Probability Theory 10 | Random Variables

[@jaimelima2420]

Thanks for putting this together. It will helpful for a lot of people. Probability theory is littered with conflicting terminologies and several times during the applied courses the concepts are not explained in full to make them consistent in mathematical terms. These strategic bridges you are building between the more abstract Measure Theory and Probability Theory are really useful to reconcile some concepts in my mind.

[@treywilkinson9989]

The connection you provide between probability theory and measure theory has helped me so much. I appreciate your effort!

[@Gouleur]

You're knocking it out of the park with this series, can't wait for the next episode!

[@brightsideofmaths]

Thank you very much :)

[@rombo8]

So clear. Italian maths books are very theoretical and I literally have headaches when I try to study on them. This video tho... SOOOOO CLEAR without it being informal! Thanks.

[@brightsideofmaths]

Thank you very much :)

[@umbranocturna6342]

Thank you so much, this is so much easier to understand than what my University is telling me.

[@molibdenum]

At 9:15, why did you say that the left side P(X element of A~) "does not make sense" by itself and is just shortcut defined as the rightmost expression? Woudn't we also be interested in the probability over the random variable also, like P(sum of throws >= 10) = P(X element of [10, 12]?

[@brightsideofmaths]

It is just formal thing. X as a map is not literally an element in A or the real numbers.

Hello Julian! As always, exceptional videos and explanations! The pace now (IMHO) is perfect - you left a small pause after you finish a concept, that is perfect. I have one question: if I want to prove that a function is measurable, is it enough to show that the pre-image of ATOMS of the sigma-algebra on the right (these "right-atoms" would be only the images of the atoms of the sigma-algebra on the left) belong to omega (the set on the left)?

[@brightsideofmaths]

Thank you very much! I also have the feeling that with some pauses, everything sounds nicer now :) Thank you for your nice input there. Regarding your question: It depends what you mean by "atoms". Indeed, your claim is true for a every collection of sets that generates the sigma-algebra. For example, for the Borel sigma algebra, the collection of intervals with rational endpoints generates the whole sigma algebra. Therefore, it is sufficient to only check the preimages of these. The rest will simply follow by unions and intersections, which the preimage can deal with.

@@brightsideofmaths Thanks for your kind answer, Julian! I mean by atoms of the sigma-algebra: 1 - A \in \script F 2 - A not equal to the empty set 3 - There are no non-empty sets B, C in \script F such that A = B U C But your answer clarifies my doubts. Thank you very much! :-)

[@skillick]

Thank you, the notation in probability theory is confusing-your videos make it much more clear.

[@brightsideofmaths]

Glad it was helpful! And thanks for your support! I also found the notations confusing and that's the reason I explain them here.

[@zhizhongpu8937]

why did you say the choice of A~ does not matter at all? at 5:28

[@jsoldi1980]

In example (a) the sigma algebra of the second set is the borel algebra, does that mean our probability function will give probabilities of intervals and not of individual sums? I was assuming if we pass the function, for example, {2, 12} we'd get 2/36, but if we use borel algebra then that'd be interpreted as an interval, so the output of the function would be just 1. Is this correct? And if so, could we have not used borel algebra on the right side if we wanted to distinguish between individual elements, for instance, if we want P({2, 4}) to be different from P({2, 3, 4})?

[@brightsideofmaths]

You should distinguish the sigma algebra and the probability measure. In the Borel Sigma algebra also singletons and finite sets are measurable. Therefore, you can define a probability measure, that gives positive probabilities to finite sets.

[@hn97754]

Could you please explain 6:09 again? Why is this trivially fulfilled? Thank you.

[@brightsideofmaths]

The power set contains all subsets :)

[@riolan1274]

There are only two kinds of elements in A tilde. One can be calculated from w1+w2, where (w1,w2) is an element in Omega. The others cannot be calculated from w1+w2, then its pre-image is empty set. Since the power set contains all the subsets of Omega, the requirements satisfied.

[@meteor3033]

Hey I've been meaning to ask, what is the techno/electronic song at the end of some of your videos? It's fire.

[@brightsideofmaths]

Oh my, it's my own composition :D I like my new song better ;)

[@dydx_mathematics2]

Im interisted on probability theory

[@teddysariel]

If I'm not mistaken, the last notation remark should involve A \in \tilde{A}, and not \tilde{A} itself. Thanks for the video, and for your work in general.

[@brightsideofmaths]

I am not sure what you mean. The name for the elements in the sigma algebra fancy-A-tilde is not important here :)

[@teddysariel]

@@brightsideofmaths You used \tilde A for the sigma-algebra, and you want to consider an element thereof.

[@brightsideofmaths]

@@teddysariel I use two different kinds of A's

[@teddysariel]

@@brightsideofmaths I see it now. My mistake.

[@brightsideofmaths]

@@teddysariel Great :)

[@junechu9701]

Thanks! It is really helpful!

[@brightsideofmaths]

Thank you for your support :)

[@sopas5918]

Thank you for the video boss

[@musicalBurr]

A possibly dumb question coming, but first, let me say how much I have enjoyed your channel over the years! Thanks so much. Ok, dumb question now: Say I have X^-1({2.5}) (or something not the sum of the two dice), then I take it that it's pre-image set is {} (i.e. empty set) which IS an element of P(omega), is this right? To continue, is this also correct (using range of real numbers): X^-1([1.5, 3.5]) = {(1,1), (1,2), (2,1)}? Thanks again!

[@brightsideofmaths]

Thank you very much. What is your X in this case?

[@musicalBurr]

@@brightsideofmaths Ah sorry that was unclear. I was continuing with your dice example at the end of the video. (You used the one point set {(1,1)} in your example to save some writing 🙂) In other words X = w1 + w2.

[@angelmendez-rivera351]

Yes, (X^(-1))[{2.5}] = {}, and yes, {} is in P(Ω). What this proves is that if p is the probability measure from the original space, then p((X^(-1))[{2.5}]) = 0, which is to say, the probability that the experiment (of throwing the two dice, and adding the results) will have a result of 2.5 is exactly 0.

[@TheXorion]

good video as always!

[@lueelee6063]

I thought one mainly used [] to denote image and pre images of functions as to avoid confusion with functions with ().

[@brightsideofmaths]

Indeed, I often do that exactly for this reason. However, most people don't do that and to avoid confusion in probability theory (where one already has a lot of strange notations), I decided against it.

[@lueelee6063]

@@brightsideofmaths I see

[@sakcee]

what is a pre-image?

[@brightsideofmaths]

My Start Learning Mathematics series explains that: tbsom.de/s/slm

[@pinklady7184]

Yoo da best!