 Solution

Initialize $0<{\theta }^{(0)}<1$. At each iteration first compute $y_3^{(t)}=x_3{\theta }^{(t)}/(2+{\theta }^{(t)})$ by the conditional expectation above. The M-step applies the complete-data maximum-likelihood estimator to the filled counts, givingRepeat until convergence. If all counts are zero, the likelihood function is constant and any parameter can be retained. For a nonempty sample an interior start makes the displayed denominator positive.

Here the three observed cell probabilities are $(1-\theta )/2$, $\theta /4$ and $(2+\theta )/4$, so the observed log-likelihood isFor a nonempty sample its second derivative on $(0,1)$ isThus the maximum-likelihood estimator is unique, allowing a boundary maximum. To verify that the iteration actually finds it, let $u(\theta )=x_3\theta /(2+\theta )$ and $F(\theta )=(x_2+u(\theta ))/(x_1+x_2+u(\theta ))$. Then $F^{\prime }(\theta )=x_1{u}^{\prime }(\theta )/(x_1+x_2+u(\theta ){)}^2\ge 0$, andIf the maximum is interior, this identity moves each iterate towards it, and monotonicity of $F$ prevents crossing it. The bounded monotone iterates converge; continuity makes their limit a fixed point, and strict concavity makes that fixed point the maximum. If the maximum is at zero or one, the sign of ${\ell }^{\prime }$ gives the same monotone convergence towards the appropriate endpoint. In the special case $x_1=0$, $F$ is identically one whenever its denominator is positive, so the maximum is reached in one step. An interior initialization avoids the possible spurious zero fixed point when $x_2=0$; starting at zero can otherwise trap the iteration even when the true maximum is positive.