20. Definitions and Inequalities

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@timdelise6326 3 жыл бұрын

How many MIT courses are there with no videos, and Prof. Strang comes through with an incredibly relevant course with full course videos. Amazing.

@georgesadler7830 3 жыл бұрын

Professor Strang thank you for explaining and analyzing Definitions and Inequalities from probability and statistics. The pictures and examples really help me to understand the quality of this lecture.

@ziqichen4654 4 жыл бұрын

For chebyshev, on the left hand side, the right bracket should close after ">=a".

@allyourcode 3 жыл бұрын

Also, at the end, he was supposed to say that P[Y >= a^2] instead of E[Y]

@1290DR 3 жыл бұрын

Mr. Professor - you are great!

@lirtowh124 5 жыл бұрын

You sir are an Inspiration:)

@SphereofTime 10 күн бұрын

4:17 variance .

@tmusic99 2 жыл бұрын

I am missing a reference to the population when explaining expected mean and expected variance. Then you sample the population for a sample mean and/or a sample variance.

@christianjimenez1877 5 жыл бұрын

Brilliant!!!

@cyberneticqualanaut7207 2 жыл бұрын

I can see and prove a covariance matrix with x and y independent is a positive definite matrix. But when x and y are dependent, I don't think the covariance matrix V is positive semidefinite. In the 2x2 case uT*V*u could be less than zero for the case when u = [1; -1] for instance.

@rusty-b3p 5 жыл бұрын

11:55 when you try to guess answer on multiply choice test

@quanyingliu7168 5 жыл бұрын

Proof of Markov Inequality: en.wikipedia.org/wiki/Markov%27s_inequality#Proof_in_the_language_of_probability_theory

@1u2t3 4 жыл бұрын

The example for Markov inequality seems not proper. For sequence 1,2,3,4,5, the mean can be 1 only if P2, P3, P4, P5 are all 0

@seamus9898 4 жыл бұрын

I agree it's a weird simplified example but the math still checks out

@strahilpastuhov1886 4 жыл бұрын

P1 need not be 1, so P2 etc. can be above zero and the mean still be 1.

@PrzemyslawSliwinski 2 жыл бұрын

@@strahilpastuhov1886 Nope. In this example you have that 2p₂ + 3p₃ + 4p₄ + 5p₅ = 1 - p₁ = p₂ + p₃ + p₄ + p₅. That is, you get p₂ + 2p₃ + 3p₄ + 4p₅ = 0.

@PrzemyslawSliwinski 2 жыл бұрын

@@seamus9898 Yup, it is almost like a dice example.

@linelendur Жыл бұрын

Yeah, I paused the video just after he collected terms and tried to move forward myself (by collecting (p1+...+p5) together) and immediately hit this issue. That said, if you replace "1" with E[x] in both example the result he's showing still falls out, as "E[x]/3" instead of "1/3"