17. Stochastic Processes II
ฝัง
- เผยแพร่เมื่อ 5 ม.ค. 2015
- MIT 18.S096 Topics in Mathematics with Applications in Finance, Fall 2013
View the complete course: ocw.mit.edu/18-S096F13
Instructor: Choongbum Lee
This lecture covers stochastic processes, including continuous-time stochastic processes and standard Brownian motion.
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
University professors, watch and learn. This is how it should be taught!! Spent a week going through my lecture notes and several book and all it took is this video to make it all stick. Thank you Choongbum Lee and thank you MIT.
RamoSFTT feeling the same!
Great lecture. When I was in Japan, one of my Japanese professors is also called Ito, and later in his fin. derivatives class, I learned that Ito's Lemma's Kiyoshi Ito was his grandpa.
respect
That's actually pretty cool! :)
Nice
Clearly the best lecturer in this course
"It looks right but Its wrong for reasons you dont know yet" - Choongbum Lee
Gonna drop this in my next argument :D
When you're scrolling through the comments on a a random course and you see math guy from wits 10+ years ago :D
@@matthewholmes2008 Lol dude what were the chances... To be fair it is a maths video :D
"It looks right but Its wrong for reasons you dont know yet".
That statement is the description of the current state of the worldwide economy.
Choongbum is a legend. The things he is making crystal clear to understand in 1 hour would take you days going over the textbooks and studying the derivations
Some notable Timestamps:
0:00:24 Recap (Lecture 5)
0:02:14 Continuous-time stochastic process
0:06:12 (Standard) Brownian Motion
0:48:08 Quadratic variation theorem
🎉🎉🎉🎉🎉😊
you are a legend for giving a very accurate timestamp for all the videos, thanks mate👍
It's really great when you can see that somebody who clearly knows what they are talking about still takes care to present his knowledge with humility and openness. Great lecturers know enough about their subjects to realize how much they don't know.
And they respect the results for the hard work they were to derive!
00:00 Stochastic processes and continuous time
08:48 Brownian motion is a continuous probability distribution
16:49 Brownian motion is the limit of simple random walks.
24:26 Brownian motion is a key model used in financial markets and other fields.
31:45 The probability that the maximum stock price is greater than a at time t is equal to 2 times the probability that the Brownian motion is greater than a.
41:21 Brownian motion has certain properties
51:11 Brownian motion has quadratic variation equal to zero
59:12 Brownian motion and Ito's calculus are used to model stock prices and estimate infinitesimal differences.
1:07:30 Ito's lemma is a highly nontrivial result that allows for calculus with Brownian motion.
1:15:38 Calculus using Brownian motion is complicated
What a profoundly simple explanation of that dB^2=dt result.
Wow! What a delightfully clear explanation of Ito's lemma! I love the way Lee steps through his explanations - at each point showing the equation, then illustrating it geometrically, then discussing it in plain English, and then discussing its broader significance. He gives three different ways to understand each step, then ties it all together into the broader mathematical context. Lee is an amazingly gifted teacher!
Dr. Lee is amazing. This is literally the best explanation I've had about Ito's Lemma.
hes a quant at millennium now
I really like Choonbum's teaching style of how he can break the topic and describe the incentives of each theory. And it is really easier for me to follow as well when professors can write them down on blackboards. I agree now that one can teach math only on blackboard.
Really enjoyed the lecture. I Made like a bandit sitting on my bed in Jakarta attending this beautiful class in mit Boston many years ago. Thanks ocw and prof choongbum lee
Thanks so much, Choongbum.
Beautiful lecture that delivers important contents, which could have been confusing, in a concise manner. Appreciate your lecture.
Absolutely the best teacher I have met so far.
An extraordinary explanation of the fundamentals.
A good lecture that clearly explains the motivation. Even good for engineering people.
its even much more useful to study with these few videos about stochastic process than to actually attend lecture at a current university
I hope MIT will upload more lessons from Dr Lee. Very pleasant to listen, you can tell he's a humble and knowledgeable person.
