# Censored Data Log Likelihood

Suppose we have a n random samples ($X_1,..., X_n$) from a negative exponential distribution. If lets say we have these n random samples are censored at t, such that ($X_1, ..., X_m$) are observed and ($X_{m+1}, ..., X_n$) exceeds t. The likelihood is the normal censored likelihood.

Question: is it possible to express the complete data's log likelihood as a function of the not complete data?

• Please explain what you mean by "complete data" and "not complete data." – whuber Nov 27 '15 at 16:10
• @Xi'an $f$($x$;$\theta$)=$\theta e^{-\theta x}$ for $x$ > 0 Likelihood function: $L$($\theta$;$x_1, ..., x_n$)=$\prod_{i=1}^m$$f(x_i;\theta)\prod_{i=m+1}^n (1-$$F$($\tau$,$\theta$)) – user96368 Dec 5 '15 at 22:09
• @whuber complete data: data not censored at $\tau$ thus likelihood is given by the first term in the likelihood function; not complete data: data censored at $\tau$ thus likelihood is given by the second term in the likelihood function – user96368 Dec 5 '15 at 22:13
• Doesn't the expression for $L$ in your comment answer the question? – whuber Dec 5 '15 at 22:20
• @Xi'an I believe the second product in the expression for $L$ in an earlier comment accommodates the $n-m$ censored values. – whuber Dec 8 '15 at 2:28