18: Chi Square Proof

More detailed explanation of why we use chi square for sample variance.

[@abdelrahmanshehata7942]

I've been looking for that explanation for too much time , Thanks a lot : )

[@GladwinNewton]

Thanks man. Brilliant explanation. Finally found a mathematical proof with clarity. Thank you very much

[@jitenkant]

Simply awesome.. loved the linearity of explanation

[@sharonchetia54]

So glad I came across this video. Cleared a lot of doubts I had

[@13statistician13]

This proof is very close, but the reasoning isn't quite right. You are missing one crucial step toward the end of your proof. You can't simply determine that (n-1)S^2/sigma^2 is Chi-squared n-1 by subtracting the Chi-squared 1 term from the Chi-squared n term because you haven't proved the independence of those two terms. You must show that the term containing S^2 is independent from the term containing Xbar before you can start subtracting or adding their respective Chi-Squared distributions. You can show this by demonstrating that the covariance of Xi-Xbar (for i =1 to n) and just Xbar itself is zero. Once you've shown independence, you can easily reason with moment generating functions that the (n-1)S^2/sigma^2 term is in fact distributed as Chi-squared n-1.

[@samtan6304]

Good catch!! but covariance = 0 alone isn't sufficient enough to show independence though, since uncorrelated ≠ independent

[@verdaarpac63]

​@@samtan6304 In the special case of the bivariate normal distribution, being uncorrelated is equivalent to independence. Both X bar and Xi - X bar are linear combinations of the independent normal observations, so they are bivariate normal.

[@mohssenify]

why books gives this as given finally i found the intuition behind it you re a life saver

[@SaipratheekKankanala]

You are brilliant bro

[@michallauer9059]

Great explanation, thanks!

[@mohssenify]

finally i found the proof i was looking for thank very much

[@niketankotadiya9542]

hi professor can u tell me what chi-square statics says? if i say chi square of any distribution is 77.23. what does it mean

[@shuoyanpei8167]

Sir can you proof chi square goodness test too please.

[@felipegomesdemelo5879]

Incredible and clear explanation! Do you recomend any books for a complete statistics study?

[@13statistician13]

Casella and Berger's Statistical Inference, 2nd edition, is a classic text.

[@charlesAcmen]

still some flaws in the progress,proof should contain matrices to transform random variables

[@Alejandro-eu9pk]

wow bruh This is dope!

[@miodraglovric5093]

(1:33) standard deviation sigma squared?

[@sharonchetia54]

Yes I see too that the denominator was left out. But anyways numerator will be 0 so entire term becomes 0 .

[@Alejandro-eu9pk]

* 5:00

[@limzijian98]

Why does it equates to 0 ?

[@donghyunlee801]

brilliant!

[@jerinfatima4999]

Tq so much

[@ygalel]

Basu's Theorem!!!

[@GladwinNewton]

epic