TDT Effect Estimates and Confidence Intervals

From TDT Evidence to an Effect Estimate

The transmission disequilibrium test (TDT) asks whether an allele is transmitted to affected offspring more often than expected under Mendelian transmission. Its null test summarizes evidence against equal transmission. An effect estimate addresses a different question: how imbalanced are the informative transmissions? This tutorial defines a conditional allelic odds ratio for that imbalance and shows how to quantify its uncertainty.

The estimate is based on informative parental transmissions, usually transmissions from a heterozygous parent: that parent could transmit either allele, so the transmitted allele can be compared with the allele not transmitted. The quantity estimated here is a conditional transmission odds ratio. It is not automatically an absolute disease-risk ratio, a population allele-frequency ratio, or a causal effect.

Definition: Let $T$ be the number of informative transmissions of the allele being studied and $U$ the number of informative nontransmissions of that allele. The estimated conditional allelic odds ratio is $$ \widehat{\theta}=\frac{T}{U}. $$ It compares the observed odds of transmitting the allele with the odds of not transmitting it, conditional on the informative parental genotypes and the sampling setup. Equal transmission corresponds to $\theta=1$.

Model and Assumptions

For the basic calculation, suppose there are $N=T+U$ informative transmissions, each classified as either transmission of the allele ($T$) or nontransmission ($U$). Use the conditional binomial model $$ T\mid N\sim\operatorname{Binomial}(N,q), $$ where $q$ is the conditional probability that an informative transmission carries the allele. The conditional transmission odds ratio is $$ \theta=\frac{q}{1-q}. $$ Under the Mendelian transmission null, $q=0.5$, so $\theta=1$. If $\theta>1$, the allele was transmitted more often than not among the informative transmissions; if $\theta<1$, it was transmitted less often.

The usual binomial standard error and Wald interval require more than just counts. Informative transmissions must be independent, or the analysis must otherwise account for their dependence; the transmissions must share a common $q$ for the simple common-effect model; and both $T$ and $U$ must be large enough for the log-odds estimate to be approximately normal. A straightforward setting is one informative affected child per independent family, with the relevant parental transmissions treated as independent under the conditional model. When several affected siblings contribute from the same family, their transmissions may be dependent; treating every transmission as independent can understate uncertainty. A cluster-aware or other appropriate method is then needed rather than mechanically applying the simple formula.

The estimate is obtained by replacing $q$ with $\hat{q}=T/N$. Therefore, $$ \widehat{\theta}=\frac{\hat{q}}{1-\hat{q}}=\frac{T/N}{U/N}=\frac{T}{U}. $$ A confidence interval is most conveniently constructed on the log scale, where the estimate has an approximately normal sampling distribution.

Worked Example: A Modest Excess of Transmissions

Suppose a hypothetical set of independent informative transmissions gives $T=36$ transmissions and $U=24$ nontransmissions. Assume the transmissions satisfy the common-$q$ binomial model and are sufficiently numerous for a Wald interval.

The estimated conditional odds ratio is $$ \widehat{\theta}=\frac{36}{24}=1.50. $$ Thus, in this hypothetical data set, the observed odds of transmitting the allele are 1.50 times the odds of not transmitting it.

The estimated standard error of the log odds ratio is $$ SE\{\log(\widehat{\theta})\}=\sqrt{\frac{1}{T}+\frac{1}{U}} =\sqrt{\frac{1}{36}+\frac{1}{24}} =\sqrt{0.0694444}=0.2635. $$ Since $\log(1.50)=0.405465$, the 95% log-scale Wald interval is $$ 0.405465\pm1.96(0.263523)=[-0.1110,\ 0.9220]. $$ Exponentiating both endpoints gives an interval of approximately $$ [\exp(-0.1110),\ \exp(0.9220)]=[0.8949,\ 2.514]. $$

Under the model, this interval is a plausible range for the conditional transmission odds ratio at the 95% confidence level. It includes 1, so these data do not rule out equal conditional transmission odds at the corresponding two-sided 5% level. The interval is fairly wide: the data are compatible with a small deficit as well as a substantial excess.

