12 exchangeability and iid

[@WestCParallel49]

This video, really helped me wrap my head around the concept of exchangeability (given that I am student in the social science area). Thanks for the upload!

[@franksheng4173]

Really nice explanation, thanks a lot

[@yulinliu850]

Excellent! Thanks a lot!

[@omkarchavan2259]

nice video

[@Felicidade101]

Is identically distributed the same as conjugate?

[@lemyul]

independent identical distribution

[@SajadSaeedi]

I disagree. Your definition of exchangeability is not clear. Let's say random variables X, Y are exchangeable. That could mean two things. Number one P(X=x,Y=y)=P(Y=y,X=x) , which is what you have considered. Number two: P(X=x,Y=y)=P(X=y,Y=x) which is what I would think of exchangability.

[@matakos22]

Yeah, and actually isn't P(X=x,Y=y)=P(Y=y,X=x) always true anyway?

[@alzero4621]

@@matakos22 Yes, that is always true. BUT, this is NOT what the video is illustrating. In your example, we implicitly assume X and Y are random variables defined on the same probability space, but in the video we don't have that; instead, in the video it would be more like P(X=x, Y=y) = P(Y'=y', X'=x'), since X is no longer "X" (underlying probability space has changed) when the position changes, and same for Y. In other words, the 'reference frame' for X is different when its 'position' is changed, so it's no longer X anymore, which invalidates the otherwise trivially true identity you mention. Remember a random variable is a function, and so when the domain of the function changes, we no longer have the same function!

[@avananana]

I get that this is a two-year-old comment but just dropping this here for future people. No, that's way off. This video is not related to joint probabilities, which is what P(X=x, Y=y) and P(Y=y, X=x) is, that's the joint probability of the two random variables X and Y. You are right that P(X=x,Y=y) = P(Y=y, X=x) but not because of exchangeability, it's just a result of the nature of how joint probabilities work. This video is related to the exchangeability of elements in a sequence of samples. P(1,1,0,0,1) is the probability that you'll get the results 1, 1, 0, 0, 1, specifically in that order, if you take five random samples from a sample space. the 0s and 1s here don't represent random variables, but they're more like sample results from a random variable, if one is allowed to say that for simplicity.

[@prashant_sinha69]

@@avananana so very well explained.