@13:30 , since zi and zj are independent (independent increments), the expectation operator can be applied to each of the terms. And not because linearity
Yes, thank you, I also got shivers down my spine when I saw this. One has to be super careful when applying this rule for the Expectation of a Product of RVs as it is only allowed for independent RVs…
10:08 I am so confused ... How is it 8 ??? Also, 10:55 I understand it is convenient to turn it into a square root. But then the Zk doesn't make any sense anymore since the root of t/n does not represent the actual increment ?
I dunno math well at all and was able to follow along with only a basic understanding of algebra/geometry and I can say that you have clearly explained this more so than anyone else I've found discussing it. While I cannot quite grasp some of the math, I think.... Conceptually what's happening was very clear.
13:45 I think maybe there is a easier way to explain this: There are only four options for Zi × Zj: assume the increment is a, then Zi × Zj = a × a = a Zi × Zj = a × -a = -a Zi × Zj = -a × a = -a Zi × Zj = -a × -a = a And the average of these are 0 ! ∴ E(Zi × Zj) = 0
thanks for the great video. but you didn't explain why 10:38 it's square root of t/n. If that's your setup then other models doesn't have to follow this scale. random walk magnitude has to be 1? thanks!
The explanation of Martingale was a bit confusing, I could not differentiate it from the Markov process as per the differentiation. This is okay that the Markov process is memoryless and the best prediction about the future is today's value irrespective of what happened in the past. While Martingale does take into the past information. My question is how a brownian can be both Markov and Martingale??
Can you please explain why you used the specific value of square_root(t/n) for k? k is just the amount by which the counter goes up or down with equal probability. It can be anything, such as a constant, or even t/n instead of it's square root.
your discrete coin flip is continous as well, you just werent supposed to connect the dots right? Then if we pass t to 0 the dots get so much squished together that we get a continous graph
Question: Why do we need a brownian motion to model asset prices if in the real world their prices don't move continuously? From what I understand they can vary within a second but it is always a discrete time increment.
For those confused as to where the t/n came from. From my understanding you are modifying the +1/-1 based on the frequency of, shall we say, "coin flips". The square root is largely for convenience. Basically if I have a 1 second interval, and I flip my coin, I can either go to +1 or -1. Now what he failed to explain properly in my view is that he didn't specify that we aren't necessarily conducting new trials with those flips. Rather we are breaking up a large one into smaller intervals. When you look at it this way, you might understand that in that case, we are bounded to stay between +1/-1. The best way I can explain this myself is imagine the +1/-1 to be a single coin flip. Then when we halve the interval, what we are really doing is flipping half a coin (difficult to imagine, but stay with me). So the result of our half coin flips would still be between -1 and 1, except now because we can get "half" results, we can end up anywhere at -1, 0, or 1 (if you don't believe me, try replacing 2 half coins in this example with 2 coins flipping heads +1 tails -1, and the prior example to the outcomes of HH/TT exclusively). This means that prior to the end of our interval, we can now take "half" flip steps, such as -0.5 and 0.5 outcomes. But notice how our time interval t = 1, and our intervals n = 2 in this case. And therefore, at any point we can move t/n. This is my intuitive explanation for this process. The square root I can't really explain on an intuitive level but consider that all the square root does is just change the scale of the movements, but the relative magnitudes are still the same. Hope this helps. PS there are some errors in this video, such as E(XY) does NOT imply E(X)E(Y) unless they are independent. This applies here since remember, the Axioms of Brownian motion imply that each time interval is independent from the other. Final note: for those asking why he calculates E[Z^2], I will give you a hint: recall the formula for Var(Z) = E[Z^2] - E[Z]^2. If we know that E[Z] = 0 then E[Z]^2 = 0. What is Var(Z) then?
Another example of where E[x_1 * x_2] =\= E[x_1] * E[x_2] If x_1 = x_2 = 1 (p=0.5) , -1 (p=0.5) Then x_1 * x_2 = 1 (p=1) So E[x_1 * x_2] = 1 =\= E[x_1] * E[x_2] = 0 * 0 Obviously they do not equal But other than that a very informative and intuitive explanation
Hey mate, I was abit lost when Z_k = t/n , I understand the need for t/n ( to slice total t into the number of steps taken) but not sure why Z_k could take that assignment, could creator or anyone share pls, thank you
the summation formula is the mean by definition, he didnt say that properly e.g if i have x data points and i take 1/n*(the sum of x_i data points from i = 1 to n) i get the mean where n is the number of points.
he just did it for convenience so when the E[Xn^2} is calculated it is t instead of t^2. Think of it as a time period going from 0 to sqrt(t) instead of t with intervals of sqrt(n) instead of n. So if we had time period of 5s as in the video with intervals of 1/4 s. We can rewrite that as time period of sqrt(25)s with interval of sqrt(1/16) which is the same thing.
i think there is a lack of explanation about the use of Zk=t/n because Zk is suppose to be -1 or +1 even if we devide the intervals : this is at least what was shown on the graphical explanation. But in the mathematical expression, Zk is reducing proportionnaly to the reduction of the intervals : this add more fractale structure to the curve as the amplitude of the oscillations of the curves are reducing at the same time we devide the intervals.
