Great stuff, but why is the impedance of free space (a vacuum) 336.7 Ohms?

A small note (given the wonderful videos) . . . this should be no 28 . . no?

For those a bit confused by the problem, I'll rewrite it here with more words - There is two tribe labelled 0 and 1. The 0 and 1 are not probability but just names. - We know with perfect accuracy if the individual are healthy or not. This is not a case were we have an imperfect test and it is possible that an individual has the disease but test negatively. In this universe, either someone is lying on the bed dying, or is running around in the jungle, with no in between. Ox Educ confusingly used the term "asymptomatic", but then say they are healthy. The rest of the problem assume we know they are healthy. - Theta = 1 does not mean that all the members are sick, it just mean "We sampled the individual from tribe 1, the one where some member are sicks" - Our model of the probability of an individual being sick is the function f, and it depend on the tribe we sample from. In tribe 0 no one is sick. so f(0) = 0 . In tribe 1, the infected tribe, half the member are sick, the other half is in good health, so f(1) = 1/2. Remember that 0 and 1 in the function parameter are just the tribe label, and not really a number. - Since all the individual we picked are healthy, we cannot be sure which tribe we picked them from. If any of them was sick, then we would have instantly knew that it was tribe 1, because it is the only on that has sick members. But with our current data, either we sampled from tribe 0 that has only healthy individual, or we sampled from tribe 1, and by chance, only picked from the healthy half of the tribe. So P(data|theta = 1,model) is 1/8, since we picked three individual, and our model say we have 1/2 chance for each individual for them to be healthy if we sample them from tribe 1

thank you for going through a complete example from scratch. helped a lot.

The videos are good information and knowledge wise but kinda boring. I loose my focus every now and then.

i needed this so much! when i started my horribly organized probability theory class i needed simple and clear explanations to all the terms that were thrown in . well i didn't get it then but luckily you exist ! wishing you all the best !

good

Do you know how complex is the induction proof?

At four minutes and 23 seconds, you started to talk about geometrically defining dual vectors. I got really excited because I was sure you were going to talk about parallel projection versus perpendicular projection which describe contra variant components and covariant components a.k.a. dual basis vectors, respectively. Is there a connection between one forms and the perpendicular projection of vectors components called dual basis vectors ?

Brutal video

where does the values of 0.25 and 0.4 comes from?

This is a triumph of excellent teaching. Thank you.

Cool video! I finaly understood it! There are not so much about this theme in the web... thank you!

I'd like to mention that we can go beyond the Monty Hall problem and touche on a fundamental issue in probability and statistics: the comparability of events with different sample spaces or magnitudes. 1. A first event with a magnitude of 3 (three doors) 2. A second event with a magnitude of 2 (two doors) While a great introduction into probability, the Monty Hall Problem only works if one accepts the comparison of two events of different magnitudes as logical. To dismiss people who cannot agree with this comparison as they do not get it is a problem if you ask me. I see the importance of highlighting both ways of thinking. In many contexts, comparing probabilities from events with different magnitudes or sample spaces is indeed problematic or even meaningless. For example, to provide another example of fundamentally different events: Comparing the probability of rolling a 6 on a die (1/6) with the probability of flipping heads on a coin (1/2) doesn't make much sense in isolation. In more complex scenarios, like comparing stock performance across different markets or time frames, not accounting for differences in magnitude can lead to serious misunderstandings. Additionally, people think if your initial choice was the car (which happens 1/3 of the time), then switching would be the wrong move. In this case, the host's reveal of a goat door doesn't help you at all. You've already won, and switching would make you lose. If one accepts the comparison of two events of different magnitudes, the Monty Hall strategy isn't about "always switch" but rather "switch because it's more likely you initially picked a goat." The host's reveal doesn't create new winning chances. The host doesn't change the fact that you probably (2/3 chance) guessed wrong at first.

Lovely video Ben! As usual.

How could you say that theta is fixed when you integrate over it’s “different” values in 1:06

Thank you very much Ox, your class is extremely outstanding and great!

How dare I found this now.. Amazing works !

I am confused by p(a,b)= 1/3 x 1/2. Does this assume A and B are independent? If they are independent, p(a|b)=p(a); no need to go through all the rest of the derivation steps.

why did you write P(\theta) that way, isn't that the density function of \theta and not the actual probability ?

First day on Bayesian statistics. Lets go!

This was useful for me in understanding why with entropy and cross entropy with KL Divergence the Cross entropy will always be greater than the entropy (you have to flip the inequality though because the function in question is a -log which is concave

Can't believe I'm learning what people learned 9yrs ago and I'm surprised what are they doing now 😳

2:26

1:48

0:13 0:14 0:14

fantastic! absolutely fantastic explanation!

Would this also imply the reverse is true for a concave function? I.e., g(E(x)) > E[g(x)] ?

Stretching is dual to squeezing -- forces are dual. The Ricci tensor (positive curvature, matter) is dual to the Weyl tensor (negative curvature, vacuum). "Always two there are" -- Yoda. The Ricci tensor is the symmetric projection of the Riemann curvature tensor. The Weyl tensor is the anti-symmetric projection of the Riemann curvature tensor. Symmetry (Bosons) is dual to anti-symmetry (Fermions). Covariant is dual to contravariant -- dual basis.

