Some notable Timestamps 0:00:33 Stochastic Process 0:10:57 (Simple) Random Walk 0:32:43 Markov Chain 0:58:41 Martingale 1:06:47 Stopping time / Optional Stopping Theorem
0:00:33 Stochastic Process 0:10:57 (Simple) Random Walk 0:32:43 Markov Chain 0:58:41 Martingale 1:06:47 Stopping time / Optional Stopping Theorem For my reference
The best teachers do a great job of introducing problems and then showing you the tools to solve them. With these teachers, you always know why you're doing something, you always have a sense of intuition for the problem, and you easily build a sense of experience having worked with these tools in future similar scenarios. There are so many instances wherein this professor does just that and it's a huge blessing to have access to this content for free
Recursive argument at 28:00: Call p the probability you hit -50 first. There’s a 50% chance you hit -50 before you hit 50, by symmetry. Once you hit 50, the game is reversed, by stationary property. Hence p = 0.5 + 0.5 * (1 - p), from which p is 1/3.
This guy is absolutely fantastic. Could not have been explained more clearly, with a sound logical structure. People complaining about him should probably try lecturing themselves before offering their criticism.
And people like you are confusing people even more when they get caught up in one of this guy's many mistakes and think that THEY are the ones who are wrong.
@@nickfleming3719 No offense, this is a free course for us, it's our own responsibility to find out wether the information is right or not when we get caught up in the instructors' mistakes. I mean the most important ability for self-taught learners like us is to be skepticism and check other information sources when we feel confused, not only in a free course but also in other paid courses. We can certainly say whatever we want in comments and I always learned a lot by some critical comments, however, I think it would be better to be grateful when we have chance to access high quality educations like this.
Let’s see you lecture, I really want to see your descriptions on such topics as: Real analysis, Complex Analysis, Functional Analysis, or Harmonic Analysis; oh please it would be delightful to see such confidence coming from you.
All you people praising this lecturer, saying how easy and simple he makes everything, are not helping. He's making tons of mistakes, and I'm thinking I must be going crazy since everybody else seems to think this is the best lecture ever.
27:00 The argument to make it work the way the intuition of the student worked is via markov chains. Set up the states -50, 0, 50 and 100, write the transition probabilities, then calculate the absorption probabilities of the two recurrent states (-50 and 100) from 0 which give 1/4 and 1/2. The probability to end up with $100 is the probability of ending up becomes 1/4 / (1/4 + 1/2) (since the two other states will eventually bleed into either one of these states we know their steady state probability will be 0) which indeed gives 1/3.
28:00 Following the thought process of the student from the audience, after the balance reaches $50, there is a 1/2 chance for the balance to reach $100 (overall probability = 1/4) or fall back to $0 (overall probability = 1/4). If the balance falls back to zero, we can consider that as the start of the second cycle, where the distribution of the conditional probability is the same as the first cycle (1/2 chance to reach $-50, 1/4 chance to reach $100, and 1/4 to reach $50 first then return to $0). Same for the third cycle, forth cycle, etc. Therefore, we can express the overall probability for the balance to reach $100 as the infinite series of 1/4 + (1/4)^2 + (1/4)^3... which gives us 1/3.
yes, and this is also consistent with Lee's solution, except that in the equation, one only needs to consider three (large) steps/grids, instead of a total of A+B steps/grids :)
Very easy solution for 28:00. P(B), P(A) be probabilities that B,A occur first respectively. Probability that we hit 50$ before -50$ is 1/2 and also probability that we hit -50$ before 50$ is 1/2. If we reach 50$ first, we see problem is flipped now, we are 50$ closer to B and -100$ closer to A. So P(B/start at 50$) = P(A/start at 0$) So we can write P(B) = 1/2(P(A)) = 1/2(1-P(B)) Solving this simple equation we get P(B) = 1/3 In fact for any A,B there is a point where we can flip the problem, so try to generalize this and come up with a proof.
48:45 In order to predict the future using transition probability matrix A, we need to use transpose of A. (A^T^3650 * [1;0]) (this is fixed at 51:41). Since the eigenvalue of A^T is equal to A, the following theorems also hold. Hope this help. Thanks for the great lecture.
Continue the reasoning from 27:22: Assume the probability of the game ends at 100 is x. As probability of the game reaches 50 is 0.5; The probability from 50 to 100 is actually (1-x). So x=0.5*(1-x) --> x=1/3
I mean i dont get all these praises. The guy gives an overview of the topic but not rigorously at all. This is not the level of depth I would have expected but it serves me well in my preparations. It feels like I have to dive deeper on my own to get real understanding of the topic.
