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.


1 Answer 1


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 $$L(\theta,\mu|X_i) = \Phi(-\mu/\sqrt{\theta})^{I(X_i=0)} (1-\Phi((1-\mu/)/\sqrt{\theta}))^{I(X_i=1)}\phi((X_i-\mu)/\sqrt{\theta})^{I(0<X_i<1)}.$$

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

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

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

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

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.

  • 1
    $\begingroup$ 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 $\mu$ $\endgroup$ Commented Apr 21, 2020 at 17:01
  • $\begingroup$ 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 $1-\Phi(-\mu/\sqrt \theta) \to \Phi((1-\mu)/\sqrt \theta)$) $\endgroup$ Commented Apr 21, 2020 at 17:12
  • 1
    $\begingroup$ 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. $\endgroup$
    – dlnB
    Commented Apr 21, 2020 at 19:50
  • $\begingroup$ 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) $\endgroup$ Commented Sep 7, 2020 at 1:42
  • $\begingroup$ It can also be written this way: $$L(\theta,\mu|X_i) = {I(X_i=0)}\Phi(-\mu/\sqrt{\theta}) + {I(X_i=1)}(1-\Phi((1-\mu/)/\sqrt{\theta}))+{I(0<X_i<1)}\phi((X_i-\mu)/\sqrt{\theta}),$$ which would mean the joint likelihood is: $$ L(\theta,\mu|X_1,...,X_n) = \prod_i ({I(X_i=0)}\Phi(-\mu/\sqrt{\theta}) + {I(X_i=1)}(1-\Phi((1-\mu/)/\sqrt{\theta}))+{I(0<X_i<1)}\phi((X_i-\mu)/\sqrt{\theta})).$$ Both representations will give the same solution, but my technique above is less messy in the algebra and derivatives. $\endgroup$
    – dlnB
    Commented Sep 10, 2020 at 1:26

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