Is an average football fan better educated (more sensitive to other cultures?), than average mathematician: kzfaq.info/get/bejne/adehptOGuN7QemQ.html

Yeah, except this is the probability of which door Monty opens, not the probability of you getting the car. The difference is that you counted only once on the either/or ⅓ side, but twice on the ⅔ side. The ⅔ side is also either/or, because the probability to get the car remains the same; Monty always opens the goat door. The chance for you to get the car remains the same with him opening either doors. So which is it, counting either/or door once, or twice? Either way it is ⅓ vs. ⅓ or ⅔ vs. ⅔ or 50/50. Guys, you’re wrong. You’ve made a mistake. Quit with your Reductio ad absurdum.You’ve allowed semantics to completely change what probability actually means. You’ve ignored the truth and continue to spread false propaganda. It’s not about debunking anything. Just get it right. 🤙

Where did the N choose X part go though?

Either I’m dumb as fuck or I’m just not paying attention like WTF if I wanted school I’ll go back for that lol more simple please

Likelihood is a probability, it can vary in the range [0 ; 1], it's a probability of "data", the sum of all those probabilities for each value of theta does not integrate to 1, but what's the point.

Dude, THIS is the kind of education worth paying for, not whatever my professors are doing

*I can prove the true nature of this 'problem'!* .. they trick you into thinking you have 'chosen' the first door. Think of it this way: the first round of the game is the fake 'choice' where you don't actually choose, you 'reserve', to protect one door and let one goat be eliminated from the game, 1st round is now done! Now they start the second game where the REAL choice is allowed, there are no longer any 'reserved' doors and the game parameters have now changed to allow 50/50 odds because you get to play this game from scratch with two doors and for the FIRST time get to actually 'choose' instead of 'reserve/protect', which is very different than the first game's round of choices..it's TWO different games NOT one, so if you do not account for this by creating two separate math problems then you are NOT accurately representing the true nature of the game, as you have to change the odds for all 'unknowns' every time you eliminate possibilities. This logic applies to different examples of this 'problem' as well. The key is understanding that the only real game is when the final choice is made, and that everything before that is just changing the parameters, you have to make the math adjustment for the new parameters as they change, it all comes down to the *state of the parameters* when you *actually choose* and NOT when you are simply 'negotiating' the parameter changes, ..thanks and you're welcome;)

Each sheep is also an independent and unique individual. This game cannot mislead players by treating every sheep as the same thing. There are three options, two of which also stand on their own. If the player can tell them apart, there is no chance of winning like 2/3

Is the reality of quantum mechanics more logical than what the math? The player never knows which goat is the one revealed.。.. If life is an endless choice, then if we face it with this way of dealing with things, then what aspect of human nature is eternal? Some people always want to control the world and act as the so-called "God"?

The sound is just awful. Seems like the presenter's microphone is 50 feet away from the presenter.

Maybe thinking this way....... Try to understand it from the perspective of the host. The player provides his prediction to the host. The host opens a pair of doors with sheep, leaving only two pairs of doors, one sheep and one car, for the player to choose. If the player predicts the door of the car, what will the host do? of couse Want the player to change their choice. The host asks you to calculate the odds with misleading you already have once choice before. you swapping your choice, 2/3 will win. On the contrary, the player predicted the door with the sheep. What will the host do to make the player still choose the door with the sheep? you can tell. ........ Dawyer's door problem, calculate the chance of the host winning.

I come back to this video so many times now really good thank you!!

Thank you! Great explanation

I have question about the result here. We find that the posterior distribution is N(thetaPrime, sigmaPrime^2) and you have given the formulas for thetaPrime and sigmaPrime^2. Now, the posterior distribution is, of course, the distribution we get when we combine the data from our observations together with our prior, so I'd expect that thetaPrime and sigmaPrime would contain the information we gained from our observations. Indeed, thetaPrime depends on the number of observations N as well as the mean of the observed values, xBar. However, sigmaPrime appears to only depend upon the number of observations N, and it seems entirely agnostic as to what the values of the observed xi's are. That is, no matter how wildly different the observed values of xi are from thetaNot (the presumed, uninformed mean prior to applying any data), the variance of the posterior distribution will be completely unaffected by the observations. As a more concrete example, suppose the measured IQs were scattered from 50 to 210 instead of being clustered in the 50s-80s. I would think sigmaPrime would need to grow to reflect the enormous variation in the observed values, but the formula you've derived implies that it does not care what the values are. Is this a consequence of your assumption that sigmaX is a "constant"? Is the result still valid?

I think we are not adding the lengths but areas of slices in the integral. For the length you'd need to integrate over square root of H' + W'

Wow,you explained it really well

Ily

What if we want the prior to carry more weight? I realise that we could do this by making it more extreme and concentrated, but what if we don't want to do that. For example, what if I have seen something happen a million times previously but never recorded the data, but I trust my memory and have a strong handle on what I believe distribution of the parameter should be. Then we receive 1000 new observations (a high number, but nowhere near the million unrecorded), we might not want the data to dominate in such a circumstance

Great video here! Thanks!

Very clear explanation, thank you!