$X_i \sim N(\mu, \sigma^2)$ (iid), $i = 1,2,...,N$, I want to estimate $\theta = (\mu, \sigma^2)$. Problem is, I don't observe $x_i$. For each $i$, I only observe $(a_i, b_i)$, and I know that $a_i < X_i < b_i$. To use maximum likelihood, I need to know the distribution of $(A, B) | \theta$. Aside from knowing that $a_i < X_i < b_i$, I can't say much about $a$ and $b$ -- their distributions are pretty complicated.

On the other hand, it's straightforward that $$\Pr(a_i < X_i < b_i | \theta, a_i, b_i) = F(b_i; \theta) - F(a_i; \theta),$$ where $F()$ is the normal cdf, correct? So $$\Pr(a_1 < X_1 < b_1, a_2 < X_2 < b_2, ... | \theta, ...) = \prod \left( F(b_i; \theta) - F(a_i; \theta) \right).$$ Intuitively, I think that I would get a good estimate of $\theta$ by finding the $\theta$ that maximizes this probability. But this is not maximum likelihood, is it?

Does this problem have a name? Does my intuitive approach have any merit? Does it have a name? Are there any references for this problem that I can study?

  • 1
    $\begingroup$ Looks related to interval censoring defined here. $\endgroup$ Commented Mar 10, 2019 at 6:55

1 Answer 1


Thanks to @StubbornAtom for pointing me in the right direction. The problem is called Interval Censored Data Analysis. My intuition about the likelihood appears to be correct -- see slide 55. The problem is solved by nonparametric MLE. The R package that does this is simply called interval.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.