L06.2 Variance

  Рет қаралды 36,289

MIT OpenCourseWare

MIT OpenCourseWare

6 жыл бұрын

MIT RES.6-012 Introduction to Probability, Spring 2018
View the complete course: ocw.mit.edu/RES-6-012S18
Instructor: John Tsitsiklis
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu

Пікірлер: 19
@adnanmohamed6517
@adnanmohamed6517 3 жыл бұрын
Thanks for explaining why concepts are true, because many people are in institutions where "teachers" only regurgitate formulas and tell you how to use them in silly problems. So, these kind of videos are our true place to learn.
@naveenchandrakumar480
@naveenchandrakumar480 5 жыл бұрын
Outstanding
@yashj8238
@yashj8238 9 ай бұрын
You are so clean man. Teachers have been ruining this subject with intimidating definitions and formulae.
@moumitasaha581
@moumitasaha581 6 жыл бұрын
excellent lecture
@artistscientist2848
@artistscientist2848 2 жыл бұрын
In the previous video, the professor said linearity of expectations can be applied only when g(x) is a linear function i.e. E(g(x)) = g(E(x)) only when g(x) is linear. But at 7:58 he uses it to factor a^2 out even though g(x) is (x-mu)^2, which is not linear. Can someone explain this?
@SprinkzMC
@SprinkzMC 2 жыл бұрын
a^2 is a constant scaling the random variable (X-mu)^2. Hence, it can come out of the expectation due to linearity. However, (X-mu)^2 remains within the expectation and is defined as the variance.
@artistscientist2848
@artistscientist2848 2 жыл бұрын
@@SprinkzMC Yes, that what I thought later on after posting this. I agree with you, thanks.
@artistscientist2848
@artistscientist2848 2 жыл бұрын
@@SprinkzMC If you jump to L07.7, around 1:25, the professor again uses linearity if expectations to say E(X^2 + 2.X.Y + Y^2) = E(X^2) + 2.E(XY) + E(Y^2) Do you have any insight for that?
@ZH59
@ZH59 2 жыл бұрын
@@artistscientist2848 You need to prove E[X+Y]=E[X]+E[Y] to make sense of these : )
@kodenejmberni
@kodenejmberni Жыл бұрын
This is a serious sin that those lectures are on the white background.
@akshatsinghal9627
@akshatsinghal9627 3 жыл бұрын
What do you mean, "a better behaved mathematical object" w.r.t. to the squared distance and not the absolute?
@artistscientist2848
@artistscientist2848 2 жыл бұрын
@Akshat The derivative is well defined at 0, which is not the case for absolute value funxt.
@avishshah2186
@avishshah2186 2 жыл бұрын
2:45
@awesomeblossom6154
@awesomeblossom6154 4 жыл бұрын
what is pmf
@saivarunreddykamatham5712
@saivarunreddykamatham5712 4 жыл бұрын
Probability Mass Function
@rodolfog2459
@rodolfog2459 4 жыл бұрын
Watch L05.3 to better understand the pmf...
@awesomeblossom6154
@awesomeblossom6154 4 жыл бұрын
@@saivarunreddykamatham5712 Thank you!
@awesomeblossom6154
@awesomeblossom6154 4 жыл бұрын
@@rodolfog2459 Definitely I will.. thanks a lot!
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