A proof of the weak law of large numbers

[@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

[@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.

[@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.

[@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.

[@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))

[@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...

[@Carterv3]

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

[@rudrakshtuwani4567]

Great video! Thanks a lot!

[@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?

[@albertkirsten8407]

Thank you!

[@samuelbassey6806]

Thanks for sharing

[@anharsaif8395]

Thank you!!

[@jsh429]

Awesome. Thank you for the video.

[@cameronmiller4144]

very helpful

[@hanxiongwang3406]

thank you very much

[@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