What Does Independence Look Like on a Venn Diagram?
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- Опубликовано: 29 сен 2024
- Usually, Venn diagrams are not very useful for illustrating independence, as the sizes of the circles and their intersections have no meaning. It can help to illustrate independence if we force the area of each region to be equal to its probability of occurring. Independence is even easier to see if we represent the events with rectangles instead of circles. I illustrate these concepts in this video.
All plots were created in R. The appropriate diameters of the circles and distance between the centres of the circles were calculated in R using the package venneuler.
Not all plots in this video are officially called Venn diagrams.
i think i'm wasting my time at university. seriously these youtube videos are soo much better. i wish there was a 'youtube university' with lecturers like these. that would make learning soo much better
Sadly, it’s kinda hard to systematically know the courses you need to learn, and information is too scattered over the internet. I do have faith in online self education, but it’s hard to know the outlined roadmap to a topic
Ohhhhh! This confirms my understanding of dependence, maybe I can finally start to work with it! 😄
Only took, what, 5 years? 😭
This is an exceptionally insightful video! Thank you so much, I have always tried to visualise this concept of independence with venn diagrams, but the rectangle method is so much simpler to make sense of. Please keep up the awesome work!
Shaun Roberts Thanks Shaun! I like this video too :)
@@jbstatistics Hi Jeremy, thank you so much for your statistics videos, they helped a lot! For this visual representation of independence, how about using 2 equilateral triangles, each with area of 1, containing smaller equilateral triangles A = 0.3 and B = 0.7 respectively (I think 0.3 and 0.7 will be easier to see than 0.3 and 0.4), A and B share 2 sides with their respective sample spaces, then we align the bases of A and B and gradually overlap them until the overlapping equilateral triangle is 0.21, at this point, all the equal ratios will be visually intuitive
@@viettran7243 To have any discussion of events A and B being independent, they must share the same sample space.
I chose 0.3 and 0.4 to have circles of different areas that could be mutually exclusive in the circle representation. It would be silly to use the same area of course, but they don’t need to be dramatically different. A circle with an area of 0.7 has a diameter of 0.9440698. That pretty much fills the square sample space, but for the corners, and you’re never fitting a circle with an area of 0.3 in one of those corners.
Squares are as simple as it gets in terms of area. Triangles would add unnecessary complication. Something as simple as “what’s the height of an equilateral triangle with a side length of 1” isn’t trivial. Neither is “what’s the side length of an equilateral triangle with an area of 1?” Sure, those are answerable, but it’s harder with triangles and thus adds unnecessary thinking and possible confusion that distracts from the topic at hand. Also, if we’re sliding two equilateral triangles whose bases share a line, then their intersection will be an equilateral triangle. I’m not sure what would be visually intuitive about the intersection corresponding to independence. And we’d have the same issue as with circles, as we’re not fitting mutually exclusive triangles with areas of 0.7 and 0.3 in any sort of sample space with a basic shape (typically rectangular).
@@jbstatistics Thank you for your explanation, eye-opening as always!
I wanted to visualize it somehow, but I couldn't figure it out by myself, I barely hoped that someone on youtube already thought about it too and tadaa! :) Thank you very much for this video!
This explains so much to me. I'm a visual person. And this video really helped me understand mutually exclusive and independent . Thanks a lot!! 😀
I'm glad to be of help!
My Ku Leuven professor recommended us to watch all ur videos haha
Great video thanx! After spending over an hour on this concept I think I finally get it. For all of those who still don't, here's how I understood it and if what I'm saying is wrong please let me know!! Consider the video's example (A=.3, B=.4, A∩B=.12). If you calculate the conditional probability of A given B you realise that it's exactly .3. Now, the original numbers (A=.3, B=.4, A∩B=.12) represent the original sample space, but when you calculate A|B you create a new sample space which is represented by B and the overlapping section of A, in other words when you calculate the conditional probability you basically find out the ratio of the overlapping A section with respect to B (.12/0.4) and ignore everything outside of B. Given that our result is 0.3, this means that A has the same ratio (i.e. same probability) both in the ORIGINAL sample space and in the NEW sample space (i.e. the one where only B and the overlapping A section exist). If you think about it you realise that A is gonna happen with the same probability regardless of whether it is inside or outside of B, or in other words its probability is not tied to that of B. The same holds for B with respect to A.
Thanks man. The video was great, but your explanation really clinched it for me. I had a vague sense that there were two different sample spaces involved, but I wasn't sure how this related to dependence vs independence.
Brilliant...thanks for sharing it!!
your reply is the best... thanks ... Understanding this concept on a Saturday
01: 39 AM (early morning), 21/05/ 2022.
Great!
Hands down the best video explaining this on RUclips. You're awesome.
Thanks so much!