I also got stuck in that part. I don't understand how or why he defines the increments (Zk) as +-sqrt(t/n). I followed that each time step is now t/n but the value of each increment is not well explained.
@@trabek123 Number of intervals in each case is the same value: n. Hence, I do not understand your line of reasoning since you say, "...intervals of sqrt(n) instead of n". Notice Xn:=Sigma(k=1)(n)Zk in each case (i.e., before and after the transformation)!
In general good lecture. The audio quality though is very poor. The lecture "Building Brownian Motion from a Random Walk", starting around minute 11 is very unclear. You suddenly move from explaining E(Zk) = 0 to E(Zk^2) = 0, without explaining what E(Zk^2) is or why you introduced it. From this point on the lecture is not clear at all. I would be good if you could clarify why the term E(Zk^2) = 0 was introduced in the lecture.
E(Zk^2) is an indicator of how far the values you can get during the brownian motion can go away from the average value (E(Zk)) and the result depends on the time. So this is very relevant to Brownian motion as you have an idea of how high the values can be.
Just to make your comment more clear, the indicator you are talking about is the variance. The formula for the variance is equal to = E(Zk^2) - (E(Zk))^2. The first term is derived as t/n and the second term is 0.
@13:30 , since zi and zj are independent (independent increments), the expectation operator can be applied to each of the terms. And not because linearity
But why is E(Zi)E(Zj) equal to zero...
because whatever is k E(Zk)=0 so E(Zi)=E(Zj)=0 but E(ZiZj) cannot be equal to 0 (except if the values Zk stay equal to 0)
Yes, thank you, I also got shivers down my spine when I saw this. One has to be super careful when applying this rule for the Expectation of a Product of RVs as it is only allowed for independent RVs…
10:08 I am so confused ... How is it 8 ???
Also, 10:55 I understand it is convenient to turn it into a square root. But then the Zk doesn't make any sense anymore since the root of t/n does not represent the actual increment ?
I dunno math well at all and was able to follow along with only a basic understanding of algebra/geometry and I can say that you have clearly explained this more so than anyone else I've found discussing it. While I cannot quite grasp some of the math, I think.... Conceptually what's happening was very clear.
"Markov is a goldfish, right?" haha, never heard that before. I like it.
Really... speaking up would make a big difference though.
13:45 I think maybe there is a easier way to explain this:
There are only four options for Zi × Zj: assume the increment is a, then
Zi × Zj = a × a = a
Zi × Zj = a × -a = -a
Zi × Zj = -a × a = -a
Zi × Zj = -a × -a = a
And the average of these are 0 !
∴ E(Zi × Zj) = 0
This needs to be upvoted more
@@robalexnat sorry when I mean a^2 for the RHS of the four lines 😂😂
wow thats a great explanation as his explanatation of this part was incorrect!
One of the best videos on the topic
He says when we flip the coin every half-second we have divided our 5s time interval into 8 time intervals?? Shouldn't it be 10 time intervals?
A 5 second interval with 0.5 second timesteps results in 8 timesteps? Does he mean 10 timesteps or am I missing something?
yeh he is kinda retarded and this video is quite crappy anyway
thanks for the great video. but you didn't explain why 10:38 it's square root of t/n. If that's your setup then other models doesn't have to follow this scale. random walk magnitude has to be 1? thanks!
The explanation of Martingale was a bit confusing, I could not differentiate it from the Markov process as per the differentiation. This is okay that the Markov process is memoryless and the best prediction about the future is today's value irrespective of what happened in the past. While Martingale does take into the past information.
My question is how a brownian can be both Markov and Martingale??
Forward to 22:45 with your headphones :D geeezfuck
Trust me
ear rape?
Geezzzz, blew up my ear drums gods sake
Can you please explain why you used the specific value of square_root(t/n) for k? k is just the amount by which the counter goes up or down with equal probability. It can be anything, such as a constant, or even t/n instead of it's square root.
I dont understand the point where z_k = t / n. No mater how smaller the interval becomes, we go up and down by 1. Am I missing something?
13:00 why is E[Z_k ^2] = t/n?
your discrete coin flip is continous as well, you just werent supposed to connect the dots right? Then if we pass t to 0 the dots get so much squished together that we get a continous graph
Question: Why do we need a brownian motion to model asset prices if in the real world their prices don't move continuously? From what I understand they can vary within a second but it is always a discrete time increment.
How do asset prices in financial markets fluctuate? Think about how often agents trade.
Thanks for the good explanations . i just understood it completely :)
For those confused as to where the t/n came from. From my understanding you are modifying the +1/-1 based on the frequency of, shall we say, "coin flips". The square root is largely for convenience. Basically if I have a 1 second interval, and I flip my coin, I can either go to +1 or -1.
Now what he failed to explain properly in my view is that he didn't specify that we aren't necessarily conducting new trials with those flips. Rather we are breaking up a large one into smaller intervals. When you look at it this way, you might understand that in that case, we are bounded to stay between +1/-1.
