Probability Theory 10 | Random Variables

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This is my video series about Probability Theory. I hope that it will help everyone who wants to learn about it.
This video is about probability theory, also known as stochastics, stochastic processes or statistics. I keep the title in this general notion because I want cover a lot of topics with the upcoming videos.
#ProbabilityTheory
#Analysis
#Calculus
#Mathematics
Here we talk about the important concept of random variables. We use the general definition between arbitrary measurable spaces, but we mostly discuss real-valued random variables as the occur in applications.
00:00 Intro/ short introduction
00:56 Example (discrete)
02:57 Definition of a random variable
04:56 Continuation of the example
07:49 Notation
09:28 Outro
(This explanation fits to lectures for students in their first year of study: Mathematics for physicists, Mathematics for the natural science, Mathematics for engineers and so on)
The Bright Side of Mathematics has whole video courses about different topics and you can find them here tbsom.de/s/start

Опубликовано:

7 авг 2024

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Комментарии : 39

@jaimelima2420 2 года назад

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 2 года назад

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

@Gouleur 2 года назад

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

@brightsideofmaths 2 года назад

Thank you very much :)

@rombo8 6 месяцев назад

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 6 месяцев назад

Thank you very much :)

@umbranocturna6342 2 года назад

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

@TheXorion 2 года назад

good video as always!

@skillick 10 месяцев назад

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

@brightsideofmaths 10 месяцев назад

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

2 года назад

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 2 года назад

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.

2 года назад

@@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! :-)

@dydx_mathematics2 3 месяца назад

Im interisted on probability theory

@junechu9701 11 месяцев назад

Thanks! It is really helpful!

@brightsideofmaths 11 месяцев назад

Thank you for your support :)

@jsoldi1980 2 года назад

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 2 года назад

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.

@pinklady7184 2 года назад

Yoo da best!

@teddysariel 2 года назад

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 2 года назад

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

@teddysariel 2 года назад

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

@brightsideofmaths 2 года назад

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

@teddysariel 2 года назад

@@brightsideofmaths I see it now. My mistake.

@brightsideofmaths 2 года назад

@@teddysariel Great :)

@molibdenum 2 года назад

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 2 года назад

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

@lueelee6063 2 года назад

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

@brightsideofmaths 2 года назад

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 2 года назад

@@brightsideofmaths I see

@musicalBurr Год назад

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 Год назад

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

@musicalBurr Год назад

@@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 Год назад

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.

@sakcee 4 месяца назад

what is a pre-image?

@brightsideofmaths 4 месяца назад

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

@hn97754 Год назад

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

@brightsideofmaths Год назад

The power set contains all subsets :)

@riolan1274 Год назад

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.