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 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.
I've been looking for that explanation for too much time ,
Thanks a lot : )
Thanks man. Brilliant explanation. Finally found a mathematical proof with clarity. Thank you very much
Simply awesome.. loved the linearity of explanation
So glad I came across this video. Cleared a lot of doubts I had
why books gives this as given finally i found the intuition behind it you re a life saver
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.
Good catch!! but covariance = 0 alone isn't sufficient enough to show independence though, since uncorrelated ≠ independent
@@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.
Great explanation, thanks!
You are brilliant bro
finally i found the proof i was looking for thank very much
wow bruh This is dope!
brilliant!
Tq so much
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
Sir can you proof chi square goodness test too please.
still some flaws in the progress,proof should contain matrices to transform random variables
Incredible and clear explanation!
Do you recomend any books for a complete statistics study?
Casella and Berger's Statistical Inference, 2nd edition, is a classic text.
Basu's Theorem!!!
(1:33) standard deviation sigma squared?
Yes I see too that the denominator was left out. But anyways numerator will be 0 so entire term becomes 0 .
* 5:00
Why does it equates to 0 ?
epic