Maximum Likelihood Estimation of Population Mean(General Normal Distribution)

Question

I am having trouble finding the estimation for the population mean of a generalized normal distribution using maximum likelihood estimation. Using the definition I have:

$$l \left(\mu,\sigma^2;x_1,\dots, x_n\right)=- \frac{n}{2} \ln 2\pi - \frac{n}{2} \ln \sigma^2 - \frac{1}{2\sigma^2} \sum_{j=1}^n \left(x_j-\mu\right)^2 $$

Taking the partial derivative with respect to $\mu$, the first two terms are $0$ as they are considered constants and we have the partial derivative of the third term equal to $0$.

Doing this I get the definition of the sample mean but without the $\frac 1 n$ part. At the end of the day I do not see how $\frac 1 n$ is possible as the third term of the above expression has no $n$ variable.

Any help regarding this matter would be appreciated.

Answer 1

$$\frac{d}{dμ}\left(-\frac{1}{2σ^2}\sum_{j=1}^n(x_j-μ)^2\right)=\frac{1}{σ^2}\sum_{j=1}^n(x_j-μ)=\frac{1}{σ^2}\sum_{j=1}^nx_j-\frac{n}{σ^2}μ$$ and setting this equal to zero you get $$μ=\frac1n\sum_{j=1}^nx_j$$ So the $n$ came from $$\sum_{j=1}^nμ=nμ$$