A proof of the weak law of large numbers

This video provides an explanation of the proof of the weak law of large numbers, using Chebyshev's inequality in the derivation. Check out ben-lambert.co... for course materials, and information regarding updates on each of the courses. Quite excitingly (for me at least), I am about to publish a whole series of new videos on Bayesian statistics on youtube. See here for information: ben-lambert.co... Accompanying this series, there will be a book: www.amazon.co....

[@elenalenaiva]

oh wow. that's the best and cleanest explanation ever, big plus for color use - it makes everything very clear and easy to understand.

[@SpartacanUsuals]

Hi, thanks for your comment - glad to hear they are useful. The reason it is N^2 is that the variance of a scalar times a random variable is that scalar squared times the variance of the random variable. This is because the variance is defined as E[(X-mu)^2] - in other words it is an operator which is quadratic in nature. Hope that helps! Ben

[@inthecillage9213]

A tonic one day before exam! Thanks sir!

[@eddie.carrera]

Nice! Thank you! I saw this at college, but forgot about how the proof was made. Chebyshev inequality 😮😮 interesting

[@kotsaris87]

00:28 That's an epsilon, not an eta

[@RealMcDudu]

could be xi as well :-)

[@lecothers]

@@RealMcDudu read as hi :D

[@lecothers]

It was short and useful thanks a lot Ben!!

[@lucasmoratoaraujo8433]

Thank you for the explanation!

[@fiazahmed5477]

Thanks, great video, very well explained.

[@adamBUFC]

Hi - thanks for these great videos, I'm finding them really useful alongside my graduate course in econometrics. I was just wondering why (at 2.20 in the video) when you take the variance of X(n)bar the denominator becomes N^2 rather than just N? I may be missing something obvious. Thanks in advance!

[@jeongsungmin2023]

I’m a bit overdue for this but I’ll answer in case other ppl see this reply. First, expand the Var(X bar) = Var(X1/N + X2/N + … + Xn/N) Then by the linearity of variances, Var(X1/N) + Var(X2/N) + … + Var(Xn/N) Each of these Xi are assumably iid so no covariances are added. If these Xi were dependent then add the 2 sigma Cov(Xi,Xj) term to the sum above. Anyways with the iid case: Var(Xi/N) = E((Xi/N)^2) - E(Xi/N)^2 from Var(A) = EX^2 - E^2X Then with some simple factorisation. Var(Xi/N) = (1/N^2)(E(Xi^2) - E^2(Xi)) = Var(Xi)/N^2 Therefore: Var(X1/N) + Var(X2/N) + … + Var(Xn/N) = (1/N^2)(N*Var(Xi))

[@takudzwamukutairi7284]

well done ................thats a good explanation

[@stathius]

Thanks a lot for the videos. I'm not sure why you assume all the samples will have variance sigma. Isn't that a r.v. in itself?

[@rudrakshtuwani4567]

Great video! Thanks a lot!

[@ben_lama]

doesn't this assume that the X_i's are L^2 RVs? The WLLN should hold even if the variance is undefined...

[@mariogonzalezsauri4632]

Can you elaborate more on how the variance of the sample mean equals 1/N^2 sum(Var(xi))=sigma^2/N. Its not clear for me.

[@samuelbassey6806]

Thanks for sharing

[@albertkirsten8407]

Thank you!

[@jsh429]

Awesome. Thank you for the video.

[@anharsaif8395]

Thank you!!

[@Carterv3]

is Khinchine's weak law of large numbers only the one with homogeneous variances? Thanks

[@miodraglovric5093]

You should read "Brave new world", to learn about alpha, beta and epsilon people! Also, in Statistics, we use lower case n to denote the sample size, not N (N is the population size). Best :)

[@askarm588]

Dear Ben, Firstly, thank you very much for your great videos Secondly, I would be very grateful if you explain last derivations. As far as I understand, Xn is a constant (sample mean). If it is constant, how you calculate the variance of Xn? I think the general question is about the nature of Xn. Best regards,

[@ivcbtg]

Xn is not a constant since it changes when the sample size n changes. you can use the property of variance, i.e. Var(aX1+bX2)=(a^2)Var(X1)+(b^2)Var(X2) there is no Covariance terms assuming independence of Xi's. Also Var(X1)=Var(X2)=......=Var(Xi) for all i assuming identical distribution of Xi's

[@cameronmiller4144]

very helpful

[@hanxiongwang3406]

thank you very much