In a text I'm reading it says that we define

$$ \begin{align} \bar{\pmb x}= \begin{bmatrix}\bar x_1 \\ \bar x_2 \end{bmatrix} \end{align} $$ And then immediately says the asymptotic covariance matrix for $\bar{\pmb x}$ is $$ \begin{align} \pmb V = N E\left[(\bar{\pmb x} - \pmb \mu)(\bar{\pmb x} - \pmb \mu)' \right] \end{align} $$

But I'm having a hard time seeing that. I'm comfortable with the covariance matrix for $\pmb x$ being $E[(\pmb x - \pmb \mu)(\pmb x - \pmb \mu)']$ but for some reason I'm having a hard time seeing how $N$ comes into the picture even though I know it gets there via the CLT.

I know that the CLT says $\bar x \sim N(\mu, \frac{\sigma^2}{N})$ but still having a hard time getting from there to the definition of the asymptotic covariance matrix.

Distributional Assumptions

$x_i$ is distributed multi-variate normal, independently, and i.i.d.


It makes sense now, the variance isn't for $\pmb x$ but for $\pmb{\bar x}$. Therefore we have:

$\bar x \sim N(\mu, \sigma^2_{\bar x})$ which when we demean and multiply by $\sqrt{N}$ we have $\sqrt{N}(\bar x-\mu) \sim N(0, N\sigma^2_{\bar x})$.


1 Answer 1


Asymptotically, $\bar{x} - \mu$ goes to zero. So the asymptotical covariance of $\bar{x}$ refers to that of $\sqrt{n}(\bar{x} - \mu)$, which is $V$, when we assume all the regularity conditions of CLT are met.

By the way, we do not really need the distributional assumption on $x$.

Edit: (Further details as requested by OP)

$V = Var(\sqrt{n}(\bar{x} - \mu)) = n Var(\bar{x}) = n E (\bar{x}-\mu)^2$ which is what you have in your textbook.

In the event that $x_t$ are i.i.d, then $n Var(\bar{x}) = Var(x)$. Maybe this is what confused you? The textbook is written down without assuming i.i.d.

  • $\begingroup$ Okay, do you think you could illustrate mathematically why we have $N$ out front for the the variance? $\bar x \sim N(\mu, \sigma^2/N$ and running the transformation described gives us $\sqrt{N}(\bar x - \mu) \sim N(0, \sigma^2)$ which still isn't $N\sigma^2$ $\endgroup$ Commented Aug 12, 2021 at 16:27
  • 1
    $\begingroup$ Please see my "edit" above. $\endgroup$ Commented Aug 12, 2021 at 21:49

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