LOD Score Derivation and Likelihood-Ratio Interpretation

Why Compare Linkage Likelihoods?

A pedigree can provide evidence about whether a marker and a disease locus are linked. The recombination fraction, $\theta$, indexes their linkage: smaller values correspond to less recombination between the loci, while $\theta=0.5$ represents no linkage. A pedigree likelihood evaluates how well a specified value of $\theta$ accounts for the observed data under the chosen model.

The LOD score turns a comparison of two likelihoods into a convenient logarithmic scale. One likelihood is evaluated at a specified recombination fraction; the other is evaluated at $\theta=0.5$. Because likelihoods from independent pedigrees multiply, their log likelihood ratios add. This makes the score useful both for interpreting a fixed comparison and for finding the recombination fraction that gives the strongest likelihood support.

Definition: For observed pedigree data $D$, let $L(\theta;D)$ be the likelihood under a specified linkage model at recombination fraction $\theta$, with the other model assumptions held fixed. The LOD score at $\theta$ is $$ Z(\theta)=\log_{10}\!\left(\frac{L(\theta;D)}{L(0.5;D)}\right). $$ The maximum LOD score is $$ Z_{\max}=\max_{0\leq \theta\leq 0.5} Z(\theta). $$

The likelihood in the numerator and denominator must refer to the same observed data and the same underlying model apart from the value of $\theta$. In particular, the value $0.5$ is the unlinked comparison, not a claim that the observed data are equally likely under every possible model. A likelihood is a function of model parameters for the observed data; it is not a probability distribution over parameter values.

For a fixed $\theta$, $Z(\theta)=0$ when $\theta=0.5$. A positive score means that the data are more likely under that specified linkage value than under the unlinked value; a negative score means they are less likely under that comparison. The likelihood ratio is $10^{Z(\theta)}$. Thus a score of $1$ means a likelihood ratio of $10$, not a 10% probability of linkage.

Deriving the Log-Likelihood-Ratio Form

The ordinary likelihood ratio for a specified $\theta$ against no linkage is $L(\theta;D)/L(0.5;D)$. Taking its base-10 logarithm gives the LOD score directly. The logarithm changes the scale but does not change which model has the larger likelihood: a ratio above $1$ has a positive logarithm, a ratio of $1$ has logarithm $0$, and a ratio below $1$ has a negative logarithm.

Z(\theta) =\log_{10} L(\theta;D)-\log_{10} L(0.5;D)

This expression also explains how evidence from separate pedigrees can be combined. If $D_1,\ldots,D_m$ are independent pedigree data sets and the model uses the same $\theta$ for each, their joint likelihood is the product of their likelihoods. Taking the log of that product gives the sum of the separate log likelihoods.

L(\theta;D_1,\ldots,D_m)=\prod_{j=1}^{m}L(\theta;D_j) \quad\Longrightarrow\quad Z_{\mathrm{total}}(\theta)=\sum_{j=1}^{m}Z_j(\theta)

This addition applies at a common specified value of $\theta$. If each pedigree has a different value of $\theta$ that maximizes its own score, adding those individual maxima need not give the maximum for the joint data: the joint analysis uses one common value and maximizes the sum at that value.

Worked Example: Calculating a LOD Score at a Specified Value

Consider hypothetical data whose likelihood, apart from a constant that does not depend on $\theta$, is $L(\theta)=45\theta^2(1-\theta)^8$. The form represents 10 independent binary outcomes, with 2 outcomes in one class and 8 in the other. Calculate the LOD score at the specified value $\theta=0.1$.

The definition compares the likelihood at $0.1$ with the likelihood at $0.5$. The factor 45 appears in both and cancels:

\frac{L(0.1)}{L(0.5)} = \frac{45(0.1)^2(0.9)^8}{45(0.5)^2(0.5)^8} = \left(\frac{0.1}{0.5}\right)^2 \left(\frac{0.9}{0.5}\right)^8 = (0.2)^2(1.8)^8 \approx 4.408

Therefore,

Z(0.1)=\log_{10}(4.408)\approx 0.6443.

For these hypothetical data, the likelihood at $\theta=0.1$ is about 4.408 times the likelihood at $\theta=0.5$. This comparison concerns the two likelihood values under the stated model; it is not the probability that $\theta=0.1$ is true.

Maximizing the LOD Score

The maximum LOD score is obtained by choosing the permitted recombination fraction that maximizes the likelihood relative to the fixed unlinked denominator. Since the denominator $L(0.5;D)$ is constant as $\theta$ varies, maximizing the ratio is equivalent to maximizing $L(\theta;D)$. The base-10 logarithm is strictly increasing, so maximizing the ratio and maximizing its logarithm also give the same $\theta$.

\underset{0\leq\theta\leq 0.5}{\arg\max}\ Z(\theta) = \underset{0\leq\theta\leq 0.5}{\arg\max}\ \frac{L(\theta;D)}{L(0.5;D)} = \underset{0\leq\theta\leq 0.5}{\arg\max}\ L(\theta;D)

The restriction to $0\leq\theta\leq0.5$ is part of the usual recombination-fraction parameterization. A maximum can occur inside this interval or at a boundary, so the likelihood must be maximized over the permitted range rather than at an unconstrained estimate if that estimate falls outside it.

