L18.3 The Chebyshev Inequality

MIT RES.6-012 Introduction to Probability, Spring 2018
View the complete course: ocw.mit.edu/RE...
Instructor: John Tsitsiklis
License: Creative Commons BY-NC-SA
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[@Dabsyboii]

I am literally crying right now cause no matter how many times I wentt to my professor or looked on google or through my notes, I could never understand this. I am so glad i found you

[@Ash-vu5vo]

You still don’t understand it unfortunately (see my comment).

[@q44444q]

The best, most intuitive, most thoughtful presentation of elementary concentration inequalities I have ever witnessed. Thank you Professor Tsitsiklis. I know it is unlikely that you see these comments, but I hope you do. Your teaching has made a world of difference to students all over the world.

[@dccn]

Thank you! The proof of markov & chebyshev inequality is easy to be understood

[@xiscc-41suhaninair94]

Love from Jadavpur university 🎉🎉

[@holly_singh]

This is great. Thank you so much Prof. Tsitsiklis

[@ns4543]

the best explanation of Chebyshev's inequality! thanks!

[@bluejimmy168]

Does anyone know why p( x - 1 >= a -1) = a - 1) is a true statement at 4:34 . How do we know that the statement is true if we dont know the distribution of the probability?

[@dsafadsddfca]

the distribution of the probability is not necessary, | x - 1| >= x-1 ie all values for x where x-1 >= a-1 also means that for the same values of x, | x - 1| >= a-1 . However, for values of x where x-1 < a-1, for those same values either | x - 1| will also be < a-1 but it may also be >= a-1 . Which means there that | x - 1| will at the very least have the same amount of values >= a-1 hence p( x - 1 >= a -1) = a - 1)

[@bluejimmy168]

@@dsafadsddfca thanks.

[@prateekpani9464]

The inequality is clearer if we see it in terms of area....say y = x-1....now y >= a-1 correspond to a certain domain-set( S1 ) and |y| >= a-1 correspond to {y | S1 union {S2 = y | y

[@q44444q]

A different explanation:
1) We know that x-1 >= a-1 implies that |x-1| >= a-1. Why? If x-1 >= 0, it's trivially true, as x-1 = |x-1|. If x-1 < 0, then a-1 < 0 too, and since any absolute value is larger or equal to zero, we know that |x-1| >= 0 > a-1, so |x-1| > a-1.
2a) We know that |x-1| >= a-1 DOES NOT ALWAYS imply that x-1 >= a-1. Here is a counterexample. Let x = -2 and a = 1. Then |x-1| = 3 >= a-1 = 0, but x-1 = -3 is not greater than a-1 = 0.
2b) Note that if x is strictly positive, then these two cases are the same: there is no counterexample.
3) Therefore, we know that there zero (case 2b) or more (case 2a) x's for which |x-1| > a-1 and x-1 >/= a-1. This is the very definition of P(|x-1| > a-1) being greater than P(x-1 > a-1).
To show this more rigorously, remember that the definition of |x| >= a is that -a

[@ylast3756]

Thank you Indian brother. The explanation was very clear.

[@zbynekba]

Amazing insight, interpretation, and intuition. Thanks

[@sandrasurendran5068]

Thank you so much.

[@imranq9241]

Wouldn't Markov be a better bound for distributions with very high variance?

[@comusic2023]

Very helpful

[@pallaviharishchandre3021]

Thank you so much

[@danielm572]

beautiful

[@mr.shanegao]

brilliant ! Great !

[@souravsidana9631]

Good explanation

[@Chankeypathak]

What is the real world usage of Chebyshev? Can someone give an example please.

[@GerardoRodriguez-cw6rj]

When the probability distribution of a data set is unknown, you can determine the probability that the random variable X falls within a certain range of values given its mean and standard deviation. By looking at normal distributions with different degrees of kurtosis, you'll see that the Chebyshev inequality matters as the same range will have different probabilities of occurring in the respective normal distributions.

[@msahasarkar590]

Very often used in the study of barren plateaus of quantum neural networks and exponential concentration of kernel methods in quantum machine learning.

[@Retinal]

@@GerardoRodriguez-cw6rj Doesn't that make the second example counter-intuitive since x is stated to have an exponential distribution?

[@xiscc-41suhaninair94]

Any west bengal students here?

[@ramanpal3908]

Yes

[@Ash-vu5vo]

This is what college tuition fees gets you - incomplete proofs from incompetent teachers. Meanwhile all the sheeple viewers are grateful thinking they understand (when they don’t). |x - m| >= c implies (x - m)^2 >= c^2, sure, that is easily seen, but we are talking about their probabilities here and this guy has just straight up claimed they are equal (and the sheeple have just accepted). The world is full of incompetent teachers who need to be held accountable - I’m simply doing my little part.

[@lynny7868]

You can do it without calling out names and sounding condescending.

[@tejasgurjar6450]

Do you know that the P(X) and P(X^2) are same?