I found the proof of the indifferentiablity of Brownian motion not very rigorous. As the differntiability is defined as: for any given \epsilon > 0, there exists \delta >0, s.t. | [B(t+\delta) - B(t)]/\delta - A | < \epsilon. The inequality derived from the hypothesis that B(t) is differentiable should be | B(t+\delta) - B(t)| < \delta*A + \epsilon. Also notice here I followed the traditional Epsilon-Delta definition here, which means the \epsilon used in the video should be the \delta I used here. Although the instructor's proof captured the intuition and the extra \epsilon term here should not effect the conclusion since \epsilon can be any given fixed positive value, which can be chosen as close to one as possible (I suppose that is what he meant by 'almost smaller or equal to'), I still feel like pointing this out in case anyone has a similar concern, like me.
Clear and Amazing intuition I got from the stochastic process. Appreciate
a combination of clarity and simplicity
23:26
I think 'motion of pollen' is better expression than 'action of pollen'.
In physics, 'action' is a terminology regarding a fundamental dynamic symmetry.
The explanation of Quadratic Variations is great
Thank you very much, great course, great explanations!
very clear explanation! so glad this is for free.
These Lectures are more attractive for those who are working in stochastic analysis.
way too good explained, thanks
This is great! very intuitive and clear.
Excellent lecturer. Smalle quibble with the explanation of Brownian motion as the limit of a simple random walk as the step-size goes to zero: you also need to scale the possible values the process can take at each step in order to recover Brownian motion in this way.
This was a very interesting lecture, thank you!
This was a great lecture on Brownian motion! Choongbum is truly really talented as a lecturer. It's too bad that he's no longer in academia.
What does he do now?
@@Anyone.c now millennial portfolio manager
This prof is a legend!
I was so messed up and had no idea what the hell my professor was talking about until I watched this video. Great lecture indeed!
Great Professor! Even his voice is confortable to be listened to.
geiiiii
AMAZING lecture! Thaks
The more I watch this series of videos on Stochastic processes, the more I realize that while they are good and give some interesting results, the sloppiness of the presentation also shows more and more. Another example is the argument made in 58:35. I thought the presenter should have made a comment about the standard deviation of the Y_i, not just the expectation of the Y_i
Thank you MIT
In the calculations shown at 39:06, why does P(Tau < t) = P(B(t) - B(Tau)>0 | Tau
really great lecture
Helpful lecture. Thank you
Great lecture
Gracias MIT ❤️
can anyone tell what will be the quadratic variation of an absolute value.or how can we apply ito's formula on skew brownian motion
very very nice lecture
38:32 Just a side note, Dr.Lee's argument is actually hinting on the strong Markov property of Wiener processes.
Thank you very much
thank you so much
excellent
Amazing lecture!
Could someone clarify my understanding. @11:48, the second point(stationary) in the theorem. Does it mean that if we take many multiple interval from one brownian motion path ~N(0,t-s) or from multiple brownian paths , same interval is taken then it has distribution ~N(0, t-s). → I think it is the latter since first proposition does not make sense, if multiple intervals are taken how to find variance?
Probability of "1" but no longer isolated intervals if the system is mapped in a finite four-dimensional tesseract where a Brownian particle moves in and out of 3 dimensional space
3:30, the stock price can't be negative, so how can it be modeled by brownian motion (or continuous unbiased random walk)?
The mathematical expectation of the product of independent random variables is equal to the product of their mathematical expectations
Oh my god. Math becomes beautiful again to me!!! First time to know that Ito lemma comes from tayler expansion after learning ito lemma for a year from my school. Thanks MIT!!!!! I should have accepted your admission. Regret.
Tank you so much!!
Around 16:06. I think it should be Z(t/n) = Y_t / (sqrt(n)). Because if Z is a standard Brownian motion, then Z(1) ~ N(0,1). if Z(t/n) = Y_t, just as what the lecturer told, then Z(1)=Z(n/n)=Y_n ~ N(0,n), not N(0,1). Is it right?
