5. Stochastic Processes I

MIT 18.S096 Topics in Mathematics with Applications in Finance, Fall 2013
Instructor: Choongbum Lee

*NOTE: Lecture 4 was not recorded.
This lecture introduces stochastic processes, including random walks and Markov chains.

License: Creative Commons BY-NC-SA

[lucastrojanowski]

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

[SeikoVanPaath]

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

[sahilsood]

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

[edwardantonian]

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.

[mattiascardecchia]

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.

[bigollameo]

This guy has the most elegant writing style and manner of presentation.

[takashikashiwase]

when you don't wanna read or write anymore but still wanna do some math, well you've got to the right place.

[AE-cjch]

Top universities have the best lecturers, making it easier for the students. It’s like a “poverty trap” for higher education.

[frasersmall]

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

[Boringpenguin]

49:03 ahh the "click" moment, seeing all the maths pieces coming together is really satisfying

[michaelcheng]

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.

[jerryzhang]

insane lecture, tried so many different online materials, this one is clear af!

[nickfleming]

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.

[DilanChecker]

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.

[ApiolJoe]

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.

[samgao]

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).

[bigollameo]

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.

[Grey_]

OMFG! This guy is genius in explaining and presenting concepts.

[vijayk]

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.

[phillipthompson]

This is great and simple stuff for students studying the particle theory and Brownian motion