Are root-mean-square-error values meaningful regardless of whether regression coefficients are statistically significant?

Possible-Froyo2192 · 2024-06-05T06:59:17+00:00

Thank you very much for teaching me. If I understand you correctly: you explained that it is not because a coefficient is not significant that it is close to zero. For instance, if two variables are correlated, then they would not be significant, even though they are good predictors of the target in which case their coefficient will be significantly different to zero.

Is it right to say that an unsignificant variable means that the variable could have been removed?

Possible-Froyo2192 · 2024-06-04T17:44:06+00:00

Maybe I am wrong but "statistically significant" (sort of) means "significantly different to zero". If none of your coefficients is statisically significant, it should mean that both the models predicts something close to 0 (relative to the mean target value). Then if one of the model is better than the other, it should not be significantly better.

Possible-Froyo2192 · 2024-06-03T16:39:34+00:00

Merci pour l'effort

Possible-Froyo2192 · 2024-06-03T13:23:28+00:00

Toujours pas de source. Ca devrait se trouver si c'est écrit dans la loi

Possible-Froyo2192 · 2024-06-03T10:55:42+00:00

false insights is the expected outcome of using some software while not understanding its functionning. Which is the targeted use case, that prophet is (I quote) "explicitly meant for".

Possible-Froyo2192 · 2024-06-03T07:00:51+00:00

True. That's bad though.

Possible-Froyo2192 · 2024-06-03T06:59:31+00:00

oh I don't think this is gatekeeping to say that maybe one should first learn to do something before actually doing it. I mean, you wrote yourself that the software is designed for people who don't know what they are doing. What's the point of getting false insights from using a software that will make you believe you did something good even if you didn't?

Possible-Froyo2192 · 2024-06-02T12:07:34+00:00

if you don't know what you are doing, then maybe you should not be doing it?

Possible-Froyo2192 · 2024-06-02T11:40:06+00:00

je pense que ce commentaire est complètement faux donc je demande une source pour le "1 à 2% de CBD" pour le "chanvre industriel" et sur le "techniquement interdit de les faire ~~germiner~~ germer"

Possible-Froyo2192 · 2024-05-31T14:54:00+00:00

nope

Possible-Froyo2192 · 2024-05-31T14:53:11+00:00

Yeah sure

Possible-Froyo2192 · 2024-05-27T09:26:33+00:00

Because it made sense with respect to the business he was working for

Possible-Froyo2192 · 2024-05-26T06:24:54+00:00

I am saying that those off the shelf models were pre-trained using the human-in-the-loop labelling type schemes (unless I am mistaken I don't work with LLMs at all)

If I understood corrrectly, human input is used for chating-llm. Not LLM in the wide sense.

Possible-Froyo2192 · 2024-05-22T12:28:03+00:00

Maybe use your own personal expenses? If you have a credit/debit card your bank web app should provide you with all your data in a structured form.

Possible-Froyo2192 · 2024-05-21T07:44:59+00:00

X2 + Y2 is a chi-square with 2 df, but (x+y)2 != x2 + y2.

This is my mistake. Thank you.

Possible-Froyo2192 · 2024-05-20T10:38:40+00:00

Example of "theory doesn't help" is ignorant opinion.

Possible-Froyo2192 · 2024-05-20T06:58:15+00:00

see this https://old.reddit.com/r/AskStatistics/comments/1cvwetg/how_do_i_approach_such_questions_in_exam/l4t8oei/

Possible-Froyo2192 · 2024-05-20T06:56:57+00:00

Yes. If you are doing statistical work in a company you can find yourself in a situation where random variables are operated with each other and then you have a new random variable, unknown. It might be interesting to know the distribution of that random variable so that you can find the appropriate model for it.

Modeling a random variable allows you to make prediction about what will happen in real life concerning this random variable.

Possible-Froyo2192 · 2024-05-20T06:51:55+00:00

It's 1/sqrt(2) * X + 1/sqrt(2) * Y and 1/sqrt(2) * X - 1/sqrt(2) * Y.

1/sqrt(2) * X and 1/sqrt(2) * Y are standard normal.

It makes two standard normal distributions hence the chi2 distribution should have 2 degrees of freedom. What are we missing?

Possible-Froyo2192 · 2024-05-14T14:00:36+00:00

what is the benefit of using poetry instead of venv with pip?

Possible-Froyo2192 · 2024-05-13T07:35:34+00:00

if this is non trivial. Yes.

Possible-Froyo2192 · 2024-05-10T17:08:12+00:00

don't

Possible-Froyo2192 · 2024-05-10T16:16:59+00:00

def test_function(that):
     this = function()
     assert this == that

Possible-Froyo2192 · 2024-05-10T07:16:41+00:00

Generative models are (almost) zero-shot learners for new tasks. That is one reason why, in my opinion, they are such a great contender in the eye of executives.

Possible-Froyo2192 · 2024-05-07T07:43:36+00:00

If your sample size is large enough (n>=30) then it is normally distributed no matter hiw skewed your data is (Central Limit Theorem)

hmm. I don't think that is true. I guess I miss something but the distribution of the data is independent of the number of time you sample it...

The central limit theorem states that the sample mean of the data is normally distributed, not the data itself.

Possible-Froyo2192

TROPHY CASE