Since this question received absolutely zero attention, here's a complete rephrase with the aim of significantly shortening it.

I have a potful of $n$ radioactive atoms. I spend $t_\text{max}$ = 1 hour recording the times when I see an atom decay. Then I'm left with $n-k$ atoms in the pot and $k$ measurements of the times: $t_1, \dots, t_k$. As you know, the distribution of these times is exponential, $p(t) = a e^{-a t}$, and I wish to find the decay rate $a$ using maximum likelihood estimation.

One way is to restrict the distribution to the interval $[0, t_\text{max}]$ and use $p(t) = a e^{-a t} / (1-e^{-a t_\text{max}})$ for the deriving the MLE.

This method doesn't use the information that there were $n-k$ atoms left after $t_\text{max}$ (which in itself would be usable for estimating $a$, where this estimate may be somewhat different from what I get from the MLE).

Question: How can I make use of the number of remaining atoms in the fitting procedure, while still using maximum likelihood estimation?

My question is in fact general, and could be asked about any distribution defined on $[0, \infty)$, not only the exponential one. The story above was just an example that made it easier to explain the problem concisely.

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    $\begingroup$ Is that any different from having censored data, where all you know for the $n-k$ atoms that didn't decay is that their decay time is greater than $t_{\mathrm{max}}$? $\endgroup$ – Scortchi - Reinstate Monica Mar 19 '14 at 15:05
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    $\begingroup$ I describe and illustrate the method of using MLE with censored data at stats.stackexchange.com/a/49456. Does that fully answer your general question? Incidentally, your specific question is also addressed using techniques of survival analysis. $\endgroup$ – whuber Mar 19 '14 at 15:16
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    $\begingroup$ Don't treat the data as truncated but as right-censored, with likelihood $[1-P(t_{\mathrm{max}})]^{n-k}\prod_{i=1}^k p(t_i)$, where $P(\cdot)$ is the cumulative distribution function. $\endgroup$ – Scortchi - Reinstate Monica Mar 19 '14 at 15:19
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    $\begingroup$ Please do not delete it! If it turns out to be a duplicate, then (because it is full of useful key words) it will most likely show up better in future searches and help lead people to the answers they seek. $\endgroup$ – whuber Mar 19 '14 at 15:23
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    $\begingroup$ @Scortchi Yesterday night I was thinking of the same what you wrote (you comment with the formula), but I made the mistake of assuming that we can't include both $1-P(t_\text{max})$ and $p(t_i)$ in the product because one is a probability and the other is a probability density, thus they are not comparable. Now I do see why they can actually both be included and this does indeed solve the problem. Thanks! (Just did a numerical experiment to verify that it works---for my peace of mind.) $\endgroup$ – Szabolcs Mar 19 '14 at 15:31