Question

This is related to Machine Learning Problem

We have talked about the fact that the sample mean estimator X = 1 , X, is an unbiased estimator of the mean u for identically distributed X1, X2, ..., Xn: E(X) = p. The sample variance, on the other hand, is not an unbiased estimate of the true variance o2: for V = 12-1(X; - X), we get that E[V] = (1 - 02. Instead, the following bias-corrected sample variance estimator is unbiased: Vo=nti Li-I(X; -X). (a) [15 MARKS) Use the fact that E[] = o to show that E[V] = (1 - 02. Hint: The proof is short, it can be done in a few lines. (b) (10 MARKS) We also discussed the variance of the sample mean estimator, and concluded that V[X] = 102, for iid variables with variance o?. We can similarly ask what the variance is of the sample variance estimator. Deriving the formula is a bit more complex for general random variables, so let's assume the X, are zero-mean Gaussian. Then we know that the following is true (though we omit the derivation): VIV] = 0 4. This variance enables us to use Chebyshev's inequality, to get a confidence estimate. After seeing 10 samples from a distribution, do you think you will have a tighter confidence estimate around the sample mean X or the sample variance V? Explain why.

Answer 1

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