Thanks
53.15 min...Quadratic variation square is always possitive..and we are summing to infinity...even if its very less initially as we add up more and more it shud be positive number right? How it will become zero for normal function? It looks like counter intuitive
@37:07 I'm slightly confused. The line he writes down then, the third line, is where I don't fully understand. I'ts a simple proof but for some reason I'm struggling here. The third line says; "probability of the process ending up above "a", at time "t", given it hits "a" before time "t"....+ the same for ending up below "a" at time "t"... basically.. if we've already hit "a" before time "t" (given that Tau
Never mind :) I see he fixes it afterwards :)
I believe the 3rd name of Brownian motion (aka Wiener process) is Levy process named after a French mathematician Paul Levy
Wondering if brownian motion can be controlled with force fields 🤔
19:44 uhhhm, pollen particles are very tiny and generally float on water. And Brownian motion is best observed at the surface of water under a microscope, where gravity is canceled out by surface tension. However, he's only considering the motion parallel to the surface of the water, so... hopefully someone has informed him of what pollen particles are by now.
this is n times better than my lecturer, as n -> \infty
Around 1:07, if Bt was differentiable, why df=(dBt/dt)dt? Doesn't it become df = dBt? I am confused about this part, can anyone explain?
Why are the 3rd order and so on terms ignored in Ito's lemma?
Emmm I got a question for the proof around 39min. What if price goes down after tau_a less that t. Then P(M(t) > a) is zero and P(tau_a < t)is 1. These two should not be equal?
If the price goes down after \tau_a and never manage to climb back before t (actually, it doesn't really matter whether the price crosses the line y=a again, since you can always find a symmetric path that ends on the other side of y=a, as long as B(t) < a, we are good), that would be the case covered by the second part on the RHS, which is P(B(t)
I think there is a "typo" in defining τ_a. But everything will go through if we define τ_a = inf {s ≥ 0 | B(s)>a}.
In the proof of the property at 39:00 he uses conditional probability P(B(t)-B(tau_a)>0 | tau_a
You're right.
53:11 where did we use the continouity of the derivative ?
Is it when we use the fact ,that a continous function is locally bounded ?
By use of mean value theorem ?
Jude is right, MVT requires the function to be continuous.
arround 1:07:00 the chaine rule is wrong. It is corrected one minute later.
Any idea what the correct statement of the chain rule would be? the video doesn't show the board when he's correcting it
I have a question on the third equation at 40:31. Why B(t)=B(\tau_a) not considered?
Because it is related to P(B(t)>a) and B(tau_a) is equal to a, so P(B(t))= B(tau_a) is not applicable. It has to be greater or less than a, whilst also crossing a
Tom Jordan why is P(M(t) > a) = P(B(t)-B(tao)>0 U tao
Because it accounts for no mass at all, if X is a continuous R.V., then P(X = x) is always 0
You can incorporate it to either the 1st or 2nd term, but it makes no difference from a distribution point of view.
1:06:24 This function f(Bt) is not smooth as it's non-differentiable at S0.
it was necessary to note the property of the sum of normal distributions
Frankly, 58:06 the proof is not very clear, espcially he didn't give the step from X~N(0,T/N) to E(X^2)=T/N. He used the formula Var(X)=E(X^2)-E^2(X) here
Var(X)=E(X^2)-E^2(X)
Now E(X) = 0
This makes Var(X) = E(X^2)
and Var(X) = T/n
This makes E(X^2) = T/n
fit to the answer
Do we really mean Brownian motion reflects the price? Brownian motion can go negative but prices cannot. Any help appreciated.
this is why actually it is stock price return and not stock price that is modelled in this way. It is explained in some way at 01:01:00 , dS/St is brownian motion
Thanks Federico...will take a look.
@@wcottee i know these comments are old but F_B is technically incorrect. dS/St is not brownian motion, it is the stochastic equivalent of derivative of brownian motion (i.e. normally distributed with a mean of 0). the answer to wcottee's question is that generally you model the log of stock prices with brownian motion, and the log can go below zero. Since brownian motion processes are invariant under shifts you can also define the process such that the brownian motion starts at the current log of the stock price. When you undo the logarithm, this effectively results in the % returns being normally distributed around 0.
