Likelihood Function for Censored Model I have a model that results in the following data generating process:
$$x=\begin{cases}\begin{array}{c}y-\theta\\0\end{array} & \begin{array}{c}if\ y>\bar{y}(\lambda_1)\\if\ y\leq\bar{y}(\lambda_1)\end{array}\end{cases}$$
Further, $y$ is a random variable with a distribution $F(\cdot|\lambda_2)$, and $\bar{y}(\cdot)$ is an increasing function.  The parameters I want to estimate are $\theta$, $\lambda_1$, and $\lambda_2$. My problem is I am having a difficult time constructing the likelihood.  It looks like a typical censured model, which would suggest the following likelihood:
$$L=\prod \left(\frac{f(x+\theta|\lambda_2)}{1-F(\bar{y}(\lambda_1)|\lambda_2)}\right) ^{I(x>0))}*F(\bar{y}(\lambda_1)|\lambda_2)^{1-I(x>0)}$$
However, this likelihood is clearly increasing in $\lambda_1$, and hence I will not be able to identify this parameter.  However, since $\lambda_1$ controls the selection process of $x$, I'm almost certain that it should be identified.  Any ideas on what the correct likelihood should be?  What am I missing.
Thanks!
 A: From what I understand, your sample is comprised of $x_i$'s, realizations of the $X_i$-random variables.  
The density function of $X_i$ is 
$$f_{X_i}(x_i) = f(x_i+\theta|\lambda_2)\cdot {I(X_i>0)}+F(\bar{y}(\lambda_1)|\lambda_2)\cdot [{1-I(X_i>0)}] $$
Note that we do not divide the first branch by $[1-F(\bar{y}(\lambda_1)|\lambda_2)]$. This happens because we do not truncate , but we  censor . Intuitively, the probability mass is not to be allocated only in the interval having lower bound $\bar{y}(\lambda_1)|\lambda_2)-\theta$, since (I assume) we observe $x_i$'s taking zero values.
The joint density function is
$$f_X(\mathbf X \mid \theta, \lambda_1,\,\lambda_2)=\prod_{i=1}^n\Big ( f(x_i+\theta|\lambda_2) \cdot {I(X_i>0)}+F(\bar{y}(\lambda_1)|\lambda_2)\cdot [{1-I(X_i>0)}]\Big) $$
When viewed as a likelihood function, i.e. a function of the parameters given a realized sample of $X_i$'s, $\{x_1,...,x_n\}$, the "ambiguity" regarding which branch of the density of each $X_i$ should be used collapses:either the one branch or the other, dependening on whether each $x_i$ is observed as zero or not. Denote the number of observed zeros $n_0$, and the number of observed non-zeros $n_1$, $n_0+n_1=n$. Then the Likelihood function becomes 
$$L\Big(\theta, \lambda_1,\,\lambda_2 \mid \{x_1,...,x_n\}\Big)=\Big (\prod_{i=1}^{n_1} f(x_i+\theta|\lambda_2)\Big) \cdot \prod_{i=1}^{n_0}\Big (F(\bar{y}(\lambda_1)|\lambda_2)\Big) $$
$$\Rightarrow L\Big(\theta, \lambda_1,\,\lambda_2 \mid \{x_1,...,x_n\}\Big)=\Big ( \prod_{i=1}^{n_1}f(x_i+\theta|\lambda_2)\Big) \cdot \Big (F(\bar{y}(\lambda_1)|\lambda_2)\Big)^{n_0}$$
Now, indeed, the likelihood is increasing in $\lambda_1$. Why? Because the distribution function of $X_i$ has a discrete jump due to the set up, since at point $0$ it concentrates non-zero probability mass. So even if just one realization in the sample is zero, the likelihood attempts to "take account" of this discrete probability part also.
