= Solution
Let $A\in\{0,1\}$ and $B\in\{0,1\}$ encode the <alleles> at two loci on a randomly chosen population <haplotype>. Write $p_{ab}=P(A=a,B=b)$, $p_A=p_{10}+p_{11}$ and $p_B=p_{01}+p_{11}$. The <linkage disequilibrium> coefficient is
$$
\boxed{D=p_{11}-p_Ap_B=p_{11}p_{00}-p_{10}p_{01}.}
$$
The determinant identity follows by substituting the marginal frequencies and $\sum p_{ab}=1$. All four <haplotype> frequencies can be recovered as
$$
\begin{aligned}
p_{11}&=p_Ap_B+D,&p_{10}&=p_A(1-p_B)-D,\\
p_{01}&=(1-p_A)p_B-D,&p_{00}&=(1-p_A)(1-p_B)+D.
\end{aligned}
$$
Thus $D=0$ is precisely <independence> of the two allelic states. Its sign depends on which <alleles> are labeled one, and its possible magnitude depends on the marginal frequencies.
One other measure is squared <allelic correlation>. For segregating loci, $0<p_A,p_B<1$, their indicator-variable <covariance> is $D$, and their <variances> are $p_A(1-p_A)$ and $p_B(1-p_B)$. Hence
$$
\boxed{r^2=\frac{D^2}{p_A(1-p_A)p_B(1-p_B)},\qquad 0\le r^2\le1.}
$$
This is the squared <correlation>, useful for measuring how well one marker predicts the other in <association mapping>. It is undefined if a locus is monomorphic. <Linkage disequilibrium> describes population <haplotype> frequencies; <genetic linkage> describes transmission within <meioses>. Physically linked loci can have $D=0$, and population mixing can produce LD even between unlinked loci.
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