MLE for Mu and SD assuming Normal Distribution

Question

In my university material I have the following summary question which I believe is broken into two parts, it goes as follows:

Define the heights of the male student population as a random variable $X\sim N(µ,\sigma)$ where $µ$ is the population mean and $\sigma$ is the population standard deviation. Demonstrate how the sample average is the maximum likelihood estimator of the mean $µ$?

My lecture material has a derivation for the MLE of $\sigma^2$ which is $\frac{1}{N}\sum_i(X_i-\bar{X})^2$

I will probably get shot down in a hail of bullets for asking but here goes: is there anything to stop me from taking the square root of the MLE of Sigma for the S.D? Can I wrap the MLE for sigma in a bracket to the power of a half and call it the S.D?

Answer 1

Let me explain you what MLE means: Given your training dataset, what is the most likely "estimate" of something. In general, you calculate this by finding a value that maximizes some probability.

Therefore, you MLE estimate of sigma^2 represents the best guess of sigma^2 given this training set. If you change the training set you will get a different value of sigma^2.

So, yes, feel free to take the square root of MLE sigma^2 and call it your MLE SD. This can be justified through the invariance property of MLE:

If $\hat{\theta}$(x) is a maximum likelihood estimate for ${\theta}$, then g($\hat{\theta}$(x)) is a maximum likelihood estimate for g(${\theta}$). For example, if ${\theta}$ is a parameter for the variance and $\hat{\theta}$ is the maximum likelihood estimate for the variance, then $\sqrt{\hat{\theta}}$ is the maximum likelihood estimate for the standard deviation.

source: Watkins: "An Introduction to the Science of Statistics"

Does that make sense?