Effective Sample Size After Imputation

Why Imputation Quality Changes Information

An imputed genotype dosage is not necessarily a known genotype. It is an expected allele count that can be fractional, such as 0.35 or 1.72, because the imputation procedure represents uncertainty about the underlying genotype. Association tests can use these dosages, but uncertainty generally reduces the information available compared with observing every genotype perfectly.

A useful summary is the effective sample size: the size of a hypothetical perfectly genotyped sample that would provide approximately the same association information as the imputed sample, under specified conditions. It is not the number of people actually analyzed. Its simplest form is the observed sample size multiplied by a dosage-quality measure. Allele frequency matters as well: even a perfectly measured variant has less genotype variation, and therefore less information, when its allele is less frequent.

Definition: Let $G$ be the true allele count for a variant, taking values 0, 1, or 2, and let $D$ be its imputed dosage. For a calibrated dosage $D=E(G\mid I)$, where $I$ denotes the imputation information, define the dosage information ratio $$ q=\frac{\operatorname{Var}(D)}{\operatorname{Var}(G)}. $$ When $0\le q\le 1$, $q$ is the fraction of genotype variance represented by the dosages. If the variant has $N$ analyzed participants, the basic effective-sample-size approximation is $N_{\mathrm{eff}}=Nq$.

In practice, an imputation-quality statistic such as dosage $R^2$ may be used as an estimate or proxy for $q$. The exact meaning depends on how that statistic was calculated; it should not be assumed that every INFO or quality score is numerically identical to the variance ratio above. The introductory material on imputation quality metrics and dosage $R^2$ provides the background for interpreting such measures.

Deriving the Effective-Sample-Size Approximation

Consider a simple additive association model in which a quantitative outcome $Y$ has mean $\mu+\beta G$, with independent errors of variance $\sigma^2$. Assume participants are independent, the dosage is calibrated, the imputation uncertainty is not related to the outcome after conditioning on the true genotype, and there are no covariates or other design features changing the genotype information. With true genotypes, the information about $\beta$ is proportional to $N\operatorname{Var}(G)/\sigma^2$. Replacing $G$ by $D$, it is proportional to $N\operatorname{Var}(D)/\sigma^2$.

The ratio of dosage-based information to perfectly observed genotype information is therefore:

\frac{N\operatorname{Var}(D)/\sigma^2} {N\operatorname{Var}(G)/\sigma^2} = \frac{\operatorname{Var}(D)}{\operatorname{Var}(G)} =q.

Thus, under these assumptions, using dosages retains approximately a fraction $q$ of the genotype information. A perfectly genotyped sample of size $Nq$ has the same proportional information at the same allele frequency. This is the basis for $N_{\mathrm{eff}}=Nq$; it is an information-equivalence approximation, not a claim that the analysis literally contains fewer participants.

The connection between dosage $R^2$ and this ratio can also be justified. Because $D=E(G\mid I)$, the conditional expectation identity gives $\operatorname{Cov}(G,D)=\operatorname{Var}(D)$. Consequently:

\operatorname{Corr}(G,D)^2 = \frac{\operatorname{Cov}(G,D)^2} {\operatorname{Var}(G)\operatorname{Var}(D)} = \frac{\operatorname{Var}(D)}{\operatorname{Var}(G)} =q.

This equality relies on calibrated conditional-mean dosages. In finite samples, or when a quality statistic is estimated differently, the reported value may not equal this population ratio exactly.

Allele Frequency and Genotype Information

Let $p$ be the frequency of the allele counted by $G$. Under Hardy–Weinberg equilibrium, the genotype variance is $2p(1-p)$. The expected dosage-based genotype information in the simple model is therefore proportional to:

N\,\operatorname{Var}(D) = Nq\,2p(1-p).

This expression separates three influences: the number of analyzed participants $N$, dosage accuracy $q$, and allele-frequency variation $2p(1-p)$. The factor $2p(1-p)$ is largest at $p=0.5$ and smaller as the frequency approaches either 0 or 1. Thus $Nq$ compares imputation information at a fixed allele frequency; it does not by itself make variants with different frequencies equally informative.

