12. Iterated Expectations

MIT 6.041 Probabilistic Systems Analysis and Applied Probability, Fall 2010
View the complete course: ocw.mit.edu/6-041F10
Instructor: John Tsitsiklis
License: Creative Commons BY-NC-SA
More information at ocw.mit.edu/terms
More courses at ocw.mit.edu

[@snehalsanghvi6122]

Those are some wonderful examples you have used professor! I've been following this entire probability series and I would like to say that I am really grateful for these videos. Thank you and keep up the good work!

[@saicharanmarrivada5077]

This is the best series of Probability on internet

[@remmymusumpuka6519]

JT is an excellent lecturer! Easy to follow his explanations on stochastic processes!

[@wolftribe66]

thinking back of my statistics course in university, i feel like i've been conned. This is so much better

[@hitashasharma2178]

I want to meet this amazing teacher and tell him how much I value this video. My stats teacher is really bad!

[@achillesarmstrong9639]

wow I watched many different sources , this is by far the best and most clearly explained video. It really help us that love mathematics and want to learn by ourselves.

[@dania_884]

Great teacher! he explained many concepts in Statistics far much better than my previous learned, cleared out many doubts. I'm following his series also. Only this Lecture 12 I'm not full understanding yet, need more practice.

[@AdiJ8]

Thank you! This is an amazing lecture

[@NehadHirmiz]

Thank you for this wonderful lecture

[@xinmingxu9940]

Amazingly explained!

[@soccergalsara]

Mental Gymnastics (Y).

[@theerawatbhudisaksang7162]

love it

[@nickiexu7259]

amazing...

[@ArsenedeBienne]

36:04 maybe it's worth mentioning that all E[Xi] are equal

[@PrakashBesra]

at 33:00, in Y=1 universe, f(x)dx=P(x) where dx=1, P(x)=1, therefore f(x)=1, now var(X|Y=1)=integral of ((x-E[X|Y=1])^2*f(x)) from 0 to 1. . that's how he got 1/12. and similarly for Y=2 universe..Is this correct?

[@beal_a]

Yes, you can compute it formally from the definition of variance, and, from what I can tell, your derivation is correct. Alternatively, you can memorize that the variance of a uniform distribution is 1/12 * (b - a)^2 for a distribution from a to b. This is what the professor was implying when he said "By now you've probably seen that formula..." at 32:50.

[@sunritroykarmakar4406]

Analogous to moi of rod about axis perpendicular to its centre.

[@sudhanshudey758]

thanks a lot

[@Scb-ef6ih]

This was great. I only need some Harvard friends I can impress.

[@mikechen3174]

Everyone here is blessed by MITocw

[@jeffreyanderson5333]

Harvard does have a series for Probability on RUclips
.

[@allandogreat]

expectation mean and var are the most difficult to be understood.

[@delYdelX]

0:21 outline

[@pratkmistry6401]

32:40 How do we write the expectation value as 1/2 and 3/2 ?

[@alikhansmt]

in case of uniform distribution, expectation is the center point of the range of values.
So for y=1, E = (1+0)/2 = 1/2
For y=2, E =(1+2)/2 = 3/2
Generally we would find by integrating x times Fx in the range of x

[@_sidvash]

Ali's response is correct. Just to elaborate on how you can do this using integration. ->
for y=1, E(X) = ∫xf(x) dx integrated over [0,1]. In the conditional universe y=1, f(x|y=1) = 1 (since area under curve has to be 1 within our current universe of y=1 and x is uniform). So E(X) = ∫x*1 dx over [0,1] which gives you 1/2.
for y=2, E(X) = ∫xf(x) dx integrated over [1,2]. In the conditional universe y=2, again, f(x|y=2) = 1 (since area under curve has to be 1 within our current universe of y=2 and x is uniform). So E(X) = ∫x*1 dx over [1,2] which gives you 3/2.
The thing to note here is that even though f(x) = {1/3 for 0

[@Roman-fb9mq]

@@_sidvash Thanks. You cleared my confusion.

[@rohtashbhall2671]

Which book sir ?

[@mitocw]

The text for this course is: Bertsekas, Dimitri, and John Tsitsiklis. Introduction to Probability. 2nd ed. Athena Scientific, 2008. ISBN: 9781886529236. For more info, see the course on MIT OpenCourseWare at: ocw.mit.edu/6-041F10.

[@empirewhite]

32:45 How come both var(X | Y=1) = 1/12 and var(X| Y=2) = 1/12
I didn't follow

[@intaemoon4356]

Hope this link helps. math.stackexchange.com/questions/728059/prove-variance-in-uniform-distribution-continuous

[@TolgaYilmaz1]

The variance of a uniform distribution on interval [a, b] is given by [(b - a)^2]/12.
So, it doesn't depend on where the interval is. It depends only on the length of the interval; in particular, it's proportional to the square of the length of the interval. So, for a uniform distribution on [0, 1], the variance is 1^2/12 = 1/12. Similarly, for [2,1], it's again 1^2/12 = 1/12. But for an interval that's twice as long, e.g. [2, 0] or [3, 1] or [4,2], it would be 2^2/12 = 1/3.

[@leodu561]

See lecture 8

[@Roman-fb9mq]

@@TolgaYilmaz1 Thanks. Really helpful.

[@forrestlin3401]

ngl, good video, but thumbnail creeeeeeep

[@leodu561]

Professor: What is the variance of *this thing* ? It's the expected value of *the thing* minus the square of the expected value of *the thing* ....
John Carpenter: Since when is my movie a random variable???!!!

[@theerawatbhudisaksang7162]

love it