Register to reply 
Help with proof of sample variance 
Share this thread: 
#1
Mar1809, 02:47 PM

P: 10

Hi,
I'm trying to prove to myself that the formulation of sample variance as [tex]s^2=\frac{1}{n1}\sum_{i=1}^{n}(y_i  \mu_y)^2[/tex] is an unbiased estimator of the population variance [tex]\sigma^2[/tex]. Of course, I proceed by checking the expected value of the sample variance, which flows smoothly until I get to [tex]\frac{n1}{n}E[s^2] = E[y_i^2]E[\mu_y^2][/tex]. Fine. Now, from the definition of variance: [tex]V[y_i^2]=E[y_i^2](E[y_i])^2[/tex]. Most of the sources I've checked insert the population variance here for [tex]V[y_i][/tex] (which is what you get to if you sub in the above). That make absolutely no sense to me. Shouldn't this be the sample variance since [tex]y_i[/tex] is the sample? Calling it the population variance seems circular to me. Can someone explain to me why this variance is the population variance and not the sample variance? Thanks. 


#2
Mar1809, 03:02 PM

P: 10

Oh, never mind. I figured it out. Above, y_i is a random draw from the population, not the sample, so by definition, its variance is the population variance, not the sample variance.



Register to reply 
Related Discussions  
Sample variance  Set Theory, Logic, Probability, Statistics  8  
Weighted verage of two variables with minimal variance  Calculus & Beyond Homework  4  
Stats  Geometric Variance Proof  Precalculus Mathematics Homework  0  
Sample mean and sample variance  Calculus & Beyond Homework  1  
Statistics: sample median, means, s.d. vs sample size  Precalculus Mathematics Homework  2 