3. Independence
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- Опубликовано: 28 авг 2024
- MIT 6.041 Probabilistic Systems Analysis and Applied Probability, Fall 2010
View the complete course: ocw.mit.edu/6-0...
Instructor: John Tsitsiklis
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
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People like this prof are a rare breed. Brilliant and yet able to articulate in a manner that everyone can understand. Thanks, MIT!
"Let's check our intuition using definition" - this should be the obvious rule on how i should learn from now on
yeah. old intuition----> math definition----> new intuition
I think it's the first I follow a course where the teacher takes 5mn at each class to review and exemplify briefly what was done the last time. This stuff is quite easy as I already saw it in 6.042 for me, but it adds a level of clarity that is just amazing!
If any teacher sees this comment: what this guy does is a GREAT idea, borrow it! We can allways spare 5mn in a 1--hour course.
John Tsitsiklis is awesome !
Clear explanation, no philosophical mumbles.
Plain old school scholar which I like very much ! Conveying idea via paper and pencil these days is rare and should be treasured !
Technology does help but in the right way.
i noticed that only when i watch MIT lectures i get not only the definitions but how to imagine the space on which the whole picture is all about!
that's crucial!!!
Prof Tsitsilis is just awesome! I have never come across such an witty scholar. Thanks MIT for making these lectures available in open.
I never thought I can understand probability easily. This professor Dr Tsitsiklis made it so easy to understand. Thank you professor, TAs and MIT.
If only probability questions had half the clarity in their information as the way Professor talks..!
As he mentioned at the end, it is very easy to get confused while solving the probability questions based on real situations because the information provided can easily have multiple interpretations if utmost clarity is not maintained.
This prof has a way of making the ideas make sense. I am grateful, because I always dread learning about anything related to stats.
Amazing! The closing example is mind-boggling: a concrete illustration of how probability applied to social sciences can be far trickier than simple coin-flipping or die throwing...
Phenomenal teacher..... He has the ability to covey concepts in a very clear and understandable way. Wow....what a joy to learn from him.
This is such a wonderful intellectual lecture i really enjoy listening your lectures.
Thanks very much. Education is the only way in my humble opinion to reduce the ignorance that tend to dominate the world in with we living today.
Professor, I bow to u. May be someday i will be this deft a teacher in what I do. Thank you
The king's sibling problem at the end is thought provoking. Thank you
IT'S ALL WOW! THANK YOU SIR WITH WONDERFUL EXPLANATION!
BEST REGARDS FROM AFGHANISTAN
"As independent as siamese twins"-priceless
The professor did a good job of explaining these concepts so that they seem intuitive :) Thanks!
Went over my head.
I have to rewatch.
I love the trees and recursive nature of the multiplication rule
Probability is nothing but the absence of information.
Oh this guy! Recognise his voice. Fantastic lecturer.
In the king example, the question is ambiguous. "What is the probability of a child being a girl?" vs "Given two children of unknown gender, if one is a boy, what is the probability the other is a girl?" It's not immediately clear which of these questions to answer from the verbiage in the presentation, and given the fact that the latter interpretation would probably not be asked in normal conversation, I don't think it's naive to answer the former interpretation. Apparently semantics is very important in probability and statistics.
He also assumed there are only 2 genders which triggered me.
In fact, that is exactly his point in presenting this example -- that you need to be careful about hidden and/or unjustified assumptions in your probability model when translating from natural language descriptions of a situation -- which he makes clear at the very end: 44:29
17:00 disjoint vs independence
Thank you, professor,
very intuitive, which is not have probability is usually taught but should be.
I think there is one more assumption, that 'GB' and 'BG' are considered sequentially on the two children context. Otherwise, it should be 50% that the sibling is a girl.
Think the answer to the questions separately.
What's your answer to probability of both siblings are boys?
What's your answer to probability of both siblings are girls?
What's your answer to probability of a boy and a girl?
After you get all answers of above questions, add them together and it should be 1. Happy to discuss if you are still not convinced.
@Ted Merrit The king is assumed to be the oldest male child. So it’s 2/3
You can have two children in 4 different ways.
The problem comes out from a student at 19:29 is the concept of conditional independence. P(A and B | B) = 0 = P(A|B) * P(B|B) = 0 * 1.
A probability paradox?
If I construct the sample space in this way: E1= king boy & king's sibling boy, E2= king boy & king's sibling girl, it would seem that the probability of E2 is 1/2.
But if I construct the sample space in this way: E1'= BB, E2'= BG, E3'= GB, E4'= GG, it seems the probability that the king's sibling is a girl is 2/3.
