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Let $X_i$ be a random variable.

Let $\{X_i\}_{i=1}^{n}$ be a sample of observations i.i.d. over $i$ with $ \mathbb{E}(X_i)=\mu$ and $Var(X_i)=\sigma^2>0$.

Let $\bar{X}_n:=\frac{1}{n}\sum_{i=1}^{n}X_i$.

Let $\hat{\sigma}_n^2\geq 0$ be a consistent estimator of $\sigma^2$ as $n \rightarrow \infty$.

By CLT, $\frac{\bar{X}_n-\mathbb{E}(\bar{X}_n)}{\sqrt{Var(\bar{X}_n)}} =\sqrt{n}\frac{\bar{X}_n-\mu}{\sigma} \rightarrow_d N(0,1)$ as $n \rightarrow \infty$.

Using $\hat{\sigma}_n^2 \rightarrow_p \sigma^2$ as $n\rightarrow \infty$ and applying the Slutsky's Theorem, we can conclude that \begin{equation} \sqrt{n}\frac{\bar{X}_n-\mu}{\hat{\sigma}_n} \rightarrow_d N(0,1) \end{equation} as $n \rightarrow \infty$.

Let $\{\epsilon_n\}_{n=1}^{\infty}$ be a sequence of positive constants approaching zero as $n \rightarrow \infty$. Hence $\tilde{\sigma}_n^2:=\hat{\sigma}^2_n+\epsilon_n$ is a consistent estimator of $\sigma^2$ as $n \rightarrow \infty$.

Therefore, \begin{equation} \sqrt{n}\frac{\bar{X}_n-\mu}{\tilde{\sigma}_n} \rightarrow_d N(0,1) \end{equation} as $n \rightarrow \infty$.

Questions:

1) Are all steps above right (in particular the final convergence in distribution result)?

2) How fast the sequence $\{\epsilon_n\}_{n=1}^{\infty}$ approaches zero is relevant somewhere? If it is relevant, how can I set that rate of convergence?

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For 1) Yes, they are both correct.

2) no, the rate is irrelevant since its a sequence of constants that disappear in the limit.

Is there a counterexample that gave you concern?

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  • $\begingroup$ In my application (which is more complex than the example I have posted here) $\hat{\sigma}^2$ can be zero and hence I need to correct it with some sequence of positive constants converging to zero. However I don't know how to set that sequence: $\frac{1}{n}$ or $\frac{1}{n^2}$ or $\frac{1}{\sqrt{n}}$ or...? $\endgroup$
    – TEX
    Oct 31 '14 at 8:45
  • $\begingroup$ Do you think that the answer does not change if I define $\tilde{\sigma}^2_n:=\max\{\hat{\sigma}_n^2, \epsilon_n\}$? $\endgroup$
    – TEX
    Oct 31 '14 at 9:07
  • $\begingroup$ @Cris That's a cool trick (I might actually use it in some of my work :-). As long as $\hat \sigma^2_n \xrightarrow{p} \sigma^2$ then your constants will cease to be important for some $n$, although if your $\epsilon$ converge slowly, then you'll be pushing this back. $\endgroup$
    – user76844
    Oct 31 '14 at 10:25

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