= Solution
Let $T_i$ be an event time and $C_i$ a potential censoring time. Under <independent censoring>, observe $Y_i=\min(T_i,C_i)$ and $\delta_i=\mathbf1\{T_i\leq C_i\}$. For an <exponential distribution> of rate $\theta$, an observed failure contributes $\theta e^{-\theta Y_i}$ to the <survival likelihood>, whereas a <right-censored> observation contributes $e^{-\theta Y_i}$. The censoring-law factors may be discarded when they are independent of $\theta$. Thus, writing $d=\sum_i\delta_i$ and $V=\sum_iY_i$ for total observed <person-time>,
$$
L(\theta)\propto\theta^d e^{-\theta V},\qquad
\ell'(\theta)=\frac d\theta-V,\qquad
\ell''(\theta)=-\frac d{\theta^2}.
$$
For $d,V>0$ the unique <maximum-likelihood estimate> is
$$
\boxed{\widehat\theta=\frac dV.}
$$
Censored times belong in $V$: they record time during which failure was not observed. With $d=0$ and $V>0$, the <likelihood> decreases with $\theta$, so the supremum is at the boundary $\theta\downarrow0$; there is no positive interior estimate. This is <exponential-rate estimation from censored exposure>.
<Uninformative censoring> means that the potential censoring time supplies no information about the latent failure time after conditioning on the modeled variables. A study ending at a fixed follow-up duration independent of prognosis is an example of <administrative censoring>. <Informative censoring> occurs, for example, when a seriously deteriorating patient withdraws because of that deterioration and the prognostic information is omitted from the model. Under <informative censoring>, those remaining in the <risk set> are selected by their future failure propensity, so the ordinary <survival likelihood> and its estimate need not target the true rate.
Using the reported durations and flags as though they were valid <right-censored> records gives
$$
\boxed{\widehat\theta_{\rm labelled}=\frac{500}{1031.6}\simeq0.48468.}
$$
Treating all durations as failures instead gives
$$
\boxed{\widehat\theta_{\rm complete}=\frac{1000}{1031.6}\simeq0.96937.}
$$
Relabelling has halved the event count without shortening any observed follow-up. The resulting estimate is therefore exactly half the complete-data estimate.
\b[This is not a simulation of uninformative ordinary right-censoring.] For a record declared censored at $Y_i$, ordinary <right censoring> requires $T_i>Y_i$; the construction instead took $Y_i$ to be the simulated failure time itself. Keeping a failure time and hiding only its flag does not produce the minimum of that time and an independent earlier censoring time. Even choosing labels at random does not fix this: independence of a flag from the recorded duration is different from <independent censoring> of a latent event time. Indeed the labelled durations still have mean one while the failure fraction is one half, so their ordinary censored estimate converges to $1/2$, not the generating rate one. Thus it is not a valid uninformative censoring scheme for the intended failure distribution; strictly, it has not implemented ordinary right-censoring of those simulated failure times at all. One could reproduce the randomly labelled record distribution with independent latent event and censoring times both of rate $1/2$, but that would be a different failure model, not the claimed rate-one model. Calling the flags random is therefore insufficient to justify the target distribution or its censored likelihood. This is the <random relabelling is not independent right censoring> issue.
A valid simulation draws independently $T\sim\operatorname{Exp}(1)$ and $C\sim\operatorname{Exp}(1)$, and records $(\min(T,C),\mathbf1\{T\leq C\})$. Since
$$
\Pr(C<T)=\int_0^\infty e^{-c}e^{-c}\,dc=\frac12,
$$
it gives <independent censoring> with the requested probability. Equivalently generate $T=-\log U$ and $C=-\log V$ from independent uniform random numbers. Another valid choice is fixed censoring at $\log2$, since $\Pr(T>\log2)=1/2$. These give a random censored fraction with expectation one half; they do not promise exactly 500 censored records. For the independent-exponential scheme, observed durations have rate two and mean one half, so $d/\sum Y_i$ correctly converges to one.
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