Help with proof of sample variance

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The discussion revolves around proving that the sample variance formula is an unbiased estimator of the population variance. The individual initially struggles with the expected value of the sample variance and questions why the variance of a random draw from the population is referred to as the population variance rather than the sample variance. They clarify that the confusion arises because y_i represents a random draw from the population, confirming that its variance is indeed the population variance. Ultimately, the individual resolves their doubt, understanding the distinction between population and sample variance. This highlights the importance of recognizing the definitions and contexts of statistical terms.
fadecomic
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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.
 
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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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