gill1109 wrote:Nice!
There are two pi's in those formulas, which one are you changing?
Terminology. In probability theory, the cumulative distribution function F of a random variable X is the function x mapsto Prob(X <= x). I don't think that your f should be called a cumulative distribution function. Your theta_0 is drawn at random from the Uniform([0, k_1]) distribution. So f(theta_0) = - 1 + 2 /sqrt(1 + 3 theta_0 /k_2) is another random variable with its own cumulative distribution function.
Here k_1 and k_2 were originally both equal to pi but you are now proposing to let one of them be different?
I double checked Pearl's computations, and I found many problems. All checks out until formula 16 (including 16 and including appendix A and B) Then it goes downhill. (17) has a typo, the correct formula is:
mu(x) = - 1/(1-x^2)^{1/2} d/dx [(1-x^2)^{1/2} / 4x d/dx g(x) x^2]
Equation (19) is messed up big time, 1/2 g(x) is the sum of the 2 integrals over [(1-x^2)^{1/2} + x] and not the first integral over (1-x^2) + the second integral over x^2. As a consequence Eq. 23 is wrong as well, but 23 does follow from the incorrect 19. Section V of the paper is based on (23) and is invalid.
I could not obtain Eq. 20 (I got a similar slightly more complex one-I am still working on it and maybe it will simplify to 20)
Eq. 21 has a typo and should read:
h(x) = 1/4 C pi 1/(1+x)^3
Eq. 22 has a typo and should read:
rho(r) r^2 = 1/3 sin (1/2 pi r) / (1+cos(1/2 pi r))^3
It was a major pain to straighten out the math on page 1421 because it is a sloppy paper and there are many false trails one need to explore to understand the intention. On Eq. 20 I got a very similar expression but I am lacking the overall 1/1-x^2 factor outside the square brackets, and in the second integral I have an additional term of (1-z^2)^{1/2}. The first integral in the square bracket is equal with g(0)/2.
gill1109 wrote:So I found that 4 pi r^2 rho(r) = 4 / 3 . 2 . sin(pi r /2) / (1 + cos pi r / 2) ^ 3
Joy Christian wrote:gill1109 wrote:So I found that 4 pi r^2 rho(r) = 4 / 3 . 2 . sin(pi r /2) / (1 + cos pi r / 2) ^ 3
This probability density is still not normalized to 1 for r in [0, 1]. It normalizes to 2/pi. It appears that the correct normalizing constant on the r.h.s. of (22) is 4pi/3.
r <- seq(from=1, to = 199, by = 2)/200 ## 100 midpoints of 100 intervals of width 1/100 between 0 and 1
length(r)
pr <- (4 * pi / 3) * sin(pi * r /2) / ((1 + cos(pi * r / 2)) ^ 3) ## A probability density?
plot(r, pr, type = "l")
sum(pr)*(1/100) ## Yes!
FrediFizzx wrote:...Good news, Joy. The new function seems to work well in the MinkweReed Mathematica simulation also.
gill1109 wrote:This illustrates the difficulty with detection loophole models.
FrediFizzx wrote:Yeah, Richard still doesn't understand about states that don't exist in the first place. He stopped asking questions about that before he finished the lesson.
gill1109 wrote:At present, there does not exist a mathematical derivation by Joy Christian of the probability distribution which we are using in the Pearle model for the second component of the hidden state. Pearle himself showed that his own choice was one of many - he just picked something simple which does the trick. Joy has opportunistically adopted Pearle's choice but has not shown that it follows from his (Joy's) physical modelling.
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