But only in the second case can I assume all events are equiprobable: E1 can happen in just one way; E2 can happen in two: E2 = E2'UE3'.
it's because the 1st set up doesnt include all the necessary information. If you also include birth order it adds an extra GB option
@@alexmallen5765 how would knowing the birth order in this particular problem be useful to identify the king' sibling? In the same way we can write BB twice
Great professor...great enthusiasm and clarity.
Answer of king question is 1/2. There is no "GB" option. it is not permutation. if you care about ordering, then you must choose another "BB" and "GG". "GG" is not important actually, but because of another "BB", the answer will be 1/2.
Right,Even I got that doubt
no it is 2/3. what ur saying wud apply if one is trying to calculate the probability of being born first or second. bt that is not wat one is trying to calculate in the example.the question is abt whether the other sibling is a girl , of which there are 2 scenarios whereas in the case of the other sibling being a boy there is only one scenario.
2/3 is correct. Condition for possible outcomes in this case is the king is from family of two, so have we consider all possible outcomes for family of 2, which makes this problem very similar to coin toss - if there is at least one head, whats the probability of the other toss being tail.
You are considering possible outcomes of “what gender could his sibling be” which is relevant to the question but not related to the way the sample space is constructed.
I am lost with calculation for last question at 29:46 ... Probability of 11th toss being H, given that first 10 tosses are heads.. If I compute using conditional probability, it comes to be same as Probability of 11th toss being H (without any condition, previous question)
x1/y1 events when first coin is picked
x1 => 11th toss is H
y1 => first 10 tosses are heads
P(x1|y1) = P(x1 intersect y1)/P(y1)
Assuming first coin => P(x1 intersect y1) => probability that first 11 tosses are H => (0.9)^11
P(x1) = (0.9)^10
Thus , P(x1|y1) = 0.9.. and multiple by 1/2 since we picked first coin.
on similar grounds P(x2|y2) = 0.1 * 1/2
final is 0.9*1/2 + 0.1*1/2 => 1/2 , which is same answer as previous question.. I might be missing something very basic.. please correct me if I am mistaken in my computation
When you´re told that the first ten tosses are heads, you don´t differentiate between coin A and B, i.e. ,using your notation, P(y) = (0.9)^10+(0.1)^10 , as both describe having tossed heads the first 10 times. Without differentiating coins at event y, you now may notice that P(x2|y) is highly unlikely, given that P(x2|y)=2.87*10^(-11). Therefore it´s reasonable to assume that P(x|y) approximately equals 0.9, because all odds are in favor of you having picked coin A.
the example of king's sibling is so profounding for me
These lectures are great
Incredible class. Thanks for the content and go on!
he has amazing grasp of the subject
Sir looks like Creed from The Office.
All in all great lectures
The subtitles are in the way. I would guess they would still be in the way if you were deaf.
All in all these lectures are great, but sadly it seems to be he doesn't really reply properly to the question the student asks @ around 20 min in - it seems to me the student is correct, and that the Venn Diagram is akward in explaining exactly what he asks about. Maybe because the event B with probability = 0 is poorly pictured using anything but a point, in a venn diagram.
Please comment on this to enlighten me of alternative views on the subject.
I didn't really get what he was trying to assert (largely because of the [inaudible]s along with my poor inference). But to me, what professor said in response, whether relevant to the student's question or not, seemed reasonable. Now that I see you say the student might be right, I'm getting curious of what he tried ask
The Venn diagram is does not directly capture the concept of independence.
to my understanding, student asked what if P(A) =1 and P(B)= 0, in that case A will be Sample space i.e, omega, only in that case P(sample space) =1 , and B = null set , then P(null set) =0. so A intersection B is also a null set, hence their Probability is 0 again .. hence P(null set) = [ P(A).P(B)]= [P(sample space).P(null set)]=[1*0]=0. so it's true
+sai trinath dubba
I am not sure if the student is right: P (A) is not 1; however, P (A) given A happened is 1.
Unrelated: What is the use of dislike button in comment section?
There seems to be a mistake in the explanation for why pairwise independence does not imply independence ... at [36:30], event C should be that 'our tosses gave different results', rather than 'our tosses gave same results'.
After checking the book, and my suspicion is confirmed. See example 1.22 on page 39.
There's no mistake there. Both examples show that pairwise independence of events does not imply total independence of events. The difference in the examples are the values of the probabilities. P(C)=1/2 in both examples, in the video P(C|A and B)=1 and in the book P(C|A and B)=0.
Great video.
