# Help with proof of sample variance

Tags: proof, sample, variance
 P: 10 Hi, I'm trying to prove to myself that the formulation of sample variance as $$s^2=\frac{1}{n-1}\sum_{i=1}^{n}(y_i - \mu_y)^2$$ is an unbiased estimator of the population variance $$\sigma^2$$. Of course, I proceed by checking the expected value of the sample variance, which flows smoothly until I get to $$\frac{n-1}{n}E[s^2] = E[y_i^2]-E[\mu_y^2]$$. Fine. Now, from the definition of variance: $$V[y_i^2]=E[y_i^2]-(E[y_i])^2$$. Most of the sources I've checked insert the population variance here for $$V[y_i]$$ (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 $$y_i$$ 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.
 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.

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