Yes, we could actually change the world literally. However, there are two sides to it one is to make available free or accessible education and thanks God the Internet and the good will and hard work of MIT, Stanford University, EdX, Coursrera, etc. have made it possible, a reality. The other side is for people to be willing to learn instead of constantly entertained or worse aroused. Alas, only a fraction of the western population seems to have the commitment to pursue such endeavor in their free time.
@@davidespano8674 I think that's a good thing from an individual's perspective. Think about it this way, say you really want to become an engineer or scientist but you're just not that smart, which is something that's not at all in your control. Well, in the past you just had to accept the fact that you weren't going to fulfill your aspirations because if you weren't smart enough to learn from reading a tersely written textbook or from a live lecture given by a lazy professor, you didn't really have any other options to make the knowledge more accessible. Nowadays, if you're committed enough to devote all of your free time to learning, there's no requirement of being smart because you can always find resources online that introduce difficult concepts at a dumbed down level, then after using those resources to gain familiarity, you can gradually work your way up to learning from more advanced resources without ever getting stuck trying to understand a lecture. However, if everyone had that level of commitment then you would no longer have any competitive edge and once again you'd find yourself too far behind the pack to keep up.
Like his explanation of how a normal distribution can be constructed by adding direction (negations), width controls (sigma^2), and nomalizers (so that integral is 1). I always had trouble remembering the normal distribution, now I don't anymore.
One question though. He mentioned that probability densities don't have to be less than 1, as long as total area under the curve is 1. (In the first ten minutes of the lecture, I didn't get the timestamp). Can someone explain how that works?
You integrate your non uniform density function if it is possible and then get your CDF . For exemple at the end of the course , the integral of the Gaussian pdf is too difficult in a closed form so We use tables
These lectures were recorded ≥ 10 years ago and it shows, so there's a nostalgiacore aesthetic involved in that feeling. It's even more pronounced if you look at Prof Strang's linear algebra lectures.
In case of normal distribution you say that when sigma smaller then that parabola will narrower but if we take sigma getter than 1 then opposite situation happened. Please explain it.
hmmm... this is interesting.I never had it in school that you cannot integrate the function exp(-x^2) analytically . And then I solved some similar integral, of exp(-x^3/3)... and got an answer and wondered why everybody told me this cannot be solved. Nope. You actually can. Here is the solution of the integral of exp(-x^2) : www.wolframalpha.com/input/?i=integral+of+exp%28-t%5E2%29+from+-inf+to+x the integral is 1/2*sqrt(pi)*(erf(x)+1) which you actually can get by yourself also with the definition of incomplete gamma functions and integration by parts. You just have to know the incomplete gamma functions. and substitute -x^2 by -t. If you do that you get and integral that is almost the very definition of the incomplete gamma function. en.wikipedia.org/wiki/Incomplete_gamma_function It seems you have to be precise what you mean with analytic. analytic is all trigonometric functions and exp plus some sqrt and all their possible combinations.... Gamma functions and all kind of other integrals not included. So no need to integrate this one numerically if you have incgamma or erf at hand. And I guess there is a reason why you call it error function. And neither do you need tables anymore This is one example where analytic math has sort of stopped before the incomplete gamma funciton and other integral defined functions and a like have been invented. I guess this is actually a case where people would vigorously disagree just because they learned it otherwise without defining precisely what means actually analytic in this case. I always wondered why on earth people think it until I came across this video by a german prof actually shading a little light on it. It seems this is only due to the historic definition of analytic th-cam.com/video/l6w868U8C-M/w-d-xo.html en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra) For an enghineer I would say this statement doesnt really make sense. Anyhow calculating gammas and gammainc functions, also hypergeometric functions is another story in itself, especially when multiplying special functions like exp and erf with each other to get them, that much I remember from calculating something also called F1. Anyway... incomplete gamma functions are already seriously cool. Since you can calculate them within standard math libraries and they are the general case of many other special funcitons like error function, exponential integral, I think even airy. For many solutions of variable coefficient second order odes it was useful in many cases. And wolfram alpha knows how to handle it pretty well ^ (try for example integral of exp(-1/3*x^3))
He said that is not have a close form. Literally, error function is an open form, because you use Taylor series to calculate, so your statement is contradictory.
The solution in the video is correct. You can verify it by doing the integral, or intuitively in that the midpoint of [a,b] is [(b-a)/2 + a], which equals (b+a)/2
(b-a)/ is not possible .Because E(X) is an Averaging operator and hence we can only expect an AVERAGE of the uniformity between b and a.Thus the answer (b+a)/2.