@@DilanCheckerits financial mathematics so it's not clearly into depth as it would be for pure mathematics or statisticians, the application is more important here
56:13 I think the confusion here comes from the fact that for the other eigenvalue, which actually is less than 1 and greater than 0, the corresponding eigenvector will converge to the 0 vector. The “sum trick” he did earlier wouldn’t work because v_1 + v_2 = \lambda (v_1 + v_2) doesn’t imply that \lambda = 1 when both v_1 and v_2 are 0. Hope I didn’t overlook anything!
I am currently working on understanding the stochastic processes, and I am very confused by the concept of “a collection of random variables”, but the trajectory thing given by the lecturer helped me understand the concept a lot easier. For a continuous random process, if I sample at very high frequency, I will get several curves in the “x(t)-t” plain (the curve depending on the setting of the random process).
58:10 Someone asked whether the algebraic manipulation led to the (seeming incorrect) conclusion that all eigenvalues lambda are 1. That was not true, since the assumption for that equation is that we are dealing with a stationary state, and therefore, the conclusion is for a stationary state, its eigenvalue must be 1, as stated by Lee.
The equation was just an eigenvalue equation for A - it didn’t assume anything about stationary state. The correct argument, against the incorrect conclusion that all eigenvalues of A is 1, is that (v1 + v2) can be 0 and hence you can’t divide that out to conclude much about lambda. The case where you can do it turns out to be when v1 and v2 are positive - thus the theorem about the unique highest eigenvalue isn’t broken.
@@eigentejas You are correct. If one assumes a stationary state (some vector (p, q) of probability that remains unchanged by further multiplying A from the left), it simply implies the existence of an eigenvalue of 1.
In 47:42 Multiplying a 2x2 matrix with a vector (1,0) will give back the p11 and p21 which stands for working today and working tomorrow(p11) and broken today but working tomorrow(p21) not the probability working and not working.
it gives the probability of the machine working tomorrow, no matter if it's broken or not today therefore p reflects the probability of the machine working in 10 years. however he should've multiply with a vector (1,1) to adjust the same for q, since if you multiply the matrix with (1,0) the value of q will be 0
N4mch3n there cannot be a vector (1,1) as they represent probabilties of the machine working and not working.The rows of the vector must add upto 1. With (1,1) it implies that the machine is working and not working at the same time.
You are right. The error is that the entrees which should sum up to one are the ones in ROWS not columns. Because he is not multiplying A^3650 by the correct vector, he had to amend the matrix A when computing the eigenvector in 52:00.
They have the audacity to call Choongbum Lee an instructor, when he can give a presentation so complete, elegant, and accessible that he could (and maybe should) teach ALL of the other professors at MIT a thing or two about how to give a lecture.and communicate ideas throughout it. This guy is @#$%ing amazing! What a beast. God, I feel stupid in comparison.
@@MrCmon113 I think they mean that he should be promoted to the position of professor. Instructors are not generally permanent positions at a university.
MIT is a top research university, and as such, professors at MIT (and other research institutions) are judged mostly according to the quality of their research, not teaching.
I have seen quite a few MIT courses and every time, the teachers were amazing. This teacher is honestly not the best, although he is very much alright.
shailesh kakkar I believe he meant the probability to be within 100 standard deviations (which is virtually 100%, not close to 90% =). And there are a lot of minor mistakes in this video and the two before, the instructor is not very well prepared. But it's still useful
+Maxim Podkolzine No, the answer is right, he means that the total area under the bell curve its 1, or 100%, but in the real word, you nead just 2 standard deviations boths sides to the total area to stay very close to 100%
+dicksonh Well if you do that, you miss 1/3 of the boundaries values and your forcast will be completely wrong, but Who Am I to change your point of view.😉
In the 1st and 2nd cases he's talking about delta hedge parity (in trading/market practice) as reflected by trend lines. In the 3rd case he's referring to the vol of vol, in this situation one must employee stochastic volitility models.
