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For the mean $\bar{X_n}$ of n iid normal observations with $EX=\mu$ and $\operatorname{Var}X=\sigma^2$, if we take $T_n=\bar{X_n}$, then $\lim \sqrt{n}\operatorname{ Var} \bar{X_n}=\sigma^2$ is the limiting variance of $T_n$.

I don't understand this. I know $\operatorname{ Var} \bar{X}=\frac{\sigma^2}{n}$. Hence if we multiply it by $\sqrt{n}$, I think the limit will tend to 0, since we have $\frac{\sqrt{n}}{n}$.

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    $\begingroup$ $Var(aX)=a^2Var(X)$. Please note the power, i.e. $^2$. $\endgroup$
    – Tan
    Commented Mar 8, 2021 at 3:33
  • $\begingroup$ @Tan I know this equation. But what does this matter with my question? $\endgroup$
    – Yao Zhao
    Commented Mar 9, 2021 at 23:57
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    $\begingroup$ I believe it should be $\lim\text{Var}(\sqrt{n}\bar{X}_n)=\sigma^2$ $\endgroup$
    – Tan
    Commented Mar 10, 2021 at 18:25

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