View Full Version : Can you really solve for an MLE given a fictitious distributi

03-25-2010, 09:01 PM
to estimate MLE, you need more than one observation, right? What if you don't have a density function? (It's a fictitious disitrubtion)

04-08-2010, 09:04 PM
In order to form a maximum likelihood estimate (MLE), you must have a likelihood function. If y_1, \ldots, y_n \sim f(y_i|\theta), where f(y_i|\theta) is a density function indexed by \theta, then the likelihood function is given by L(\theta|y_1,\ldots,y_n) = \prod_{i=1}^n f(y_i|\theta). Of course, without an explicit density function, we can't go much farther than this in computing an MLE.

In general, no more than one observation is required in order to compute a MLE. However, the MLE may not be unique. This problem is called unidentifiability. That is, when there are few observations, the MLE may be a poor estimate of \theta.

04-08-2010, 10:54 PM
Just to be totally confusing I wan to point out that there is
non parametric maximum liklihood (MELE) that leads to k-m and empirical cdf. Not alot to say about one observation though!! :yup:

04-09-2010, 09:09 AM
BioStatMatt is correct.
Without disitrubtion function , cann't get maximum likelihood estimate .
But for k-m , using empirical distribution , it would be better with more observations.