9. Multiple Continuous Random Variables
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- Опубліковано 8 лис 2012
- MIT 6.041 Probabilistic Systems Analysis and Applied Probability, Fall 2010
View the complete course: ocw.mit.edu/6-041F10
Instructor: John Tsitsiklis
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu
Excellent lecture, great visual and conceptual explanations, much appreciated!
The best visualization on conditional continuous r.v. I've ever seen (even after I've already taken some courses on probability).
The way he explained the Buffon Needle Experiment.
Wish I knew about this before.
Thanks MIT opencourseware for these lectures
The needle example is a great one that illustrates the usefulness of Statistics well :)
Wonderful explanation for Buffon's needle problem. Thanks a lot.
This guy gets it. Intuition is a must. Its actually the first time I saw a teacher teaching a subject the exact way I would.
Amazing material. Thank you.
20:33 standard 4-step procedure of calculating probability.
Amazing lecture!
Thank you so much, sir. From India.
lovely experiment
guys, for the triangle doubt, people are confusing this with fx,y = 1/lx.
its not that, it's the distribution of y and x.
which means 0
Thank you!
The number of paper clips that intersected the lines were 3. So the probability is 2/3 and hence pi therefore is 1.5.
I noticed that too, thx for working out the math!
Shouldn't it be 3/5 though?
There is a grid of lines and he counted that lowest paper clip crossed the bottom line that was not drawn on the paper.
in 47:38 what's the point of limiting the boundaries of the integral to exclude zero regions? wouldn't integrating over those areas (over the entire x) just "add" zeroes, thus yielding the same result?
Best Explanation
I have a few questions regarding the needle experiment.
1. Why did he act like the bottom line didn't exist ? I feel like we only did the reasoning for the top line.
2. If the paperclip falls outside of the two lines we are we still in the univers we defined considering x can be greater than d/2.
Very efficient
If X is in [0, d/2] we shouldn't be counting the paper clips outside two lines, right?
How do we get the picture (triangle) in the stick-breaking example which is later used to get the limits of the integral
Because Y
The line y=x represents the values where both y and x are equal. Anything below this line are values where y < x and this is the region which we want. After first break, we are left with a length of x. the second break length must be less than x.
why did he take range of theta as 0 to pi/2? This will be valid if needle falls like '/' but what if needle falls like '\', then theta would be greater than pi/2 right? in my opinion theta should be between 0 and pi.
I calculated by assuming theta between 0 to pi, and answer came out to be the same, but i cant understand the intuition behind taking 0
He said that theta should be the acute angle formed between the line and the needle, so of course the range would be from 0 to pi/2 only.
I am interested in interpreting the case where theta = 0 (ie parallel needle): In this case, it seems to me -- based on the resulting, solved probability formula, P = 2L/Pi*d -- that the only thing mattering is the ratio L/d, but in the case of theta = 0, L is not 0, but the probability goes to 0 as a result of the 0 theta value. Obviously this leads to a zero probability, which we might expect of any individual point, however, according to the formula, there is no way to account for this scenario. The parallel needle crossing is certainly an option, and of course each of these probabilities should be zero, but why does its contribution not show up in the given formula if it is technically an option?
Wait, in the needle problem, you're calculating the probability for the needle to crosse only one of the lines (the top one). But in your paper clip experiment, you counted all those that intersected both lines. isn't that a flaw? How would you count for the needle intersecting the other line which would be (d-x) away from the center of the needle. Do we do l/2sin(theta)>=(d-x)?
Youcef Yahiaoui I noticed that too. But I think you would just have to multiply the final answer by two, since the needle cannot intersect both lines at once. So, the probability would just be 4*l/pi*d? Which would mean the value of pi from the random experiment with paper clips would actually by 4, which is a little closer to pi than 2!
+Nishaad Rao +Youcef Yahiaoui
The number of lines is irrelevant. Let's say you did this with 10 lines instead of 2. You'd get a higher number of needles that crossed a path by chance, but you'd also get a correspondingly higher number of needles that did not cross a path because you have proportionally extra empty space (i.e. empty space between the lines) a needle could land in. So in the end, the ratio of #needles crossing a path / total # of needles (which is P) would be unchanged. So P remains the same whether you have 2 lines or 10 lines (or any number of lines).
Of course, if you just had a single line, the concept of d (distance between lines) would be irrelevant and the 2l/d*pi wouldn't apply.
You get a much better value for pi if you do this sort of experiment using a computer simulation rather than manually for obvious reasons.
Youcef here X is the random variable which takes the distance from the centre of needle to the "nearest" line..the catch here is the word "nearest"...hope you got my point...
it was always both lines, he just used one to work out the geometry, but it could also be done on the lower line.
could someone explain how to do the integral at 28:00?
