Isnt it possible to prove the statement without using axiom 2? Once we find P(S)=P(S)+P(⍉), it directly folows P(⍉)=0 (you do'nt even have to use S, any set works).
Love your work. Helps me keep going with my learning :) Thanks the recent splurge in proof math it’s really been lifting my interest in different areas.
In Finland we use the notation m.o.t. (= "mikä on todistettavaa") as a sign of finishing a proof, which is the same as the latin q.e.d. (quod erat demonstrandum). Alternatively we may use a small empty box (without the cross).
Hi! I think there is a problem in this proof if we are very careful. Axiom three doesn't say that you can count the probability of P(A U B)=P(A)+P(B) even if A and B are mutually exclusive. Axiom three talks about infinite sequence of events... You can proof of cause that P(A U B)=P(A)+P(B) is indeed true, but it does not follow directly from axiom three.
Actually, you cannot use finite additivity to prove the probability of the empty set is 0 at all, so the video is totally wrong. That's because the derivation of finite additivity from infinite additivity requires the knowledge that the probability of the empty set is 0 since the proof is basically you have disjoint events labelled A of i from i = 1 to infinity that we define is the empty set for all i>n. Then the sum from n+1 to infinity is the infinite sum of the probabilities of empty set, which is 0. Since the proof of finite additivity relies on knowing the probability of the empty set is 0, we cannot use it to prove the thing it depends on. In fact, it is the other way around. We need to prove the probability of the empty set is 0 to get finite additivity. Hence, the proof in the video is circular reasoning.
@@Exachad "Yes, it seems that your statement is correct. I thought that if one first proves that P(A ∪ B) = P(A) + P(B) (which you can certainly do) and then proceeds as he did in this video, without considering myself what proving P(A ∪ B) = P(A) + P(B) requires. It does require (at least the way I am thinking) knowing the probability of the empty set. So, here he used that knowledge in his proof. My original statement was about the incorrect use of axiom 3, which is still valid, I suppose. I mean, you could prove that the probability of the empty set is 0, then that P(A ∪ B) = P(A) + P(B), and proceed as he did, but why bother if you have already proved that the probability of the empty set is 0."
Intuitively obvious , but the proof is an another problem, isn't it? Meanwhile I'm enjoying a cup of cardamom coffee now, just a smidgen cardamom into coffee ,then you can experience the depth of flavor of coffee ! Please try it.
Probability is the branch of math I dislike the most. I passed the one obligatory course in second time, I was only 1 point above the corner :) unfortunately 100% of job orders related with Mathematics are Probability,Statistics and its applications (Finance, Data analysis, machine learning) so its not easy for me to find a Job.
Just another observation: Sample space is defined as (wikipedia): "In probability theory, sample space (also called sample description space[1] or possibility space[2]) of an experiment or random trial is the set of all possible outcomes or results of that experiment.[3]" But impossible event (outcome) CAN'T be the member of this set! It it were it would be possible. This is a paradox. So axiom 2 fails here. I think Wittgenstein would agree the words are not defined according Logical positivism norm. LOL REGARDS
Isnt it possible to prove the statement without using axiom 2? Once we find P(S)=P(S)+P(⍉), it directly folows P(⍉)=0 (you do'nt even have to use S, any set works).
Love your work. Helps me keep going with my learning :) Thanks the recent splurge in proof math it’s really been lifting my interest in different areas.
❤️
@@TheMathSorcerer But sir empty set fi is a subset of every set so how S and fi are mutually independent ? pls answer sir
Just beautiful!! It is a nice feeling of gratitude when after a time to understand the proofs in math you get also in love with math!
In Finland we use the notation m.o.t. (= "mikä on todistettavaa") as a sign of finishing a proof, which is the same as the latin q.e.d. (quod erat demonstrandum). Alternatively we may use a small empty box (without the cross).
I was thinking if you shouldn't prove before that the axiom 3 is also true for finite sequences of disjoint sets
Hi! I think there is a problem in this proof if we are very careful. Axiom three doesn't say that you can count the probability of P(A U B)=P(A)+P(B) even if A and B are mutually exclusive. Axiom three talks about infinite sequence of events... You can proof of cause that P(A U B)=P(A)+P(B) is indeed true, but it does not follow directly from axiom three.
Actually, you cannot use finite additivity to prove the probability of the empty set is 0 at all, so the video is totally wrong. That's because the derivation of finite additivity from infinite additivity requires the knowledge that the probability of the empty set is 0 since the proof is basically you have disjoint events labelled A of i from i = 1 to infinity that we define is the empty set for all i>n. Then the sum from n+1 to infinity is the infinite sum of the probabilities of empty set, which is 0. Since the proof of finite additivity relies on knowing the probability of the empty set is 0, we cannot use it to prove the thing it depends on. In fact, it is the other way around. We need to prove the probability of the empty set is 0 to get finite additivity. Hence, the proof in the video is circular reasoning.
@@Exachad "Yes, it seems that your statement is correct. I thought that if one first proves that P(A ∪ B) = P(A) + P(B) (which you can certainly do) and then proceeds as he did in this video, without considering myself what proving P(A ∪ B) = P(A) + P(B) requires. It does require (at least the way I am thinking) knowing the probability of the empty set. So, here he used that knowledge in his proof. My original statement was about the incorrect use of axiom 3, which is still valid, I suppose. I mean, you could prove that the probability of the empty set is 0, then that P(A ∪ B) = P(A) + P(B), and proceed as he did, but why bother if you have already proved that the probability of the empty set is 0."
Intuitively obvious , but the proof is an another problem, isn't it? Meanwhile I'm enjoying a cup of cardamom coffee now, just a smidgen cardamom into coffee ,then you can experience the depth of flavor of coffee ! Please try it.
why mathematicians ate like that!!!!
The probability for an impossible event to happen is zero, I think it is an axiom
stunning video The Math Sorcerer. I crushed the thumbs up on your video. Maintain up the outstanding work.
Probability is the branch of math I dislike the most. I passed the one obligatory course in second time, I was only 1 point above the corner :) unfortunately 100% of job orders related with Mathematics are Probability,Statistics and its applications (Finance, Data analysis, machine learning) so its not easy for me to find a Job.
math is a hope giver
Just another observation: Sample space is defined as (wikipedia):
"In probability theory, sample space (also called sample description space[1] or possibility space[2]) of an experiment or random trial is the set of all possible outcomes or results of that experiment.[3]"
But impossible event (outcome) CAN'T be the member of this set! It it were it would be possible. This is a paradox. So axiom 2 fails here. I think Wittgenstein would agree the words are not defined according Logical positivism norm. LOL
REGARDS
you should have read more of that wikipedia page. Sample space is not the set of events.
What I don't get is why would we even need to prove such an obvious thing. This is basically an axiom right?
Great and intuitive.
thank you:)
Nice proof and Thanks for your efforts I wonder if you can make a video in your opinion what is the most Efficient way to read a maths book.
Nice
OR we can use m/n
m is no of favourable outcomes which is equal to zero
n is total no of possible outcomes
required probability = 0/n=0
But sir empty set fi is a subset of every set so how S and fi are mutually independent ? pls answer sir
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When the imposter is SU∅
it's obvious
sir i want to contact you ,can i get your email ?
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