For our sample $w_1, w_2, \cdots, w_n$, the [likelihood function](https://en.wikipedia.org/wiki/Likelihood_function) is given by

$$

\begin{aligned}

L(\mu, \sigma | w_i) = \prod_{i=1}^{n} f(w_i, \mu, \sigma) \\

\end{aligned}

L(\mu, \sigma | w_i) = \prod_{i=1}^{n} f(w_i, \mu, \sigma)

$$

The likelihood function can be viewed as both

Let's first find the maximum likelihood estimate (MLE) of $\mu$

$$

\begin{aligned}

\frac{\delta \ell}{\delta \mu}

= - \frac{1}{2\sigma^2} \times 2 \sum_{i=1}^n (\ln w_i - \mu) = 0 \\

\implies \sum_{i=1}^n \ln w_i - n \mu = 0 \\

\implies \hat{\mu} = \frac{\sum_{i=1}^n \ln w_i}{n}

\end{aligned}

$$

Now let's find the MLE of $\sigma$

$$

\begin{aligned}

\frac{\delta \ell}{\delta \sigma^2}

= - \frac{n}{2\sigma^2} + \frac{1}{2\sigma^4}

\sum_{i=1}^n (\ln w_i - \mu)^2 = 0 \\

\implies \frac{n}{2\sigma^2} =

\frac{1}{2\sigma^4} \sum_{i=1}^n (\ln w_i - \mu)^2 \\

\implies \hat{\sigma} =

\left( \frac{\sum_{i=1}^{n}(\ln w_i - \hat{\mu})^2}{n} \right)^{1/2}

\end{aligned}

$$

Now that we have derived the expressions for $\hat{\mu}$ and $\hat{\sigma}$,