These whole series is brilliant. Excellent job by Prof. Tsitsiklis. My respect and admiration.

Clear, concise, efficient, and most importantly doesn't make you take 1000 notes. Love his reviews to start off every lecture... sadly many professors are either too lazy, inept, or inefficient to spend time on review each class. I've watched many many math videos, and this Professor is probably the best math lecturer I am yet too hear. (Though Professor Leonard's calc series is wonderful as well.)

I have the property 'memorylessness of math concepts' Whatever happened in the lecture has no bearing on my exam.

So if anyone is interested for an explanation of 38:00-38:31 roughly. x takes on values which are all the Natural numbers 1,2,3,4.... etc. Now if we want to find the probability values after some number of trials we'd obviously have to start after some variable, lets say 'n' amount of numbers. Expressing where we start by x, given the values that x takes would mean we'd have to write x+n whatever n is. In the video it is 1. This implies that the expectation value given that x is all values greater than some n: E[x|x>n] notice how the specification of 'greater than n' is only nontrivial insofar as x takes on values from all the natural numbers. We specify this since we dont change the domain of values that x itself takes. Our problem would be solved if we just suggested that x take on values greater than n in itself, however we do not. Now given this, we basically have to restrict the expectation values to when x>n, we'd have to start calculating expectation values at x+n. So suppose we start at the point n=5, ie. x is always greater than 5 (again, in the video it starts at n=1), then we'd do add 5 to natural number in x and thus we could attain the distribution we we're trying to attain by adjusting the random variables this way from the initial distribution. In other words for any E[x+n] (note it is implicit here however this means E[x+n|initial sample space]). This is a calculation of the expectation values of x greater than n. We can express it in terms of the actual formula to make it more evident. (x+n)p_X(x+n)=E[x+n] again, where x takes on the value of the natural numbers. Meaning that thus for every natural number from 1, we are adding n, making the formula apply on the values that are n greater than what would otherwise be in just E[x]. So as a consequence then E[x|x>n]=E[x+n|initial sample space]=E[x+n].

memorylessness can be expressed as P(X > m+n | X > m) = P(X > n)

I really like the explanations of memorylessness.

Great explanation Kudos professor I initially had doubt at why E(x-1/x-1>0)=E(x)+1 but i did not understand there.But then u explained that he wasted one toss + he started again that make sense to me. Really i'm loving probability because of u Last but not least Thanks for mit for providing such excellent videos

I like to hear Prof. Tsitsiklis say "equally likely".

"In a previous life" lmaooo #mathGoneSpiritual

What are the chances that the joint pmf is 4/20 ?

OCW really helps me with my studies. I will donate once I have money again :,D

Thank you! :)

Great lecture!

Thank you!

OCW is the Best!

At 33:00 I'm confused how X-2 is the remaing number of heads, it should be Tails right as we only have one head and then the experiment stops.

in conditional PMF, what if the probability of each random variable is not same? Then how do we scale each probability?

Whats the "cheat sheat for exam" thing?

Great lecture Dr. Tsitsiklis. I have a question where you were applying divide and conquer to solve the Geometric example where A1: X=1 and A2: X>1. You said that P(X>1) = (1-p). But, this is the probability that the first toss is a T. The case where the Head appears in the 5th toss for example also corresponds to P(X>1) and this probability is (1-p)^(4).q. It seems to me that its the geometric sum of all the cases corresponding to X>1... May be I am wrong in my thinking or just a little confused as to why you set P(X>1) to be the same as getting T in the first toss.

I dont understand why individuals dont always make the sample space explicit. That way, a nonconditional probability is conditional on the given sample space. It shows that probability is dependant on sample space and there can be varying specification of it.

How does the +1 end up coming in

it's good to see in 1.25x speed

"Whatever happened in the past it happened, but has no bearing what's going to happen in future" i guess this is true for life also 🙂

Great lecture! But was the memorylessness property correctly formulated? I think it should be P(X+2|X>2). Or maybe I understood his notation incorrectly.

At 40:40 why is P(X>1) = (1-p) ? Are we using total probability concept here?

I don't understand why E[X|X>1]=1+E[X] if the first try fails? if the first try fails then how come there is +1 because +1 means that you obtain the HEAD at the first try? And if so how can we add 1 to that?

am confused how come memorylessness didn't work in the example of kings sibling example. clearly getting a girl or boy is independent. P( female sibling ) should be equal to 1/2

1/16

Cheat sheet was amazing 😂😂

Awesome lectures however I call bullshit to the following: At 32:03 he says the following: "you can make it formal (the argument about the memoryless property) by just manipulating PMFs formally". And then proceeds up to 34:04 by which time he's supposedly "proven formally" the memoryless property of the geometric distribution and declares victory. Except of course he's actually proven absolutely nothing at all. Either that or I am plain thick and have failed to see the "proof".

24:23 THUG LIFE 😎😎😎