Wonderful explanation!! Thank you for helping me :) I understand alot better now, wish teachers in my school could teach like you :D
Thank you for this great illustration.
You are very welcome, and thank you for the compliment!
This was amazing!! 👏 I even started imagining inter-dependence of 3 events in 3D blocks.
excellent illustrations and concept are conceived , makes the subject practically sound. Traditional books by authors are simply based for exam purpose. Amazing presentation by the Professor.
Thanks so much for the very kind words!
I like to think that independent events belongs to different sample space. Thus two circles (events) belong to two different rectangle.
That doesn't make sense. When we speak of independent events, we are discussing events in the same sample space. Events A and B are independent if and only if P(A n B) = P(A)P(B). If they are not in the same sample space, then what is A n B?
Is it possible to make a trigonometrical relationship in independence variables?
I did some homework with no results
- C=A.B
- I tried to make a trigonometric relation, using the cosene law with gama opposite angle of C
- Having took the video example A=0.3 B=0.4 C=0.12. Of course gama doesn’t change with A=0.6 B=0.8 C=0.24 but independence its no more.
- From the same example, with x=1.25, 1.5 etc (xA.xB=x.x.C) and so, gama generally changes aprox 6 degres with low tendence until B reaches 1
- With A=B goes from a rect to an equi (60º), making A 0 to 1(nothing)
The only thing was cos gama = ½(A/B+B/A-AB) ; C=A.B nothing again
And with A=B cos gama=1/2(2-A^2)
Thanks for posting this!
If only I had watched this video earlier! This explaination has cleared almos all of my confusions. Keep up the good work 👍
Thankyou for this! Literally couldnt find anything more meaningful than this!
Can you make a video on the question
Two persons A and B throw a die alternately till one of them gets a three, and wins the game. Find probability of A, if A begins the game
Simply the best lecture on these basic concepts. Clear and simple. Most university classes or online lectures are too deep down on the math and not intuitive at all.
I’m only watching this because I feel asleep and in class and missed that part
The rectangles make so much sense! Thank you so much for the insight.
another excellent video, thanks so much man.
Sir you are a legend
Love the author for making this awesome video!!!
Sir, your videos are truely amazing.. I have a request if you get some time can you put some tutorials on R too. Thanks. Cheers!
+Sunil Bhatia Thanks for the compliment! R tutorials are part of the plan, but I need to find a little time first :)
This guy is the best statistics professor in the internet. Period.
Amazing explanation! I got it it forever!
Thank u for all great videos you help me a lot
Very nice explanation, thank you
Great video
THANK YOU FOR MAKING THIS VIDEO
Please quit your awful singing career, JB, this is your true calling!
In all seriousness though, I hope you eventually get the time to continue uploading videos, these are an asset to the Internet.
Thanks for the compliment -- I'll get back to video production in the near future. (And c'mon -- JB's got a couple of good songs!)
Awesome videos. I am a statistician that graduated years ago, but never had the opportunity to apply these techniques. I am reviewing the theory in order to work with statistics in a near future. I was wondering if it is possible to know what are the next videos that you are planning to upload? Is it possible to give any suggestions or are you following your own criteria?
Hi Fabio. I'm glad you like my videos! I'm happy to hear suggestions, but I've got a backlog of a few years worth of video ideas in my head I'm trying to find some time for more production, but time is tight these days. Cheers.
Thank you, this absolutely satisfies me, you're great.
Thanks! I'm glad to be of help.
Holy crap my mind is blown 😍
I finished the stats course and got a C+ because i didn't study all the material due to procrastination.. Nevertheless, without your channel i would have got an F, seriously.
So THANK YOU.
I'm glad I could help. I suggest that in future courses you give it your best effort and shoot for that A+ :)
jbstatistics You're absolutely right, I need to start preparing for exams earlier than I have.
Great illustration, really insightful for a beginner like me.
+Sam Chan Thanks! I'm glad you found it helpful.
Perfect.. thanks a lot for the explanation!
You are very welcome!
Very nice!
Great video, thanks for posting it ;)
+Gang Fang You are very welcome!
Smart!
Quite insightful. Thanks...
You're welcome!
This is seriously awesome!
Thanks!
The best, not one of. Thanks a lot!
Thanks!
why did you stop making videos?
I've just been very busy for a few years. Getting back to video production has always been part of the plan, and I hope to start making videos again later this year. I have a long list of topics I'd like to cover.
i hope so, you have great videos by the way
Thanks Roland!
does that means if the final square has got all squares inscribed in it then the events are independent. But what if p(A∩B)
If the events are independent, then the final square will look like the 2 individual squares with one laid on top of the other (with the lines in the same places). If the events are not independent, then the final square will not look like this.
If P(A n B) does not equal P(A)*P(B), then the events are dependent.