@@robertvolgman6590 Thanks for the reply, Robert.
Can someone explain why P(M(t) > a) = P(B(t)-B(tao)>0 and tao
Let's denote the time range as 0 = a}, but it makes no difference from a distribution point of view)
If you think about it, B(s) at time t can really only be either above the level a, or below (Again I am ignoring the equality case here)
You can choose an arbitrary reference level and the statement will still be true. Just so happens choosing a in this case is the most natural thing to do, as demonstrated by the calculations in the video.
So, these are the only two possibilities no matter what conditional event you impose on them, that's why we can break P(\tau_a < t) into the sum above.
Let A,B be two sets. Is it easy to see that P(A)=P(A and B)+P(A and B^{c})?
Then if you set A={t:tao0} does it works?
At 1:13:15 mark, I still don't understand why are we working out the taylor expansion up until the second derivative only? why not up until 5 or 6 or even infinity?
because the rest of the terms goes to 0 as t approaches 0. See it now?
@@lazywarrior hmmm and so this happens due to the Quadratic variation concept, right?
@@ninmarwarda5154 no. This is just evaluating the limit. My previous reply contain typos. I should say as delta t approaches 0. Remember that big O notation in some form of Taylor expansion? That remainder = zero at the limit.
Thank you !
This stuff is his right hand stuff eh. It's technical but he seems to know it though. He just has to look it up in a book if he needs it. I think he was playing. Reality has boundaries. It's very small. The rest is just cooking. But it can be studied.
Can someone please explain why (dB(t))2=dt . Thank you
Because of quadratic variation being non-zero.
when he says we take the infinitesimal difference of the LHS, he just ignores the sum. How does one get around this? Thank you
@@MiniNinjaUK1 Note that we can rewrite the RHS into something like RHS = sum_{i = 1}^{n} [t_{i} - t_{i-1}], this is a telescoping sum which evaluates to T.
Then very informally speaking, when n is crazily large, we can treat t_{i} - t_{i-1} as super small time increment, which is like dt
By the same heuristic, we can also treat (B_{t_{i}} - B_{t_{i-1}})^2 as a very small change of B(t) all squared (whatever that is), and we named it (dB(t))^2
At first, it may be kind of confusing why we even adopted such a naming, since we were using calculus notation for something that is inherently not differentiable everywhere a.s.
However, if you accept the "abusing of notation" here and move on to the next lecture, you will see why we restate the quadratic variation in this "differential" form and how powerful it is.
34:13, "tau_a = min_t { B(t)=a }" This notation looks wrong.
I like to see examples in the world to make theoretical more clear. Printing presses operate by printing color dot patterns. You need a magnifying glass to see the dots. In the printing field there is an occasionally used technique of computer generated (yes, an oxymoron) random dot pattern of C, M, Y, K (basically primary colors + black) (as opposed to standard evenly spaced dots) called stochastic dot pattern. It is used for a high end fine art look and to prevent moire patterns.
Can someone help me out with something, whether I'm think about this correctly? He seems to prove that for each t, the brownian motion is non-differentiable wp1, which is different from the third property he writes on the board that brownian motions are nowhere differentiable wp1. What he proves is that after fixing a t, the collection of paths that are not differentiable at that particular t is measure 1. If this collection of paths were the same for each t, then you'd get a full measure set of paths that are nowhere differentiable. But if for each t you get a different measure zero set of paths that are differentiable at t, since there are uncountably many t, the union of these measure 0 sets is not necessarily measure 0. This would imply that the set of nowhere differentiable paths is not measure one... so it seems like he proved something different from what he wrote. Is that right?
I think your confusion partly stems from the fact that he is not actually proving the exact version of the 3rd property. As it is just a heuristic "proof", inevitably there will be some logical flaws/holes hidden within.
If you have a strong measure-theoretic probability background, then maybe skimming through the original proof made by Dvoretsky, Erdos, and Kakutani would be a good choice.
