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The negative loglikelihood is as follows:

$$\dfrac{nd}{2} \log 2\pi + \dfrac{n}{2} \log |\Sigma| + \dfrac{1}{2}\sum_{i=1}^n(x_i-\mu)^T\Sigma^{-1}(x_i-\mu) \tag{1}$$

If I take differentiation with respect to $\mu$ on $(x_i-\mu)^T\Sigma^{-1}(x_i-\mu)$, the result becomes as follows: $$2\Sigma^{-1}\mu - 2\Sigma^{-1} = 2\Sigma^{-1}(\mu-x_i) \tag{2}$$

So

$$\dfrac{\partial l(u, \Sigma)}{\partial \mu} = \dfrac{1}{2}\sum_{i=1}^n2\Sigma^{-1}(\mu-x_i) = \Sigma^{-1}\sum_{i=1}^n(\mu-x_i) \tag{3}$$

But what I can't do with the algebraic steps for the next step to get the following result:

$$\mu_{MLE}^{*} = \dfrac{1}{n}\sum_{i=1}^n x_i \tag{4}$$

How can I go from $(3)$ to $(4)$? Hope to get algebraic steps for it.

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1 Answer 1

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It's quite easy. Just equate the equation 3 to zero and solve for mu. Have a try and lets see what you get.

Have a look at this https://en.m.wikipedia.org/wiki/Maximum_likelihood_estimation

Example part. I think you get some mistake in your equations.

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  • $\begingroup$ I have missed some part in the equation. 3. I have included $\sum_{i=1}^n$. I am horrible at doing algebraic steps... $\endgroup$
    – user122358
    Commented Mar 2, 2017 at 8:32
  • $\begingroup$ Hope it is ok. $$\Sigma^{-1}\sum_{i=1}^n (\mu-x_i) = 0$$ $$n\mu-\sum_{i=1}^n x_i = 0$$ $$n\mu = \sum_{i=1}^n x_i$$ $$\mu = \dfrac{\sum_{i=1}^n x_i}{n}$$ $$\therefore \mu_{MLE}^{*} = \dfrac{1}{n}\sum_{i=1}^n x_i$$ $\endgroup$
    – user122358
    Commented Mar 2, 2017 at 8:42
  • $\begingroup$ The last equation is correct and once you equate it to zero you will gt the answer, but I need to check all you equation to make sure they are correct. $\endgroup$
    – user151245
    Commented Mar 2, 2017 at 8:44
  • $\begingroup$ You have some typo in third step. You miss x_i $\endgroup$
    – user151245
    Commented Mar 2, 2017 at 8:45
  • $\begingroup$ Thank you for having me think about the problem. Even though I don't get the straight answer, I still get the support and it feels a lot better than being alone facing the wall. Thank you for taking your time, too. $\endgroup$
    – user122358
    Commented Mar 2, 2017 at 8:50

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