I will really apreaciate your help with this question. I have a variable y + a that follows a log-normal distribution. I want to find proof of whether the MLE estimator for a can or can't be found. However, I am having problems to do it.
As far, I have the following:
$$f(y|a) \sim \frac{1}{\sigma\sqrt(2\pi)[{y + a}]} \exp \Big[- \frac{[\ln(y + a) - \mu]^2}{2\sigma^2}\Big]$$
The log MLE is given by:
$$L(a) = -\frac{n}{2}\ln(2\pi) - n\ln \sigma - \sum^N_{i=1}\frac{1}{2\pi\sigma^2}\Big(\ln(y_{i} + a) - \mu\Big)^2 - \sum^N_{i=1}\ln(y_{i} + a)$$
And then:
$$\frac{d\ln L(a)}{da} = \frac{1}{\pi\sigma^2}\sum^N_{i=1}\frac{\ln(y_{i} + a) + \mu}{y_{i} + a} - \sum^N_{i=1}\frac{1}{y_{i} + a}$$
However, I am not sure how to extract the constant a (even if that is possible) from:
$$\frac{1}{\pi\sigma^2}\sum^N_{i=1}\frac{\ln(y_{i} + a) + \mu}{y_{i} + a} - \sum^N_{i=1}\frac{1}{y_{i} + a} = 0$$
If you could help me with that and check if I did not make any mistake, I will really appreciate it.
Thanks
Should be $$\frac{1}{ \sigma^2} \sum_{i=1}^N \frac{-\ln(y_i+a)+\mu}{y_i+a} - \sum_{i=1}^N \frac{1}{y_i + a} = 0$$ It's not going to be possible to solve for $a$ in "closed form", but numerical methods can be used.