To get variance, applied variance to both sides, var(sum(yi) over i). because yis are iid variance becomes sum(var(yi)). Var of each yi is one, and so variance is t. Var of each yi is one by computational formula of variance, E[yi^2]-E[yi]=1
I'm confused about the machine working/broken example. At 0:49:09 I believe it should be [1 0]*A^3650 = [p q]. Then for eigenvector at 1:17:40 it should be A(transpose)*[v1,v2] = [v1,v2], as you can see he modified the matrix from A to A transpose. With the way it is shown here p, q should have different meaning.
Would be a honor to be part of your class, professor. Your content is just awesome and your care with the understanding of the students can be noticed by your looks. Thank you.
And I bet if you ask them a week later what they learned, 99% will not remember a thing from the lecture. Unless this material is just review for them, math like this needs to be savored and digested for complete understanding
@@legendariersgaming Not necessarily. I suppose those who say it are probably boasting. But sometimes, it isn't difficult to get everything even I you watch in 2x
Amazing Lecture, I think at 57:56 , the equation v1 + v2 = lambda(v1+v2) only holds for lambda = 1(the only case where both v1 and v2 can be positive) , for the other eigenvalue v1+ v2 =0. This Should extend to any dimension.
I think that I don't understand the independence property of random walks, given around 21:00. His verbal explanation sounds a lot like the Markov property, but I doubt that he would define the same thing two different ways without saying that they're equivalent. Are there any systems with the independence property, but not the Markov property, or vice-versa?
Yes, I think this wouldn't have happened if he wrote down the matrix in form of the conditional probabilities, like P(w|f) before filling out A with numbers.
@@Grassmpl Here my mind is getting fucked because the probabilities across the rows are supposed to add to 1, then he switches them to columns out of nowhere and I'm like where was that even supposed to happen?
Amazing lecture. Made it A LOT easier to understand the concepts and applications. Books on subject don't usually give examples, which makes it that much harder to understand.
@48:24 "probability distribution of day 3651 and day 3650 are the same." @54:04 if av=v, day 3651=day3650, then the machine of his example last forever?
Interesting to see a proof that the simple random walk is expected to take t steps in order to move sqrt(t), which is relevant in Markov chain Monte Carlo theory.
Also, if the Riemann Hypothesis is true, then it means the variance of the number of prime numbers up to x compared to the expected number given by the Prime Number Theorem is proportional to sqrt(x), which is connected to this as well.
Lectures 4 and 22 are not available. The lectures were done by guest speakers. Most common reason they are not available: they didn't sign the IP forms giving us permission to publish their lecture. Lecture 4's topic was "Matrix Primer" and Lecture 22's topic was "Calculus of Variations and its Application in FX Execution". See the course materials at: ocw.mit.edu/18-S096F13. Best wishes on your studies!
the last corollary is neat indeed, but the assumption of the theorem seems not be fulfilled. there does not exist T>tau, since it's possible for the random walker to bump between the lines -50 and 100 as long as it likes... can sb clarify?
Here are the prerequisites for this course: 18.01 Single Variable Calculus, 18.02 Multivariable Calculus, 18.03 Differential Equations, 18.05 Introduction to Probability and Statistics or 18.440 Probability and Random Variables, 18.06 Linear Algebra. We did a quick search of our videos and maybe this video would help? ua-cam.com/video/7CYXy9J4Aao/v-deo.html See the course on MIT OpenCourseWare for more info and materials at: ocw.mit.edu/18-S096F13. Best wishes on your studies!
Can’t the derivative be taken until 0 or check the order and negativity/positivity of the function be assessed to shortcut how many options to search for in each axis?
This is a good video. just that there is a little mistake under the transition matrix. With the matrix provided, the last entry under the first column should have been P subscript 3m and not 2m.
It is a homogeneous equation. So we only look for a solution to the homogeneous equation. Let w(k) = b*w(k-1) then w(k) = b^k * w(0) (by plugging it into itself k times). Plugging this into the equation and dividing through by w(0) we get 0.5*b^2 -b - 0.5 = 0. Solving this we get only a single root b = 1. So we try a simpler form w(k) = Ck + D. Using the boundary conditions w(100) = 1 and w(-50) = 0 we get C = 1/150 and D = 1/3. The original question asked for w(0) = C*0 + D = D = 1/3. Search google for "Solving linear recurrence relations". Note that if the probability of going up is not the same as going down, we may not end up with a single root and the trial form of w(k) would not be so simple.
I think he meant the increment was stationary, but yeah I agree he was sloppy in words here, which really should not happen in an introduction course since wrong first impression is harder to be corrected.
