following. Suppose X(n) is the population size at time n. Let N₁, N₂, ... be iid with generating function A(z). Then

X(n+1) = N₁ + N₂ + ... + N_X(n).

This can be used to derive a relationship between the pgf of X(n) and the pgf of X(n+1).

Thanks. I'll go do some reading.

So I spent today trying to learn how to do the normal distribution stuff, but I can't even do the first question in the textbook:

"Suppose 90% of the patients with a certain disease can be cured with a certain drug. What is the approximate probability that, of 50 such patients, at least 45 can be cured with the drug?"

The back of the book says .5948, however I get .5.

Here is my working

The mean is 50*.9 = 45.
The std. deviation is sqrt(50*.9*.1) = sqrt(4.5)

So we are looking for P(X >= 45) = P(Z >= 0) = 1 - phi(0) = .5

I've asked two people and they don't know what's happening.

Assistance is greatly appreciated.

Jason, what is your take on Jaynes' book on probability. Many claim it is the most influential book on probability theory.

Using the Poisson aproximation to the Binomial for the number who don't get cured, n=50, p=.1, lamda=np=5, I get about 61.5% for the probability of 5 or fewer non-cures. I got this by calculating,

Sum[k=0..5] ( exp(-5) (5^k) / k! )

There may be some Finagaling Factor that's recommended for n=50 so small and p=.1 so large. The Poisson aproximation is best for large n, say greater than 100, and very small p. I'm not sure how good the fit would be here. Ok but not great I'm guessing.

A better informed reply can probably shed more light.

PairTheBoard

how does one calculate pot odds??

The lower boundary for the value 45 starts from 44.49 not from 45.0. Using 45.0 misses 1/2 of the distribution that is at the value 45.
For example go to http://davidmlane.com/hyperstat/z_table.html and put in the mean , sd, and 44.49 and see what p is.

I always thought that technique was just good for fine tuning a little extra accuracy. It makes a big difference in this case though doesn't it. A quarter of a standard deviation away from the mean chops off about 10% worth of probability.

Now, which is the better aproximation? The 59.48% given by the Normal or the 61.5% given by the Poisson?

PairTheBoard

Thanks for linking that site, but as far as I can see it only supports my results that it is 0.5. Does anyone know where the 0.5948 came from?

I manually computed the exact answers using a binomial RV and recall I got .61 something like what you approximated with the poisson, although that doesn't make any intuitive sense.

I don't think it's correct to use 44.49 as the lower bound. The normal graph is continuous, which means the probability of any particular value occuring is zero. Therefore the probability that it is >= 45 is the same as the probability that it is > 45.

Can someone please explain how 0.5948 was calculated?

So the Poisson apx is better. I thought it might be.

Pokerlogist is correct. If you use the 44.5 for the lower bound you get the 0.5948, at least in my Normal Table. The 44.5 reaches down about 0.24 standard deviations below the mean for the standard normal to pick up the extra 0.0948. Thus my comment that a quarter of a standard deviation near the mean chops off about 10% worth of probability.

Yes, the Normal is continuous but you are using it to aproximate a discrete random variable - the binomial. So you're getting an aproximation for exactly 45 cures from the Normal's probability in the continuous range from 44.5 to 45.49. Or normalized, from 0.24 standard deviations below the mean to 0.24 standard deviations above the mean.

Note, in that range (-0.24,0.24) for the standard normal the probability is 18.96%. ie. The Normal aproximates a 18.96% probability there will be exactly 45 cures.

The Poisson aproximates exactly 5 non-cures by

exp(-5)(5^5/5!) = 17.5%

You can see more of the difference in how the two distributions are fitting for exactly 46 cures. For the range (0.24,0.71) the standard Normal gives a probability of about 0.7611-0.5948 = 16.63%

while the Poisson aproximates exactly 4 non-cures by the same 17.5%

exp(-15)(5^4/4!) = 17.5%

PairTheBoard

Okay, got it!

Thanks

It looks like your question was answered, so I will just reiterate what others have said. When approximating a discrete random variable with a continuous one, we often use a "histogram correction", which means replacing the integer n with the interval [n - 1/2, n + 1/2]. In this case, it means calculating P(X ≥ 44.5) instead of P(X ≥ 45).

For a binomial distribution, when λ = np is of "moderate" size, then the Poisson(λ) distribution is usually a better approximation than the normal. Note that for large λ, the Poisson(λ) distribution is approximately normal anyway.

As a mathematician, I find the book fairly useless and, in some places, just plain wrong. For example, Chapter 18 (or at least the first part of it) desperately needs to be rewritten using σ-algebras and measure theory. Equation (18.1) is nonsense, since we can take E = A. Also, Jaynes' argument in Section 12.4.3, which justifies his choice of the Haldane prior, is flawed since a (defined by (12.45)) depends on X, and therefore depends on θ. This flaw is irreparable in my opinion. In addition, the book does not seem to add anything new at all to the mathematics of probability theory.

However, as someone who is interested in the philosophical foundations of probability, I find it fascinating and extremely valuable. This book puts forward an interpretation of probability which is coherent, useful, and generally consistent with how we think of probability intuitively. His treatment of the indifference principle is outstanding. It is the only treatment I have ever seen which is able to be precise, while at the same time avoiding contradictions. His coverage of various so-called "paradoxes" in probability is also outstanding. What is influential and important about this book is the way Jaynes thinks about probability, philosophically. From that perspective, I think the book is fantastic.