Questions regarding the scale of MTAG input and output statistics

[LouChen-med]

Dear developers,

Thank you very much for developing and maintaining the MTAG software.

I have a few questions regarding the scale of the input and output summary statistics in MTAG:

1. As MTAG only takes Z-scores as input, are both binary and continuous traits effectively treated as phenotypes with mean = 0 and standard deviation = 1 in the MTAG framework?
2. For the MTAG output GWAS results, are the reported mtag_beta and mtag_se also on the standardized phenotype scale (mean = 0, standard deviation = 1)?
3. Is there any recommended method to transform the MTAG output mtag_beta and mtag_se from this standardized scale back to log(OR) for binary traits, or to the original measurement units for continuous traits?

Thank you very much for your time and help.

Best regards,
Chen Lou

[paturley]

Hello Chen, 1) Because MTAG works with z-scores, it doesn't matter whether the inputs are based on continuous or binary traits in terms of MTAG's p-values and z-scores. 2) Yes, the beta and the SE are reported as if the input phenotype standard deviation = 1. (The mean shoudn't affect the betas or SEs.) 3) I think just standard approaches should work fine. For continuous traits, if you multiply by the standard deviation of the original phenotype, that should get you back to the right units. For binary traits where the GWAS was done with logistic regression, the latent variable assumed by a logistic has variance 1 already, so I don't think any transformation is necessary. Best, Patrick

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On Sat, Feb 7, 2026 at 9:04 AM Chen Lou ***@***.***> wrote: *LouChen-med* created an issue (JonJala/mtag#250) <#250> Dear developers, Thank you very much for developing and maintaining the MTAG software. I have a few questions regarding the scale of the input and output summary statistics in MTAG: 1. As MTAG only takes Z-scores as input, are both binary and continuous traits effectively treated as phenotypes with mean = 0 and standard deviation = 1 in the MTAG framework? 2. For the MTAG output GWAS results, are the reported mtag_beta and mtag_se also on the standardized phenotype scale (mean = 0, standard deviation = 1)? 3. Is there any recommended method to transform the MTAG output mtag_beta and mtag_se from this standardized scale back to log(OR) for binary traits, or to the original measurement units for continuous traits? Thank you very much for your time and help. Best regards, Chen Lou — Reply to this email directly, view it on GitHub <#250>, or unsubscribe <https://github.com/notifications/unsubscribe-auth/AFBUB5JWSCQM7MQHPBZP3LL4KXWIBAVCNFSM6AAAAACUKNKGPCVHI2DSMVQWIX3LMV43ASLTON2WKOZTHEYTAMZUHEYDIMA> . You are receiving this because you are subscribed to this thread.Message ID: ***@***.***>

[LouChen-med]

Hello Patrick,
Thank you very much for your kind explanation and for taking the time to respond.

However, I am still not fully clear on what is meant by the statement: “For binary traits where the GWAS was done with logistic regression, the latent variable assumed by a logistic has variance 1 already.”

Below, I outline a series of definitions and transformations for regression coefficients in the binary-trait setting. However, I am not fully confident that my interpretation is correct, and I would greatly appreciate your help in checking it. It also seems that discussions in the GWAS literature/software documentation on this point are relatively limit
For binary traits, the MTAG output (MTAG_beta) appears to correspond to case II②. If I want to transform the MTAG_beta to the log(OR) scale, I believe this corresponds to case III②. Since MTAG uses the effective sample size, multiplying MTAG_beta by 2 would yield the estimate in log(OR) units.

let
K = case/samplesize
n_total+z^2 ≈ n_total （often z^2 << n_total in MTAG paper）
If n_total in the following formulas is replaced by n_eff = as.integer(4casecontrol/(case + control)) then K = 0.5,1/sqrt(K (1 − K ))=2

I: standardized regression coefficients （The derivation of the formulas can be referred to 'GWAS of 126,559 Individuals Identifies Genetic Variants Associated with Educational Attainment', Science 2013,sup note p4）

se_standard = σy / sqrt(2 * freq * (1 − freq) * (n_total + z^2))
b_standard = σy * z * (1 / sqrt(2 * freq * (1 − freq) * (n_total + z^2)))
① For continuous traits, σy = 1 (equivalent to unit = SD);
② For binary traits, σy = sqrt(K (1 − K )), where y ∈ {0,1}. such as bolt-lmm (https://doi.org/10.1534/genetics.117.300360)

II：standardized regression coefficients (SD fixed to 1)

se_sd1 = 1 / sqrt(2 * freq * (1 − freq) * (n_total + z^2))
b_sd1 = 1* z * (1 / sqrt(2 * freq * (1 − freq) * (n_total + z^2)))
① For continuous traits, σy = 1 (equivalent to unit = SD); no difference from above.
② For binary traits, using σy = 1 is equivalent to further scaling y ∈ {0,1} such that mean = 0 and SD = 1, where the original SD is 'sqrt(K (1 − K ))'.

