Is square of sample mean a consistent estimator?

Is square of sample mean a consistent estimator?

is a random sample drawn from a finite population then it is not correct to accept the sample mean square as a consistent estimator of the population variance. However, it can be mathematically shown that in this case it is correct to accept the sample variance as a consistent estimator of the population variance.

Is the square of also an unbiased estimator of the square of the population mean?

It is an unbiased estimator of the population mean. The square-root of S2 is a statistic commonly used to estimate the standard deviation of a population. S2 is an unbiased estimator of the square of the population standard deviation; in general, (S2)½ is a biased estimator of the population standard deviation.

Is the sample mean squared an unbiased estimator?

Plainly, this is a biased estimator of population variance.

Which is the best estimator for mean and variance?

Thus, we may replace μ by our estimate of the μ, the sample mean, to obtain the following estimator for σ 2 : S ¯ 2 = 1 n ∑ k = 1 n ( X k − X ¯) 2. S ¯ 2 = 1 n ( ∑ k = 1 n X k 2 − n X ¯ 2). Let X 1, X 2, X 3., X n be a random sample with mean E X i = μ, and variance V a r ( X i) = σ 2.

How to calculate Sample variance as an average?

Here’s a general derivation that does not assume normality. Let’s rewrite the sample variance S2 as an average over all pairs of indices: S2 = 1 (n 2) ∑ { i, j } 1 2(Xi − Xj)2. Since E[(Xi − Xj)2 / 2] = σ2, we see that S2 is an unbiased estimator for σ2.

How to prove the square of the sample mean?

But, how can i prove that the square of the sample mean is an biased (or maybe unbiased) estimator of the variance? E [ X ¯ 2] = E [ ( ∑ i = 1 n X i n) 2] = E [ ∑ i = 1 n X i n × ∑ i = 1 n X i n] = 1 n 2 E [ ∑ i = 1 n X i × ∑ i = 1 n X i] =…..

Is the square of the sample mean an unbiased estimator?

It’s trivial to show that the square of the sample mean is neither a consistent nor unbiased estimator in the general case. The sample mean is 2, no matter what. The population variance is 0. The sample mean squared is 4. I’d bet though this isn’t what the homework is asking for.