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I am wondering the correct way to calculate the covariance matrix found by using Maximum Likelihood Estiamtion. The following equation is the result after having followed algebraic steps:

$\Sigma_{MLE}^*=\dfrac{1}{n}\sum_{i=1}^n(x_i-\mu_{MLE}^*)(x_i-\mu_{MLE}^*)^T$

And I interpreted the above equation into matrix form as follows:

$\Sigma_{MLE}^* = \dfrac{1}{n}\begin{bmatrix}X_1-\mu_1^* \\ \\ X_2-\mu_2^*\\ \\ \vdots \\ \\ X_n-\mu_n^*\end{bmatrix}\begin{bmatrix}X_1-\mu_1^* & X_2-\mu_2^* & \dots &X_n-\mu_n^*\end{bmatrix}$

$= \dfrac{1}{n}\begin{bmatrix}(X_1-\mu_1^*)(X_1-\mu_1^*) & (X_1-\mu_1^*)(X_2-\mu_2^*) & \dots & (X_1-\mu_1^*)(X_n-\mu_n^*) \\ (X_2-\mu_2^*)(X_1-\mu_1^*) & (X_2-\mu_2^*)(X_2-\mu_2^*) & \dots & (X_2-\mu_2^*)(X_n-\mu_n^*) \\ & \vdots \\ (X_n-\mu_n^*)(X_1-\mu_1^*) & (X_n-\mu_n^*)(X_2-\mu_2^*) & \dots & (X_n-\mu_n^*)(X_n-\mu_n^*)\end{bmatrix}$

I wonder if I have understood correctly, hope to hear some advice if I am wrong.

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  • $\begingroup$ Your very first equation is already the maximum-likelihood estimator of the population covariance. $\endgroup$ – SmallChess Mar 13 '17 at 9:28
  • $\begingroup$ Do you want to prove why your first equation is the MLE estimator? Is this what you're asking? $\endgroup$ – SmallChess Mar 13 '17 at 9:28
  • $\begingroup$ If possible, I'd like to look at the proof in detail as long as if I can follow the math. But at least, I'd like to know if I have denoted the matrix form equation correctly for maximum likelihood estimation case. $\endgroup$ – user122358 Mar 13 '17 at 9:31
  • $\begingroup$ Notice that what you are calculating is covariance matrix using the maximum likelihood estimate of $\mu$. It does not prove that your estimate is a MLE since you have not proven that it actually maximizes the likelihood! Moreover, you didn't even stated what likelihood you are maximizing (supposedly it's multivariate normal distribution, but it needs to be said explicitly, since there is no such a thing as maximizing any likelihood). $\endgroup$ – Tim Mar 13 '17 at 9:35
  • $\begingroup$ I assume a multivariate normal distribution. $\endgroup$ – user122358 Mar 13 '17 at 9:38

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