= Solution
Put $N=n_0+n_1+n_2$ and $s=2x=2\theta(1-\theta)$. The usual <recombination fraction> range $0\le\theta\le1/2$ gives $0\le s\le1/2$, and the preceding <likelihood function> becomes
$$
\boxed{\mathcal L(s)\propto s^{\,2n_0+n_1}(1-s)^{\,2n_2+n_1}.}
$$
To see its relation to <nonparametric linkage analysis>, use <Bayes' theorem> and the constant disease <probability> $1/16$ to obtain the affected-pair marker-sharing <probabilities>
$$
(z_0,z_1,z_2)=\left(\frac{L_0/4}{1/16},\frac{L_1/2}{1/16},\frac{L_2/4}{1/16}\right)=\bigl(s^2,\,2s(1-s),\,(1-s)^2\bigr).
$$
Their <multinomial likelihood> has exactly the same $s$-dependent kernel, since its extra factor $2^{n_1}$ is constant. Set $A=2n_0+n_1$ and $B=2n_2+n_1$. Then $A+B=2N$, and $B$ is the total number of shared ancestral copies. At fixed $N$, the <Fisher-Neyman factorization theorem> makes $B$ a <sufficient statistic>. The observed <mean> shared proportion is $\widehat p=B/(2N)$, and the <binomial likelihood> kernel gives
$$
\boxed{\widehat s=\min\!\left(\frac A{2N},\frac12\right),\qquad \widehat\theta=\frac{1-\sqrt{1-2\widehat s}}2.}
$$
The unlinked null is $s=1/2$, whereas linkage corresponds to $s<1/2$, or expected shared proportion $1-s>1/2$. If $\widehat p\le1/2$, the constrained <likelihood> is maximized at the unlinked null. For $\widehat p>1/2$, its log <likelihood> ratio is
$$
2N\left\{\widehat p\log(2\widehat p)+(1-\widehat p)\log\bigl(2(1-\widehat p)\bigr)\right\},
$$
whose derivative in $\widehat p$ is $2N\log[\widehat p/(1-\widehat p)]>0$. Hence it orders samples exactly by excess <IBD> sharing. \b[Under this fully penetrant recessive model, parametric linkage testing and a one-sided test of the overall IBD-sharing proportion use the same information.] This is <allele-sharing sufficiency for recessive linkage>, not a general equivalence for all disease models. Use limiting values for terms such as $0\log0$, and assume $N>0$ for the estimators.
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