# Maximum Likelihood Estimator for Censored Data

Let $$X^n=(X_1,X_2,...,X_n)$$ denote a sample where

(1) $$X_i=\mathbf 1_{(\epsilon_i + \mu \geq 0)}(\mu+\epsilon_i)+\mathbf 1_{(\epsilon_i + \mu \leq 1)}(\mu+\epsilon_i)+\mathbf 1_{(\epsilon_i + \mu > 1)}$$, $$i=1,2,...,n$$;

(2) $$\epsilon_i \sim F(\cdot\,;\,\theta)$$, iid, where $$F$$ admits a smooth density $$f(\cdot\,;\,\theta)$$;

(3) $$\mu \in M$$, $$M$$ being a known closed interval on the real line.

For concreteness, let's assume $$\epsilon_i \sim N(0,\theta)$$, $$\theta >0$$, and $$M=[0,1]$$.

In sum, $$X_i$$ is censored data; in this example, censored in the unit interval. This is not the same thing as estimating the parameters of a truncated distribution because there are (potentially, depending on $$F$$ and $$M$$) two mass points at the limits of the censoring.

I want to estimate $$\mu$$ and $$\theta$$ using a ML estimator, knowing $$f$$ and given $$X^n$$.

It seems like a way to do it is to focus on $$X_i \in (0,1)$$ in which case $$X_i$$ is distributed according to a truncated normal, for which it is fairly straightforward to estimate the parameters.

However, this estimating procedure may leave out quite a bit of observations, so I was wondering how to deal with the mass points arising from the censoring in the data.

Other estimation suggestions (not MLE-based) are also welcome.

First, this is censored data, not truncated data. Throughout I will use $$\Phi(\cdot)$$ and $$\phi(\cdot)$$ to denote the standard normal distribution and density functions, respectively. We need to consider three possibilities.

First is lower censoring (observing zero), which occurs with probability $$P(\epsilon+\mu \leq 0) = P(\epsilon \leq -\mu) = \Phi(-\mu/\sqrt{\theta})$$.

Second is upper censoring (observing 1), which occurs with probability $$P(\epsilon+\mu \geq 1) = P(\epsilon \geq 1-\mu) = 1-\Phi((1-\mu/)/\sqrt{\theta})$$.

Third is observing $$X_i$$ in the interval $$(0,1)$$, for which we will just use the normal pdf $$\phi((X_i-\mu)/\sqrt{\theta})$$.

The likelihood function for observation $$i$$ is therefore $$L(\theta,\mu|X_i) = \Phi(-\mu/\sqrt{\theta})^{I(X_i=0)} (1-\Phi((1-\mu/)/\sqrt{\theta}))^{I(X_i=1)}\phi((X_i-\mu)/\sqrt{\theta})^{I(0

To get the joint likelihood, assuming independence, we take the product over all observations:

$$L(\theta,\mu|X_1,X_2,...,X_n) =$$ $$\prod_{i=1}^N\Phi(-\mu/\sqrt{\theta})^{I(X_i=0)} (1-\Phi((1-\mu/)/\sqrt{\theta}))^{I(X_i=1)}\phi((X_i-\mu)/\sqrt{\theta})^{I(0

The log-likelihood is obtained by taking the natural log of both sides:

$$\ln L(\theta,\mu|X_1,X_2,...,X_n) = \sum_{i=1}^N I(X_i=0) \ln \Phi(-\mu/\sqrt{\theta})+ I(X_i=1)\ln(1-\Phi((1-\mu/)/\sqrt{\theta}))+ I(0

Then take the partial derivatives with respect to $$\mu$$ and $$\theta$$ and solve for the system of two equations and two unknowns. This will give you your solution and does not 'omit' the censored observations.

• There are some mistakes in the answer: upper censoring $P(\epsilon + \mu \geq 1)$ not $\geq 0$; $\phi((X_i-\mu)/\sigma)$ replacing $\theta$ by $\sigma$ or mentioning somewhere that you are using $\theta$ as the std. dev. instead of the var. as in the original question; and I would also suggest making explicit the dependence of $L$ on $\mu$ – user_newbie10 Apr 21 at 17:01
• Fixed, thank you. – dlnB Apr 21 at 17:06
• But isn't $P(\epsilon+\mu\geq 1)=\Phi((1-\mu)/\sqrt \theta)\ne 1-\Phi(-\mu/\sqrt \theta)$ (i.e. would imply just changing $1-\Phi(-\mu/\sqrt \theta) \to \Phi((1-\mu)/\sqrt \theta)$) – user_newbie10 Apr 21 at 17:12
• Not quite. It should be $P(\epsilon+\mu \geq 1) = P(\epsilon \geq 1-\mu)=1 - \Phi((1-\mu)/\sqrt{\theta})$. I have corrected the typo. – dlnB Apr 21 at 19:50
• In your likelihood function for observation $i$, why is a product of three terms given rather than a sum? (They are disjoint events, so to me it seems intuitive to have a sum) – cwindolf Sep 7 at 1:42