Exact and Asymptotic Calibration of the TDT

Why Calibration Matters for the TDT

The transmission disequilibrium test (TDT) compares how often a particular allele is transmitted to an affected child with how often it is not transmitted. The statistic is only part of the test: calibration specifies how unusual that statistic would be under the null hypothesis. With few informative transmissions, the null distribution is discrete, so a large-sample approximation can produce a noticeably different p-value from an exact calculation.

This tutorial compares the exact binomial test with the usual uncorrected chi-squared TDT. Both use the same pair of counts, but they answer the calibration question differently: one evaluates probabilities under the finite-sample null distribution, while the other uses a continuous large-sample approximation.

Counts and the Conditional Null Model

Definition: For the basic TDT, let \(b\) be the number of informative transmissions of the allele being tested to affected offspring, and let \(c\) be the number of corresponding nontransmissions from heterozygous parents. The total number of informative transmissions is \(m=b+c\). Under the null hypothesis of equal transmission, and assuming the informative transmissions are independent, \(B\), the number transmitted, has distribution \(B\sim\operatorname{Binomial}(m,1/2)\). Thus \(C=m-B\).

An informative transmission occurs when a heterozygous parent can transmit either of two alleles, so the transmitted allele can be compared with the allele that was not transmitted. The basic calibration assumes that, conditional on the informative parental genotypes and the relevant family design, each transmission has probability \(1/2\) of being the allele being tested. It also assumes independent informative transmissions. For the simple calculations here, families are independent and each counted transmission contributes one independent Bernoulli trial. If the data include dependent transmissions, the simple binomial and chi-squared calibrations may not be appropriate without adjustment.

Under the null, \(E(B)=m/2\) and \(\operatorname{Var}(B)=m/4\). A value \(b\) much larger or smaller than \(m/2\) is evidence against equal transmission. A two-sided test is appropriate when departures in either direction count as evidence.

Worked Example: An Exact Test With Few Informative Transmissions

Suppose a hypothetical TDT has \(b=9\) transmissions and \(c=1\) nontransmission, so \(m=10\). Under the stated null model, \(B\sim\operatorname{Binomial}(10,1/2)\). Calculate the two-sided exact p-value.

The observed count is nine, so outcomes at least as far from the null center of five are \(B\geq 9\) or \(B\leq 1\). By symmetry of the binomial distribution at probability \(1/2\), these two tails have equal probability:

p_{\text{exact}} =P(B\leq 1)+P(B\geq 9) =2P(B\leq 1) =2\sum_{k=0}^{1}\binom{10}{k}(1/2)^{10}

Substituting the binomial coefficients gives

p_{\text{exact}} =2\left(\frac{\binom{10}{0}+\binom{10}{1}}{2^{10}}\right) =2\left(\frac{1+10}{1024}\right) =\frac{22}{1024} =0.02148.

At a two-sided significance level of \(0.05\), this exact test rejects equal transmission. The p-value is the probability, under the null model, of an outcome at least this extreme; it is not the probability that the null hypothesis is true.

The Exact Binomial Calibration

For a two-sided exact binomial test with null transmission probability \(1/2\), the usual equal-tailed p-value is twice the probability of the smaller tail, capped at one. Since \(b+c=m\), the smaller count is \(\min(b,c)\), giving

p_{\text{exact}} =\min\left\{1,\; 2\sum_{k=0}^{\min(b,c)} \binom{m}{k}(1/2)^m \right\}.

This calculation uses the full finite-sample distribution, so it does not require a large \(m\). The p-value is discrete: only certain values can occur for a given \(m\). Consequently, a nonrandomized exact test may be conservative at a chosen significance level. In this setting, “exact” means that the null probabilities are calculated from the specified binomial model, not that the test has exactly its nominal rejection probability for every possible sample size.

A two-sided test convention should be stated when reporting results. The equal-tailed definition above is standard for the symmetric null probability \(1/2\). A mid-p calculation, which subtracts part of the probability at the observed outcome, is a different procedure and is not the exact p-value defined here.

The Asymptotic Chi-Squared TDT

The usual uncorrected TDT statistic compares the difference \(b-c\) with its null variability. Under the binomial null, \(B-m/2\) has variance \(m/4\), and \(b-c=2b-m\). Standardizing the difference therefore gives \((b-c)/\sqrt{m}\). Squaring removes the direction of departure:

X^2_{\text{TDT}}=\frac{(b-c)^2}{b+c} =\frac{(b-c)^2}{m}.

For sufficiently large \(m\), this statistic is compared with a chi-squared distribution with one degree of freedom. The resulting p-value is the upper-tail probability \(P(\chi^2_1\geq X^2_{\text{TDT}})\). This is an approximation, not a finite-sample identity. Unless specified otherwise, “chi-squared TDT” here refers to this uncorrected statistic; a continuity-corrected version is a separate approximation.

Justification: Under the null, \(B\sim\operatorname{Binomial}(m,1/2)\). The central limit theorem implies that \(Z=(B-m/2)/\sqrt{m/4}\) is approximately standard normal for large \(m\). Since \(B=b\) and \(m=b+c\), \(Z=(2b-m)/\sqrt{m}=(b-c)/\sqrt{m}\). Squaring gives \(Z^2=(b-c)^2/m\). The square of a standard normal variable has a chi-squared distribution with one degree of freedom, which justifies the large-sample reference distribution.

