Let $X_1, X_2, \ldots, X_n$ be i.i.d exponential random variables with parameter $\lambda$, where the form of the distribution for each $x_i$ is
$$f(x_i) =\lambda e^{-\lambda x_i}.$$
The m.l.e of $\lambda$ when there isn't censoring is $\frac{n}{x_1+x_2+\dots+x_n}$.
The expectation of this estimate ($=\frac{n\lambda}{n-1}$) can be directly calculated using the fact that the sum of iid exponentials is distributed as gamma (with parameters $n$ and $\lambda$ in this example).
Now, suppose only the first $r$ events will be collected and the remaining $n-r$ will be censored. The form of the m.l.e of $\lambda$ will be
$\frac{r}{(x_1+\dots+x_r) + (x_{r+i}+\dots+x_n)}$, where the observations $x_1,\ldots,x_n$ are observed event times and the $x_{r+1},\ldots,x_n$ are the censored times. Is there a direct way to calculate the expectation of this estimate?
EDIT - I was able to find information about exponential order statistics that enabled me to calculate the expectation directly (Cox and Oakes, "Analysis of Survival Data"...page 38)