WHY do teachers have to write the definition on the chalkboard…”Collection of random variables…” I heard that from the get go and writing it down is not going to help me (memorize it) !
B and -A are two levels of the posible balances one can have. We are measuring the probablity that, starting at a certain level k, our balance reaches B before it reaches -A. If we start at k=B, then we are already in B and so the probability that we reach B before -A is 1. If we start at k=-A, then we already lost the bet, and so the probability of reaching B before -A is 0.
***** Lecture 4 was not recorded. The topic was "Matrix Primer". See the MIT OpenCourseWare site for more course information and materials at ocw.mit.edu/18-S096F13
At 16:53, the variance is corrected to ' t ' using the central limit theorem. I did watch the last video. Still, how did this variance for X(t)'s sum go to t?
Some notable Timestamps
0:00:33 Stochastic Process
0:10:57 (Simple) Random Walk
0:32:43 Markov Chain
0:58:41 Martingale
1:06:47 Stopping time / Optional Stopping Theorem
Thanks pal!
Thank you
thank you
thanks!
Thanks!
0:00:33 Stochastic Process
0:10:57 (Simple) Random Walk
0:32:43 Markov Chain
0:58:41 Martingale
1:06:47 Stopping time / Optional Stopping Theorem
For my reference
49:03 ahh
This is a really useful comment!
The best teachers do a great job of introducing problems and then showing you the tools to solve them. With these teachers, you always know why you're doing something, you always have a sense of intuition for the problem, and you easily build a sense of experience having worked with these tools in future similar scenarios. There are so many instances wherein this professor does just that and it's a huge blessing to have access to this content for free
Recursive argument at 28:00:
Call p the probability you hit -50 first. There’s a 50% chance you hit -50 before you hit 50, by symmetry. Once you hit 50, the game is reversed, by stationary property.
Hence p = 0.5 + 0.5 * (1 - p), from which p is 1/3.
Thank you!
Ahh, yeah idk why I didn't get that the first time
p=2/3 whuch gives us the the required answer of hitting 100 first is 1/3
This guy is absolutely fantastic. Could not have been explained more clearly, with a sound logical structure. People complaining about him should probably try lecturing themselves before offering their criticism.
And people like you are confusing people even more when they get caught up in one of this guy's many mistakes and think that THEY are the ones who are wrong.
@@nickfleming3719 No offense, this is a free course for us, it's our own responsibility to find out wether the information is right or not when we get caught up in the instructors' mistakes. I mean the most important ability for self-taught learners like us is to be skepticism and check other information sources when we feel confused, not only in a free course but also in other paid courses. We can certainly say whatever we want in comments and I always learned a lot by some critical comments, however, I think it would be better to be grateful when we have chance to access high quality educations like this.
@@nickfleming3719 aaaaaaaaaaaaaaaaaaa_aaa$zzzzq xzxzzxxzxaa$zzz azxaaaa_x¢
Let’s see you lecture, I really want to see your descriptions on such topics as: Real analysis, Complex Analysis, Functional Analysis, or Harmonic Analysis; oh please it would be delightful to see such confidence coming from you.
@@maxpopkov1432 Easy game
This guy has the most elegant writing style and manner of presentation.
when you don't wanna read or write anymore but still wanna do some math, well you've got to the right place.
All you people praising this lecturer, saying how easy and simple he makes everything, are not helping.
He's making tons of mistakes, and I'm thinking I must be going crazy since everybody else seems to think this is the best lecture ever.
What mistakes?
@@faisalajin491 47:02
@@nickfleming3719 please explain further what is the mistake
@@lucasgarcia78matrix values not in the right position
49:03 ahh the "click" moment, seeing all the maths pieces coming together is really satisfying
27:00 The argument to make it work the way the intuition of the student worked is via markov chains.
Set up the states -50, 0, 50 and 100, write the transition probabilities, then calculate the absorption probabilities of the two recurrent states (-50 and 100) from 0 which give 1/4 and 1/2.
The probability to end up with $100 is the probability of ending up becomes 1/4 / (1/4 + 1/2) (since the two other states will eventually bleed into either one of these states we know their steady state probability will be 0) which indeed gives 1/3.
