It is important to reiterate the information given in this short lecture about the conditions that exists in the problem 1st condition: Independent Events,moving in a certain direction does not affect the next choice movement event i.e. unbiased decision [P(BF)= P(B)xP(F)] 2nd condition: Mutually exclusive event,If moved to the direction of F it exclude the possibility of moving to the direction of B i.e one decision eliminates the other [P(FB,BF) = P(FB)+P(BF)] The last part was specific for the problem,while the above are general 3rd condition: Conditional probability,the movement was forward given that the end point is position 1. This is one of those problem where there is a combination of 3 fundamental concepts of probability at the same time. Great work MIT
no you guys don't. All he did was go over the probability of ending up in position 0 or 1, which is just summing up probabilities of each path taken. That's nothing. He barely went into conditional probability which is the essence of random walks, represented as Markov chains. Do you know why? Nah. He was able to compute exactly the probability distribution of starting forward. What happens when the probability distribution is high dimensional? How do you do probabilistic inference of a high dimensional model which is an intractable problem?
Are the last step of each possibilities independent from the previous one ,such as if I take F then taking the B step is inevitable .then the p(B) must be 1 provided we have to end up at that same position we started ....I have some confusion in this step.
It is important to reiterate the information given in this short lecture about the conditions that exists in the problem
1st condition: Independent Events,moving in a certain direction does not affect the next choice movement event i.e. unbiased decision [P(BF)= P(B)xP(F)]
2nd condition: Mutually exclusive event,If moved to the direction of F it exclude the possibility of moving to the direction of B i.e one decision eliminates the other
[P(FB,BF) = P(FB)+P(BF)]
The last part was specific for the problem,while the above are general
3rd condition: Conditional probability,the movement was forward given that the end point is position 1.
This is one of those problem where there is a combination of 3 fundamental concepts of probability at the same time.
Great work MIT
I don't even know how i ended up here, but I understood most of this stuff.
no you guys don't. All he did was go over the probability of ending up in position 0 or 1, which is just summing up probabilities of each path taken. That's nothing. He barely went into conditional probability which is the essence of random walks, represented as Markov chains. Do you know why? Nah. He was able to compute exactly the probability distribution of starting forward. What happens when the probability distribution is high dimensional? How do you do probabilistic inference of a high dimensional model which is an intractable problem?
I found it super relaxing to watch! Thank you
This is a easy problem on a line isn’t it. Because if X is the position after n time steps, the probability that X=k is kind of just a Binomial RV.
clear
Kuang Xu is a genius.
你是他朋友吧
i did some check he is his student in standford.
@@achillesarmstrong9639 get a life
what about if you have another, 1-p-q, a circulation
when q=1-p
Thank you so muchh
Bro is handsome
Are the last step of each possibilities independent from the previous one ,such as if I take F then taking the B step is inevitable .then the p(B) must be 1 provided we have to end up at that same position we started ....I have some confusion in this step.
And btw what was the question,sir ?
Easy peasy
Is DZ video wt iam looking for 😏
the sound is not good I am quite disappointed
what was the problem?
You can find more info on this problem on MIT OpenCourseWare in the Unit 1, Lecture 3 section of this course: ocw.mit.edu/6-041SCF13.
@@mitocw I'm getting the 'Page not found' error.