It's probably late, but if anyone sees this, please help me out.
In the example he gave, the experiment entails flipping a coin 3 times. Each outcome of the experiment consists of the results of each of the 3 coin tosses. When we say each coin flip is independent of each other, I get confused, since the results each coin flip is not an outcome of the entire experiment(an outcome consists of the results of 3 tosses).
Where's my flaw?
Also related is: how can we be sure that the probability of getting a head on the first coin toss is "p". The probability of that evet is the sum of all outcomes that have heads in the first toss.
How can we be sure that sum is p?
the lecture is awesome but my head started spinning now!!!
Awesome, thanks a lot for this wonderful lesson 💗
Excellent Lecture!
(completed)
Given one of the children became a king, we know it's a boy. So, now there are 4 possibilities.
1. Male, King
2. Female, King
3. King, Male
4. King, Female
(First born first)
In this, it's safe to assume '1' didn't happen, as in that case the king wouldn't be the king. So, we got three possibilities with 2 of them having a female sibling. Thus probability is = 2/3.
Doesn't this make more sense?
Amazing professor with the looks and passion of Walter White 😄
Given that you are somehow eager to watch the next episode, what's the probability that this is a good lecture 😂
so if my probability of a event B ocurs doesn't change if A ocurrs, A and B are independent?
16:35 - why is it not intuitive that if P(A) = 0, then A is independent with any other event?
If the probability of it's raining on Mars is 0, then is it still 0 even if any other event happened.
I think that he means the independence between A and B holds, even if P(A)=0.
It's difficult to imagine that an event has no probability, but you've got an example by yourself already.
at 37:15 ,why P(A∩B∩C)=1/4 isnt supposed to be 1/8 as each one of event have a probability of 1/2
Help !
P(A∩B∩C)=P(A)P(B)P(C) only if A,B and C are independent.But they are not in the example which is being discussed at 37:15
Visually, you can see that only one of the four possibilities falls within each of A, B and C, so P(A∩B∩C) = 1/4.
Mathematically, recall that the Multiplication Rule is P(A∩B) = P(A)*P(B|A), *not to be confused with* P(A∩B) = P(A)*P(B) (which is the definition of independence).
In the example, P(A) = 1/2 and P(B|A) = 1/2, so P(A∩B) = (1/2)*(1/2) = 1/4
Therefore, P(A∩B∩C) = P((A∩B)∩C) = P(A∩B)*P(C|A∩B). Since the sample space of A∩B consists of only one possible outcome in the sample space, and that one possible outcome is one in which the first and second toss give the same result, P(C|A∩B) = 1/1 =1. So, P(A∩B∩C) = (1/4)*(1) = 1/4.
awesome， your reply is really helpful！
Clear explanation！
I never knew that probability could help with curing my OCD
At 31:50 , why do A2 and A6 have a little c on top to make it look like A2 complement, A6 complement ?
They are complements. It's simply a random choice of event. What matters is that the probability of an event concerning the first three flips is unaffected by information regarding the 5th and 6th coin flip.
How is P(AnB)=0 but P(A)*P(B)=1/12?
He was trying to show that the two were in fact not the same! They would only be the same if A and B were independent, but he's making the point that in this case they are not.
Really nice explanation
Independent means not related or connected or attached or involved.
In probability theory though, the term 'independent' only refers to this mathematical relationship. Just like a 'negative' number only means that when you add it to its 'positive' counterpart you get exactly zero, not that it is 'pessimistic' or 'has an excess of electrons' or any other meaning that 'negative' might have.
In the king's example in the end, shouldn't bg and gb mean the same thing since it does not matter how you have ordered them? So that according to that logic you still get 1/2 like how you had got intuitively.
Snehal Sanghvi From what I understand, you have to care about the ordering, cause "bg" is not a atom event, it's a sequence of two events. Thus "gb" is a different sequence. Otherwise if you only count "a girl and a boy" as an atom event, you have underestimated that probability by half: P(a girl and a boy) = P(2nd = boy| 1st = girl) + P(2nd = girl | 1st = boy).
+Snehal Sanghvi i thought the same ... the question was sibling, not older or younger sibling..
+Bo Zhang I don't think the order matters. If we add the information the king was born first or second here, then the probability of the second child being a girl is 1/2, while if we don't state anything it's 2/3. Does that make sense? I don't think so..
why is the probability of P(1head) is 3p(1-p)^2...? Shouldnt it be 2p(1-p)^2, because only 2 probability it would happen instead of 3.... or am i not understanding it clearly? its in 9:14 please help
Hi.. no, it's 3p(1-p)^2. There are three different ways it could happen: (Head, Tails, Tails), (Tails, Head, Tails) and (Tails, Tails, Head)
ah yeah I miss that, thanks
Thank you Sir.