Constructing a Wald Confidence Interval

The construction uses the estimated log odds ratio and its standard error. For $T>0$ and $U>0$, the estimate is $\log(\widehat{\theta})=\log(T/U)$. The estimated standard error is $$ SE\{\log(\widehat{\theta})\}=\sqrt{\frac{1}{T}+\frac{1}{U}}. $$ For a two-sided confidence level $1-\alpha$, the approximate interval on the log scale is $\log(\widehat{\theta})\pm z_{1-\alpha/2}SE\{\log(\widehat{\theta})\}$, where $z_{1-\alpha/2}$ is the corresponding standard normal quantile. Exponentiating the two endpoints returns the interval to the odds-ratio scale.

Calculate the estimate.
Use $\widehat{\theta}=T/U$, and record what allele and informative transmissions the counts describe.

Calculate uncertainty on the log scale.
Use $SE=\sqrt{1/T+1/U}$, provided the independent-binomial and large-sample conditions are reasonable.

Form and transform the interval.
Calculate $\log(\widehat{\theta})\pm z_{1-\alpha/2}SE$, then exponentiate both endpoints.

The standard error follows from a short approximation. Under the binomial model, the estimated transmission probability is $\hat q=T/N$, with estimated variance approximately $\hat q(1-\hat q)/N$. The derivative of the logit function $\log\{q/(1-q)\}$ with respect to $q$ is $1/[q(1-q)]$. The delta method therefore gives $$ \operatorname{Var}\{\log(\hat q/(1-\hat q))\} \approx \left\{\frac{1}{\hat q(1-\hat q)}\right\}^{2} \frac{\hat q(1-\hat q)}{N} =\frac{1}{N\hat q(1-\hat q)} =\frac{1}{T}+\frac{1}{U}. $$ The last equality uses $T=N\hat q$ and $U=N(1-\hat q)$. Taking the square root yields the stated estimated standard error.

Worked Example: More Informative Transmissions

In a second hypothetical data set, $T=72$ and $U=48$. Assume independent informative transmissions and a common conditional transmission probability, with enough observations for the normal approximation.

The estimated conditional odds ratio is $$ \widehat{\theta}=\frac{72}{48}=1.50,\qquad \log(\widehat{\theta})=\log(1.50)=0.405465. $$ Although the estimate matches the first example, the larger counts should produce a smaller standard error.

Indeed, $$ SE=\sqrt{\frac{1}{72}+\frac{1}{48}} =\sqrt{0.0347222}=0.186339. $$ The 95% log-scale interval is $$ 0.405465\pm1.96(0.186339)=[0.04024,\ 0.77069]. $$ After exponentiation, the 95% interval for the conditional odds ratio is approximately $$ [\exp(0.04024),\ \exp(0.77069)]=[1.041,\ 2.161]. $$

The point estimate is again 1.50, but the interval is narrower because there are more informative transmissions. This interval lies above 1, indicating that under the stated model it is inconsistent with equal transmission odds at the two-sided 5% level. More observations increased precision; they did not change the estimated effect in this example.

Effect Estimation Is Not the TDT Null Test

A confidence interval describes uncertainty about an effect parameter; a null test evaluates a specified hypothesis. For the simple TDT, the null is $H_0:q=0.5$, equivalently $H_0:\theta=1$. The familiar one-degree-of-freedom asymptotic TDT statistic can be written $(T-U)^2/(T+U)$ under this null. Its p-value measures how surprising a statistic at least this extreme would be if the null model held; it is not the probability that the null is true. The odds ratio and its interval retain the direction and magnitude of the estimated imbalance, which the test statistic alone does not convey.

For a Wald interval, whether 1 is included corresponds approximately to whether the two-sided Wald test of $\log(\theta)=0$ rejects at the matching level. But this is not a guarantee that every test and interval will agree exactly: exact and asymptotic procedures can differ, and different variance assumptions can produce different results. Most importantly, the interval is not a substitute for specifying the estimand and the assumptions that make the calculation meaningful.

