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Maximum Likelihood Estimator for Log Normal Distribution

Maximum Likelihood Estimator for log normal distribution, MLE, Likelihood Function
If   $X_{1}, X_{2},......X_{n} ~  LogN(\mu, \sigma^2) \tag{1}$

The pdf of log normal distribution is given by

`f(x; \mu, \sigma^2) = \frac{1}{x \sigma \sqrt{2 \pi}} \exp \left( -\frac{1}{2} \left( \frac{\log(x) - \mu}{\sigma} \right)^2 \right), \quad x > 0 \tag{2}`

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)`

this is mle for $\mu$ when $\sigma^2$ is known.

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`




Log Normal Density Curve


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