Worked Example: Effective Sample Size for One Variant

Suppose a hypothetical association analysis includes $N=12{,}000$ unrelated participants. A variant has minor-allele frequency $p=0.30$, and its dosage-quality estimate is $q=0.82$. For this calculation, assume the quality estimate represents the dosage variance ratio, the dosages are calibrated, and the simple model conditions above are a reasonable approximation.

The effective sample size is:

N_{\mathrm{eff}}=Nq=12{,}000(0.82)=9{,}840.

So the dosages provide approximately the same genotype information, at this allele frequency, as perfect genotypes in 9,840 participants. This is an information comparison, not a revised count of the study participants.

Under Hardy–Weinberg equilibrium, the true genotype variance is:

\operatorname{Var}(G)=2p(1-p)=2(0.30)(0.70)=0.42.

The dosage variance is approximately $q\operatorname{Var}(G)=0.82(0.42)=0.3444$. The proportional association information is therefore $12{,}000(0.3444)=4{,}132.8$, in units proportional to genotype information. Using perfect genotypes in 9,840 participants gives $9{,}840(0.42)=4{,}132.8$, the same quantity.

Worked Example: Why Sample Size Alone Is Not Enough

Compare two hypothetical variants analyzed in the same type of simple study. Variant A has $N=8{,}000$, $p=0.25$, and $q=0.90$. Variant B has $N=12{,}000$, $p=0.10$, and $q=0.60$. Assume Hardy–Weinberg equilibrium and that each $q$ is a valid dosage variance ratio.

Their effective sample sizes are:

N_{\mathrm{eff,A}}=8{,}000(0.90)=7{,}200, \qquad N_{\mathrm{eff,B}}=12{,}000(0.60)=7{,}200.

The effective sample sizes match, but the allele-frequency terms do not. For A, $\operatorname{Var}(G)=2(0.25)(0.75)=0.375$, so the proportional information is $8{,}000(0.90)(0.375)=2{,}700$. For B, $\operatorname{Var}(G)=2(0.10)(0.90)=0.18$, so it is $12{,}000(0.60)(0.18)=1{,}296$. Variant B has fewer genotype-based information units in this comparison despite the same $N_{\mathrm{eff}}$, because its allele frequency produces less genotype variation.

These calculations compare information on the genotype scale. They do not by themselves determine a p-value or power: effect size, outcome variability, covariates, and the study design also matter.

Using the Approximation for Planning and Interpretation

For a single variant at a fixed allele frequency, the $Nq$ calculation is useful for summarizing how imputation quality changes the scale of information. If a study needs the information corresponding to $N_{\mathrm{target}}$ perfectly genotyped participants and expects quality $q>0$, the corresponding imputed sample size is approximately $N_{\mathrm{target}}/q$. This is a planning approximation, not a guarantee of a particular power or significance result.

Worked Example: Planning for a Target Effective Sample Size

A hypothetical study aims to obtain approximately the information of 10,000 perfectly genotyped participants for a variant. Suppose the expected dosage-quality ratio is $q=0.75$, the allele frequency is held fixed for the comparison, and the simple-model assumptions are reasonable. Solve $Nq=10{,}000$ for the required analyzed sample size:

N=\frac{10{,}000}{0.75}=13{,}333.3.

Because participants are counted in whole numbers, approximately 13,334 analyzed participants are needed to reach or slightly exceed that effective sample-size target. The calculation does not include additional loss from missing phenotypes, covariate adjustment, relatedness, or other design features; if those reduce information, the required recruitment total may be higher.

In the simple model, standard errors scale inversely with the square root of information. At the same allele frequency and with other factors held constant, the standard-error multiplier relative to perfect genotypes is approximately $1/\sqrt{q}$. For example, $q=0.64$ corresponds to $N_{\mathrm{eff}}=0.64N$ and a standard error about $1/\sqrt{0.64}=1.25$ times as large. This is a relative approximation, not a prediction of a particular study's confidence interval.

