A^c and A are disjoint, and their union is equal to Ω. Therefore, P(A^c) + P(A) = P(Ω). P(Ω) = 1 by axiom 0, hence P(A^c) + P(A) = 1. Thus P(A^c) = 1 - P(A). Q. E. D.
technically, you don't have P(A) + P(B) = P(A u B), you would need to establish it first. you could do this by taking a sequence seq(A_i, i=1, oo) = (A, B, ∅, ∅, ...), then uni(A_i, i=1, oo) = A u B u uni(A_i, i=3, oo) = A u B u uni(∅, i=3, oo) = A u B u ∅ = A u B, and so P(uni(A_i, i=1, oo)) = P(A u B) similarly, sum(P(A_i), i=1, oo) = P(A) + P(B) + sum(P(A_i), i=3, oo) = P(A) + P(B) + sum(P(∅), i=3, oo) = P(A) + P(B) + sum(0, i=3, oo) = P(A) + P(B) + 0 = P(A) + P(B). this proves that P(A u B) = P(A) + P(B). i used here fact that P(∅) = 0, which creator has stated as an axiom. this is not needed. it's enough to assume that P(Ω) = 1 and use the same approach first with the following sequence: seq(A_i, i=1, oo) = (Ω, ∅, ∅, ∅, ...). from first part we will get: P(Ω u ∅) = P(Ω) and the second part we have to do carefully: sum(P(A_i), i=1, oo) = P(Ω) + P(∅) + sum(P(∅), i=3, oo) = P(Ω) + P(∅) + R, where R is some number from [0,1]. combining we have: P(Ω) = P(Ω) + P(∅) + R, so 0 = P(∅) + R >= P(∅). we know that P(∅) is some number from [0,1] hence P(∅) = 0.
i'm solving it backward so it is obvious to solve it forward. assume P(A^c) = 1 - P(A) then P(A^c) + P(A) = 1 using the properties of probability measures of (b) P( A^c) + P(A) = P(A^c ∪ A) P( A^c ∪ A ) = 1 but A^c = Ω/A. therefore P( Ω/A ∪ A ) = 1 ~> p(Ω) = 1. i have solved it backward now i can do it forward similarly. Q.E.D
First of all, thank you SO MUCH for these videoes. Your series on measure theory just helped me pass my exam in the course :-) Could you ever consider doing a series on Markov Chains? I am currently doing a course revolving around these and find them a bit difficult to comprehend, and your channel has truly been a godsend to me during my other courses.
Is it true to say that the measure theoretic approach to probability is firmly rooted in the ‘frequentist’ interpretation of probability? Can the measure theoretic approach be reconciled with a Bayesian interpretation of probability?
The measure theoretic approach is needed for the modelling and formalisation. The interpretation is not needed yet. Therefore it is compatible to both interpretations in my understanding.
This is really nice explanation of the probability measure. Besides, the notebook application of your lecture is really nice, can I have the name of it?
First of all thanks for videos. In definition of sigma algebra (a) we can assume either empty set or Ω contained in curly(A), we don't need both of them contained in curly(A). Because axiom (b) implies each other.
Thank you very much. Sure but we don't have to choose the axioms to be minimal. I want property (a) in this way because it is easy to check in examples.
The thing I still don’t understand is why we have to explicitly say that this definition holds in the limit of an infinite number of subspaces (at 3:50). I’ve never taken real analysis, but can’t we use the first 3 axioms to show that since the disjoint union of any infinite number of subsets of the sample space must still be a subset of the sample space. And the probability of any subset must be between 0 and 1. Then doesn’t the limit for the probability of the disjoint union of an infinite number of subsets already have to exist, be between 0 and 1, and be the infinite sum of each disjoint subsets probability. Or is there something about limiting processes that I don’t understand correctly?
@@brightsideofmaths Aren’t sets “closed” under countably infinite union? In the sense that if you take the union of countably infinite of subsets, and call it S, S will still be a subset of omega. Because of this, you should be able to find the probability of that S. So, shouldn’t the infinite sum of the probabilities just converge to whatever the probability of S is?
@@ryanlangman4266 The infinite union is still a subset of Omega. That is correct. However the question is still, how will you be able to use the third bullet point in the infinite union on the left-hand side? I could give you counterexamples, but for probability measure I only know complicated ones. For a measure (with value infinity), counterexamples are very simple.
Your explanation helped me a lot to get my head around this topic. Thanks :) What's still puzzling me, however, is that in our lecture we defined the function s.t. P: A --> [0, +infinity] (in contrast to you restricting it to 1, i.e. P: A --> [0, 1]). Is both correct? Does it make a difference?
Thanks very much for making this video. It's super helpful! I am wondering why the curved A is a subset of power set of omega? Specifically, I am curious in what cases the curved A is not equal to the power set of omega? An example to demonstrate this is much appreciated. Thank you!
Hello I have a question we were tasked to prove that the probability measure is countably additive. In my proof, I consider two cases wherein all A_i 's are nonempty (case 1) and there are some A_i's that are empty (case 2). I don't have any idea how to start the proof in case 1. Do you have any idea how to prove it using the assumption in case 1? Thanks for the reply
Hi, thanks for the video! How often will you upload videos to this playlist? I studying this at university right now and would love to watch your videos throughout the course.
