Probabilty Bounds

MIT 6.041SC Probabilistic Systems Analysis and Applied Probability, Fall 2013
View the complete course: ocw.mit.edu/6-0...
Instructor: Kuang Xu
License: Creative Commons BY-NC-SA
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[@scottferson170]

Maybe it's relevant to mention the probability ~0.0134 can be estimated numerically in R with the code: many = 10000000; s = 0; for (i in 1:10) s = s + runif(many); sum(s>=7)/many

[@sabrinarahmanlifestyle]

Do you have any video about chernoff bounds?
Please reply I need that urgent

[@asdfqwer2988]

10:21 If CLT provides the best upper bound, do we still need to know chebyshev and markov inequality? If the answer is yes, when to use each of them?

[@sairishi]

The CLT is a limiting case and not an upper bound

[@austintalbot5258]

The central limit theorem only applies asymptotically converging at a rate 1/sqrt n. This is a misleading video in my opinion

[@cu7695]

At 4:21 How do you get from 1/2*(Var(X)/2^2) to the 1/8*(10*1/12) ?

[@nathansherrard4111]

The 1/8 is the 1/2 * 1/(2^2). The 10*1/12 is Var(X) = n*Var(Xi). Here n = 10 and Xi is a uniform RV, which have variance (b-a)^2 / 12, or in our case (1-0)^2 / 12 = 1/12.

[@ajayshaan8573]

Sorry for the noobish question but what formula did you use to calculate variance? Is (b-a)^2 / 12 a standard formula for variance? Can I calculate variance through E(X) too?

[@nathansherrard4111]

Yes, for a uniform R.V. on an interval [a,b], the variance is known to be (b-a)^2 / 12. You can easily find the derivation via google. One method indeed does use E[X], as Var(X) = E[X^2] - (E[X])^2.

[@asdf-wv7dp]

This way it's: E[X^2] = E[X] = 1/2; E[X]^2 = 1/4; E[X^2] - E[X]^2 = 1/2 - 1/4 = 1/4. Therefore the bound should be 5/16.

[@asdf-wv7dp]

nevermind, we've a continuous rv.

[@emilyhuang2759]

There is some concept about the Expectation value that I am not getting and how you are allowed to move around the numbers like that for Cheyshevs. What is it?

[@intoeleven]

I still don't get it 2:53 why P(x - 5 >=2) == P(x - 5 =2) == P(5 - X

[@jaysonzhou5786]

Your equation is true without any assumption on X. But here X~\sum_i=1^10 Uni(0,1),which has mean 5. So the pdf of (X-5) is symmetric regarding 0. Therefore the first equation holds.

[@edwardk8477]

very clear explanation! appreciate the mini lecture 👍

[@jier65]

This is really really great. Thank you very much!

[@LeCoolCroco]

Very nice explanation thank you

[@MrCigarro50]

Dear Professor Xu, Is there a distribution for which this inequality becomes an equality?

[@scottferson170]

Both the Markov and Chebyshev inequalities are best-possible, which means that there are distributions that touch the limits. But they are discrete distributions that touch the limits at points.

[@gib1262]

How can u get 5?

[@mehdihachimi9624]

Xi is uniform [0,1] -> E[Xi] is 1/2 or 0.5

[@emilyhuang2759]

@@mehdihachimi9624 But, it is not 0.5 is is 5. How do they just take away the decimal point? Why multiply by 10?

[@belibelizo]

its E(X) not E(Xi), E(Xi)=0.5 and E(X)=E(sum of Xi's) which are independent
@@emilyhuang2759

[@emilyhuang2759]

So E(X) is like the cumulative density function? And the E(Xi) is the probability density function?

[@shadowgodprolood3387]

you invoke a basic property of the expectation, its linearity so if you have E(x1+....+xn)=E(x1)+....E(xn) in this case each Xi has an expectation of 1/2 because it's a uniform(0,1) RV and when you add 10 of them you get 1/2*10=5