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
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
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.
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.
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 $100 as the infinite series of 1/4 + (1/4)^2 + (1/4)^3... which gives us 1/3.
yes, and this is also consistent with Lee's solution, except that in the equation, one only needs to consider three (large) steps/grids, instead of a total of A+B steps/grids :)
I am currently working on understanding the stochastic processes, and I am very confused by the concept of “a collection of random variables”, but the trajectory thing given by the lecturer helped me understand the concept a lot easier. For a continuous random process, if I sample at very high frequency, I will get several curves in the “x(t)-t” plain (the curve depending on the setting of the random process).
Very easy solution for 28:00. P(B), P(A) be probabilities that B,A occur first respectively. Probability that we hit 50$ 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.
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
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.
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.
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.
48:45 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.
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
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 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.
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.
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
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.
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.
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'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.
Exercises 4.1. Use Itˆo’s formula to write the following stochastic processes Xt on the standard form dXt = u(t, ω)dt + v(t, ω)dBt for suitable choices of u ∈ Rn, v ∈ Rn×m and dimensions n, m: a) Xt = B2 t , where Bt is 1-dimensional b) Xt = 2 + t + e Bt (Bt is 1-dimensional) c) Xt = B2 1 (t) + B2 2 (t) where (B1, B2) is 2-dimensional d) Xt = (t0 + t, Bt) (Bt is 1-dimensional) e) Xt = (B1(t)+B2(t)+B3(t), B2 2 (t)−B1(t)B3(t)), where (B1, B2, B3) is 3-dimensional
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?
this video really puts in perspective how much shittier my actual professor is in comparison... Not to be rude or anything but geesus this lecture is on another level completely.
The part he couldn't explain at all is that in the matrix product, you need not only to get aXb, it implies multiply and sum. The outcome allways be one that is the result of the Perron- Frobenius theorem.
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!
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?
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.
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!
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
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.
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
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
when you don't wanna read or write anymore but still wanna do some math, well you've got to the right place.
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?
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?
insane lecture, tried so many different online materials, this one is clear af!
49:03 ahh the "click" moment, seeing all the maths pieces coming together is really satisfying
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.
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
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 $100 as the infinite series of 1/4 + (1/4)^2 + (1/4)^3... which gives us 1/3.
yes, and this is also consistent with Lee's solution, except that in the equation, one only needs to consider three (large) steps/grids, instead of a total of A+B steps/grids :)
Day traders need courses like this, OMG!
i am a trader i am also trying to collect this kind of content for leverage trading
I am currently working on understanding the stochastic processes, and I am very confused by the concept of “a collection of random variables”, but the trajectory thing given by the lecturer helped me understand the concept a lot easier. For a continuous random process, if I sample at very high frequency, I will get several curves in the “x(t)-t” plain (the curve depending on the setting of the random process).
Very easy solution for 28:00.
P(B), P(A) be probabilities that B,A occur first respectively. Probability that we hit 50$ 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.
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.
Don't spend your time for another channels. It is the best one!
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
Thanks for ur efforts, I was just preparing for my first class about stochastic.
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.
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.
The lecture is so calming.
WORLDS BEST PROFESSOR HANDS DOWN!!!!
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.
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.
Bravo for the stopping time definition . Very helpful
a completely different level can not be compared with the first lectures
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.
This guy is amazing. His explanation is clear.
I wish my lecturers could lecture in such a well structured way :(
the best intuition behind stochastic processes !, really good
OMG you are a genius stochastic process never looked this simple and intuitive
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
Absolutely fantastic video, presented with such clarity. Extremely helpful. Thank you.
Who is this guy?
His explanation on the subject is awesome
Choongbum Lee
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 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.
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.
At 41:35, it should be P_m1 instead of P_2m.
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
you sir are a gift! Thankyou for your clear lecturing!
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
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.
choong bum the humble korean god
@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
Awesome lecture. Just found out he went to the same college for his undergrad as me
it's a shame for MIT to have such a teacher.
this is some darn good stochastic processin', i tell u wat
thank so much for MIT...it very helpful for my short time study at University.
this guy is a genius
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).
This is Dr Ahmed i am referring your lecture to my students
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!!
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.😉
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.
He is sooo good!
This is next level.
1:02:00
Randomwalk is a martingale
Thanks a lot. Very clear explanation.
such a teacher can work only in a technical school
is this a compliment?
Great !! video. Always grateful for this content.
Great lecture. Learned a lot.
simply amazing
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.
WOW... THANKS FOR THIS....
The first time I see a teacher who rewrites everything
Very good explanation.
great job
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!
Exercises
4.1. Use Itˆo’s formula to write the following stochastic processes Xt on the
standard form
dXt = u(t, ω)dt + v(t, ω)dBt
for suitable choices of u ∈ Rn, v ∈ Rn×m and dimensions n, m:
a) Xt = B2
t
, where Bt is 1-dimensional
b) Xt = 2 + t + e
Bt (Bt is 1-dimensional)
c) Xt = B2
1
(t) + B2
2
(t) where (B1, B2) is 2-dimensional
d) Xt = (t0 + t, Bt) (Bt is 1-dimensional)
e) Xt = (B1(t)+B2(t)+B3(t), B2
2
(t)−B1(t)B3(t)), where (B1, B2, B3)
is 3-dimensional
can anyone help me understand why f(k) = 0.5*f(k+1) + 0.5*f(k-1) holds?
Stochastic Processes Point Processes Total = P
Good presentation
very good presentation, enjoyed it!
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
確率方程式＝Stochastic Processes I
I think instead of multiplying (A^3560)[1;0] he should do [1 0] * (A ^ 3650)
I believe it has to do with the eigenvector relationship... Av = (lamba)(v)
@@kaydenwoodsmusic I'll have to rewatch the video.. :-D
This guy is the best; makerere shd employ him
I think it should be N(0, 1/4) at 17:13
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?
Show me the lectures of the Poisson process
Shouldn't it be [1 0] * A^3650 = [p q] ?
I believe it has to do with the eigenvector relationship... Av = (lamba)(v)
49:10 blew my mind!
this video really puts in perspective how much shittier my actual professor is in comparison...
Not to be rude or anything but geesus this lecture is on another level completely.
Bruno Borgatti it’s pronounced Jesus js
the matrix at 0:49:09 was wrong. Also, the transition matrix is (p_{1j},p_{2j}....), not (p_{k1},p_{k2},....).
Thanks a lot!!! Very good teacher :)
The part he couldn't explain at all is that in the matrix product, you need not only to get aXb, it implies multiply and sum. The outcome allways be one that is the result of the Perron- Frobenius theorem.
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!
stopping time 개념이 헷갈렸었는데, 정말 직관적으로 이해가 가네요. 감사합니다!
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
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?
Reasonable course = not stupid copyright terror.
Very interesting
درس رائع جدا
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.