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If I have a vector $\mathbf{\bar{X}}=\frac{1}{N}\sum_{i=1}^N \mathbf{X}_i$ such that it is known by CLT that $\sqrt{N}(\mathbf{\bar{X}}-\bar{\mu})$ (here $\bar{\mu}$ is the mean vector) converges in distribution to a multivariate normal $N(\mathbf{\bar{0}},\Sigma)$ what are the parameters for the limiting distribution for just $\mathbf{\bar{X}}$ if its possible to use this fact? Im not sure how to adjust for the scaling in the multivariate case

fejz1234
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  • What do you mean by parameters of limiting distribution? – StubbornAtom Jun 03 '19 at 20:59
  • @StubbornAtom I assume that it will be normally distributed, and consider its distribution a "limiting distribution". So what I mean with the parameters, is the parameters for the multivariate normal (mean vector, covariance matrix) – fejz1234 Jun 03 '19 at 21:01
  • When you say $\sqrt{N}(\mathbf{\bar{X}}-\bar{\mu})$ converges in distribution to a multivariate normal, that itself is the (non-degenerate) limiting distribution of $\mathbf{\bar{X}}$ (with proper scaling as usual). – StubbornAtom Jun 03 '19 at 21:08
  • @StubbornAtom Would you mind explain what you mean with degenerate ? – fejz1234 Jun 03 '19 at 21:13
  • Degenerate means constant. What I am trying to say is that you cannot distinguish the scaled limiting distribution and the limiting distribution of $\mathbf{\bar{X}}$. They are the same. What you can only say is that for large $N$ (i.e. without the limit), $\mathbf{\bar{X}}$ has a $N(\bar{\mu},\frac{1}{N}\Sigma)$ distribution as mentioned in the answer below. – StubbornAtom Jun 03 '19 at 21:16

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If $$\sqrt{N}(\mathbf{\bar{X}}-\bar{\mu}) \rightarrow N(\mathbf{\bar{0}},\Sigma)$$ then multiplying both sides by $\frac{1}{\sqrt{N}}$ (this is very dirty) and using $\operatorname{var}(\alpha X) = \alpha^2 \operatorname{var}(X)$,then $$(\mathbf{\bar{X}}-\bar{\mu}) \rightarrow N(\mathbf{\bar{0}},\frac{1}{N}\Sigma)$$ Add $\bar{\mu}$ on both sides, we get$$\mathbf{\bar{X}}\rightarrow N(\bar{\mu},\frac{1}{N}\Sigma)$$

As @StubbornAtom notes and i agree, the first expression is sufficient to characterize the limiting behavior.

Ahmad Bazzi
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    In the last expression you run the risk of saying the limiting distribution depends on $N$, which is why the first expression is perfectly adequate in every sense. – StubbornAtom Jun 03 '19 at 20:49
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    Yes exactly what i write is “sloppy”, that is why CLT is elegant as it is, namely the first expression, as you say @StubbornAtom – Ahmad Bazzi Jun 03 '19 at 20:51