= Solution
For $y=0,1,\ldots$,
$$
\mathbb P_\theta(Y_i=y)=e^{-\theta y}(1-e^{-\theta}),
$$
so $Y_i$ is geometric with success probability $1-e^{-\theta}$. The log likelihood is
$$
n\log(1-e^{-\theta})-n\theta\overline Y.
$$
Its score equation gives
$$
e^{-\theta}=\frac{\overline Y}{1+\overline Y},
$$
and hence
$$
\widetilde\theta_n=\log\left(\frac{1+\overline Y}{\overline Y}\right).
$$
When $\overline Y=0$, the extended MLE is $+\infty$; this event has probability tending to zero.
Back to article page