28:00 Following the thought process of the student from the audience, after the balance reaches $50, there is a 1/2 chance for the balance to reach $100 (overall probability = 1/4) or fall back to $0 (overall probability = 1/4). If the balance falls back to zero, we can consider that as the start of the second cycle, where the distribution of the conditional probability is the same as the first cycle (1/2 chance to reach $-50, 1/4 chance to reach $100, and 1/4 to reach $50 first then return to $0). Same for the third cycle, forth cycle, etc. Therefore, we can express the overall probability for the balance to reach $100 as the infinite series of 1/4 + (1/4)^2 + (1/4)^3... which gives us 1/3.
yes, and this is also consistent with Lee's solution, except that in the equation, one only needs to consider three (large) steps/grids, instead of a total of A+B steps/grids :)
Very easy solution for 28:00.
P(B), P(A) be probabilities that B,A occur first respectively. Probability that we hit 50$ before -50$ is 1/2 and also probability that we hit -50$ before 50$ is 1/2.
If we reach 50$ first, we see problem is flipped now, we are 50$ closer to B and -100$ closer to A. So P(B/start at 50$) = P(A/start at 0$)
So we can write
P(B) = 1/2(P(A)) = 1/2(1-P(B))
Solving this simple equation we get P(B) = 1/3
In fact for any A,B there is a point where we can flip the problem, so try to generalize this and come up with a proof.
insane lecture, tried so many different online materials, this one is clear af!
48:45 In order to predict the future using transition probability matrix A, we need to use transpose of A. (A^T^3650 * [1;0]) (this is fixed at 51:41). Since the eigenvalue of A^T is equal to A, the following theorems also hold. Hope this help. Thanks for the great lecture.
Continue the reasoning from 27:22: Assume the probability of the game ends at 100 is x. As probability of the game reaches 50 is 0.5; The probability from 50 to 100 is actually (1-x). So x=0.5*(1-x) --> x=1/3
There is a reason he teaches at MIT this guy explains things so clearly and with ease! Im in H.S and can understand this! Absolutely amazing
Stop the cap
No you don’t stop lying.😂
@@jackg2630honestly you see random variables in high school you don’t need much more than that to understand here
@@jackg2630 why wouldn't he understand this?
@@jackg2630I don’t understand your skepticism either. This lecture is meant to be more of an overview and taste of the deeper math anyhow
I mean i dont get all these praises. The guy gives an overview of the topic but not rigorously at all. This is not the level of depth I would have expected but it serves me well in my preparations. It feels like I have to dive deeper on my own to get real understanding of the topic.
It's a class for finance people. Did you expect a graduate course?
@@Nikifuj908 To me it seems it's more taylored towards Math Majors who want to specialize in quantitative finance.
@@DilanCheckerits financial mathematics so it's not clearly into depth as it would be for pure mathematics or statisticians, the application is more important here
Top universities have the best lecturers, making it easier for the students. It’s like a “poverty trap” for higher education.
Luckily the best ones (MIT, Stanford) recognize that and release things like this OCW
So you're talking about the boot theory in higher education?
This lecturer isn’t all that great, in fact.
56:13 I think the confusion here comes from the fact that for the other eigenvalue, which actually is less than 1 and greater than 0, the corresponding eigenvector will converge to the 0 vector. The “sum trick” he did earlier wouldn’t work because v_1 + v_2 = \lambda (v_1 + v_2) doesn’t imply that \lambda = 1 when both v_1 and v_2 are 0. Hope I didn’t overlook anything!
OMFG! This guy is genius in explaining and presenting concepts.
I am currently working on understanding the stochastic processes, and I am very confused by the concept of “a collection of random variables”, but the trajectory thing given by the lecturer helped me understand the concept a lot easier. For a continuous random process, if I sample at very high frequency, I will get several curves in the “x(t)-t” plain (the curve depending on the setting of the random process).
Wow ! What a clear and concise lecturer. His ability in minimizing excess data to keep to the pure path of understanding is excellent. He is a star.
58:10 Someone asked whether the algebraic manipulation led to the (seeming incorrect) conclusion that all eigenvalues lambda are 1. That was not true, since the assumption for that equation is that we are dealing with a stationary state, and therefore, the conclusion is for a stationary state, its eigenvalue must be 1, as stated by Lee.
The equation was just an eigenvalue equation for A - it didn’t assume anything about stationary state.
The correct argument, against the incorrect conclusion that all eigenvalues of A is 1, is that (v1 + v2) can be 0 and hence you can’t divide that out to conclude much about lambda. The case where you can do it turns out to be when v1 and v2 are positive - thus the theorem about the unique highest eigenvalue isn’t broken.
