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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?

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  • $\begingroup$ Please explain what you mean by "complete data" and "not complete data." $\endgroup$ – whuber Nov 27 '15 at 16:10
  • $\begingroup$ @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$)) $\endgroup$ – user96368 Dec 5 '15 at 22:09
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    $\begingroup$ @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 $\endgroup$ – user96368 Dec 5 '15 at 22:13
  • $\begingroup$ Doesn't the expression for $L$ in your comment answer the question? $\endgroup$ – whuber Dec 5 '15 at 22:20
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    $\begingroup$ @Xi'an I believe the second product in the expression for $L$ in an earlier comment accommodates the $n-m$ censored values. $\endgroup$ – whuber Dec 8 '15 at 2:28
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Literally, the answer to your question "is it possible to express the complete data's log likelihood as a function of the not complete data? " is NO. For censored data, there will be many (infinitely) many ways to "complete" them. Maybe you could ask what is the expected complete data likelihood, when conditioning with the censored data. That way of thinking leads to the EM-algorithm.

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