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
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
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.
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
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.
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
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 :)
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).
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 $100 is the probability of ending up becomes 1/4 / (1/4 + 1/2) (since the two other states will eventually bleed into either one of these states we know their steady state probability will be 0) which indeed gives 1/3.
48:45 In order to predict the future using transition probability matrix A, we need to use transpose of A. (A^T^3650 * [1;0]) (this is fixed at 51:41). Since the eigenvalue of A^T is equal to A, the following theorems also hold. Hope this help. Thanks for the great lecture.
They have the audacity to call Choongbum Lee an instructor, when he can give a presentation so complete, elegant, and accessible that he could (and maybe should) teach ALL of the other professors at MIT a thing or two about how to give a lecture.and communicate ideas throughout it. This guy is @#$%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.
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
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.
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.
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!
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.
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.
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
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.
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
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.
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.😉
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.
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!
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.
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.
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?
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.
@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?
A question about Markov Chains: Would I be right to say that Perron-Frobenius guarantees that 1 will be an eigenvalue of a positive square matrix, with a positive eigenvector? And the fact that 1 is an eigenvalue guarantees that the probability distribution will converge to the eigenvector?
[00:00]([00:00](ruclips.net/video/TuTmC8aOQJE/видео.html)) Stochastic processes are collections of random variables indexed by time. - Discrete-time stochastic processes have random variables that can be discrete or continuous. - Continuous-time stochastic processes have random variables that can jump and be continuous. [03:31]([03:31](ruclips.net/video/TuTmC8aOQJE/видео.html)) Stochastic processes are random processes that can be described mathematically. - There are different ways to describe a stochastic process. - Stochastic processes can have different levels of randomness and dependency. [09:06]([09:06](ruclips.net/video/TuTmC8aOQJE/видео.html)) Stochastic processes involve predicting future stock prices and understanding the long-term behavior of sequences. - Stochastic processes help answer questions about the occurrence of boundary events, such as extreme stock price drops or the probability of idle phones in a call center. - The focus of this video is on discrete-time stochastic processes, with continuous-time processes covered later in the course. [12:37]([12:37](ruclips.net/video/TuTmC8aOQJE/видео.html)) A one-dimensional simple random walk is a sequence of random variables that can either go up or go down at each time step. - Plotting the values of X_t over time on a line forms a random trajectory. - Over a long period of time, the distribution of X_t converges to 0. [18:08]([18:08](ruclips.net/video/TuTmC8aOQJE/видео.html)) The simple random walk stays close to the curves square root of t and minus square root of t. - Even though theoretically it can take extreme values of t and -t, in reality it stays close to the curves. - The simple random walk will mostly stay within the area between the two curves. [20:36]([20:36](ruclips.net/video/TuTmC8aOQJE/видео.html)) Stochastic processes with independent increment and stationary properties - The difference between random variables at different times is mutually independent - Same amount of time intervals have the same distribution, and non-overlapping intervals are independent [27:19]([27:19](ruclips.net/video/TuTmC8aOQJE/видео.html)) The probability of hitting line B first is A / (A + B). - The probability of hitting line A first is B / (A + B). - This can be proven by fixing B and A, and for each k between -A and B, we define f(k) as the probability of hitting line A first when starting at k. [30:15]([30:15](ruclips.net/video/TuTmC8aOQJE/видео.html)) Simple random walk is a powerful and easy-to-control stochastic process - The probability of hitting a specific point is determined by recursive formulas - Simple random walk can be approximated in many cases and computed by hand [35:47]([35:47](ruclips.net/video/TuTmC8aOQJE/видео.html)) A Markov chain is a discrete-time stochastic process where the probability of the next value depends only on the current value - The probability of X at t+1 is equal to a specific value, given the history up to time t - The probability of X at t+1 is the same as the value at time t+1, given only the last value [40:03]([40:03](ruclips.net/video/TuTmC8aOQJE/видео.html)) The transition probability matrix is crucial and contains all the information about the Markov chain. - The transition probability matrix sums up to 1 over all possible states. - The probability of a future