Recent content by bpet

In order to determine if the set is a semiring you first need to specify exactly what are the addition and multiplication operations.

It's not quite a beta-binomial distribution. However you can do a similar integral $$\int_0^1 P(A_{wins}\cap y|\alpha,\beta)Beta(p|\alpha,\beta)$$. Note also that the expression for P further simplifies before you do the integral.

Thanks for writing up the details. Note that only the one S-polynomial needs to be computed if we do the reductions on all polynomials immediately.

Ok so I used Buchberger’s algorithm https://en.m.wikipedia.org/wiki/Buchberger%27s_algorithm, only one polynomial was added to the basis as described in the previous post and this reduced to x+1, then the original polynomials reduced to 0 and y^2-1 trivially. As for the irreducible components...

Which bit would you like me to explain further? Keep in mind I’m tapping this out on a phone here. Cheers

ok so the reduced Groebner basis is (x+1) and (y^2-1) obtained from computing y(x^2y+xy)-x(xy^2-1) and reducing several times. That would mean, if I understand the definitions correctly, there are 3 irreducible components corresponding to y=1, y=-1 and x=-1.

This page has some theory for this type of equation: Quadratic residue - Wikipedia

There is a more general formula for when the variable can be either discrete, continuous, or a mixture of the two (or even singular if you wish). We have P[X+Y\le x] = P[X\le x-Y] = E[F_X(x-Y)] = \int F_X(x-y)dF(y) The integral above is a Stieltjes integral so we recover the standard...

Substitute $$\tan(x)=\frac{2\sin^2(x)}{\sin(2x)}$$ so the product telescopes down to $$\frac{2^{\frac{1}{2}+\frac{1}{4}+\ldots}|\sin(1)|}{\lim_{n\to\infty}\sqrt[2^{n+1}]{|\sin(2^{n+1})|}} = 2\sin(1)\approx 1.68294196962$$ For a proof that the limit in the denominator is 1 see Agnew, R. P. and...

Let p_k(n) be the probability that case k has occurred after n dice have been thrown. It is not hard to show that these probabilities satisfy a recurrence relation \begin{pmatrix}p_1(n+1) \\ p_2(n+1) \\ p_3(n+1) \end{pmatrix} = \begin{pmatrix} \frac{1}{6} & 0 & 0 \\ \frac{2}{3} & \frac{1}{3} & 0...

I would have thought that since one player has two cards of type A then the answer would simply be (1/6)*(1/5)*(1/4)*(1/3)

That doesn't quite sound right, can you give an example?

I can't say much without seeing more details but the transition matrix of the modified process should be the same except with a row of zeros for the absorbing state - that shouldn't break the law of conservation of probability.

Using the Negative Binomial CDF this can be expressed in terms of the regularised incomplete Beta function 1-I_{P_t}(100,100)=I_{P_h}(100,100)

Standard convolution formulas are not likely to be of much use for this approach because X and Y are dependent. Also don't worry about the PDF just yet, it's trivial to calculate (if it exists) once you've got the CDF. A CDF can be written as the expected value of a Boolean indicator...