Such courses are merely basic motions of mathemstics, but not any sort of elite courses. Additionally, there are obvious flaws in the notion of probability distribution for continuous rv, such as : P(a
Just "nice sets" instead of the definition of a sigma-algebra? No mention that one must use Lebesgue integrals and not Riemann integrals in general case? I understand it's not a "theoretical" course but I didn't expect they cut corners this way at MIT too...
Yea, pretty much all the intro-ish MIT courses I have seen so far have been at pretty much a business school level, though they tend to be taught superbly.
![[UNCUT] The Loyal Pin ปิ่นภักดิ์ EP.15 (3/4)](http://i.ytimg.com/vi/IVwR87OH0_A/mqdefault.jpg)
You can change the world by providing free education for elite courses that most of the people can not pay for. Thanks so much, amazing explanation.
Yes, we could actually change the world literally. However, there are two sides to it one is to make available free or accessible education and thanks God the Internet and the good will and hard work of MIT, Stanford University, EdX, Coursrera, etc. have made it possible, a reality. The other side is for people to be willing to learn instead of constantly entertained or worse aroused. Alas, only a fraction of the western population seems to have the commitment to pursue such endeavor in their free time.
@@davidespano8674 I think that's a good thing from an individual's perspective. Think about it this way, say you really want to become an engineer or scientist but you're just not that smart, which is something that's not at all in your control. Well, in the past you just had to accept the fact that you weren't going to fulfill your aspirations because if you weren't smart enough to learn from reading a tersely written textbook or from a live lecture given by a lazy professor, you didn't really have any other options to make the knowledge more accessible. Nowadays, if you're committed enough to devote all of your free time to learning, there's no requirement of being smart because you can always find resources online that introduce difficult concepts at a dumbed down level, then after using those resources to gain familiarity, you can gradually work your way up to learning from more advanced resources without ever getting stuck trying to understand a lecture. However, if everyone had that level of commitment then you would no longer have any competitive edge and once again you'd find yourself too far behind the pack to keep up.
This guy is truly amazing , what an outstanding quality of teaching!
His choice of words are excellent and he makes the whole lecture so fluid. Thank you MIT
He is literally teaching me my whole class.....maymester is no joke
this guy is amazing
It blows my mind how good he is at teaching!
Like his explanation of how a normal distribution can be constructed by adding direction (negations), width controls (sigma^2), and nomalizers (so that integral is 1). I always had trouble remembering the normal distribution, now I don't anymore.
If only teachers were this elaborate, we didn't have to memorize all the horrible formulas
@@sreenjaysen927 Most teachers don't grasp the idea as well as he does.
Love this guy. I'm teaching a course like this, and am taking notes. These lectures are a pleasure to listen to! :)
Noble and invaluable teaching
This guy is an absolute genius.
Excellent relationship between parabola and normal curve.
while listening to lecture, one has some general concerns, excellent part of the professor is he addresses most of my concerns without me asking!!
Excellent lecture. Different way to look at the Normal RV.
I can fully understand why MIT is the best technological university in our universe after watching some videos from MIT OpenCourseWare.
Excellent teaching styles.
Excellent professor and lectures!! Thank you!
Beautiful lecture!
Really clear lecture. Awesome job.
Really
One question though. He mentioned that probability densities don't have to be less than 1, as long as total area under the curve is 1. (In the first ten minutes of the lecture, I didn't get the timestamp). Can someone explain how that works?
真的是讲得好，虽然看英文视频有点儿费劲，但是值了
周睿 其实考虑与其去理解没讲好的中文的时间算下来还是可以的
看了这个视频觉得至少我的大学老师都是渣渣
He definitely says it backwards at 49:18. I love this guy though
"nice"=measurable ;)
Thank you sir!
I don't agree that discrete are equivalent to continuous - what does he mean analogous?
The probability that an IQ score is a continuous measurement is 0 at all scores (points) got it
at 24.45 u explain cdf abt uniform continuous r.v the distribution goes linearly... what abt the case if it is non-uniform??
You integrate your non uniform density function if it is possible and then get your CDF . For exemple at the end of the course , the integral of the Gaussian pdf is too difficult in a closed form so We use tables
It's a weird feeling to be a sophomore in college and be nostalgic for college lecture halls...