state depends only on the current state. [46:18]([46:18](ruclips.net/video/TuTmC8aOQJE/видео.html)) The probability distribution after a long time can be approximated using matrix A to the power of 3651. - The example involves two states, working and broken. The transition probabilities between the states are given by the matrix A. - After 3650 days, the probability distribution on that day can be represented by A to the power of 3650 multiplied by [1, 0]. - The values of p and q represent the probabilities of the system being in the working or broken state respectively after a long period of time. [49:01]([49:01](ruclips.net/video/TuTmC8aOQJE/видео.html)) The matrix has a dominant eigenvalue with a positive eigenvector - The matrix [p, q] represents the probability distribution over time - The Perron-Frobenius theorem guarantees the existence of a dominant eigenvalue and positive eigenvector [55:15]([55:15](ruclips.net/video/TuTmC8aOQJE/видео.html)) A stationary distribution exists and is the same as the initial distribution. - The Perron-Frobenius theorem states that there is exactly one eigenvector corresponding to the largest eigenvalue, which is 1. - If all the entries in the transition probability matrix are positive, there will be a unique stationary distribution. [58:09]([58:09](ruclips.net/video/TuTmC8aOQJE/видео.html)) A stochastic process is a martingale if the expectation of the next value equals the current value. - A martingale models a fair game in which the expected value remains unchanged over time. - The random walk is an example of a martingale. [1:05:48]([1:05:48](ruclips.net/video/TuTmC8aOQJE/видео.html)) A martingale is a fair game where the expected value is fixed. - A random walk is an example of both a Markov chain and a martingale, but they are different concepts that should not be confused. - There are Markov chains that are not martingales and martingales that are not Markov chains. - A martingale game ensures that your expected value cannot be positive or negative, regardless of the strategy you use. - In a martingale game, your expected value is fixed and you are not supposed to win or lose, at least in expectation. - A stopping time in a stochastic process is a non-negative integer-valued random variable. [1:09:04]([1:09:04](ruclips.net/video/TuTmC8aOQJE/видео.html)) A stopping time is a non-negative integer valued random variable that represents the time to stop in a strategy based on the values of the stochastic process up to a certain point. - The decision to stop depends only on the events up to a specific time index. - Strategies that depend on future values are not considered stopping times. [1:14:49]([1:14:49](ruclips.net/video/TuTmC8aOQJE/видео.html)) No matter what strategy you use, if you're a mortal being, then you cannot win. - The expectation of your value at the stopping time is always equal to the balance at the beginning. - The corollary of this theorem is that the expectation of X_at_tau is equal to 0. [1:17:16]([1:17:16](ruclips.net/video/TuTmC8aOQJE/видео.html)) Understanding the concept of martingales and how they relate to winning. - The content of this video is interesting and thought-provoking. - The application of martingales in modeling and the implication of not being able to win if it fits the mathematical formulation of a martingale.
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?
***** 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
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
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!
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
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!
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.
49:03 ahh the "click" moment, seeing all the maths pieces coming together is really satisfying
insane lecture, tried so many different online materials, this one is clear af!
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
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
OMFG! This guy is genius in explaining and presenting concepts.
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
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.
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.
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 :)
Day traders need courses like this, OMG!
i am a trader i am also trying to collect this kind of content for leverage trading
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).
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 $100 is the probability of ending up becomes 1/4 / (1/4 + 1/2) (since the two other states will eventually bleed into either one of these states we know their steady state probability will be 0) which indeed gives 1/3.
48:45 In order to predict the future using transition probability matrix A, we need to use transpose of A. (A^T^3650 * [1;0]) (this is fixed at 51:41). Since the eigenvalue of A^T is equal to A, the following theorems also hold. Hope this help. Thanks for the great lecture.
i got confused there, and this really helped me !
They have the audacity to call Choongbum Lee an instructor, when he can give a presentation so complete, elegant, and accessible that he could (and maybe should) teach ALL of the other professors at MIT a thing or two about how to give a lecture.and communicate ideas throughout it.
This guy is @#$%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.
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
The lecture is so calming.
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.
Don't spend your time for another channels. It is the best one!
Thanks for ur efforts, I was just preparing for my first class about stochastic.
Bravo for the stopping time definition . Very helpful
WORLDS BEST PROFESSOR HANDS DOWN!!!!
a completely different level can not be compared with the first lectures
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.
