The pdf of log normal distribution is given by
The likelihood function of log Normal pdf is given by
`L(\mu, \sigma^2) = \prod_{i=1}^{n} f_X (x_i) \tag{3}`
`L(\mu, \sigma^2) = \prod_{i=1}^{n} \frac{1}{x_i \sigma \sqrt{2 \pi}} \exp \left[ -\frac{1}{2} \left( \frac{\log(x_i) - \mu}{\sigma} \right)^2 \right] \tag{4}`
`L(\mu, \sigma^2) = \frac{1}{(\sigma \sqrt{2 \pi})^n \prod_{i=1}^{n} x_i} \exp \left( -\frac{1}{2} \sum_{i=1}^{n} \left( \frac{\log(x_i) - \mu}{\sigma} \right)^2 \right) \tag{5}`
Now taking the log both sides
`\log L(\mu, \sigma^2) = -n \log(\sigma \sqrt{2 \pi}) - \sum_{i=1}^{n} \log(x_i) - \frac{1}{2} \sum_{i=1}^{n} \left( \frac{\log(x_i) - \mu}{\sigma} \right)^2 \tag{6}`
Now for finding the mle for $\mu$ when $\sigma^2$ is known we partially differentiate it with respect to $\mu$ we get.
`\frac{\partial}{\partial \mu} \log L(\mu, \sigma^2) = \frac{1}{\sigma^2} \sum_{i=1}^{n} \left( \mu - \log(x_i) \right) \tag{7}`
now putting it equal to zero
`\frac{1}{\sigma^2} \sum_{i=1}^{n} \left( \mu - \log(x_i) \right) = 0\tag{8}`
`\hat {\mu}_{MLE} = \frac{1}{n} \sum_{i=1}^{n} \log(x_i)`
Now for finding the mle for $\sigma^2$ when $\mu$ is known we partially differentiate it with respect to $\sigma^2$ we get.
`\log L(\mu, \sigma^2) = -n \ln(\sigma \sqrt{2 \pi}) - \sum_{i=1}^{n} \ln(x_i) - \frac{1}{2} \sum_{i=1}^{n} \left( \frac{\ln(x_i) - \mu}{\sigma} \right)^2 \tag{10}`
`\frac{\partial}{\partial \sigma^2} \log L(\mu, \sigma^2) = \frac{-n}{2\sigma^2} - \sum_{i=1}^{n} \frac{\ln(x_i) - \mu}{\sigma^3} \tag{11}`
now putting it equal to zero to
`\frac{-n}{2\sigma^2} - \sum_{i=1}^{n} \frac{\ln(x_i) - \mu}{\sigma^3} = 0 \tag{12}`
`\hat{\sigma}_{MLE}^2 = \frac{1}{n} \left( \sum_{i=1}^{n} (\ln(x_i) - \mu) \right)^2 `
this is he MLE for $\sigma^2$ when $\mu$ is known but when $\mu$ is unknown the mle of $\sigma^2$ is given by
`\hat{\sigma}_{MLE}^2 = \frac{1}{n} \left( \sum_{i=1}^{n} (\ln(x_i) -\overline{\log(x_{n})}) \right)^2`