Estimating the population variance from a sample - part two

[@박수민-b8j4z]

Wow! This is an astonishing explanation for the sample variance having (1/(n-1))!! Thank you Ben!

[@1982sadaf]

@ 2:35, why? Why the inequality holds?

[@SportsManVegetal]

I could imagine a case in which the sample variance is greater than the population variance. Where am I wrong in thinking that the inequality sign is not always less than or equal to.

[@SavannahSilver]

At 3:25, can anyone connect me to a video explaining how to show that the blue estimator would be an unbiased estimator? I wish that was worked out in the video

[@lastua8562]

how can we "show" at 3.15 is an unbiased estimator for sigma squared?

[@kottelkannim4919]

Hopefully, I am not misleading you. 0. Take an expectation of both sides 1. Replace Xi by the "model" Xi=mu+epsilon_i 2. Assume homoscedasticity, V (epsilon_i ) = sigma^2, and E (epsilon_i | Xi ) = 0

[@meetmehta3385]

In the last video, didn't you conclude that it N/(N-1)? Then why is it 1/N-1 now?

[@user-lm9ny2kh8s]

In the last video N/(N-1) was multiplying sigma^2. Sigma^2= (∑(xi-xbar)^2)/N so sigma(hat)^2 = (N/N)*((∑(xi-xbar)^2)/(N-1)) The N cancels out. sigma(hat)^2 = (∑(xi-xbar)^2)/(N-1) which is what you see in this video.