Maximum Likelihood Estimator for Non Absolutely Continuous Distributions Let $\theta\in[0,1]$ be a parameter.  Suppose that $Y$ is a random variable that takes value $\theta$ with probability $1/2$ and is uniformly distributed on $[0,1]$ with probability $1/2$.  What is the maximum likelihood estimator of $\theta$ given independent observations $y_1,...,y_n$?
Stepping back from formalism, it seems very clear how to guess $\theta$ given a list of observations of $Y$.  However, I don't see how a likelihood function can be defined for this family of distributions.  
 A: Let's start by stating the intuitive answer: 
Clearly the observation value which appears most often is the maximum likelihood estimate for $\theta$.  There will be such a value with probability $1-\frac{n+1}{2^n}$ if $n \gt 1$ and with probability $1$ if $n=1$; otherwise, each observation value will appears only once with probability $\frac{n+1}{2^n}$ if $n \gt 1$ and they are each equally likely to be $\theta$ even though there is a positive probability that none of them are
Now let's try to express this in likelihood terms, noting that the probability a single observation is $\theta$ is $\frac12$ and the density for any observation is not $\theta$ is $\frac12$, though there is a fundamental distinction between probabilities and densities:
In a sense the likelihood is proportional to the product of the $n$ probability densities and probabilities associated with the observations.  But providing that none of these are zero, any non-zero probability should conceptually lead to a higher likelihood than a probability density, e.g. combining $k$ probabilities and $n-k$ probability densities has an intrinsically higher likelihood than $k-1$ probabilities and  $n-k+1$ probability densities.  You cannot quite say that a probability corresponds to an infinite density or that a finite density corresponds to a zero probability, since you may have several of each, but this does provide a weak analogy to how probabilities dominate densities 
So if you have a value $x_m$ which appears $k \gt 1 $ times together $n-k$ other distinct observations, if $\theta = x_m$ then the likelihood is proportional to $\frac1{2^n}$ with $k$ probabilities and $n-k$ densities while if $\theta \not = x_m$  then the likelihood is proportional to $\frac1{2^n}$ with $1$ or $0$ probabilities and $n-1$ or $n$ densities.  The $\theta = x_m$ case then has a higher intrinsic likelihood than any $\theta \not = x_m$, making $x_m$ the maximum likelihood estimator
On the other hand if you have $n$ distinct observations $\{x_i\}$  then if any one of them is $\theta$ then the likelihood is proportional to $\frac1{2^n}$ with $1$ probability and $n-1$ densities while if $\theta \not \in \{x_i\}$  then the likelihood is proportional to $\frac1{2^n}$ with $0$ probabilities and $n$ densities.  The former cases then have a higher intrinsic likelihood than the latter cases making each of the $\{x_i\}$ maximum likelihood estimators
