L18.7 Convergence in Probability Examples

MIT RES.6-012 Introduction to Probability, Spring 2018
View the complete course: ocw.mit.edu/RE...
Instructor: John Tsitsiklis
License: Creative Commons BY-NC-SA
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[@Celdorsc2]

This is too abstract for me! We clearly have two states of r.v.: Y=0 and Y=n^2. Why if n-> infinity Y=n^2 disappears?

[@jaydenou6818]

in this case the probability of Y=n^2 happening is depending on the value of n, which take the probability 1/n . As n goes larger, the probability of 1/n goes even smaller , thus the chance of seeing Y takes value of n^2 is smaller. It doesn't necessary means Y=n^2 disappears. (If you think on this way you might have misunderstood of concept of 'convergence in probability')
FYI: Base on my understanding, 'convergence in probability' means "as the series of Y_n ({Y_1,Y_2,Y_3.....}) takes more values, more proportion of Y within the series are going to take a value that approaches to a certain number". In this case, as n -> infinity, more of Y will take value of 0 , since the probability of getting 0 is 1-(1/n). For example, if n is 1000, the probability of getting Y=n^2=1000^2=1,000,000 is 1/1000 , yet the probability of getting a 0 is 999/1000. Thus, as this trend follow on, more proportion of Y in series Y_n are suppose to be 0 , then we could say Y_n 'convergence in probability' to 0 .

[@matiasbajac]

0
I have a question, it is okay prove the range xn-x1 of the uniform (0,1) converge in probability to 1 showing the difference between both with definition of probability?

[@mincheolshin5241]

Thank you for a good lecture.

[@bronsonvance1344]

instaBlaster...

[@nileshkharat236]

0:52 why is the probability 1/n?

[@felixsch5078]

simply for illustrative purposes. Note, that in this discrete case, the probability of observing an outcome that has value n^2 is 1/n. The probability of observing an outcome of 0 is 1-1/n.

[@yasminaama9466]

@@felixsch5078 I think he meant why does the probability of Yn greater than or equal to epsilon equal 1/n

[@Heshammehrem]

you have two outcomes either 0 or n². The pobability of this event P(|Y - 0| ≥ Ɛ) [which equal P(|Y| ≥ Ɛ) ] is the probability that the the 2nd outcome(Yn= n²) to happen.
Or in other words:
P(|Y - 0| ≥ Ɛ) = P(|Y| ≥ Ɛ) = P(|Y | = n²) = 1∕n

[@trinity832]

@@Heshammehrem Thanks lad !!

[@naren2412]

@@Heshammehrem That helped ...Thanks !!!

[@chtibareda]

those Xi rv are the same, so the min of them should be equal to any of them no ?

[@MohammadrezaParsa-k7p]

Very Nice 👌😎

[@mariosmag2918]

Thank you