Worked Example: An Estimate Below One and a Nonsignificant TDT

Suppose a third hypothetical study has $T=18$ and $U=30$ independent informative transmissions. Use the same common-$q$ binomial model and large-sample Wald approximation.

The conditional odds-ratio estimate is $$ \widehat{\theta}=\frac{18}{30}=0.6000,\qquad \log(\widehat{\theta})=\log(0.6000)=-0.510826. $$ The log-scale standard error and 95% interval are $$ SE=\sqrt{\frac{1}{18}+\frac{1}{30}}=\sqrt{0.0888889}=0.298142, $$ $$ -0.510826\pm1.96(0.298142)=[-1.09518,\ 0.07353]. $$ Exponentiating gives an interval of approximately $[0.3345,\ 1.076]$.

The estimate indicates that the observed odds of transmitting the allele are 0.60 times the odds of not transmitting it in this hypothetical sample. The interval includes 1, so it does not establish a difference from equal transmission at the two-sided 5% Wald level. For comparison, the asymptotic TDT statistic is $$ \frac{(T-U)^2}{T+U}=\frac{(18-30)^2}{48}=3.000, $$ which has an approximate chi-square p-value of 0.0833 with one degree of freedom. This test result is consistent with the Wald interval's inclusion of 1. The estimated direction is still informative, but the data do not give sufficiently precise evidence to distinguish it from equal transmission at that threshold.

Limits and Common Pitfalls

The confidence interval quantifies uncertainty in the conditional transmission odds ratio under its model. It does not by itself establish that the allele causes disease, and it should not be relabelled as a general population disease odds ratio without additional modeling and assumptions. Its scope is defined by which families and transmissions are informative and by how the conditional sampling model represents them.

A second limit is the Wald approximation. It can perform poorly when counts are small or one count is zero: the log estimate or the standard error formula may be undefined, and the normal approximation may be unreliable even with a very large-looking odds ratio. In such cases, a confidence interval obtained by inverting the appropriate binomial test, or another justified method, may be preferable. Any alternative still needs to match the conditional model and handle dependence appropriately.

Pitfall: calling $T/U$ the probability of transmission. It is an odds ratio. The estimated probability is $T/(T+U)$; the odds are that probability divided by its complement.
Pitfall: interpreting an odds ratio of 1.5 as a 50% increase in disease risk. Here it means 1.5 times the conditional odds of transmission versus nontransmission in the specified informative set. It is not, without further assumptions, a 50% increase in disease risk.
Pitfall: treating every child as independent by default. Multiple transmissions from a family may be correlated. If the independence assumption is not defensible, the simple standard error can be too small.
Pitfall: treating a p-value as an effect size. A small p-value is evidence against the null under the test assumptions; it does not state how large the conditional transmission imbalance is. Report the estimate and interval as well.
Pitfall: saying an interval gives a 95% probability that the fixed parameter lies inside it. In the frequentist interpretation, the method produces intervals that cover the fixed parameter in 95% of repeated samples under its assumptions. The calculated interval is not a 95% posterior probability statement.

Key takeaway: For informative transmission counts $T$ and $U$, the conditional allelic odds-ratio estimate is $T/U$. Its standard Wald interval is constructed on the log scale using $\sqrt{1/T+1/U}$, then exponentiated. Interpret both the estimate and interval as conditional transmission quantities, and check independence, common-effect, and large-sample assumptions before relying on them.

Check Your Understanding

Try these questions using the definitions and assumptions in this tutorial.

If $T=40$ and $U=20$, what is the estimated conditional allelic odds ratio, and what does its value mean in words?
Why is the standard error for the log odds ratio calculated from both $T$ and $U$, rather than from $T+U$ alone?
What does it mean, in context, if a confidence interval for the conditional odds ratio includes 1?
Why can multiple affected siblings from the same family make the simple binomial standard error inappropriate?
Explain why an odds ratio of 1.5 in this analysis is not automatically a 50% increase in disease risk.