Assumptions, Diagnostics, and Alternatives

The calculation is most interpretable when the quality value refers to the variant and analyzed sample in question, dosages are calibrated, the participants are independent, and the analysis is close to the simple additive setting. Hardy–Weinberg equilibrium is needed for the convenient frequency formula $2p(1-p)$, but not for the general variance ratio $q=\operatorname{Var}(D)/\operatorname{Var}(G)$. If equilibrium is not a suitable approximation, use the relevant genotype variance rather than substituting $2p(1-p)$.

In a regression with covariates, the information depends on the variation in genotype dosage remaining after accounting for those covariates, not necessarily the unadjusted dosage variance. Case-control imbalance, relatedness, missing outcomes, and other design features can also change information. Therefore, $Nq$ is a useful summary under stated conditions, not a universal replacement for the association model's standard error or information calculation.

Several checks help reveal when a simple summary may mislead:

Confirm that the quality statistic is for the same variant and sample used in the association test, and check how that statistic is defined.
Compare the allele frequency calculated from dosages with the expected or reference frequency, after ensuring that alleles are harmonized. A discrepancy may indicate a frequency mismatch or other data problem, rather than merely low $q$.
Check whether dosage variance is plausible relative to the genotype variance expected at that frequency. A finite-sample estimate can fluctuate; an impossible or extreme value merits investigation rather than automatic interpretation as true information.
Inspect whether quality varies across relevant groups or covariate strata. A single overall quality value can conceal uneven information, especially if dosage accuracy differs across groups.

Dosage-based association testing retains uncertainty and is generally preferable to replacing each dosage with a single hard-call genotype when the analysis supports dosages. Hard calls discard fractional information and can lose additional information, depending on the call threshold and uncertainty. Filtering low-quality variants instead removes them from analysis, which may be appropriate when quality is inadequate, but does not recover the information they lack. These are different choices: dosage analysis represents uncertainty, hard calling simplifies it, and filtering excludes the variant.

A frequent mistake is to say, “The study has 12,000 participants, so the variant has an effective sample size of 12,000,” or to multiply by $q$ and then claim that this is the actual number of people analyzed. The correction is to state the approximation precisely: for a calibrated dosage with $q=0.82$, the variant has information comparable to perfect genotypes in $12{,}000(0.82)=9{,}840$ participants, under the comparison's assumptions. If allele frequencies differ, also compare the $2p(1-p)$ term rather than treating equal $Nq$ values as equal total information.

Key takeaway: For a calibrated dosage, $Nq$ summarizes the effective sample size at a fixed allele frequency under a simple association model. The broader genotype-information scale also depends on allele frequency through genotype variance; study design and covariate adjustment can further change the information available.

Check Your Understanding

Work through each question, then compare your reasoning with the solutions.

A hypothetical variant is analyzed in 5,000 participants with dosage variance ratio $q=0.72$. What is its approximate effective sample size?
For $N=10{,}000$, $p=0.20$, and $q=0.80$, calculate the genotype variance under Hardy–Weinberg equilibrium and the proportional dosage information $Nq\,2p(1-p)$.
Two variants have the same $Nq$, but one has $p=0.50$ and the other has $p=0.10$. Under Hardy–Weinberg equilibrium, which has the larger genotype-based information scale, and why?
At fixed allele frequency and otherwise equal conditions, how does $q=0.49$ affect the approximate standard error relative to perfect genotypes?

Solutions. (1) $N_{\mathrm{eff}}=5{,}000(0.72)=3{,}600$; this is an information-equivalent count, not the number actually analyzed. (2) The genotype variance is $2(0.20)(0.80)=0.32$, and the proportional dosage information is $10{,}000(0.80)(0.32)=2{,}560$. (3) The $p=0.50$ variant has the larger scale: $2(0.50)(0.50)=0.50$, compared with $2(0.10)(0.90)=0.18$. With the same $Nq$, its genotype-based information is larger. (4) The standard-error multiplier is approximately $1/\sqrt{0.49}=1/0.70=1.43$. Thus the dosage-based standard error is about 1.43 times as large under the stated comparison.