Given a probability triple (Ω, F, P) then for A ∈ F with P(A) is not equal to zero, P(·|A) : F → [0, 1]. is a probability measure over (Ω, F). How to proof it? Can you recommend an article or book where i can read about it?
@@brightsideofmaths Thanks. But please can you at least tell me what this "dot" is , i want to read about it. I never saw this symbol before? Or any resource i can read about this symbol and proof? I shall be thankful
I here the term countable a lot in books and your videos. Why is it that we require unions of subsets or other things to be countable? The series are already infinite so why do we put a big emphasis on countable?
Countable is the "smallest" infinity. A lot of things break with infinite many things. However, some things can be saved for countable many things. That is the rough idea and therefore I always emphasise it when it is used.
The most confusing part is that some people say random variable is a mapping from sigma-algebra set to borel set. ruclips.net/video/olYxcBhR9yM/видео.html
A^c and A are disjoint, and their union is equal to Ω. Therefore, P(A^c) + P(A) = P(Ω). P(Ω) = 1 by axiom 0, hence P(A^c) + P(A) = 1. Thus P(A^c) = 1 - P(A). Q. E. D.
😂😂OK
technically, you don't have P(A) + P(B) = P(A u B), you would need to establish it first.
you could do this by taking a sequence seq(A_i, i=1, oo) = (A, B, ∅, ∅, ...), then
uni(A_i, i=1, oo) = A u B u uni(A_i, i=3, oo) = A u B u uni(∅, i=3, oo) = A u B u ∅ = A u B, and so P(uni(A_i, i=1, oo)) = P(A u B)
similarly, sum(P(A_i), i=1, oo) = P(A) + P(B) + sum(P(A_i), i=3, oo) = P(A) + P(B) + sum(P(∅), i=3, oo) = P(A) + P(B) + sum(0, i=3, oo) = P(A) + P(B) + 0 = P(A) + P(B).
this proves that P(A u B) = P(A) + P(B).
i used here fact that P(∅) = 0, which creator has stated as an axiom.
this is not needed.
it's enough to assume that P(Ω) = 1 and use the same approach first with the following sequence: seq(A_i, i=1, oo) = (Ω, ∅, ∅, ∅, ...).
from first part we will get: P(Ω u ∅) = P(Ω)
and the second part we have to do carefully:
sum(P(A_i), i=1, oo) = P(Ω) + P(∅) + sum(P(∅), i=3, oo) = P(Ω) + P(∅) + R, where R is some number from [0,1].
combining we have:
P(Ω) = P(Ω) + P(∅) + R, so 0 = P(∅) + R >= P(∅).
we know that P(∅) is some number from [0,1] hence P(∅) = 0.
@ All correct. Thank you for your correction.
Just started studying this in my graduate studies.. really enjoying your thorough explanation of the idea behind each concept.
Your videos are super helpful. Please keep making them!!
These videos are great. I've never seen any formal course on probability and I'm really enjoying it. Thank you. What book could you recommend me?
Well that was faster than expected, but it's a welcomed surprise.
great video your channel is severely underrated
Thank you :)
Thank you sir for sharing knowledge 😊
It's my pleasure! Thank you for your support!
This was very helpful, thanks!
Will you make a series on stochastic calculus?
i'm solving it backward so it is obvious to solve it forward.
assume
P(A^c) = 1 - P(A)
then
P(A^c) + P(A) = 1
using the properties of probability measures of (b)
P( A^c) + P(A) = P(A^c ∪ A)
P( A^c ∪ A ) = 1
but A^c = Ω/A.
therefore P( Ω/A ∪ A ) = 1 ~> p(Ω) = 1.
i have solved it backward now i can do it forward similarly. Q.E.D
First of all, thank you SO MUCH for these videoes. Your series on measure theory just helped me pass my exam in the course :-)
Could you ever consider doing a series on Markov Chains? I am currently doing a course revolving around these and find them a bit difficult to comprehend, and your channel has truly been a godsend to me during my other courses.
Is it true to say that the measure theoretic approach to probability is firmly rooted in the ‘frequentist’ interpretation of probability? Can the measure theoretic approach be reconciled with a Bayesian interpretation of probability?
The measure theoretic approach is needed for the modelling and formalisation. The interpretation is not needed yet. Therefore it is compatible to both interpretations in my understanding.
This is really nice explanation of the probability measure. Besides, the notebook application of your lecture is really nice, can I have the name of it?
Thanks! Yeah, it's Xournal.
First of all thanks for videos. In definition of sigma algebra (a) we can assume either empty set or Ω contained in curly(A), we don't need both of them contained in curly(A). Because axiom (b) implies each other.