I was curious to know where you are from.Does anybody know?
Greece
Legendary..!
superb lecture 👌
Wouldn't the fact that the king is actually a king lower the probability of the other sibling being a boy? Because if he were a boy and older, the current king would not be king, but his brother would be instead! Could someone please elucidate this?
That's the problem with conditional probability. It is stated that HE IS THE KING. The whole question is already conditioned to that fact. The probability of having a brother = 1/3 is already lesser than the probability in the case where the second boy could be king, in which case it would be = 1/2. You need to consider all alternatives to see that, i'll show how:
[Brackets means crown]
[B]B [B]G B[B] B[G]
[G]B [G]G G[B] G[G]
Now we filter all alternatives that exist:
- King is a boy.
- King is the first boy.
I'll do that with an asterisk:
[B]B* [B]G* B[B] B[G]
[G]B [G]G G[B]* G[G]
So we get: [B]B [B]G G[B] alternatives. Only 1/3 of them have a king with a brother.
If we removed the condition King is the first boy, we would get:
[B]B* [B]G* B[B]* B[G]
[G]B [G]G G[B]* G[G]
Result: [B]B [B]G B[B] G[B]... now 2/4 of the alternatives express a king with a brother.
We could even consider more prior possibilities if we didn't knew they were false, for example:
- a situation with no crowns at all
- a situation with two crowns
In fact, these situations are not false and have all happened already in human history. A royal family giving rise to two kings of different countries. Or a situation where the first king dies. Or when the king is deposed. Given the complexities of the reality, none of the answers is correct, because we would have to know the probability of each of these special situations to calculate the real probability of the event given by the original question.
super nice lecture
37.50 ??
The instructor looks like the "Tom Yorke" of statistics to me!
I didn't know teachers of STEM could speak in clear, logical English (or any other natural language).
In the last example, aren't GB and BG are same? Does order matter?
Yes, order matters. Consider a slight variation on the question: "What is the probability that a family with two children has a) two boys, b) two girls, or c) one boy and one girl. Intuitively, it may be tempting to give each possible outcome equal probabilities of 1/3, but there are actually 4 possible outcomes, not 3: boy then boy, boy then girl, girl then boy, and girl then girl. So, P(a) = P(b) = 1/4, and P(c) = 1/4 + 1/4 = 1/2.
The sample space for the problem has four elements -- {Boy, Boy}, {Boy, Girl}, {Girl, Boy}, {Girl, Girl} -- each of which has an equal probability of 1/4 of happening on its own. The event c represents the outcome where a family of two children has one son and one daughter. Since c does not specify the order that the boy and girl were born in, both {Boy, Girl} and {Girl, Boy} are outcomes that count toward c. Therefore, P(c) = P(Boy, Girl) + P(Girl, Boy) = 1/4 + 1/4 = 1/2.
Probability questions can be very counter-intuitive, and I've found that it can help to understand a problem by just doing the experiment yourself. So if you still have a hard time believing or agreeing with what I'm saying, try flipping a fair coin twice, 20 times, and record the results of each pair of flips. Both coins should be heads in about 5 of the double flips, both should be tails about 5 times, and you should get one each of heads and tails about 10 times. Increase the number of flips to make the experiment more accurate, if you like.
What about this:
The king could be born first and then the family has a boy.
The family could have a boy first and then the king is born.
Why doesn't order matter in the BB case?
They’re the same in that you end up with a girl or a boy. But they are not the same EVENT. You have a girl first and then a boy, and you have a boy first and then a girl. Because it is assumed that children are born one at a time, there are therefore 2 distinct ways to have a boy and a girl.
TY
where are lectures 4,5?
The Scholar playlist: ruclips.net/video/j9WZyLZCBzs/видео.html. For more info, see the course on MIT OpenCourseWare at: ocw.mit.edu/6-041SCF13. Best wishes on your studies!
LOL the king example is hilarious
6
MI who? :)
42:40 every childbirth is basically a coin flip. yeah, right.
except for reproductive practices
check how great is this lecturer. I am just from YEMEN>...
mit.edu/~jnt/home.html
The context of the problem on 43:24 is so rude! Why does
only a king can be born in this family??? Do you have something against female's rights to become a queen?? By the way, the formulation of this problem is so ridiculous. We can not say that there was born a boy or a girl, because a child have not identified its gender yet!!