= Solution
Write $\theta=e^\beta$ for the group-1 to group-0 hazard ratio. At $t_1$, all four subjects are at risk, so subject 1 contributes $1/(2+2\theta)$. At $t_2$, subjects 2, 3, and 4 are at risk, so subject 2 contributes $\theta/(1+2\theta)$.
If $t_2<c<t_4$, subject 3 leaves before $t_4$ and subject 4, if it fails, is alone in its risk set. Hence
$$
L_p(\theta)=\frac1{2+2\theta}\frac{\theta}{1+2\theta},
$$
independently of $k$. Its log derivative vanishes when $1-2\theta^2=0$, giving
$$
\widehat\theta=\frac1{\sqrt2}.
$$
If $c=t_4$, subject 3 remains in the risk set at a failure of subject 4. Therefore
$$
L_p(\theta)=\frac1{2+2\theta}\frac{\theta}{1+2\theta}
\left(\frac1{1+\theta}\right)^k.
$$
For $k=0$, $\widehat\theta=1/\sqrt2$. For $k=1$, the score equation is $1-\theta-4\theta^2=0$, so
$$
\widehat\theta=\frac{\sqrt{17}-1}{8}\simeq0.39.
$$
Thus $\widehat\theta$ is constant at $1/\sqrt2$ for $t_2\leq c<t_4$. At $c=t_4$ it remains there when $k=0$ and drops to approximately $0.39$ when $k=1$, because only then does the censoring time change an event's risk set.
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