Let $U_1 \dots U_n $ be a i.i.d sample form an exponential distribution with parameter $\lambda$. We only observe: $$Z_i = \mathbf{1}_{(U_i<c)}$$ Find the likelihood for $\lambda$.
I thought the following:
$$Z_i = \begin{cases} 1 & \mbox{if }\; U_i < c \\ 0 & \mbox{if }\; U_i \geq c \end{cases}$$ implies that $P(Z_i=0)=P(U_i <c) $ and $P(Z_i=1)=P(U_i \geq c) $ and from this I can compute the likelihood. Since $Z_i$ can only take values $0$ or $1$, if $n_0$ is the number of times in which $z_i$ take value $0$ and $n_1=n-n_0$, I can write the likelihood for $\lambda $ as: $$L(Z,\lambda)= \prod^{n_1}_ {i=1}\left(1-e^{-\lambda c}\right) \prod^{n_0}_ {i=1}\left(e^{-\lambda c}\right)$$
Is it correct?
As Did laconically remarked, the answer is "Yes."
Note that you can also write the product as
$$ L(Z,\lambda)=\left(\mathrm e^{-\lambda c}\right)^{n_0}\left(1-\mathrm e^{-\lambda c}\right)^{n_1}\;. $$