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Let $X_1, \ldots, X_n$ be an i.i.d sample with known mean $\mu$ and unknown variance $\sigma^2$.
Is it true that sample variance $\displaystyle \widehat{\sigma}^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \mu)^2$ is asymptotically normal estimator of $\sigma^2$ ?

In other words is it true that $\sqrt{n}(\widehat{\sigma}^2 - \sigma^2) \overset{d}{\longrightarrow} \mathcal{N}(0, V)$ for some $V$ ?

(I only found a theorem which claims that if mean $\mu$ is unknown then estimator $S^2 = \frac{1}{n} \sum_{i=1}^n (X_i - \overline{X})^2$ of $\sigma^2$ is asymptotically normal)

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This is true, if you have some additional integrability condition. For example, the result holds if the fourth moment of $X$ exists. To see this, you just need to apply the central limit theorem and note that $(X-\mu)^2$ is a sequence of independent, square-integrable random variables.

The integrability condition can still be slightly weakened, as there are many formulations of the CLT (see e.g. Kallenberg).

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  • $\begingroup$ Thanks. Using CLT I got the following formula: $\sqrt{n}(\widehat{\sigma}^2 - \sigma^2) \overset{d}{\longrightarrow} \mathcal{N}(0, \mathrm{Var}((X_1 - \mu)^2) )$. How do you think is it correct? $\endgroup$ – Rodvi Feb 28 '18 at 8:48
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    $\begingroup$ That's correct. $\endgroup$ – Dominik Feb 28 '18 at 8:49

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