Worked Example: Comparing Both Calibrations at a Moderate Count

Suppose a hypothetical TDT has \(b=14\) and \(c=6\), so \(m=20\). Assume the 20 informative transmissions satisfy the independent equal-transmission null model. Calculate each two-sided p-value.

For the exact test, the smaller count is six. By symmetry, the two tails are \(B\leq 6\) and \(B\geq 14\):

p_{\text{exact}} =2\sum_{k=0}^{6}\binom{20}{k}(1/2)^{20} =\frac{2(1+20+190+1140+4845+15504+38760)}{2^{20}} =\frac{120920}{1048576} =0.1153.

For the uncorrected chi-squared test, substitute the same counts:

X^2_{\text{TDT}} =\frac{(14-6)^2}{14+6} =\frac{64}{20} =3.200, \qquad p_{\chi^2}=P(\chi^2_1\geq 3.200)=0.07364.

Both p-values exceed \(0.05\), so neither calibration rejects equal transmission at that level. They are not numerically interchangeable: the asymptotic p-value is smaller here, and the difference is material enough to affect interpretation near a decision threshold.

Worked Example: A Larger Count and an Approximation Near the Threshold

Now suppose a hypothetical study has \(b=60\), \(c=40\), and \(m=100\), with independent informative transmissions under the null model. Calculate the exact and asymptotic results.

The smaller count is 40, so the exact two-sided calculation is

p_{\text{exact}} =2\sum_{k=0}^{40}\binom{100}{k}(1/2)^{100} =0.05689.

The chi-squared calculation is

X^2_{\text{TDT}} =\frac{(60-40)^2}{100} =\frac{400}{100} =4.000, \qquad p_{\chi^2}=P(\chi^2_1\geq 4.000)=0.04550.

The exact p-value is just above \(0.05\), while the asymptotic p-value is just below it. The large count does not guarantee that two p-values will agree closely enough to give the same threshold-based decision in every data set. When a result is near the chosen threshold, reporting the exact calculation alongside the approximation makes the calibration difference visible.

Diagnosing Calibration in Practice

The key diagnostic is not merely whether \(b+c\) seems “large.” The chi-squared reference distribution comes from an approximation to the binomial null, and its adequacy depends on how well that approximation represents the tail probability relevant to the observed imbalance. The exact test is straightforward for a single binomial count, so comparing the two p-values is often a useful direct check.

The contrast can be especially noticeable with small counts because the binomial distribution has substantial discreteness. For example, the first worked example gives an exact p-value of \(0.02148\) and an asymptotic p-value of approximately \(0.0114\). Both are below \(0.05\), but the asymptotic value makes the evidence appear stronger than the exact calculation indicates. For a result close to a decision threshold, even a moderate discrepancy can change the conclusion at that threshold.

A practical reporting sequence is:

Check the count definition.
Confirm that \(b\), \(c\), and \(m=b+c\) count the intended informative transmissions, and that the assumed independence and equal-transmission null are appropriate for the basic test.

Calculate the exact result.
For a two-sided test under probability \(1/2\), evaluate the two binomial tails using the smaller of \(b\) and \(c\).

Calculate the asymptotic result.
Compute \((b-c)^2/(b+c)\) and its upper-tail probability from a chi-squared distribution with one degree of freedom.

Compare and report transparently.
Describe the chi-squared value as an approximation. If the p-values differ meaningfully, especially near a decision threshold, report that difference rather than treating the methods as identical.

These calibrations address the single-test null distribution. They do not by themselves adjust for testing many markers, account for nonindependent transmissions, or establish that the assumptions hold in a more complex family design. Those issues require additional methods beyond this basic comparison.

Common Mistake

Wrong reasoning: “The chi-squared p-value is close to the exact p-value in one example, so it is the exact TDT p-value.”
Correction: The chi-squared result comes from a large-sample approximation. Its accuracy can vary with the count and the observed imbalance. The exact binomial p-value uses the finite-sample null probabilities, while the chi-squared p-value uses a continuous reference distribution. State which calibration was used, and do not replace one with the other without checking the approximation.

Key takeaway: Under the basic independent equal-transmission null, \(B\sim\operatorname{Binomial}(m,1/2)\). The exact two-sided binomial test directly uses this distribution; the uncorrected TDT statistic \((b-c)^2/(b+c)\) uses a one-degree-of-freedom chi-squared approximation. Compare their p-values when calibration matters.

Check Your Understanding

Use the equal-transmission binomial null and distinguish exact probabilities from large-sample approximations.

If \(b=7\) and \(c=3\), what are \(m\) and the null expectation of \(B\)?
Write the equal-tailed exact two-sided p-value formula for general \(b\) and \(c\) under the null probability \(1/2\).
For \(b=12\) and \(c=8\), calculate the uncorrected TDT chi-squared statistic.
Why does the standardization of \(B\) lead to a one-degree-of-freedom chi-squared approximation after squaring?
Why can an exact p-value and an asymptotic p-value lead to different decisions when the result is near \(0.05\)?