Worked Example: Deriving the Maximum LOD Score

Use the same hypothetical likelihood $L(\theta)=45\theta^2(1-\theta)^8$, now to find the maximum over $0\leq\theta\leq0.5$. The constant 45 does not affect the maximizing value. For an interior maximum, differentiate the log likelihood:

\ell(\theta)=\log L(\theta) =\log 45+2\log\theta+8\log(1-\theta), \qquad \frac{d\ell}{d\theta}=\frac{2}{\theta}-\frac{8}{1-\theta}.

Setting the derivative to zero gives

\frac{2}{\theta}-\frac{8}{1-\theta}=0 \quad\Longrightarrow\quad 2(1-\theta)=8\theta \quad\Longrightarrow\quad \hat{\theta}=0.2.

This candidate is inside the permitted interval. The second derivative is $-2/\theta^2-8/(1-\theta)^2$, which is negative throughout the interior, so the log likelihood is concave there and the stationary point is a maximum. The likelihood is zero at $\theta=0$, while the positive interior likelihood at $0.2$ exceeds the value at the other endpoint, $\theta=0.5$. Thus the constrained maximum is at $0.2$.

The maximum likelihood ratio is

\frac{L(0.2)}{L(0.5)} = \left(\frac{0.2}{0.5}\right)^2 \left(\frac{0.8}{0.5}\right)^8 = (0.4)^2(1.6)^8 \approx 6.872.

Consequently, $\hat{\theta}=0.2$ gives $Z_{\max}=\log_{10}(6.872)\approx0.8370$. The data are about 6.872 times as likely at the best-fitting permitted value of $\theta$ as at $\theta=0.5$, under this example’s likelihood model.

Combining Likelihood Contributions

The same calculation applies when a likelihood is built from multiple independent components. The numerator for the joint data is the product of the component likelihoods at the chosen $\theta$, and its unlinked denominator is the product at $0.5$. This allows a fixed-$\theta$ score to be found either from the joint likelihood or by adding component LOD scores.

Worked Example: A Common Recombination Fraction Across Two Data Sets

Suppose two hypothetical independent data sets contribute likelihoods proportional to $\theta^2(1-\theta)^8$ and $\theta(1-\theta)^4$, respectively. Constants independent of $\theta$ are omitted because they cancel in the likelihood ratio. Find the maximum LOD score for the combined data, assuming one common $\theta$.

Multiplication gives the joint likelihood up to a constant:

L_{\mathrm{joint}}(\theta)\propto \theta^2(1-\theta)^8\theta(1-\theta)^4 =\theta^3(1-\theta)^{12}.

Its log likelihood, up to an additive constant, has derivative

\frac{d}{d\theta} \left[3\log\theta+12\log(1-\theta)\right] = \frac{3}{\theta}-\frac{12}{1-\theta}.

Setting this derivative to zero gives $3(1-\theta)=12\theta$, so the maximizing value is $\hat{\theta}=3/15=0.2$. The second derivative is negative in the interior, and this value is within the permitted interval. The combined likelihood ratio is

\frac{L_{\mathrm{joint}}(0.2)}{L_{\mathrm{joint}}(0.5)} = (0.4)^3(1.6)^{12} \approx 18.01, \qquad Z_{\max}=\log_{10}(18.01)\approx1.256.

Thus the best-fitting common recombination fraction is $0.2$, and the joint likelihood at that value is about 18.01 times the likelihood at $0.5$. The two data sets contribute jointly through their product likelihood; the maximization is performed for that joint likelihood.

What the Score Does and Does Not Say

A LOD score is a scale for comparing likelihoods under specified model assumptions. It is not a p-value, a posterior probability, or a direct measure of the probability that two loci are linked. Its interpretation is conditional on the pedigree data and the model used to compute the likelihood. If model assumptions or nuisance parameters change, the likelihoods and resulting score can change as well.

The maximum score also answers a particular optimization question: which permitted $\theta$ gives the largest likelihood ratio against the unlinked value? It does not itself establish that the model is correct, and maximizing over $\theta$ does not turn the score into a probability statement. Keep the specified-$\theta$ score distinct from the maximum score: the former compares one chosen value with $0.5$, while the latter reports the strongest such comparison over the allowed range.

Key takeaway: The LOD score is the base-10 logarithm of the likelihood at a specified $\theta$ divided by the likelihood at $\theta=0.5$. Maximizing it over the permitted interval is equivalent to maximizing the linkage likelihood, because the denominator is fixed and the logarithm is increasing.

Common Pitfall

A common error is to read a LOD score of $1$ as a 10% probability that linkage is present, or to call it a p-value. That reasoning confuses a likelihood ratio with a probability about a hypothesis. A score of $1$ means only that the likelihood at the specified $\theta$ is 10 times the likelihood at $\theta=0.5$, under the stated model. It does not give the probability of either model, and it does not by itself quantify a sampling error rate.

Check Your Understanding

Answer each question using the likelihood-ratio definition and the stated model assumptions.

If $L(\theta)/L(0.5)=0.1$, what is the sign of the LOD score, and what does the comparison say?
Why does maximizing $Z(\theta)$ give the same maximizing value as maximizing $L(\theta)$ over $0\leq\theta\leq0.5$?
A LOD score is $0.5$. What is the corresponding likelihood ratio, to three significant figures?
For independent pedigree data sets with one common specified $\theta$, why can their LOD scores be added?
Why is a LOD score not the probability that linkage is present?