This guy is amazing. His explanation is clear.
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!
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.
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.
I wish my lecturers could lecture in such a well structured way :(
you sir are a gift! Thankyou for your clear lecturing!
He is sooo good!
Who is this guy?
His explanation on the subject is awesome
Choongbum Lee
the best intuition behind stochastic processes !, really good
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.
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
At 41:35, it should be P_m1 instead of P_2m.
Great !! video. Always grateful for this content.
1:02:00
Randomwalk is a martingale
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.
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
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.
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.😉
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).
@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
Absolutely fantastic video, presented with such clarity. Extremely helpful. Thank you.
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
great job
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
This is next level.
simply amazing
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.
OMG you are a genius stochastic process never looked this simple and intuitive
Thanks a lot. Very clear explanation.
WOW... THANKS FOR THIS....
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!
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.
Great lecture. Learned a lot.
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!
choong bum the humble korean god
The first time I see a teacher who rewrites everything
it's a shame for MIT to have such a teacher.
the matrix at 0:49:09 was wrong. Also, the transition matrix is (p_{1j},p_{2j}....), not (p_{k1},p_{k2},....).
this guy is a genius
Very good explanation.
this is some darn good stochastic processin', i tell u wat
Good presentation
very good presentation, enjoyed it!
49:10 blew my mind!
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?
This is Dr Ahmed i am referring your lecture to my students
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.
Show me the lectures of the Poisson process
Thanks a lot!!! Very good teacher :)
partial differential equations pde
Very interesting
Awesome lecture. Just found out he went to the same college for his undergrad as me
@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?
I think it should be N(0, 1/4) at 17:13
A question about Markov Chains: Would I be right to say that Perron-Frobenius guarantees that 1 will be an eigenvalue of a positive square matrix, with a positive eigenvector? And the fact that 1 is an eigenvalue guarantees that the probability distribution will converge to the eigenvector?
such a teacher can work only in a technical school
is this a compliment?
thank so much for MIT...it very helpful for my short time study at University.
Thanks man
[00:00]([00:00](ruclips.net/video/TuTmC8aOQJE/видео.html)) Stochastic processes are collections of random variables indexed by time.
- Discrete-time stochastic processes have random variables that can be discrete or continuous.
- Continuous-time stochastic processes have random variables that can jump and be continuous.
[03:31]([03:31](ruclips.net/video/TuTmC8aOQJE/видео.html)) Stochastic processes are random processes that can be described mathematically.
- There are different ways to describe a stochastic process.
- Stochastic processes can have different levels of randomness and dependency.
[09:06]([09:06](ruclips.net/video/TuTmC8aOQJE/видео.html)) Stochastic processes involve predicting future stock prices and understanding the long-term behavior of sequences.
- Stochastic processes help answer questions about the occurrence of boundary events, such as extreme stock price drops or the probability of idle phones in a call center.
- The focus of this video is on discrete-time stochastic processes, with continuous-time processes covered later in the course.
[12:37]([12:37](ruclips.net/video/TuTmC8aOQJE/видео.html)) A one-dimensional simple random walk is a sequence of random variables that can either go up or go down at each time step.
- Plotting the values of X_t over time on a line forms a random trajectory.
- Over a long period of time, the distribution of X_t converges to 0.
[18:08]([18:08](ruclips.net/video/TuTmC8aOQJE/видео.html)) The simple random walk stays close to the curves square root of t and minus square root of t.
- Even though theoretically it can take extreme values of t and -t, in reality it stays close to the curves.
- The simple random walk will mostly stay within the area between the two curves.
[20:36]([20:36](ruclips.net/video/TuTmC8aOQJE/видео.html)) Stochastic processes with independent increment and stationary properties
- The difference between random variables at different times is mutually independent
- Same amount of time intervals have the same distribution, and non-overlapping intervals are independent
[27:19]([27:19](ruclips.net/video/TuTmC8aOQJE/видео.html)) The probability of hitting line B first is A / (A + B).
- The probability of hitting line A first is B / (A + B).
- This can be proven by fixing B and A, and for each k between -A and B, we define f(k) as the probability of hitting line A first when starting at k.