@@eigentejas You are correct. If one assumes a stationary state (some vector (p, q) of probability that remains unchanged by further multiplying A from the left), it simply implies the existence of an eigenvalue of 1.
In 47:42 Multiplying a 2x2 matrix with a vector (1,0) will give back the p11 and p21 which stands for working today and working tomorrow(p11) and broken today but working tomorrow(p21) not the probability working and not working.
it gives the probability of the machine working tomorrow, no matter if it's broken or not today
therefore p reflects the probability of the machine working in 10 years. however he should've multiply with a vector (1,1) to adjust the same for q, since if you multiply the matrix with (1,0) the value of q will be 0
N4mch3n there cannot be a vector (1,1) as they represent probabilties of the machine working and not working.The rows of the vector must add upto 1. With (1,1) it implies that the machine is working and not working at the same time.
You are right. The error is that the entrees which should sum up to one are the ones in ROWS not columns. Because he is not multiplying A^3650 by the correct vector, he had to amend the matrix A when computing the eigenvector in 52:00.
At 41:35, it should be P_m1 instead of P_2m.
They have the audacity to call Choongbum Lee an instructor, when he can give a presentation so complete, elegant, and accessible that he could (and maybe should) teach ALL of the other professors at MIT a thing or two about how to give a lecture.and communicate ideas throughout it.
This guy is @#$%ing amazing! What a beast.
God, I feel stupid in comparison.
What is your problem with the word "instructor"?
"How dare they call him a teacher! He is too good at teaching for that!"
@@MrCmon113 I think they mean that he should be promoted to the position of professor. Instructors are not generally permanent positions at a university.
MIT is a top research university, and as such, professors at MIT (and other research institutions) are judged mostly according to the quality of their research, not teaching.
I have seen quite a few MIT courses and every time, the teachers were amazing.
This teacher is honestly not the best, although he is very much alright.
@42:00 Isn't the transition prob matrix incorrect. Where the lower left corner should be P_{m,1} instead of P_{2,m}
Yes
The lecture is so calming.
There is a mistake at 1;15:23 . An Expectation of a random variable is a number not a random variable. So E(Xtau)=E(X0).
Day traders need courses like this, OMG!
i am a trader i am also trying to collect this kind of content for leverage trading
a completely different level can not be compared with the first lectures
At 19:12 , the probability of a N(0,1) to be between -1 and 1 is ~68%, not close to 90% or more as said. Otherwise, great lecture.
shailesh kakkar I believe he meant the probability to be within 100 standard deviations (which is virtually 100%, not close to 90% =). And there are a lot of minor mistakes in this video and the two before, the instructor is not very well prepared. But it's still useful
+Maxim Podkolzine No, the answer is right, he means that the total area under the bell curve its 1, or 100%, but in the real word, you nead just 2 standard deviations boths sides to the total area to stay very close to 100%
+serrjosl p(-1
+dicksonh Well if you do that, you miss 1/3 of the boundaries values and your forcast will be completely wrong, but Who Am I to change your point of view.😉
In the 1st and 2nd cases he's talking about delta hedge parity (in trading/market practice) as reflected by trend lines. In the 3rd case he's referring to the vol of vol, in this situation one must employee stochastic volitility models.
Don't spend your time for another channels. It is the best one!
To get variance, applied variance to both sides, var(sum(yi) over i). because yis are iid variance becomes sum(var(yi)). Var of each yi is one, and so variance is t. Var of each yi is one by computational formula of variance, E[yi^2]-E[yi]=1
1:15:16 you might want to say that E(X_\tau) = E(X0). Remember that X0 is a random variable too.
Or condition on the value of X_0
I'm confused about the machine working/broken example. At 0:49:09 I believe it should be [1 0]*A^3650 = [p q]. Then for eigenvector at 1:17:40 it should be A(transpose)*[v1,v2] = [v1,v2], as you can see he modified the matrix from A to A transpose. With the way it is shown here p, q should have different meaning.
i understand now😂 the matrix A at 0:49:09 is wrong😂
Yeah I was having this exact same confusion, what you've said seems to be perfectly right, now it all makes sense to me. Thanks a lot!
1:02:00
Randomwalk is a martingale
I don't understand how 2 and 3 are different? They seem same to me. 6:00
Uhm one is 2 paths and the other is infinite paths
Thanks for ur efforts, I was just preparing for my first class about stochastic.
Would be a honor to be part of your class, professor. Your content is just awesome and your care with the understanding of the students can be noticed by your looks. Thank you.