These lectures were recorded ≥ 10 years ago and it shows, so there's a nostalgiacore aesthetic involved in that feeling. It's even more pronounced if you look at Prof Strang's linear algebra lectures.
thanks a lot. great job!
In case of normal distribution you say that when sigma smaller then that parabola will narrower but if we take sigma getter than 1 then opposite situation happened. Please explain it.
The explanation is quite simple; if sigma is greater than 1, then the value of x^2/sigma changes slowly, hence the wide parabola.
22:38 That's the camera man
No wonder MIT students ace every competition they participate in
@ 30:01, in the loose graph, should the height of the rectangle based on [0,1] be 1/2 which is as tall as the arrowed upright line in the middle?
you are right
hmmm... this is interesting.I never had it in school that you cannot integrate the function exp(-x^2) analytically .
And then I solved some similar integral, of exp(-x^3/3)... and got an answer and wondered why everybody told me this cannot be solved. Nope. You actually can. Here is the solution of the integral of exp(-x^2) :
www.wolframalpha.com/input/?i=integral+of+exp%28-t%5E2%29+from+-inf+to+x
the integral is 1/2*sqrt(pi)*(erf(x)+1)
which you actually can get by yourself also with the definition of incomplete gamma functions and integration by parts. You just have to know the incomplete gamma functions. and substitute -x^2 by -t. If you do that you get and integral that is almost the very definition of the incomplete gamma function.
en.wikipedia.org/wiki/Incomplete_gamma_function
It seems you have to be precise what you mean with analytic. analytic is all trigonometric functions and exp plus some sqrt and all their possible combinations....
Gamma functions and all kind of other integrals not included.
So no need to integrate this one numerically if you have incgamma or erf at hand. And I guess there is a reason why you call it error function. And neither do you need tables anymore
This is one example where analytic math has sort of stopped before the incomplete gamma funciton and other integral defined functions and a like have been invented. I guess this is actually a case where people would vigorously disagree just because they learned it otherwise without defining precisely what means actually analytic in this case.
I always wondered why on earth people think it until I came across this video by a german prof actually shading a little light on it. It seems this is only due to the historic definition of analytic
th-cam.com/video/l6w868U8C-M/w-d-xo.html
en.wikipedia.org/wiki/Liouville%27s_theorem_(differential_algebra)
For an enghineer I would say this statement doesnt really make sense. Anyhow calculating gammas and gammainc functions, also hypergeometric functions is another story in itself, especially when multiplying special functions like exp and erf with each other to get them, that much I remember from calculating something also called F1.
Anyway... incomplete gamma functions are already seriously cool. Since you can calculate them within standard math libraries and they are the general case of many other special funcitons like error function, exponential integral, I think even airy. For many solutions of variable coefficient second order odes it was useful in many cases. And wolfram alpha knows how to handle it pretty well ^ (try for example integral of exp(-1/3*x^3))
He said that is not have a close form. Literally, error function is an open form, because you use Taylor series to calculate, so your statement is contradictory.
what a guy!
Is there someone similar for Data structures and algorithms?Please let me know.Cuz this guy is bomb.
search somebody Abdul Bari
Did you find something nice??
there is Algorithm from prof. Srinivasan
Thanks a lot sir :-)
@19:20 i think the E(X) sud be = b-a/2
The solution in the video is correct. You can verify it by doing the integral, or intuitively in that the midpoint of [a,b] is [(b-a)/2 + a], which equals (b+a)/2
(b-a)/ is not possible .Because E(X) is an Averaging operator and hence we can only expect an AVERAGE of the uniformity between b and a.Thus the answer (b+a)/2.
pls answer anyone
Such courses are merely basic motions of mathemstics, but not any sort of elite courses. Additionally, there are obvious flaws in the notion of probability distribution for continuous rv, such as :
P(a
@45:00 . How does the professor conclude that variance is 1 if the random variable is divided by standard deviation?
var(X/a) = 1/a^2 * var(X). In this case: var(X/sigma) = 1/sigma^2 * var(X) = 1/sigma^2 * sigma^2 = 1.
Just "nice sets" instead of the definition of a sigma-algebra? No mention that one must use Lebesgue integrals and not Riemann integrals in general case? I understand it's not a "theoretical" course but I didn't expect they cut corners this way at MIT too...
I hope that moment generating functions are covered eventually though.
lol "trivial"!
Even the mere existence of non measurable sets is non trivial...
Yea, pretty much all the intro-ish MIT courses I have seen so far have been at pretty much a business school level, though they tend to be taught superbly.