[30:15]([30:15](ruclips.net/video/TuTmC8aOQJE/видео.html)) Simple random walk is a powerful and easy-to-control stochastic process
- The probability of hitting a specific point is determined by recursive formulas
- Simple random walk can be approximated in many cases and computed by hand
[35:47]([35:47](ruclips.net/video/TuTmC8aOQJE/видео.html)) A Markov chain is a discrete-time stochastic process where the probability of the next value depends only on the current value
- The probability of X at t+1 is equal to a specific value, given the history up to time t
- The probability of X at t+1 is the same as the value at time t+1, given only the last value
[40:03]([40:03](ruclips.net/video/TuTmC8aOQJE/видео.html)) The transition probability matrix is crucial and contains all the information about the Markov chain.
- The transition probability matrix sums up to 1 over all possible states.
- The probability of a future state depends only on the current state.
[46:18]([46:18](ruclips.net/video/TuTmC8aOQJE/видео.html)) The probability distribution after a long time can be approximated using matrix A to the power of 3651.
- The example involves two states, working and broken. The transition probabilities between the states are given by the matrix A.
- After 3650 days, the probability distribution on that day can be represented by A to the power of 3650 multiplied by [1, 0].
- The values of p and q represent the probabilities of the system being in the working or broken state respectively after a long period of time.
[49:01]([49:01](ruclips.net/video/TuTmC8aOQJE/видео.html)) The matrix has a dominant eigenvalue with a positive eigenvector
- The matrix [p, q] represents the probability distribution over time
- The Perron-Frobenius theorem guarantees the existence of a dominant eigenvalue and positive eigenvector
[55:15]([55:15](ruclips.net/video/TuTmC8aOQJE/видео.html)) A stationary distribution exists and is the same as the initial distribution.
- The Perron-Frobenius theorem states that there is exactly one eigenvector corresponding to the largest eigenvalue, which is 1.
- If all the entries in the transition probability matrix are positive, there will be a unique stationary distribution.
[58:09]([58:09](ruclips.net/video/TuTmC8aOQJE/видео.html)) A stochastic process is a martingale if the expectation of the next value equals the current value.
- A martingale models a fair game in which the expected value remains unchanged over time.
- The random walk is an example of a martingale.
[1:05:48]([1:05:48](ruclips.net/video/TuTmC8aOQJE/видео.html)) A martingale is a fair game where the expected value is fixed.
- A random walk is an example of both a Markov chain and a martingale, but they are different concepts that should not be confused.
- There are Markov chains that are not martingales and martingales that are not Markov chains.
- A martingale game ensures that your expected value cannot be positive or negative, regardless of the strategy you use.
- In a martingale game, your expected value is fixed and you are not supposed to win or lose, at least in expectation.
- A stopping time in a stochastic process is a non-negative integer-valued random variable.
[1:09:04]([1:09:04](ruclips.net/video/TuTmC8aOQJE/видео.html)) A stopping time is a non-negative integer valued random variable that represents the time to stop in a strategy based on the values of the stochastic process up to a certain point.
- The decision to stop depends only on the events up to a specific time index.
- Strategies that depend on future values are not considered stopping times.
[1:14:49]([1:14:49](ruclips.net/video/TuTmC8aOQJE/видео.html)) No matter what strategy you use, if you're a mortal being, then you cannot win.
- The expectation of your value at the stopping time is always equal to the balance at the beginning.
- The corollary of this theorem is that the expectation of X_at_tau is equal to 0.
[1:17:16]([1:17:16](ruclips.net/video/TuTmC8aOQJE/видео.html)) Understanding the concept of martingales and how they relate to winning.
- The content of this video is interesting and thought-provoking.
- The application of martingales in modeling and the implication of not being able to win if it fits the mathematical formulation of a martingale.
Very clear!
a layman explanation for the last theorem: if you're trying play a fair game, don't even bother to start. it's just a waste of your time
That's not true, if you are risk seeking, then a fair game is free real estate for you!!
Thanks
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?
18:40 Mark
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
The teacher rewrites the material from the sheet that he took
確率方程式＝Stochastic Processes I
This guy is the best; makerere shd employ him
ank you for lecture
My cities transit system is stochastic- the busses arrive at random stops at random times smh