WORLDS BEST PROFESSOR HANDS DOWN!!!!
i think all people that writes "I changed video´s speed to 2" is trying to say: "i am more brilliant than anyone here".
And I bet if you ask them a week later what they learned, 99% will not remember a thing from the lecture. Unless this material is just review for them, math like this needs to be savored and digested for complete understanding
@@legendariersgaming Not necessarily. I suppose those who say it are probably boasting. But sometimes, it isn't difficult to get everything even I you watch in 2x
I changed video speed to x4
Bravo for the stopping time definition . Very helpful
I wish my lecturers could lecture in such a well structured way :(
This is great and simple stuff for students studying the particle theory and Brownian motion
In 1996 I took the most mathematical advanced course I have ever taken: RANDOM VIBRATIONS.
This course reminded me of that great course.
Amazing Lecture, I think at 57:56 , the equation v1 + v2 = lambda(v1+v2) only holds for lambda = 1(the only case where both v1 and v2 can be positive) , for the other eigenvalue v1+ v2 =0. This Should extend to any dimension.
Brilliant! Thank you.
This guy is amazing. His explanation is clear.
the matrix at 0:49:09 was wrong. Also, the transition matrix is (p_{1j},p_{2j}....), not (p_{k1},p_{k2},....).
I think that I don't understand the independence property of random walks, given around 21:00. His verbal explanation sounds a lot like the Markov property, but I doubt that he would define the same thing two different ways without saying that they're equivalent. Are there any systems with the independence property, but not the Markov property, or vice-versa?
Who is this guy?
His explanation on the subject is awesome
Choongbum Lee
partial differential equations pde
the best intuition behind stochastic processes !, really good
can anyone help me understand why f(k) = 0.5*f(k+1) + 0.5*f(k-1) holds?
At time 47:30, the interpretation of p,q are incorrect, unless A is transposed
Its a bit too late to flip at time 51:55. Students have already copied down incorrect notes.
Yes, I think this wouldn't have happened if he wrote down the matrix in form of the conditional probabilities, like P(w|f) before filling out A with numbers.
Can you explain that with a little bit more information?
@@Grassmpl Here my mind is getting fucked because the probabilities across the rows are supposed to add to 1, then he switches them to columns out of nowhere and I'm like where was that even supposed to happen?
you sir are a gift! Thankyou for your clear lecturing!
17:15 if the variance is t, how's std equal to square root of t. Isn't supposed to be just 1 since you'd divide variance with t first?
Amazing lecture. Made it A LOT easier to understand the concepts and applications. Books on subject don't usually give examples, which makes it that much harder to understand.
@48:24 "probability distribution of day 3651 and day 3650 are the same." @54:04 if av=v, day 3651=day3650, then the machine of his example last forever?
47:52
Shouldn't we premultiply here,.
i.e [1 0](A^3650) = [p q]
pre-multiply (with [1,0] as 1x2 vector) instead of post-multiply.
Great !! video. Always grateful for this content.
Show me the lectures of the Poisson process
Interesting to see a proof that the simple random walk is expected to take t steps in order to move sqrt(t), which is relevant in Markov chain Monte Carlo theory.
Also, if the Riemann Hypothesis is true, then it means the variance of the number of prime numbers up to x compared to the expected number given by the Prime Number Theorem is proportional to sqrt(x), which is connected to this as well.
Absolutely fantastic video, presented with such clarity. Extremely helpful. Thank you.
19:14 what's the name of that theorem?
Chanson reflection principle of Wiener process / brownian motion
and it is useful for pricing barrier options and lookback options
@23:00 how can simple random walk be stationary when variance grows with time ? Did he mean increments are stationary ?
I think it should be N(0, 1/4) at 17:13
OMG you are a genius stochastic process never looked this simple and intuitive
Where are lectures 4 and 22?
Lectures 4 and 22 are not available. The lectures were done by guest speakers. Most common reason they are not available: they didn't sign the IP forms giving us permission to publish their lecture. Lecture 4's topic was "Matrix Primer" and Lecture 22's topic was "Calculus of Variations and its Application in FX Execution". See the course materials at: ocw.mit.edu/18-S096F13. Best wishes on your studies!
the last corollary is neat indeed, but the assumption of the theorem seems not be fulfilled. there does not exist T>tau, since it's possible for the random walker to bump between the lines -50 and 100 as long as it likes... can sb clarify?
choong bum the humble korean god
what is the prerequisite for this course, does anywhere i can find a detail simplified version of all the explanation relating to this topic.
