Recent content by camillio

For an example, let's have a probability space (\Omega, \mathcal{F}, \mathbb{P}) and a stochastic process (coin tossing) X adapted to a filtration (\mathcal{F}_t)_{t=1,2}, \mathcal{F}_t\subset \mathcal{F}. Then, in the beginning, one knows nothing but possible results, \mathcal{F}_0 =...

Start with 0 -> -1. This gives you relation between b and d. Then simply think in limits, when f(z) -> inf, f(z) -> 1?

I wouldn't implicitly consider it normal unless stated. Anyway, one can exploit some of available probabilistic inequalities. Since the Chebyshev's inequality is usually taught, try this one, maiad. Most likely, you will succeed to get your "correct" value :-)

That smart, 0xDEADBEEF! I'd however add one more thing - choose the indices of the deviations list in a random way, otherwise the result will be biased due to shrinkage of the the sequence of the distributions' supports. Other (easier) possibility could be to shuffle the deviations with...

Well, immediately I'm not sure how to make your values don't exceed the limits. Still, there are some possibilities how to generate them in the way that they do not with high probability: 1) Notice, that you have 4 classes X_1, ..., X_4 with parameters \alpha_1,...,\alpha_4, each with mean value...

I think so. You even don't need the above fractions, the ratio follows directly from 5938/8387. But be careful about its interpretation, especially if you want to talk about probability!

Regarding the OP: you're right from one point of view and not necessarily right from another :-) If your problem is viewed in the frequency framework, then OK. However, from the Bayesian viewpoint, it may be wrong. It's the same thing like the dispute that the sun will rise tomorrow. According...

Hi Dan, I suggest following the Bayesian procedure with multinomial model and Dirichlet prior. You can set your prior for Dirichlet in terms of shaping parameters. This prior can correspond to your [.2, .5, .2, .1] vector with parameters set to suit this need. Then, you update the Dirichlet...

In addition to chiro's reply, I'd add that the basic reasons why frequentist methods dominate are (i) historical - they were generally accepted quite recently, (ii) they are much harder to learn and understand for non-mathematicians (non-statisticians), (iii) they do not provide a simple bunch...

Well, the Bayesian approach is already a well established counterpart of the traditional frequentist statistics. The reason why it took so much time (although it's far older then frequentism) consisted mainly in the enormous denial from the traditional Fisher's and Neyman-Pearson's schools...

I know it's not online, but anyway, I strongly recommend this book (any edition, of course): Gelman, A., Carlin, J. B., Stern, H. S. Rubin, D. B., Bayesian Data Analysis, Second Edition (Chapman & Hall/CRC Texts in Statistical Science), 2nd edn. (Chapman and Hall/CRC, 2003). Its main...

Damn, you're right! I'm deeply sorry, my trivial fault :-(

I know, but what then with the following? <Qx, y> = \int_H <x,z> <y,z> \mu(dz) If I understand correctly, <Qe_n, e_n> = \int <e_n, x> <e_n, x> \mu(dx) = \int <e_n, x>^2 \mu(dx) which doesn't coincide with abs. value for complex numbers.

Hello, I'm reading Gaussian measures on Hilbert spaces by S. Maniglia (available via google) and I have the following issue, regarding the proof. He states in Lemma 1.1.4: Let μ be a finite Borel measure on H. Then the following assertions are equivalent: (1) \int_H |x|^2 \mu(dx) < \infty (2)...

Thank you Christoff :-) Probably for the irrational points it follows directly from their property of being accumulation points of sequences of rationals.