You take (l/2) to the left side of the integral since it is constant (l/2) (4/d pi) integral[sin (theta) dtheta] from 0 to pi/2 which is equal to 1. You are left with (l/2)(4/d pi)(1) = (2l)/(d pi)
49:15 - how can this curve "blow up" as you get close to zero? It's a probability density function so surely it has to integrate to 1?
I thought the same, replying here in case someone known the answer
the integral is the probability not the pdf. the individual pdf can blow up it's fine as long as the pdf (marginal in this case) integrates to 1. math.stackexchange.com/a/2304102
You can certainly integrate some functions which blow up at the origin. For example, try integrating the function 1/sqrt(x) from 0 to 1. Or look up the Delta function.
I forgot how to integrate that final integral on stick breaking example. LOL
what is book he is referring as readings ?
bertsekas probability
I take the risk of looking stupid. The professor said at 32:10 that density is probability of little intervals. But didn't he defined in the earlier lecture that density is probability of unit intervals? Am I getting anything wrong here?
nt unit,,,very small interval
a year late, but velocity can be considered distance over unit time intervals, but you can still calculate how far someone has travelled in 0.01 seconds, even though that may not be the unit you originally intended for your calculation.
Both interpretations are correct yet insufficient in light of measure theory
@@Isaac668 3blue1brown eh ;)
Can anyone be awesome enough to explain how he gets (2/d) * (2/pi) as the density of X and theta respectively? He is saying it has to integrate to 1 which I kind of understand as the PDF but not sure how he gets there... @25:07
ua-cam.com/video/KSrPJe7y9oA/v-deo.html
Uniform => 1/(d/2)=2/d
If l > d, would the probability of crossing greater than 1?
No, the needle can still be parallel with the line(s)
But this hypothetical violates the assumption that we’re making (that l < d), and it would cause the probability to be > 1
mark zuckelburg on front seat!
The notion of continuous is an outcome of misunderstanding of the roles of human -invented number systems. Statistics is more like a sort of guessing game than facts-based science because in true science, there can be only one correct interpretation of a phenomenon, like in solving math- problems, in which there can be a correct solution to a problem, and in order for everyone to able to arrive at the same correct solution, everyone has to interpret or understand the problem in the same way, although they can have different approaches to get to the same final destination. But in statistics, there can be different subjective interpretations of the same set of data. It depends on each individual's purposes how the same set of data is interpreted, which means that there are merely or more subjective and arbitrary truths than any objective understanding in such invented notions. There are numerous nonsensical statements in statistics, such as that an area under a probability distribution or density curve which goes to infinity is equal to 1, which obviously nonsense because that area keeps extending forever to the unknowable unknown or infinity, and thus only ignorant human mathematicians claim that the area can be measured as to be equal to 1, The reason why when you apply such human invented notions of mathematics and their formulas , you can get the number 1 is that there are numerous flaws in such notions. Since the space on here is limited, i will not explain and give examples to show you guys the absurdity or irrationally of such subjective and arbitrary motions. The notion that
f(X = c) = 0 is another typical example of being ignorant of factual facts of the real physical world, in which such notions of mathematics never exist
If you say that since you are interested in only the probability of an interval of data points or values, and any single data point will be ignored, then it will be acceptable. But when you claim that such a single point of data or value is equal to nought or 0, you have not only exposed your poor knowledge of mathematics, but also unconscoiusly irresponsibly spread nonsense or distorted facts to promote your fundamental misunderstanding and fanatical subjective beliefs in false claims. Such single points of data or values are the indispensable parts of any interval of values. Without them, there will be no definable interval. Hence, it is important to include or exclude the 2 ending points which define the boundary or the upper bound and lower bound of an interval of data points or values. Let's take a simple example to see how inclusion and exclusion of the ending points can change the total value of an interval.
Let a and b be the lower bound and upper bound of an interval of probability which a continuous rv X can take on .
If you express the probability distribution forntge random rv X as
P(a
liked this first.
30:10 But only 3 needles cross the line?!?
there's also one in the bottom
@@leodu561 Only 3 cross the line: 2 on top and 1 on the bottom. The rest - 5 - does not cross one of the two lines. Or maybe I am going crazy.... :)
@@ostihpem On the bottom of the sheet there is a third line: the paperclip on the bottom is crossing that line.
11:50 guy on left is watching other lectures
people having trouble with the triangle can look at this video. It's the same instructor and same course just edX version of it. ua-cam.com/video/aXFbBcabaQA/v-deo.html
I wonder how he writes 'e' aside 'l'😂
坏！
The notion of continuous rv is based on misunderstanding of the roles of numbers. Statistics is more like a guessing game than a sort of facts- based science because there can be different subjective interpretations of the same data set, and there are numerous nonsensical statements in statistics, such as the are under a density probability curve which goes infinity is equal to 1, and negative infinity and positive infinity are used as specific values to evaluate subjective mathemstical statements. Blindly ignorantly following such nonsense is obviously not any way to have any improvement in human extremely limited knowledge