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Given $\{X_i\}$ a sample i.i.d. observations of a $J$-dimensional random variable with $E(X_i)=\mu\in\mathbb{R}^J,V(X_i)=\Sigma\in\mathbb{R}^{J\times J}$.

Is it true that: $$ E\left[(X_i-\bar{X}_n)(X_i-\bar{X}_n)'\right]=\Sigma\equiv E\left[(X_i-\mu)(X_i-\mu)'\right] $$

Is there a compact proof for this in the case it's true? Thanks for your help! :D

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    $\begingroup$ I think you still need the correction($\frac{1}{n-1}$) as in univariate case to get the unbiased estimation. $\endgroup$ – Deep North Oct 7 '15 at 5:35
  • $\begingroup$ @DeepNorth is correct, because this is the univariate case--repeated $J\times J$ times. $\endgroup$ – whuber Oct 7 '15 at 15:38
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No.

As a very simple counter example, consider the univariate case with a single sample. Then the left side of your equation reduces down to

$E[(x - x)^2] = 0$

The right side of your equation would then reduce down to

$E[(x - \mu)^2] = \sigma^2$

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  • $\begingroup$ I believe you mean "one-sample" rather than univariate. $\endgroup$ – whuber Oct 7 '15 at 15:38
  • $\begingroup$ @whuber: actually, I do mean univariate, as in $J = 1$. But how the OP stated the problem implies one sample, as I understand it. $\endgroup$ – Cliff AB Oct 7 '15 at 16:23
  • $\begingroup$ I think most people would understand the presence of an index on "$X$", as written in the question, as indicating a sample of more than one. $\endgroup$ – whuber Oct 7 '15 at 17:02
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    $\begingroup$ @whuber: to help clarify, I added the note of a single sample $\endgroup$ – Cliff AB Oct 7 '15 at 17:51

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