How do I take the log, and then the derivative, of this liklihood function?

Question

I have $X_1 ... X_n \sim N(\mu, \sigma^2)$, the overall goal is to find MLE for $\sigma$.

I made this Liklihood function: $$L(\sigma) = \prod_{i=1}^{n} \frac{1}{\sqrt{2\pi\sigma^2}} e^{-\frac{(x_i-\mu)^2}{2\sigma^2}}$$

Now I want to find what $\ell(\sigma)$ which is taking the log of the liklihood function.

The log of the second part of the function is gonna be simply $$- \frac{1}{2\sigma^2} \sum_{i=1}^2 (x_i - \mu)^2$$

What about the first part?

And then, how do I take the derivative of $\ell '(\sigma)$ ?

Answer 1

Since $$\mathcal L(\sigma \mid \boldsymbol x, \mu) = \prod_{i=1}^n (2\pi \sigma^2)^{-1/2} e^{-(x_i - \mu)^2/(2\sigma^2)} = (2\pi)^{-n/2} \sigma^{-n} e^{-\sum (x_i - \mu)^2/(2\sigma^2)},$$ then $$\ell(\sigma \mid \boldsymbol x, \mu) = \log \mathcal L(\sigma \mid \boldsymbol x, \mu) = -\frac{n}{2} \log 2\pi - n \log \sigma - \frac{1}{2\sigma^2} \sum_{i=1}^n (x_i - \mu)^2.$$ Then $$\frac{\partial \ell}{\partial \sigma} = - \frac{n}{\sigma} + \frac{1}{\sigma^3} \sum_{i=1}^n (x_i - \mu)^2.$$ The rest of the computation I leave as an exercise.