The best way I can explain this myself is imagine the +1/-1 to be a single coin flip. Then when we halve the interval, what we are really doing is flipping half a coin (difficult to imagine, but stay with me). So the result of our half coin flips would still be between -1 and 1, except now because we can get "half" results, we can end up anywhere at -1, 0, or 1 (if you don't believe me, try replacing 2 half coins in this example with 2 coins flipping heads +1 tails -1, and the prior example to the outcomes of HH/TT exclusively).
This means that prior to the end of our interval, we can now take "half" flip steps, such as -0.5 and 0.5 outcomes. But notice how our time interval t = 1, and our intervals n = 2 in this case. And therefore, at any point we can move t/n.
This is my intuitive explanation for this process. The square root I can't really explain on an intuitive level but consider that all the square root does is just change the scale of the movements, but the relative magnitudes are still the same. Hope this helps.
PS there are some errors in this video, such as E(XY) does NOT imply E(X)E(Y) unless they are independent. This applies here since remember, the Axioms of Brownian motion imply that each time interval is independent from the other.
Final note: for those asking why he calculates E[Z^2], I will give you a hint: recall the formula for Var(Z) = E[Z^2] - E[Z]^2. If we know that E[Z] = 0 then E[Z]^2 = 0. What is Var(Z) then?
Another example of where E[x_1 * x_2] =\= E[x_1] * E[x_2]
If x_1 = x_2 = 1 (p=0.5) , -1 (p=0.5)
Then x_1 * x_2 = 1 (p=1)
So E[x_1 * x_2] = 1 =\= E[x_1] * E[x_2] = 0 * 0
Obviously they do not equal
But other than that a very informative and intuitive explanation
one still needs the central limit theorem to actually make sense of the limit here...
Hey mate, I was abit lost when Z_k = t/n , I understand the need for t/n ( to slice total t into the number of steps taken) but not sure why Z_k could take that assignment, could creator or anyone share pls, thank you
the summation formula is the mean by definition, he didnt say that properly e.g if i have x data points and i take 1/n*(the sum of x_i data points from i = 1 to n) i get the mean where n is the number of points.
thank you!! helped alot
Hi
struggling on the reasoning why is Xi=t/n then Xi =\sqrt (t/n).
Any help thankful
he just did it for convenience so when the E[Xn^2} is calculated it is t instead of t^2. Think of it as a time period going from 0 to sqrt(t) instead of t with intervals of sqrt(n) instead of n.
So if we had time period of 5s as in the video with intervals of 1/4 s. We can rewrite that as time period of sqrt(25)s with interval of sqrt(1/16) which is the same thing.
i think there is a lack of explanation about the use of Zk=t/n because Zk is suppose to be -1 or +1 even if we devide the intervals : this is at least what was shown on the graphical explanation. But in the mathematical expression, Zk is reducing proportionnaly to the reduction of the intervals : this add more fractale structure to the curve as the amplitude of the oscillations of the curves are reducing at the same time we devide the intervals.
I also got stuck in that part. I don't understand how or why he defines the increments (Zk) as +-sqrt(t/n). I followed that each time step is now t/n but the value of each increment is not well explained.
@@trabek123 Number of intervals in each case is the same value: n. Hence, I do not understand your line of reasoning since you say, "...intervals of sqrt(n) instead of n". Notice Xn:=Sigma(k=1)(n)Zk in each case (i.e., before and after the transformation)!
In general good lecture. The audio quality though is very poor. The lecture "Building Brownian Motion from a Random Walk", starting around minute 11 is very unclear. You suddenly move from explaining E(Zk) = 0 to E(Zk^2) = 0, without explaining what E(Zk^2) is or why you introduced it. From this point on the lecture is not clear at all. I would be good if you could clarify why the term E(Zk^2) = 0 was introduced in the lecture.
Var(Zk) = E(Zk^2)-(E(Zk))^2
but the reasoning at 12:17 for E(Zk^2) is a bit unclear
A little step is that since Zi, Zj with i≠j are independent then E[Zi * Zj] =E[Zi]*E[Zj] = 0 * 0 = 0.
Hi, nice video. clear and helpful. What software are you using?
So why are we getting E(Zk)2?? How does this have anything to do w/ Brownian motion?
E(Zk^2) is an indicator of how far the values you can get during the brownian motion can go away from the average value (E(Zk)) and the result depends on the time. So this is very relevant to Brownian motion as you have an idea of how high the values can be.
Just to make your comment more clear, the indicator you are talking about is the variance. The formula for the variance is equal to = E(Zk^2) - (E(Zk))^2. The first term is derived as t/n and the second term is 0.
(As answered to medhi shiriz)Var(Zk) = E(Zk^2)-(E(Zk))^2
but the reasoning at 12:17 for E(Zk^2) is a bit unclear
The video with the lowest volume in TH-cam ever!
install some volume up extension
addons.mozilla.org/en-US/firefox/addon/soundfixer/ worked well for me on Firefox :)
Who all are watching the video at x0.75 :)
You are jimmy two times
Speak up!
Hopeless lacks clarity