# MLE for Mu and SD assuming Normal Distribution

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?

• Your last sentence doesn't make sense -- it's $\sigma^2$ you would be wanting to take square root of, not $\sigma$. Please fix Commented May 8, 2016 at 6:27

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.

Does that make sense?

• You haven't said why it's o.k. to "feel free to take the square root of MLE sigma^2 and call it your MLE SD." To the OP: look up invariance property of maximum likelihood estimation. I won't give you a link because you'll learn more by tracking it down and understanding it yourself, rather than having it handed to you on a platter. Commented May 7, 2016 at 22:16
• Thanks @MarkL.Stone. I suppose after sometime we should edit the answer to reflect the invariance property. Commented May 7, 2016 at 22:21