I don't understand the proof for P(M(t) > a). Isn't P( B(t) - B(tau_a) >0 | tau_a
Just goes to show a good teacher can make complex content a lot more accessible. My professor was awful in comparison.
Ich küss' Dein Auge. Ehrenmann.
wish i had his mind
Am I the only one who thinks at th-cam.com/video/PPl-7_RL0Ko/w-d-xo.html P(\tau_a
0:27 black board is clean as shit
shit is not the right word to use for such situations.
can't tell if these comments are sarcastic or not...
greyreynyn why would they be sarcastic?
Do you mean stochastic?
sen adamsin
sorry for my sexist language
Is this lecture for math students or financial students?
From the course description: "The purpose of the class is to expose undergraduate and graduate students to the mathematical concepts and techniques used in the financial industry. Mathematics lectures are mixed with lectures illustrating the corresponding application in the financial industry. MIT mathematicians teach the mathematics part while industry professionals give the lectures on applications in finance." See the course on MIT OpenCourseWare for more information and materials at ocw.mit.edu/18-S096F13.
For quants
0.5/0.1 =0.5/0.1 =5 but don't be equal 0.5
i have a vague feeling, that the 37:02 equality: P(M(t)>a) = P(tau_a < t) might not always be true. Namely M(t) > a means, that in the span of time interval (0,t), at some time s, we had B(s) > a, whereas tau_a < t means, that in the span of time interval (0,t), at some time s=tau_a, we had B(tau_a) = B(s) = a. what if in the latter case in the span of time interval (tau_a, t) we had B(s)
michał botor why should M(t)
oh, boy. i just realized, that this is even more complicated then i originally thought, so let me restate the above again, and hopefully this will help you to resolve your doubts as well. unless my reasoning is flawed, of course. ;)
so..
M(t) records the maximum value of brownian motion in the time interval (0, t).
tau_a records *the first time* brownian motion attains the value of a.
moreover, since (by definition of brownian motion) B(0) = 0 and t -> B(t) is a continuous function, then brownian motion starts from 0 and cannot jump in value, i.e. (1) to reach the value of a it must go through all of the values between 0 and a first, and (2) to reach the value greater than a it must reach the value of a first, which i can summarize as follows if 0< b < a < c, then tau_b < tau_a < tau_c.
what i'm trying to say is that because of the two mentioned defining properties of brownian motion tau_a not only records for you the moment when brownian motion first attained the value of a, but also implicitly tells you, that in the past it never exceeded the value of a, and so it must be, that M(tau_a) = a.
now, to answer your question, if we assume that tau_a < t, then from what i said above at time t maximum value of browinan motion M(t) is at least as big as it was at time tau_a, namely M(tau_a) = a, and if it is greater than a, then it must have reached this greater value after time tau_a. however if at time tau_a brownian motion will bounce off of the value of a and never reach it again in the span of time interval (tau_a, t), then its maximum value will be the one that was recored at time tau_a, i.e. M(t) = M(tau_a) = a.
on the contrary if we assume, that M(t) > a, then well M(t) > a.
an this is the reason, why i'm worrying, that the equality P(M(t)>a) = P(tau_a
michał botor i have this same confusion. Have you figured it out?
@@michalbotor That's why a more "careful" treatment will be using the set {M(t) >= a} instead of simply ignoring the equality case
In distributional sense they are both the same, but the argument seems to be more solid if we directly consider P(M(t) >= a)
@@michalbotor Because the process is continuous the probability that M(t) is any arbitrary constant is 0, so P(M(t) = a)=0, so the statement is still true. You are right that the process could "bounce" off a and never reach it again, but in a brownian motion process this is infinitely unlikely to happen.
I don't think he proves Brownian motion is nowhere differentiable correctly.
Quadratic variation theorem he's confused by the wrong proof
57:50 I cannot believe he made that silly mistake...
What mistake?
1:04:12 f is a smooth function. 1:05:03, f is not smooth at S_0
I didn't come here for a lecture.
f'(s) in square