Here are the prerequisites for this course: 18.01 Single Variable Calculus, 18.02 Multivariable Calculus, 18.03 Differential Equations, 18.05 Introduction to Probability and Statistics or 18.440 Probability and Random Variables, 18.06 Linear Algebra. We did a quick search of our videos and maybe this video would help? ua-cam.com/video/7CYXy9J4Aao/v-deo.html See the course on MIT OpenCourseWare for more info and materials at: ocw.mit.edu/18-S096F13. Best wishes on your studies!
He is sooo good!
Shouldn't it be [1 0] * A^3650 = [p q] ?
I believe it has to do with the eigenvector relationship... Av = (lamba)(v)
Can’t the derivative be taken until 0 or check the order and negativity/positivity of the function be assessed to shortcut how many options to search for in each axis?
Can someone explains me whats the difference of the stochastic processes number 2 and 3 defined at minute 4:30
? Thank you so much
why did he take the transpose, at 51:50 ?
This is a good video. just that there is a little mistake under the transition matrix. With the matrix provided, the last entry under the first column should have been P subscript 3m and not 2m.
I believe it should have been m1.
The first time I see a teacher who rewrites everything
32:30... that f(k) formula thing...how would that be solved?
It is a homogeneous equation. So we only look for a solution to the homogeneous equation. Let w(k) = b*w(k-1) then w(k) = b^k * w(0) (by plugging it into itself k times). Plugging this into the equation and dividing through by w(0) we get 0.5*b^2 -b - 0.5 = 0. Solving this we get only a single root b = 1. So we try a simpler form w(k) = Ck + D. Using the boundary conditions w(100) = 1 and w(-50) = 0 we get C = 1/150 and D = 1/3. The original question asked for w(0) = C*0 + D = D = 1/3. Search google for "Solving linear recurrence relations". Note that if the probability of going up is not the same as going down, we may not end up with a single root and the trial form of w(k) would not be so simple.
My cities transit system is stochastic- the busses arrive at random stops at random times smh
Is this course part of masters program in Statistics?
I'm not sure, over here in the UK Stochastic Processes is part of an undergraduate degree, but this might be more advanced!
this is some darn good stochastic processin', i tell u wat
Can technically everything be a markov chain if the history is included in the current state?
great job
Thanks a lot. Very clear explanation.
16:53, how does the variance equal t?
18:40 Mark
Awesome lecture. Just found out he went to the same college for his undergrad as me
Around 32 mins time mark, why f(B) = 1 and f(-A) = 0?
Thanks for the help.
Isn’t discrete the same as continuous, at the limit?
Great lecture. Learned a lot.
But random walk is not a stationary process...
I think in his definition it is stationary.
Random walk with or without drift is not a stationary process.
I think he meant the increment was stationary, but yeah I agree he was sloppy in words here, which really should not happen in an introduction course since wrong first impression is harder to be corrected.
49:10 blew my mind!
WHY do teachers have to write the definition on the chalkboard…”Collection of random variables…” I heard that from the get go and writing it down is not going to help me (memorize it) !
very good presentation, enjoyed it!
@32:25 why f(b)=1,f(-A)=0?
Please can someone help
B and -A are two levels of the posible balances one can have. We are measuring the probablity that, starting at a certain level k, our balance reaches B before it reaches -A. If we start at k=B, then we are already in B and so the probability that we reach B before -A is 1. If we start at k=-A, then we already lost the bet, and so the probability of reaching B before -A is 0.
What was discussed in lecture 4? It goes directly from lecture 3 to lecture 5
***** Lecture 4 was not recorded. The topic was "Matrix Primer". See the MIT
OpenCourseWare site for more course information and materials at ocw.mit.edu/18-S096F13
i dont understand the difference between 2 and 3 at 4:39
At 16:53, the variance is corrected to ' t ' using the central limit theorem. I did watch the last video. Still, how did this variance for X(t)'s sum go to t?
Could anyone explain this? thanks!
@@finalpurez hey
So the way BM is defined is drift times dt plus sigma times sqrt dt. So the Var is sqrt dt squared which makes it T
@@ruchirsharma6415 sorry, are u able to explain it via the variance formula?
@@finalpurezmake a table of t vs outcome and sum. Calculate variance for each t till atleast t=4 and observe the pattern