#statstab #501 Prediction intervals for GLMs
Thoughts: Sometimes the prediction of the next dat point can be [0,1]. Not very useful.
#prediction #uncertainty #predictionintervals #glm #binomial #probability
https://fromthebottomoftheheap.net/2017/05/01/glm-prediction-intervals-i/
One of my more popular answers on StackOverflow concerns the issue of prediction intervals for a generalized linear model (GLM). My answer really only addresses how to compute confidence intervals for parameters but in the comments I discuss the more substantive points raised by the OP in their question. Lately...
I'm doing some symmetric monoidal algebra that involves careful counting of some signs of certain permutations. The work depends on a couple of combinatorial identities that I haven't seen anywhere else—do you recognize these (below)?!
The identities involve the "choose two" binomial coefficients.
For ease of typing, I'll use this notation:
[a;2] = binomial(a,2) = a·(a-1)/2
(read "a choose 2")
The two identities are
(I1):
[a+b;2] = [a;2] + [b;2] + ab
and
(I2):
[ab;2] = a[b;2] + b[a;2] + 2[a;2][b;2]
In particular, (I2) means there is a mod 2 congruence
[ab;2] ≡ a[b;2] + b[a;2]
and that's the form that has been particularly useful for me.
Neither of these identities are hard to prove directly from the definition, and they hold for positive *and negative* integers a and b. (That extension to all integers is important for my applications too.)
I've done some internet searching (wikipedia [1,2] and other general references), but I haven't found mention of these particular identities. So, I'm wondering if anyone here recognizes them. (Boosts appreciated!)
Note: These particular binomial coefficients [a;2], for positive a, are also called *triangular numbers*. I'll rewrite (I1) and (I2) in terms of triangular numbers in the next post, in case people will recognize that alternate form (but I doubt it).
[1] https://en.wikipedia.org/wiki/Binomial_coefficient
[2] https://en.wikipedia.org/wiki/Triangular_number
(1/2)
One day, one decomposition
A138389: Binomial primes: positive integers n such that every i not coprime to n and not exceeding n/2 does not divide binomial(n-i-1,i-1)
3D graph, threejs - webGL ➡️ https://decompwlj.com/3Dgraph/Binomial_primes.html
2D graph, first 500 terms ➡️ https://decompwlj.com/2Dgraph500terms/Binomial_primes.html
#decompwlj #math #mathematics #sequence #OEIS #javascript #php #3D #numbers #binomial #primes #PrimeNumbers #coprime #graph #threejs #webGL
i'm up to 67 posts of binomial, my blog/newsletter. the focus has drifted since i started it, but it has remained a place for reflections and play.
someone i met 10+ years ago sent me a message saying they read and enjoyed the stuff so far. perhaps you will, too!
One day, one decomposition
A121943: Numbers k such that the central binomial coefficient C(2k,k) is divisible by k^2
3D graph, threejs - webGL ➡️ https://decompwlj.com/3Dgraph/A121943.html
2D graph, first 500 terms ➡️ https://decompwlj.com/2Dgraph500terms/A121943.html
#decompwlj #math #mathematics #sequence #OEIS #javascript #php #3D #numbers #central #binomial #coefficient #divisible #graph #threejs #webGL
Try to prove the following two results that relate the harmonic numbers to the golden ratio. Have an excellent weekend.
\[\displaystyle\sum_{n=1}^\infty\binom{2n}n\dfrac{H_n}{5^n}=2\sqrt5\ln\varphi\]
\[\displaystyle\sum_{n=1}^\infty\binom{2n}n\dfrac{H_n}{5^nn}=\frac{2\pi^2}{15}-2\ln^2\varphi\]
where \(\varphi=\frac{1+\sqrt5}2\) is the golden ratio; and \(H_n=\left(1+\frac12+\frac13+\ldots+\frac1n\right)\) is the \(n\)-th harmonic number.
#GoldenRatio #HarmonicNumbers #HarmonicNumber #Logarithm #Pi #Summation #Math #Sum #InfiniteSum #Binomial #BinomialCoefficient #Maths #WeekendChallenge
In the process I also thought of a #binomial #coefficient interpretation for choosing k things out of a bag of n objects when the objects can be put back and picked again. In probability problems, this is referred to as picking "with replacement".
Usually, when we say, n choose k, we do not allow repeated choices, every chosen object has to be chosen once. In this case, I want to allow replacements but at the same time, I want to keep using my trusted n choose k idea.
Here's my way around this: We'll still be picking k things, but we'll pick them not from 1 to n but from the set {1, 2, ..., 𝑛, 𝑟₁, 𝑟₂, ..., 𝑟ₖ₋₁}. That is, in addition to the n objects, we add what I'm calling "replacement tokens" 𝑟ₓ. If any of the 𝑟ₓ gets picked, then it is interpreted as the 𝑥th choice was put back and you are now choosing to pick that again. Since the 𝑘th choice is not put back, we only need replacement tokens for choices 1 to k-1.
With these replacement tokens, the problem becomes a standard choose k things out of this set, which we can resolve using the binomial coefficient to get: \({ n+k-1 \choose k }\).
I believe the standard approach to this is via #StarsAndBars but I liked the idea of this "replacement token". Admittedly, I didn't want to use stars and bars here and made up some stuff which I happened to like. :)
Krzysztof Sakrejda
Dr Mircea Zloteanu 🌺🌞🍃
Niles Johnson
mushey
Rémi Eismann
Pustam | पुस्तम | পুস্তম🇳🇵
Akshar Varma
lagomoof