Thank you very much. Sure but we don't have to choose the axioms to be minimal. I want property (a) in this way because it is easy to check in examples.
thank you so much sir. Now i have clarity of it. Thank you so much sir.🙂
The thing I still don’t understand is why we have to explicitly say that this definition holds in the limit of an infinite number of subspaces (at 3:50). I’ve never taken real analysis, but can’t we use the first 3 axioms to show that since the disjoint union of any infinite number of subsets of the sample space must still be a subset of the sample space. And the probability of any subset must be between 0 and 1. Then doesn’t the limit for the probability of the disjoint union of an infinite number of subsets already have to exist, be between 0 and 1, and be the infinite sum of each disjoint subsets probability. Or is there something about limiting processes that I don’t understand correctly?
How do you go from the finite union to the infinite union?
(Of course, I highly recommend that you watch my real analysis course :)
@@brightsideofmaths Aren’t sets “closed” under countably infinite union? In the sense that if you take the union of countably infinite of subsets, and call it S, S will still be a subset of omega. Because of this, you should be able to find the probability of that S. So, shouldn’t the infinite sum of the probabilities just converge to whatever the probability of S is?
@@ryanlangman4266 The infinite union is still a subset of Omega. That is correct. However the question is still, how will you be able to use the third bullet point in the infinite union on the left-hand side?
I could give you counterexamples, but for probability measure I only know complicated ones. For a measure (with value infinity), counterexamples are very simple.
For the last sentence, you can see it here:
math.stackexchange.com/questions/929528/additive-but-not-sigma-additive-function
Your explanation helped me a lot to get my head around this topic. Thanks :)
What's still puzzling me, however, is that in our lecture we defined the function s.t. P: A --> [0, +infinity] (in contrast to you restricting it to 1, i.e. P: A --> [0, 1]). Is both correct? Does it make a difference?
Good question! Did you define a measure or a probability measure?
Thank you for this video!
Thanks very much for making this video. It's super helpful! I am wondering why the curved A is a subset of power set of omega? Specifically, I am curious in what cases the curved A is not equal to the power set of omega? An example to demonstrate this is much appreciated. Thank you!
Good question! For this discussion I have a whole Measure Theory series tbsom.de/s/mt
Hello I have a question we were tasked to prove that the probability measure is countably additive. In my proof, I consider two cases wherein all A_i 's are nonempty (case 1) and there are some A_i's that are empty (case 2). I don't have any idea how to start the proof in case 1. Do you have any idea how to prove it using the assumption in case 1? Thanks for the reply
Does the countable union of subsets property not follow inductively from the union of two subsets property and vice versa?
No, it doesn't :)
Probability Theowy :3
Finally, I got it
Hi, thanks for the video! How often will you upload videos to this playlist? I studying this at university right now and would love to watch your videos throughout the course.
Thank you very much! I don't want to make promises but a rough estimate would be one video per week.
Yoo da best, best and bestest. 👏
for property (c) in your definition of sigma algebra, does each A_j has to be disjoint or not?
For the sigma-algebras, there is no restriction to disjoint sets.
@@brightsideofmaths I see, so some A_j can overlap in cursive A
Can you give me an example where the sigma-algebra is not equal to the power set?
Borel Sigma algebra on R.
Quickly: why does “fancy A” map to R? Should it not map to [0, 1]?
The codomain does not have to be equal to the image :)
Thanks, Love from Saudi Arabia 🫡
Given a probability triple (Ω, F, P) then for A ∈ F with
P(A) is not equal to zero,
P(·|A) : F → [0, 1].
is a probability measure over (Ω, F).
How to proof it?
Can you recommend an article or book where i can read about it?
We will discuss this soon :)
@@brightsideofmaths Thanks. But please can you at least tell me what this "dot" is , i want to read about it. I never saw this symbol before? Or any resource i can read about this symbol and proof? I shall be thankful
@@AKhan-3 The dot in your expression is just an empty space for the independent variable.
Great video! But What do you mean with “ total mass” ?
The full measure :)
Gutes Video!
I here the term countable a lot in books and your videos. Why is it that we require unions of subsets or other things to be countable? The series are already infinite so why do we put a big emphasis on countable?
Countable is the "smallest" infinity. A lot of things break with infinite many things. However, some things can be saved for countable many things. That is the rough idea and therefore I always emphasise it when it is used.
Could you teach something about product measure in probability space?
Sure. This will definitely come!
@@brightsideofmaths thanks
love from china nice video
Thanks!
Perfect!!!!
what is the name of the music at the end of your videos?
It does not have a name yet :)
@@brightsideofmaths So did you do it yourself? :D
@@Hold_it Yes, I do everything here myself :)
Nice
The most confusing part is that some people say random variable is a mapping from sigma-algebra set to borel set.
ruclips.net/video/olYxcBhR9yM/видео.html
I still don't get it💔😭😭
I have a community forum where you can ask questions :)
p(a^c)=1-p(a),p(a^c)=p(Ω)-p(a)=p(-ΩUa),therefore the probability is 100%
I don't really understand why you write sigma-algebra = P(Ω) ; sigma-algebra is just a set of all subsets (or no?) and P(Ω) by definition is 1...
Please note the different P's here :)
@@brightsideofmaths ah ok, thank you !
You
i thought this course is supposed to be for beginners.
Yes, for beginners in probability theory :) If you need more mathematical knowledge check out the dependencies: tbsom.de/startpage/
1st
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