III：unstandardized regression coefficients

se_unstandard = σy / sqrt(2 * freq * (1 − freq) * (n_total + z^2))
b_unstandard = σy * z * (1 / sqrt(2 * freq * (1 − freq) * (n_total + z^2)))
① For continuous traits, σy equals the actual SD of the phenotype (e.g., height with SD = 20 cm, beta unit = cm);
② For binary traits, σy = 1 / sqrt(K (1 − K )),
b_unstandard = b_standard / [K (1 − K )] = b_sd1/ sqrt(K (1 − K )), with unit = log(OR).

Best,
Chen

[LouChen-med]

Hello Patrick,

Regarding the standardization of binary traits, I further examined the GenomicSEM formulation and conducted a comparison with the MTAG framework. I would like to discuss this further with you.

Background

In Genomic structural equation modelling provides insights into the multivariate genetic architecture of complex traits (Nature Human Behaviour, 2019), the logistic regression coefficient (log-odds ratio) was standardized as follows.

The original beta from logistic regression is expressed in log(OR) units:

$$\sigma^2_{\mathrm{SNP}} = 2\,\mathrm{MAF}\,(1-\mathrm{MAF})$$ $$b^{*}_{\text{logit},\text{SNP},P} = \frac{Z}{\sqrt{\,v(1-v)\,N_{\text{total}}\,\sigma^2_{\text{SNP}}\,}}.$$ $$\mathrm{se}\!\left(b^{*}_{\text{logit},\text{SNP},P}\right) = \frac{b^{*}_{\text{logit},\text{SNP},P}}{Z} = \frac{1}{\sqrt{\,v(1-v)\,N_{\text{total}}\,\sigma^2_{\text{SNP}}\,}}.$$

The standardized SNP effect ($b_{\mathrm{std}}$) is obtained by scaling
($b^{*}_{\mathrm{logit},\mathrm{SNP},P}$) to the unit-variance liability scale; that is, by dividing by the square root of the total liability variance:

$$b_{\mathrm{std}} = \frac{b^{*}_{\mathrm{logit},\mathrm{SNP},P}} {\sqrt{\sigma^2_{\mathrm{SNP}}\!\left(b^{*}_{\mathrm{logit},\mathrm{SNP},P}\right)^2 + \frac{\pi^2}{3}}} \qquad (1)$$

and the corresponding standard error is

$$\mathrm{se}(b_{\mathrm{std}}) = \frac{\mathrm{se}\!\left(b^{*}_{\mathrm{logit},\mathrm{SNP},P}\right)} {\sqrt{\sigma^2_{\mathrm{SNP}}\!\left(b^{*}_{\mathrm{logit},\mathrm{SNP},P}\right)^2 + \frac{\pi^2}{3}}}.$$

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Alternative formulation (balanced binary phenotype)

In Pervasive Downward Bias in Estimates of Liability-Scale Heritability in GWAS Meta-analysis: A Simple Solution (Biological Psychiatry), $b^*$ is defined as the linear regression coefficient of a standardized binary phenotype (assuming a balanced case–control design; The 0/1 phenotype was standardized to have mean 0 and variance 1) and used for meta-analysis.

It is unclear to me whether this $b^*$ is already on the unit-variance liability scale.

$$b^* = \frac{Z}{\sqrt{4v(1-v)\,n_{\text{total}}\sigma^2_{\text{SNP}}\,}} \qquad (2)$$ $$SE(b^*) = \frac{1}{\sqrt{\,4v(1-v)\,n_{\text{total}}\,\sigma^2_{\text{SNP}}\,}}$$

with the approximation

$$b_{\text{logit}} \approx 2 b^* \quad\Rightarrow\quad b^*\approx \frac{1}{2}b_{\text{logit}}.$$

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Questions

1. Between Eqs. (1) and (2), MTAG appears to standardize the binary phenotype to mean 0 and variance 1 and adopts Eq. (2) (with $4v(1-v)n$ supplied to MTAG as $N_{\text{eff}}$), whereas GenomicSEM uses Eq. (1). In both cases, the resulting beta seems to be interpreted as a standardized effect (i.e., per unit variance). Could you clarify what the practical difference is between these two formulations, or whether they are in fact equivalent under the usual GWAS assumptions?

2. Conceptually, how do Eqs. (1) and (2) differ? They appear numerically close. In Eq. (2), does $b^*$ already correspond to the unit-variance liability scale? I have not yet conducted a comprehensive empirical comparison.

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Best regards,

Chen Lou
Wenzhou Medical University